D}\phi\ e^{\int\limits_{0}^{\beta}\,d(it)\int\,d^{3}{x}\,{\cal
L}(t\rightarrow-i\tau)}$ $\displaystyle=$
$\displaystyle\int\limits_{\phi(\bm{\vec{x}},0)=\pm\phi(\bm{\vec{x}},\beta)}{\cal
D}\phi\ e^{\int\limits_{0}^{\beta}\,d\tau\int\,d^{3}{x}\,{\cal
L}(t\rightarrow-i\tau)}.$

One can compute the partition function in Euclidean time $\tau$ and discrete
frequency $i\omega_{n}$ directly using (206).

###  6.2 Scalar Field Partition Function

####  6.2.1 Partition function for free real scalar field

A real non-interacting scalar field Lagrangian is given in (166) by

$\displaystyle{\cal L}_{0}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}m^{2}\phi^{2}=\frac{1}{2}\left[\left(\frac{\partial\phi}{\partial
t}\right)^{2}-({\mathbf{\nabla}}\phi)^{2}-m^{2}\phi^{2}\right],$ (207)

in Minkowski space-time. This can be written only in Euclidean time Kapusta ;
Schmitt ; Yang ; Inga  as

$\displaystyle{\cal L}_{0}(t\rightarrow-i\tau)$
$\displaystyle{=\atop{t\rightarrow-i\tau}}$
$\displaystyle\frac{1}{2}\left[\left({\partial_{t}\phi}\right)^{2}-({\mathbf{\nabla}}\phi)^{2}-m^{2}\phi^{2}\right]=-\frac{1}{2}\left[\left({\partial_{\tau}\phi}\right)^{2}+({\mathbf{\nabla}}\phi)^{2}+m^{2}\phi^{2}\right]\,.$
(208)

The Fourier transform of the field $\phi(X)$ can be written as

$\displaystyle\phi(X)=\phi(x,\tau)$
$\displaystyle{=\atop{t\rightarrow-i\tau}}\atop{k_{0}\rightarrow i\omega_{n}}$
$\displaystyle\frac{1}{\sqrt{V\beta}}\sum_{K}\,e^{-iK\cdot
X}\,\phi(K)=\frac{1}{\sqrt{V\beta}}\sum_{K}\,e^{i\bm{\vec{k}\cdot\vec{x}}}\,\,e^{-i(i\omega_{n})(-i\tau)}\,\phi(\omega_{n},\bm{\vec{k}})$
(209) $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{V\beta}}\sum_{K}\,e^{i\bm{\vec{k}\cdot\vec{x}}}\,e^{-i\omega_{n}\tau}\,\phi(\omega_{n},\bm{\vec{k}}),$

where $\sum_{K}=\sum_{n,\bm{\vec{k}}}$ and $V$ is the three volume.

Using (208) in (206), one can write the partition function for free scalar
field as

$\displaystyle{\cal Z}_{0}$ $\displaystyle=$
$\displaystyle\,\int\limits_{\phi(\bm{\vec{x}},0)=\phi(\bm{\vec{x}},\beta)}{\cal
D}\phi\
e^{-\frac{1}{2}\int\limits_{0}^{\beta}\,d\tau\int\,d^{3}{x}\,\left[\left({\partial_{\tau}\phi}\right)^{2}+({\mathbf{\nabla}}\phi)^{2}+m^{2}\phi^{2}\right]},$
(210)

where scalar field obeys periodicity condition.

Now we calculate explicitly the terms in the exponential of (210):

  1. First term:

$\displaystyle\int\limits_{0}^{\beta}d\tau\int
d^{3}{x}\,\left(\partial_{\tau}\phi\right)^{2}=\int
d^{4}X\,\frac{1}{V\beta}\sum_{K}\,\sum_{K^{\prime}}\,\partial_{\tau}\left[e^{-i\omega_{n}\tau}\,e^{i\bm{\vec{k}\cdot\vec{x}}}\phi(\omega_{n},\bm{\vec{k}})\right]$
(211)
$\displaystyle\times\partial_{\tau}\left[e^{-i\omega_{m}\tau}\,e^{i\bm{\vec{k}^{\prime}\cdot\vec{x}}}\phi(\omega_{m},\bm{\vec{k}^{\prime}})\right]$
$\displaystyle=$ $\displaystyle\int
d^{4}X\,\frac{1}{V\beta}\sum_{K}\,\sum_{K^{\prime}}\,e^{i(\bm{\vec{k}}+\bm{\vec{k}^{\prime}})\cdot\bm{\bm{x}}}e^{-i(\omega_{n}+\omega_{m})\tau}\phi(\omega_{n},\bm{\vec{k}})(-i\omega_{n})(-i\omega_{m})\phi(\omega_{m},\bm{\vec{k}^{\prime}})$
$\displaystyle=$
$\displaystyle\frac{1}{V\beta}\sum_{K}\sum_{K^{\prime}}\left[V\delta^{3}(\bm{\vec{k}}+\bm{\vec{k}^{\prime}})\right]\left[\beta\delta(-\omega_{n}-\omega_{m})\right]\phi(\omega_{n},\bm{\vec{k}})(-i\omega_{n})(-i\omega_{m})\phi(\omega_{m},\bm{\vec{k}^{\prime}})$
$\displaystyle=$ $\displaystyle\sum_{K}\
\,\phi(\omega_{n},\bm{\vec{k}})\,\omega_{n}^{2}\phi(-\omega_{n},-\bm{\vec{k}})=\sum_{K}\,\omega_{n}^{2}\,\phi(K)\,\phi(-K)$
$\displaystyle=$
$\displaystyle\sum_{K}\,\omega_{n}^{2}\,\phi^{*}(K)\,\phi(K),$

where we have used $\phi(K)=\phi^{*}(-K)$ since $\phi(X)$ is real in (209). We
have also used of the following relations:

$\displaystyle\int\limits_{0}^{\beta}d\tau\ e^{i(-\omega_{n}-\omega_{m})\tau}$
$\displaystyle=$ $\displaystyle\beta\delta(-\omega_{n}-\omega_{m})\,,$ (212)
$\displaystyle\int d^{3}x\ e^{i(\bm{\vec{k}}+\bm{\vec{k}^{\prime}})\cdot x}$
$\displaystyle=$ $\displaystyle
V\delta^{3}(\bm{\vec{k}}+\bm{\vec{k}^{\prime}})\,,$ (213)
$\displaystyle\omega_{-n}$ $\displaystyle=$ $\displaystyle-\frac{2\pi
n}{\beta}=-\omega_{n}\,.$ (214)

  2. Second term:

$\displaystyle\int\limits_{0}^{\beta}d\tau\int
d^{3}{x}\,\left(\mathbf{\nabla}\phi\right)^{2}$ $\displaystyle=$
$\displaystyle\,\sum_{K}{k}^{2}\phi^{*}(K)\,\phi(K).$ (215)

  3. Third term:

$\displaystyle\int\limits_{0}^{\beta}d\tau\int d^{3}{x}\,\,m^{2}\phi^{2}$
$\displaystyle=$ $\displaystyle\,\sum_{K}\,m^{2}\,\phi^{*}(K)\,\phi(K)\,.$
(216)

Using (211), (215) and (216) one can write

$\displaystyle{-\frac{1}{2}\int\limits_{0}^{\beta}\,d\tau\int\,d^{3}{x}\,\left[\left({\partial_{\tau}\phi}\right)^{2}+({\mathbf{\nabla}}\phi)^{2}+m^{2}\phi^{2}\right]}$
$\displaystyle=$
$\displaystyle-\,\frac{1}{2}\,\sum_{K}\,\,\phi^{*}(K)\Big{[}k_{0}^{2}+{k}^{2}+m^{2}\Big{]}\phi(K)$
(217) $\displaystyle=$
$\displaystyle-\,\frac{1}{2}\,\sum_{K}\,\,\phi^{*}(K)\Big{[}k_{0}^{2}+\omega_{k}^{2}\Big{]}\phi(K)$
$\displaystyle=$ $\displaystyle-\frac{1}{2}\,\sum_{K}\,\phi^{*}(K)\,{\cal
G}^{-1}_{0}(K)\phi(K)\,,$

where ${\cal G}^{-1}_{0}=(\omega_{n}^{2}+\omega_{k}^{2})$ is the free inverse
propagator in Euclidean time with energy $\omega_{k}=\sqrt{k^{2}+m^{2}}$.

Using (217) in (210) one can write the free scalar field partition function as

$\displaystyle{\cal Z}_{0}\,\,$ $\displaystyle=$
$\displaystyle\,\int\limits_{\phi(\bm{\vec{x}},0)=\phi(\bm{\vec{x}},\beta)}{\cal
D}\phi\ \exp\left[-\frac{1}{2}\,\sum_{K}\,\phi^{*}(K)\,[{\cal
G}^{-1}_{0}(K)]\phi(K)\right]$ (218) $\displaystyle=$
$\displaystyle\prod_{K}\int\limits_{\phi(\bm{\vec{x}},0)=\phi(\bm{\vec{x}},\beta)}{\cal
D}\phi\ \exp\left[-\frac{1}{2}\,\phi^{*}(K)\,\,[{\cal
G}^{-1}_{0}(K)]\,\,\phi(K)\right].$

We also note that inverse propagator is in general a diagonal matrix.

The integral can be performed using the standard identity555This identity is a
generalisation of the one dimensional Gaussian integral
$\int\limits_{-\infty}^{\infty}dx\ \ \exp(-\frac{1}{2}ay^{2})=\sqrt{2\pi/a}$
and can be shown by expressing the bilinear
${\mathbf{y}}\cdot{\widehat{A}}{\mathbf{y}}$ in terms of eigenvalues of
$\widehat{A}$.

$\displaystyle\int
d^{D}y\,\,e^{-\frac{1}{2}\,{\mathbf{y}}\cdot{\widehat{A}}{\mathbf{y}}}=(2\pi)^{D/2}({\mbox{det}}{\widehat{A}})^{-1/2},$
(219)

for hermitian positive definite matrix $\widehat{A}$ and $D$ is the dimension
of the system. The partition function can now be written as

$\displaystyle{\cal Z}_{0}$ $\displaystyle=$ $\displaystyle
N\left[{\mbox{det}}{\cal G}^{-1}_{0}\right]^{-\frac{1}{2}},$ (220)

where the determinant is taken over momentum space as ${\cal G}^{-1}_{0}(K)$
is diagonal. We also note that the constant factor is absorbed in $N$, which
is irrelevant as it is temperature independent. Now the logarithm of a
partition function up to a constant, becomes

$\displaystyle\ln{\cal Z}_{0}\,$ $\displaystyle=$
$\displaystyle\ln\left[{\mbox{det}}[{\cal
G}^{-1}_{0}(K)]\right]^{-\frac{1}{2}}$ (221) $\displaystyle=$
$\displaystyle-\frac{1}{2}\ln\prod_{K}[{\cal
G}^{-1}_{0}(K)]=-\frac{1}{2}\sum_{K}\ln[{\cal G}^{-1}_{0}(K)]$
$\displaystyle=$
$\displaystyle-\frac{1}{2}\sum_{n,{k}}\ln(\omega_{n}^{2}+\omega_{k}^{2})=-\frac{1}{2}\,V\int\frac{d^{3}{k}}{(2\pi)^{3}}\,\sum_{n}\ln(\omega_{n}^{2}+\omega_{k}^{2}).$

Below we perform the sum integral as

$\displaystyle
T\sum_{n}\frac{1}{T}\ln\left(\omega_{n}^{2}+\omega_{k}^{2}\right)=T\sum_{k-0}\frac{1}{T}\ln\left(-k_{0}^{2}+\omega_{k}^{2}\right)=T\sum_{k_{0}}\frac{1}{T}\left[\ln\left(\omega_{k}-k_{0}\right)+\ln\left(\omega_{k}+k_{0}\right)\right],$
(222)

where $\omega=i\omega_{n}=k_{0}$.

Now concentrating on the first term: using (134) we can write

$\displaystyle T\sum_{k_{0}}\frac{1}{T}\ln\left(\omega_{k}-k_{0}\right)$
$\displaystyle=$ $\displaystyle\frac{1}{2\pi iT}\oint
dk_{0}\ln\left(\omega_{k}-k_{0}\right)\frac{1}{2}\ \coth\left(\frac{\beta
k_{0}}{2}\right)$ (223) $\displaystyle=$ $\displaystyle-\frac{1}{4\pi iT}\oint
dk_{0}\left(\frac{-1}{\omega_{k}-k_{0}}\right)\int
dk_{0}\coth\left(\frac{\beta k_{0}}{2}\right)$ $\displaystyle=$
$\displaystyle\frac{1}{4\pi iT}\oint
dk_{0}\left(\frac{1}{\omega_{k}-k_{0}}\right)\int dk_{0}\coth\left(\frac{\beta
k_{0}}{2}\right)$ $\displaystyle=$ $\displaystyle\frac{1}{4\pi iT}\oint
dk_{0}\left(\frac{1}{\omega_{k}-k_{0}}\right)\frac{2}{\beta}\ln\left[\sinh\left(\frac{\beta
k_{0}}{2}\right)\right]$ $\displaystyle=$ $\displaystyle\frac{1}{2\pi i}\oint
dk_{0}\left(\frac{1}{\omega_{k}-k_{0}}\right)\ln\left[\sinh\left(\frac{\beta
k_{0}}{2}\right)\right]$ $\displaystyle=$
$\displaystyle\ln\left[\sinh\left(\frac{\beta\omega_{k}}{2}\right)\right].$

Similarly for the second term: choosing the opposite contour as the first one
we get,

$\displaystyle T\sum_{k_{0}}\ln\left(\omega_{k}+k_{0}\right)$ $\displaystyle=$
$\displaystyle\ln\left[\sinh\left(\frac{\beta\omega_{k}}{2}\right)\right],$
(224)

we note here that there will be a term of $i\pi$ in (224), which we neglect as
it would make partition function imaginary. So, the results of frequency sum
in (222) can be written as,

$\displaystyle
T\sum_{n}\frac{1}{T}\ln\left(\omega_{n}^{2}+\omega_{k}^{2}\right)$
$\displaystyle=$ $\displaystyle
2\ln\left[\sinh\left(\frac{\beta\omega_{k}}{2}\right)\right]$ (225)
$\displaystyle=$ $\displaystyle\beta\omega_{k}-2\ln
2+2\ln\left(1-e^{-\beta\omega_{k}}\right).$

We used

$\displaystyle\sinh\left(\frac{\beta\omega_{k}}{2}\right)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left(e^{\frac{\beta\omega_{k}}{2}}-e^{\frac{-\beta\omega_{k}}{2}}\right)=\frac{1}{2}e^{\frac{\beta\omega_{k}}{2}}\left(1-e^{-\beta\omega_{k}}\right)$
$\displaystyle\therefore\ln\left[\sinh\left(\frac{\beta\omega_{k}}{2}\right)\right]$
$\displaystyle=$ $\displaystyle\frac{\beta\omega_{k}}{2}-\ln
2+\ln\left(1-e^{-\beta\omega_{k}}\right).$ (226)

Now using (225) in (221) one gets logarithm of the partition function

$\displaystyle\ln{\cal Z}_{0}\,$ $\displaystyle=$
$\displaystyle-V\int\frac{d^{3}{k}}{(2\pi)^{3}}\,\left[\frac{\beta\omega_{k}}{2}-\ln
2+\ln\left(1-e^{-\beta\omega_{k}}\right)\right],$ (227)

where again $\ln 2$ can be neglected as it is independent of temperature. This
agrees with the result obtained in (16).

The free energy density for free scalar field can be obtained from (6a) as

$\displaystyle F_{0}$ $\displaystyle=$ $\displaystyle-T\ln{\cal
Z}_{0}\,=VT\,\int\frac{d^{3}{k}}{(2\pi)^{3}}\,\left[\frac{\beta\omega_{k}}{2}+\ln\left(1-e^{-\beta\omega_{k}}\right)\right],$
(228)

In the infinite volume limit the pressure for free scalar field can be
obtained from (6b) as

$\displaystyle{\cal P}_{0}$ $\displaystyle=$
$\displaystyle-\frac{{\Omega}_{0}}{V}=T\,\int\frac{d^{3}{k}}{(2\pi)^{3}}\,\left[-\frac{\beta\omega_{k}}{2}-\ln\left(1-e^{-\beta\omega_{k}}\right)\right],$
(229)

where the first term is the zero temperature part.

####  6.2.2 Partition function for interacting scalar field

The interaction Lagrangian density as given in (166) as

$\displaystyle{\cal L}_{I}$ $\displaystyle=$
$\displaystyle-\frac{\lambda}{4!}\phi^{4},$ (230)

The partition function in first order $\lambda$ as given in (199)

$\displaystyle\left(\ln{\cal Z}_{I}\right)_{1}$ $\displaystyle=$
$\displaystyle\left\langle{\cal F}_{I}\right\rangle_{0}=\frac{1}{{\cal
Z}_{0}}\,{\rm{Tr}}\left[\,e^{-\beta{\cal H}_{0}}\ {\cal
T}\left(-\int\limits_{0}^{\beta}{\cal H}^{\prime}\ d\tau\right)\right].$ (231)

Using path integral it becomes Kapusta

$\displaystyle\left(\ln{\cal Z}_{I}\right)_{1}$ $\displaystyle=$
$\displaystyle\frac{\int{\cal D}\phi\ {\cal S}_{I}\,e^{i{\cal
S}_{0}[\phi]}}{\int{\cal D}\phi\ e^{i{\cal S}_{0}[\phi]}}$ (232)
$\displaystyle=$ $\displaystyle\,\frac{\int_{\tiny{\mbox{\sf periodic}}}{\cal
D}\phi\ \left(-\int\limits_{0}^{\beta}\,d\tau\int\,d^{3}{x}\,{\cal
L}_{I}({t\rightarrow-i\tau})\right)\,e^{-\int\limits_{0}^{\beta}\,d\tau\int\,d^{3}{x}\,{\cal
L}_{0}({t\rightarrow-i\tau})}}{\int_{\tiny{\mbox{\sf periodic}}}{\cal D}\phi\
e^{-\int\limits_{0}^{\beta}\,d\tau\int\,d^{3}{x}\,{\cal
L}_{0}({t\rightarrow-i\tau})}},$

where free partition function is already calculated in (218) as

$\displaystyle{\cal Z}_{0}\,$ $\displaystyle=$
$\displaystyle\,\int_{\tiny{\mbox{\sf periodic}}}{\cal
D}\phi_{n,\bm{\vec{k}}}\
\exp\left[-\frac{1}{2}\,\sum_{n,k}\,[\omega^{2}_{n}+\omega_{k}^{2}]\left|\phi_{n,\bm{\vec{k}}}\right|^{2}\right].$
(233)

Now we need to compute the numerator of (232). Using the same method as the
free case one can proceed as Kapusta ; Yang

$\displaystyle\left(\ln{\cal Z}_{I}\right)_{1}$ $\displaystyle=$
$\displaystyle\frac{1}{{\cal Z}_{0}}\,\int_{\tiny{\mbox{\sf periodic}}}{\cal
D}\phi_{n,\bm{\vec{k}}}\
\left(\lambda\int\limits_{0}^{\beta}\,d\tau\int\,d^{3}{x}\,\phi^{4}\right)\,\exp\left[-\frac{1}{2}\,\sum_{n,k}\,[\omega^{2}_{n}+\omega_{k}^{2}]\left|\phi_{n,\bm{\vec{k}}}\right|^{2}\right].$
(234)

Using the Fourier decomposition of fields in (209), one can get

$\displaystyle\lambda\int\limits_{0}^{\beta}\,d\tau\int\,d^{3}{x}\,\phi^{4}$
$\displaystyle=$
$\displaystyle\frac{\lambda}{(V\beta)^{2}}\,\sum_{m_{1},\bm{\vec{q}_{1}}}\sum_{m_{2},\bm{\vec{q}_{2}}}\sum_{m_{3},\bm{\vec{q}_{3}}}\sum_{m_{4},\bm{\vec{q}_{4}}}\,\phi_{m_{1},\bm{\vec{q}_{1}}}\,\phi_{m_{2},\bm{\vec{q}_{2}}}\,\phi_{m_{3},\bm{\vec{q}_{3}}}\,\phi_{m_{4},\bm{\vec{q}_{4}}}$
(235)
$\displaystyle\times\int\limits_{0}^{\beta}\,d\tau\int\,d^{3}{x}\,\,e^{i(\bm{\vec{q}_{1}}+\bm{\vec{q}_{2}}+\bm{\vec{q}_{3}}+\bm{\vec{q}_{4}})\cdot{x}}\,\,e^{-i(\omega_{m_{1}}+\omega_{m_{2}}+\omega_{m_{3}}+\omega_{m_{4}})\tau}$
$\displaystyle=$
$\displaystyle\frac{\lambda}{(V\beta)^{2}}\,\sum_{m_{1},\bm{\vec{q}_{1}}}\sum_{m_{2},\bm{\vec{q}_{2}}}\sum_{m_{3},\bm{\vec{q}_{3}}}\sum_{m_{4},\bm{\vec{q}_{4}}}\,\phi_{m_{1},\bm{\vec{q}_{1}}}\,\phi_{m_{2},\bm{\vec{q}_{2}}}\,\phi_{m_{3},\bm{\vec{q}_{3}}}\,\phi_{m_{4},\bm{\vec{q}_{4}}}$
$\displaystyle\times\beta\delta(-\omega_{m_{1}}-\omega_{m_{2}}-\omega_{m_{3}}-\omega_{m_{4}})\
V\
\delta^{3}(\bm{\vec{q}_{1}}+\bm{\vec{q}_{2}}+\bm{\vec{q}_{3}}+\bm{\vec{q}_{4}}).$

We have used two delta functions following (212) and (213) from $\tau$ and $x$
integrations, respectively and they guarantee the energy momentum conservation
in the interaction vertex. Now, the non-zero contribution comes when
$\omega_{m_{1}}=-\omega_{m_{2}}$, $\omega_{m_{3}}=-\omega_{m_{4}}$ and
$\bm{\vec{q}_{1}}=-\bm{\vec{q}_{2}}$, $\bm{\vec{q}_{3}}=-\bm{\vec{q}_{4}}$.
Obviously, there are another two combination that would give non-zero
contributions. This allows 3 non-zero permutations among $\omega_{m_{i}}$ and
${q}_{i}$. Then one can write

$\displaystyle\lambda\,\int\limits_{0}^{\beta}\,d\tau\int\,d^{3}{x}\,\phi^{4}$
$\displaystyle=$ $\displaystyle 3\lambda\,(\beta
V)\,\left(\frac{1}{V\beta}\right)^{2}\,\sum_{m,\bm{\vec{q}}}\sum_{l,\bm{\vec{p}}}\,\left|\phi_{m,\bm{\vec{q}}}\right|^{2}\,\left|\phi_{l,\bm{\vec{p}}}\right|^{2}.$
(236)

Using (236) in (234), one gets

$\displaystyle\left(\ln{\cal Z}_{I}\right)_{1}$ $\displaystyle=$
$\displaystyle 3\lambda\beta
V\left(\frac{1}{V\beta}\right)^{2}\sum_{m,\bm{\vec{q}}}\sum_{l,\bm{\vec{p}}}\frac{{\prod\atop{n,\bm{\vec{k}}}}\int_{\tiny{\mbox{\sf
periodic}}}{\cal D}\phi_{n,\bm{\vec{k}}}\
\left|\phi_{m,\bm{\vec{q}}}\right|^{2}\,\left|\phi_{l,\bm{\vec{p}}}\right|^{2}\,\exp\left[-\frac{1}{2}\,(\omega^{2}_{n}+\omega_{k}^{2})\left|\phi_{n,\bm{\vec{k}}}\right|^{2}\right]}{{\prod\atop{n,\bm{\vec{k}}}}\int_{\tiny{\mbox{\sf
periodic}}}{\cal D}\phi_{n,\bm{\vec{k}}}\
\exp\left[-\frac{1}{2}\,(\omega^{2}_{n}+\omega_{k}^{2})\left|\phi_{n,\bm{\vec{k}}}\right|^{2}\right]}$
(237) $\displaystyle=$ $\displaystyle 3\lambda\beta
V\left(\frac{1}{V\beta}\right)^{2}\left[\sum_{n\bm{\vec{,}k}}\frac{\int_{\tiny{\mbox{\sf
periodic}}}{\cal
D}\phi_{n,\bm{\vec{k}}}\left|\phi_{n,\bm{\vec{k}}}\right|^{2}\,\exp\left[-\frac{1}{2}(\omega^{2}_{n}+\omega_{k}^{2})\left|\phi_{n,\bm{\vec{k}}}\right|^{2}\right]}{\int_{\tiny{\mbox{\sf
periodic}}}{\cal
D}\phi_{n,\bm{\vec{k}}}\exp\left[-\frac{1}{2}\,(\omega^{2}_{n}+\omega_{k}^{2})\left|\phi_{n,\bm{\vec{k}}}\right|^{2}\right]}\right]^{2},$

if $m=l=n$ and $\bm{\vec{q}}=\bm{\vec{p}}=\bm{\vec{k}}$, the integral over
$\phi_{n,\bm{\vec{k}}}$ gets factorised as above. When $m\neq l\neq n$ and
$\bm{\vec{q}}\neq\bm{\vec{p}}\neq\bm{\vec{k}}$, the $\phi_{n,\bm{\vec{k}}}$
integral in the numerator and denominator are identical and they cancel out,
thus the integral disappears.

Using Gaussian integral $\int dy\ y^{2n}\,e^{-\alpha
y^{2}}=\Gamma(n+1/2)/\alpha^{n+1/2}$, one gets

$\displaystyle\left(\ln{\cal Z}_{I}\right)_{1}$ $\displaystyle=$
$\displaystyle 3\,\lambda\,\beta V\
\left(\frac{1}{V\beta}\right)^{2}\left[\frac{\Gamma(1+\frac{1}{2})}{\alpha^{3/2}}\times\frac{\alpha^{1/2}}{\Gamma(\frac{1}{2})}\right]^{2}=3\,\lambda\,\beta
V\ \left(\frac{1}{V\beta}\right)^{2}\left[\frac{1}{2\alpha}\right]^{2}$ (238)
$\displaystyle=$ $\displaystyle 3\,\lambda\,\beta V\
\left(\frac{1}{V\beta}\right)^{2}\left[\sum_{n,k}\frac{1}{(\omega_{n}^{2}+\omega_{k}^{2})}\right]^{2}=3\,\lambda\,\beta
V\left[\frac{1}{V\beta}\sum_{n,k}\frac{1}{\omega_{n}^{2}+\omega_{k}^{2}}\right]^{2}.$

Taking continuum limit $\sum_{k}\rightarrow V\,\int d^{3}{k}/(2\pi)^{3}$, one
gets

$\displaystyle\left(\ln{\cal Z}_{I}\right)_{1}$ $\displaystyle=$
$\displaystyle 3\,\lambda\,\beta
V\left[\frac{1}{\beta}\sum_{n}\int\frac{d^{3}k}{(2\pi)^{3}}\,\frac{1}{\omega_{n}^{2}+\omega_{k}^{2}}\right]^{2}.$
(239)

where the $\sum_{n}$ is Euclidean. Casting this into Minkowski
$\omega_{n}=k_{4}=-ik_{0}$, one can write

$\displaystyle\left(\ln{\cal Z}_{I}\right)_{1}$ $\displaystyle=$
$\displaystyle-\ 3\,\lambda\,\beta
V\left[\frac{1}{\beta}\sum_{k_{0}}\int\frac{d^{3}k}{(2\pi)^{3}}\,\frac{1}{k_{0}^{2}-\omega_{k}^{2}}\right]^{2},$
(240)

where one can use contour integration in (131) vis-a-vis (134) and (136). We
also note the following points:

  1. $\bullet$

The term inside the square braces are the sum-integral that comes from the
loop integral $\int d^{4}K/(2\pi)^{4}$ and the scalar propagator
($1/(k_{0}^{2}-{k}^{2}-m^{2})=1/(k_{0}^{2}-\omega_{k}^{2})$ with energy
$\omega_{k}=\sqrt{{k}^{2}+m^{2}}$.

  2. $\bullet$

$\lambda$ is the interaction vertex.

  3. $\bullet$

$\beta V$ is the left out factor that comes from the energy-momentum
conservation (e.g., (235)) in the vertex.

  4. $\bullet$

The square, $[\cdots]^{2}$, indicates that two loops connected to one
interaction point $\lambda$.

  5. $\Rightarrow$

All these together correspond to a topologically distinct diagram [].

  6. $\bullet$

$3$ is the symmetry factor that comes from 3 different permutation of
contraction allowed by the energy-momentum conservation.

Now the first order correction to the scalar partition function in (240) can
be represented in Feynman diagram as

$\displaystyle\left(\ln{\cal Z}_{I}\right)_{1}$ $\displaystyle=$
$\displaystyle-\ 3\,\beta
V\times{\includegraphics[height=14.22636pt,width=28.45274pt]{vaccum_bubble.pdf}}.$
(241)

The frequency sum in (240) is exactly similar to that of tadpole diagram as
done in Subsec. 5.1 but with different symmetry factor ($1/2$). Excluding this
$1/2$ factor and $\lambda$ and the zero temperature part, the result of the
sum integral can be obtained from (171) as $T^{2}/12$ . Thus, the first order
correction to the scalar partition function in (240) becomes

$\displaystyle\left(\ln{\cal Z}_{I}\right)_{1}$ $\displaystyle=$
$\displaystyle-\ 3\,\lambda\,\beta
V\left[\frac{1}{\beta}\sum_{n}\int\frac{d^{3}k}{(2\pi)^{3}}\,\frac{1}{k_{0}^{2}-\omega_{k}^{2}}\right]^{2}=-\
3\,\lambda\,\beta V\ \frac{T^{4}}{144}=-\ \lambda\,\frac{VT^{3}}{48}.$ (242)

####  6.2.3 Pressure

The logarithm of the scalar partition function up to first order in coupling
can now be written using (194), (227) and (242) as

$\displaystyle\ln{\cal Z}(\beta)$ $\displaystyle=$ $\displaystyle\ln{\cal
Z}_{0}+\left(\ln{\cal Z}_{I}\right)_{1}$ (243) $\displaystyle=$
$\displaystyle-V\int\frac{d^{3}{k}}{(2\pi)^{3}}\,\ln\left(1-e^{-\beta\omega_{k}}\right)-\
\lambda\,\frac{VT^{3}}{48}+{\cal O}(\lambda^{2}),$

where we have also dropped the $T=0$ contribution in $\ln{\cal Z}_{0}$. In the
infinite volume limit the pressure up to first order can be obtained as

$\displaystyle{\cal P}=-\frac{\Omega}{V}$ $\displaystyle=$
$\displaystyle-T\,\int\frac{d^{3}{k}}{(2\pi)^{3}}\,\ln\left(1-e^{-\beta\omega_{k}}\right)-\
\lambda\,\frac{T^{4}}{48}+{\cal O}(\lambda^{2})$ (244) $\displaystyle=$
$\displaystyle\frac{\pi^{2}T^{4}}{90}-\ \lambda\,\frac{T^{4}}{48}+{\cal
O}(\lambda^{2}),$

where $\Omega$ is the thermodynamic potential. The other thermodynamic
quantities can be obtained from pressure. One can also compute the higher
order corrections to the partition function following (200) and (201), and
thus higher order thermodynamic quantities.

###  6.3 Fermion Field

Until now we have discussed partition function for real scalar field and it’s
pressure up to first order in coupling. In this section we will compute the
partition function for free fermionic fields. The computation of interacting
fermionic partition function is postponed until we introduce gauge theory,
quantum electrodynamics (QED). Since fermions are anticommuting fields, they
are Grassmann variables. We start this section by reviewing some of the
properties of Fermionic fields.

####  6.3.1 Fermionic Lagrangian and conserved charge

The Lagrangian density that describes the non-interacting fermion is given in
Minkowski space-time pbpal ; Peskin  as

$\displaystyle{\cal L}$ $\displaystyle=$
$\displaystyle{\bar{\psi}}\left(i\partial\\!\\!\\!/-m\right)\psi,$ (245)

where $m$ is the mass of the fermion and the fields $\psi$ and $\bar{\psi}$
are to be treated independently. The $\gamma$-matrices in Dirac-Pauli
representation are given as

$\displaystyle\gamma^{0}=\left(\begin{array}[]{ll}I&0\\\
0&-I\end{array}\right)\hskip 21.68121pt{\mbox{and}}\hskip
21.68121pt\gamma^{i}=\left(\begin{array}[]{ll}0&\sigma^{i}\\\
-\sigma^{i}&0\end{array}\right)\,,$ (250)

where $I$ is a $2\times 2$ unit matrix and the Pauli matrices $\sigma^{i}$’s
are

$\displaystyle\sigma^{1}=\left(\begin{array}[]{ll}0&1\\\
1&0\end{array}\right)\hskip
14.45377pt\sigma^{2}=\left(\begin{array}[]{ll}0&-i\\\
i&0\end{array}\right)\hskip 14.45377pt{\mbox{and}}\hskip
14.45377pt\sigma^{3}=\left(\begin{array}[]{ll}1&0\\\
0&-1\end{array}\right)\,.$ (257)

Using the Euler-Lagrange equation for $\bar{\psi}$ field

$\displaystyle\frac{\partial{\cal L}}{\partial\bar{\psi}}$ $\displaystyle=$
$\displaystyle{\partial_{\mu}}\left(\frac{\partial{\cal
L}}{\partial(\partial^{\mu}\bar{\psi})}\right)\Rightarrow\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi=\partial_{\mu}[0]=0,$
(258)

one gets the Dirac equation for $\psi$ field. The Hamiltonian density is given
as

$\displaystyle{\cal H}_{d}$ $\displaystyle=$
$\displaystyle\pi\partial_{0}\psi+\partial_{0}\bar{\psi}\bar{\pi}-{\cal L}.$
(259)

The conjugate momenta can be obtained as

$\displaystyle\pi$ $\displaystyle=$ $\displaystyle\frac{\partial{\cal
L}}{\partial(\partial_{0}\psi)}=i\psi^{\dagger},$ (260)
$\displaystyle{\bar{\pi}}$ $\displaystyle=$ $\displaystyle\frac{\partial{\cal
L}}{\partial(\partial_{0}{\bar{\psi}})}=0.$ (261)

Using (260) and (261) in (259), the Hamiltonian density becomes

$\displaystyle{\cal H}_{d}$ $\displaystyle=$ $\displaystyle
i\psi^{\dagger}\partial_{0}\psi-{\bar{\psi}}\left(i\partial\\!\\!\\!/-m\right)\psi=i{\bar{\psi}}\gamma^{0}\partial_{0}\psi-i{\bar{\psi}}\gamma^{0}\partial_{0}\psi+i{\bar{\psi}}\gamma^{i}\partial_{i}\psi+{\bar{\psi}}m\psi$
(262) $\displaystyle=$ $\displaystyle
i\bar{\psi}\left(\gamma^{i}\partial_{i}+m\right)\psi.$

This is the Hamiltonian density for canonical ensemble. However, allowing a
local transformation, i.e., $\alpha$ depends on $X$, one gets

$\displaystyle{\cal L}\rightarrow{\cal L}^{\prime}$ $\displaystyle=$
$\displaystyle\bar{\psi}e^{i\alpha(X)}\left(i\partial\\!\\!\\!/-m\right)\psi
e^{-i\alpha(X)}=\bar{\psi}\left(i\partial\\!\\!\\!/-m\right)\psi\,+\,\bar{\psi}\partial\\!\\!\\!/\alpha(X)\psi$
(263) $\displaystyle=$ $\displaystyle{\cal
L}\,+\,\bar{\psi}\partial\\!\\!\\!/\alpha(X)\psi.$

As seen if $\alpha(X)=\alpha$, is a constant, then ${\cal L}^{\prime}={\cal
L}$, the Lagrangian density is invariant under global symmetry. This symmetry
will lead to a conserved current according to Noether’s theorem. By solving
the equation of motion for $\alpha$, one gets

$\displaystyle{\partial_{\mu}}\left(\frac{\partial{\cal
L}^{\prime}}{\partial(\partial^{\mu}\alpha)}\right)$ $\displaystyle=$
$\displaystyle\frac{\partial{\cal L}^{\prime}}{\partial\alpha}$
$\displaystyle\partial_{\mu}\left[\bar{\psi}\gamma^{\mu}\psi\right]$
$\displaystyle=$ $\displaystyle\partial_{\mu}j^{\mu}=0,$ (264)

where the conserved current is found as

$\displaystyle j^{\mu}$ $\displaystyle=$
$\displaystyle\bar{\psi}\gamma^{\mu}\psi\,.$ (265)

Now the temporal component
$j^{0}=\bar{\psi}\gamma^{0}\psi=\bar{\psi}^{\dagger}\psi=\rho$ is associated
with conserved number density. The conserved number can be obtained as

$\displaystyle N$ $\displaystyle=$ $\displaystyle\int d^{3}x\,j^{0}=\int
d^{3}x\,\,\bar{\psi}^{\dagger}\psi=\int d^{3}x\,\,\rho\,.$ (266)

Now, the new Hamiltonian density in presence of chemical potential associated
with a conserved number becomes

$\displaystyle{\cal H}_{d}-\mu\rho$ $\displaystyle=$ $\displaystyle
i\bar{\psi}\left(\gamma^{i}\partial_{i}+m\right)\psi-\mu\rho=\bar{\psi}\left(i\gamma^{i}\partial_{i}+m-\mu\gamma^{0}\right)\psi.$
(267)

The corresponding new Lagrangian density becomes

$\displaystyle{\cal L}$ $\displaystyle=$
$\displaystyle\bar{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m+\mu\gamma^{0}\right)\psi,$
(268)

which indicates that the presence of the chemical potential is like changing
the zeroth component of the gauge field (external field), through the
substitution $\partial_{0}-i\mu$ in the Lagrangian.

####  6.3.2 Partition function and pressure for free fermions

The partition function for free fermionic field reads from (206) as

$\displaystyle{\cal Z}_{0}$ $\displaystyle=$
$\displaystyle\,\int\limits_{\psi(\bm{\vec{x}},0)=-\psi(\bm{\vec{x}},\beta)}{\cal
D}[{\bar{\psi}}]\ {\cal D}[{\psi}]\
e^{\int\limits_{0}^{\beta}\,d\tau\int\,d^{3}{x}\,{\cal
L}({t\rightarrow-i\tau})}.$ (269)

As before, the Fourier transform of the fermionic fields $\psi(X)$ and
$\bar{\psi}(X)$ can be written as

$\displaystyle\psi(X)=\psi(\bm{\vec{x}},\tau)$
$\displaystyle{=\atop{t\rightarrow-i\tau}}\atop{k_{0}\rightarrow i\omega_{n}}$
$\displaystyle\frac{1}{\sqrt{V\beta}}\sum_{K}\,e^{-iK\cdot
X}\,\psi(K)=\frac{1}{\sqrt{V\beta}}\sum_{n,\bm{\vec{k}}}\,e^{i\bm{\vec{k}\cdot\vec{x}}}\,\,e^{-i\omega_{n}\tau}\,\psi(\omega_{n},\bm{\vec{k}})\,,$
$\displaystyle\bar{\psi}(X)=\bar{\psi}(\bm{\vec{x}},\tau)$
$\displaystyle{=\atop{t\rightarrow-i\tau}}\atop{k_{0}\rightarrow i\omega_{n}}$
$\displaystyle\frac{1}{\sqrt{V\beta}}\sum_{K}\,e^{iK\cdot
X}\,\bar{\psi}(K)=\frac{1}{\sqrt{V\beta}}\sum_{n,\bm{\vec{k}}}\,e^{-i\bm{\vec{k}\cdot\vec{x}}}\,\,e^{i\omega_{n}\tau}\,\bar{\psi}(\omega_{n},\bm{\vec{k}})\,.$
(270)

where $V$ is the three volume. The Lagrangian density in (268) can now be
written in Euclidean time as

$\displaystyle{\cal L}({t\rightarrow-i\tau})$
$\displaystyle{=\atop{t\rightarrow-i\tau}}$
$\displaystyle\bar{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m+\mu\gamma^{0}\right)\psi{=\atop{t\rightarrow-i\tau}}\bar{\psi}\left(i\gamma^{0}\partial_{0}-i\gamma^{i}\partial_{i}-m+\mu\gamma^{0}\right)\psi$
(271) $\displaystyle=$
$\displaystyle-\bar{\psi}\left(\gamma^{0}\partial_{\tau}+i\gamma^{i}\partial_{i}+m-\mu\gamma^{0}\right)\psi\,.$

Using the Fourier transformed of fermionic fields in (270) we compute

$\displaystyle\int\limits_{0}^{\beta}\,d\tau\int\,d^{3}{x}\,{\cal
L}({t\rightarrow-i\tau})$ $\displaystyle=$
$\displaystyle-\frac{1}{V\beta}\int\limits_{0}^{\beta}\,d\tau\int\,d^{3}{x}\,\sum_{n,\bm{\vec{k}}}\sum_{m,\bm{\vec{k}^{\prime}}}e^{-i\bm{\vec{k}\cdot\vec{x}}}\,\,e^{i\omega_{n}\tau}\,\bar{\psi}(\omega_{n},\bm{\vec{k}})$
(272) $\displaystyle\hskip
14.45377pt\left[\gamma^{0}\partial_{\tau}+i\gamma^{i}\partial_{i}+m-\mu\gamma^{0}\right]e^{i\bm{\vec{k}^{\prime}\cdot\vec{x}}}\,\,e^{-i\omega_{m}\tau}\,\psi(\omega_{m},\bm{\vec{k}^{\prime}})$
$\displaystyle=$
$\displaystyle-\frac{1}{V\beta}\int\limits_{0}^{\beta}\,d\tau\int\,d^{3}{x}\,\sum_{n,\bm{\vec{k}}}\sum_{m,\bm{\vec{k}^{\prime}}}e^{-i\bm{\vec{k}\cdot\vec{x}}}\,\,e^{i\omega_{n}\tau}\,\bar{\psi}(\omega_{n},\bm{\vec{k}})$
$\displaystyle\hskip
14.45377pt\left[\gamma^{0}(-i\omega_{m})+i\gamma^{i}(ik^{\prime}_{i})+m-\mu\gamma^{0}\right]e^{i\bm{\vec{k}^{\prime}\cdot\vec{x}}}\,\,e^{-i\omega_{m}\tau}\,\psi(\omega_{m},\bm{\vec{k}^{\prime}})$
$\displaystyle=$
$\displaystyle-\frac{1}{V\beta}\sum_{n,\bm{\vec{k}}}\sum_{m,\bm{\vec{k}^{\prime}}}\bar{\psi}(\omega_{n},\bm{\vec{k}})\left[\gamma^{0}(-i\omega_{m})+i\gamma^{i}(ik^{\prime}_{i})+m-\mu\gamma^{0}\right]\psi(\omega_{m},\bm{\vec{k}^{\prime}})$
$\displaystyle\times\beta\delta(\omega_{n}-\omega_{m})\,V\delta^{3}(\bm{\vec{k}^{\prime}}-\bm{\vec{k}})$
$\displaystyle=$
$\displaystyle-\sum_{n,\bm{\vec{k}}}\bar{\psi}(\omega_{n},\bm{\vec{k}})\left[-\gamma^{0}(i\omega_{n}+\mu)-\gamma^{i}k_{i}+m\right]\psi(\omega_{n},\bm{\vec{k}})$
$\displaystyle=$
$\displaystyle-\sum_{n,\bm{\vec{k}}}\bar{\psi}(\omega_{n},\bm{\vec{k}})\left[S_{0}^{-1}((i\omega_{n}+\mu),\bm{\vec{k}})\right]\psi(\omega_{n},\bm{\vec{k}}),$

where $S_{0}^{-1}(i\omega_{n}+\mu,\bm{\vec{k}})$ is the inverse of free
fermionic propagator in presence of chemical potential $\mu$. The partition
function in (269) becomes

$\displaystyle{\cal Z}_{0}$ $\displaystyle=$
$\displaystyle\prod_{n,\bm{\vec{k}}}\int\limits_{\tiny\mbox{\sf
antiperiodic}}{\cal D}[{\bar{\psi}}]\ {\cal D}[{\psi}]\
e^{-\bar{\psi}(\omega_{n},\bm{\vec{k}})\left[S_{0}^{-1}((i\omega_{n}+\mu),\bm{\vec{k}})\right]\psi(\omega_{n},\bm{\vec{k}})}$
(273) $\displaystyle=$
$\displaystyle\prod_{n,{k}}\mbox{det}\left[S_{0}^{-1}((i\omega_{n}+\mu),\bm{\vec{k}})\right],$

where we have used the result of the functional integral in (81) involving
Grassman variables. The logarithm of ${\cal Z}_{0}$ becomes

$\displaystyle\ln{\cal Z}_{0}$ $\displaystyle=$
$\displaystyle\sum_{n,k}\ln\,\mbox{det}\left[S_{0}^{-1}((i\omega_{n}+\mu),\bm{\vec{k}})\right]$
(274) $\displaystyle=$
$\displaystyle\sum_{n,k}\ln\mbox{det}\left[-\gamma^{0}(i\omega_{n}+\mu)-\gamma^{i}k_{i}+m\right]=\sum_{n,k}\ln\mbox{det}[Y].$

Now one can write

$\displaystyle\mbox{det}[Y]$ $\displaystyle=$
$\displaystyle\mbox{det}\left\\{\left(\begin{array}[]{ll}-(i\omega_{n}+\mu)&\hskip
28.90755pt0\\\
0&(i\omega_{n}+\mu)\end{array}\right)+\left(\begin{array}[]{ll}0&-\bm{\vec{\sigma}\cdot\vec{k}}\\\
\bm{\vec{\sigma}\cdot\vec{k}}&\hskip
14.45377pt0\end{array}\right)+\left(\begin{array}[]{ll}m&0\\\
0&m\end{array}\right)\right\\}$ $\displaystyle=$
$\displaystyle\mbox{det}\left\\{\left(\begin{array}[]{ll}-(i\omega_{n}+\mu)+m&\hskip
21.68121pt-\bm{\vec{\sigma}\cdot\vec{k}}\\\
\bm{\vec{\sigma}\cdot\vec{k}}&(i\omega_{n}+\mu)+m\end{array}\right)\right\\}\,,$
(283)

where each element is a $2\times 2$ matrix. Using the identity
$(\bm{\vec{\sigma}\cdot\vec{k}})^{2}=k^{2}$, one can compute the determinant
as

$\displaystyle\mbox{det}[Y]$ $\displaystyle=$
$\displaystyle\left[k^{2}+m^{2}-(i\omega_{n}+\mu)^{2}\right]^{2}.$ (284)

Combining (284) and (274), one can write

$\displaystyle\ln{\cal Z}_{0}$ $\displaystyle=$ $\displaystyle
2\,\sum_{n,{k}}\ln\left[k^{2}+m^{2}-(i\omega_{n}+\mu)^{2}\right]$ (285)
$\displaystyle=$ $\displaystyle
2V\sum_{n}\int\frac{d^{3}k}{(2\pi)^{3}}\left[\ln((\omega_{k}-\mu)-i\omega_{n})+\ln((\omega_{k}+\mu)+i\omega_{n})\right]\
,$

where $\omega_{k}=\sqrt{k^{2}+m^{2}}$ and $\sum_{k}$ is replaced by
$V\int\frac{d^{3}k}{(2\pi)^{3}}$.

Now we will perform the frequency sum in (285):

  1. First term: using (144), one can write

$\displaystyle\sum_{n}\ln[(\omega_{k}-\mu)-i\omega_{n}]$ $\displaystyle=$
$\displaystyle T\sum_{n}\frac{1}{T}\ln[(\omega_{k}-\mu)-k_{0}]$
$\displaystyle=$ $\displaystyle\frac{1}{2\pi iT}\oint
dk_{0}\ln[(\omega_{k}-\mu)-k_{0}]\,\frac{1}{2}\,\tanh\left(\frac{\beta
k_{0}}{2}\right)$ $\displaystyle=$ $\displaystyle-\frac{1}{4\pi iT}\oint
dk_{0}\left(\frac{-1}{(\omega_{k}-\mu)-k_{0}}\right)\int
dk_{0}\tanh\left(\frac{\beta k_{0}}{2}\right)$ $\displaystyle=$
$\displaystyle\frac{1}{4\pi iT}\oint
dk_{0}\left(\frac{1}{(\omega_{k}-\mu)-k_{0}}\right)\int
dk_{0}\tanh\left(\frac{\beta k_{0}}{2}\right)$ $\displaystyle=$
$\displaystyle\frac{1}{4\pi iT}\oint
dk_{0}\left(\frac{1}{(\omega_{k}-\mu)-k_{0}}\right)\frac{2}{\beta}\ln\left[\cosh\left(\frac{\beta
k_{0}}{2}\right)\right]$ (286) $\displaystyle=$ $\displaystyle\frac{1}{2\pi
i}\oint
dk_{0}\left(\frac{1}{(\omega_{k}-\mu)-k_{0}}\right)\ln\left[\cosh\left(\frac{\beta
k_{0}}{2}\right)\right]$ $\displaystyle=$
$\displaystyle\ln\left[\cosh\left(\frac{\beta(\omega_{k}-\mu)}{2}\right)\right]$
$\displaystyle=$
$\displaystyle\ln\left[\frac{1}{2}\left(e^{{\beta(\omega_{k}-\mu)}/{2}}+e^{{-\beta(\omega_{k}-\mu)}/{2}}\right)\right]$
$\displaystyle=$
$\displaystyle\ln\left[\frac{1}{2}e^{{\beta(\omega_{k}-\mu)}/{2}}\left(1-e^{-\beta(\omega_{k}-\mu)}\right)\right]$
$\displaystyle=$ $\displaystyle\left[\frac{\beta(\omega_{k}-\mu)}{2}-\ln
2+\ln\left(1-e^{-\beta(\omega_{k}-\mu)}\right)\right]$ $\displaystyle=$
$\displaystyle\left[\frac{\beta(\omega_{k}-\mu)}{2}+\ln\left(1-e^{-\beta(\omega_{k}-\mu)}\right)\right],$

where again $\ln 2$ is neglected as it is independent of temperature.

  2. Second term:

$\displaystyle\sum_{n}\ln[(\omega_{k}+\mu)+i\omega_{n}]$ $\displaystyle=$
$\displaystyle
T\sum_{n}\frac{1}{T}\ln[(\omega_{k}-\mu)-k_{0}]=\ln\left[\cosh\left(\frac{\beta(\omega_{k}+\mu)}{2}\right)\right]$
(287) $\displaystyle=$ $\displaystyle\left[\frac{\beta(\omega_{k}+\mu)}{2}-\ln
2+\ln\left(1-e^{-\beta(\omega_{k}+\mu)}\right)\right]$ $\displaystyle=$
$\displaystyle\left[\frac{\beta(\omega_{k}+\mu)}{2}+\ln\left(1-e^{-\beta(\omega_{k}+\mu)}\right)\right].$

Now using (286) and (287) in (285), one gets logarithm of the partition
function

$\displaystyle\ln{\cal Z}_{0}\,$ $\displaystyle=$ $\displaystyle
2V\int\frac{d^{3}{k}}{(2\pi)^{3}}\,\left[{\beta\omega_{k}}+\ln\left(1+e^{-\beta(\omega_{k}-\mu)}\right)+\ln\left(1+e^{+\beta(\omega_{k}+\mu)}\right)\right],$
(288)

which agrees with that obtained in quantum statistical mechanics in (24) and
(26).

The free energy density for free fermionic field can be obtained as

$\displaystyle F_{0}$ $\displaystyle=$ $\displaystyle-\frac{T\ln{\cal
Z}_{0}}{V}\,=-2T\,\int\frac{d^{3}{k}}{(2\pi)^{3}}\,\left[{\beta\omega_{k}}+\ln\left(1+e^{-\beta(\omega_{k}-\mu)}\right)+\ln\left(1+e^{+\beta(\omega_{k}+\mu)}\right)\right].$
(289)

In the infinite volume limit the pressure for fermionic field can be obtained
as

$\displaystyle{\cal P}_{0}$ $\displaystyle=$ $\displaystyle-
F_{0}=2T\,\int\frac{d^{3}{k}}{(2\pi)^{3}}\,\left[{\beta\omega_{k}}+\ln\left(1+e^{-\beta(\omega_{k}-\mu)}\right)+\ln\left(1+e^{+\beta(\omega_{k}+\mu)}\right)\right],$
(290)

where the first term is the zero temperature part which should be dropped as
it only shifts the vacuum energy. After performing the integration of
temperature dependent part, one obtains pressure for free massless fermion

$\displaystyle{\cal P}_{0}$ $\displaystyle=$
$\displaystyle\frac{7\pi^{2}T^{4}}{180}+\frac{\mu^{2}T^{2}}{6}+\frac{\mu^{4}}{12\pi^{2}}\,.$
(291)

####  6.3.3 A reverse way: first number density and then pressure and entropy
density for fermions

The partition function in Minkowski space time can be written

$\displaystyle{\cal Z}(\beta,[\mu])=\int{\cal D}[\bar{\psi}]{\cal
D}[{\psi}]e^{i\int d^{4}X\mathcal{L}(\psi,{\bar{\psi}};[\mu])}.$ (292)

The pressure can be written as

${\cal P}(\beta;[\mu])=\frac{1}{\cal V}\ln{\cal
Z}(T;[\mu])=\frac{T}{V}\ln{\cal Z}(T;[\mu])\ ,$ (293)

where the four-volume, ${\cal V}=\beta V$ with $V$ is the three-volume.

Figure 22: The Feynman diagram for number density originating from $\mu$
derivative of the pressure diagram that brings a $\gamma_{0}$.

Now the number density $\rho$ can be obtained HM ; HMT  as

$\displaystyle\\!\\!\\!\\!\\!\rho$ $\displaystyle\equiv$
$\displaystyle\frac{\partial{\cal P}(\beta;[\mu])}{\partial\mu}=\frac{i}{{\cal
V}{\cal Z}[\beta;j]}\ {\int{\cal D}[\bar{\psi}]{\cal D}[\psi]\int d^{4}x\
{\bar{\psi}(x)}\gamma_{0}[\mu]\psi(x)}\,e^{\left({i\int d^{4}x{\cal
L}(\psi,{\bar{\psi}};[\mu])}\right)}\ .$ (294)

The full fermionic propagator in presence of uniform $\mu$ can be written as

$\displaystyle i{\cal S}_{\alpha\sigma}[\mu](x,x^{\prime})$ $\displaystyle=$
$\displaystyle\frac{\int{\cal D}[\bar{\psi}]{\cal
D}[\psi]\psi_{\alpha}(x){\bar{\psi}_{\sigma}(x^{\prime})}\exp\left({i\int
d^{4}x{\cal L}(\psi,{\bar{\psi}};[\mu])}\right)}{\int{\cal D}[\bar{\psi}]{\cal
D}[\psi]\exp\left({i\int d^{4}x{\cal L}(\psi,{\bar{\psi}};[\mu])}\right)}\ .$
(295)

Now using (295) and performing the traces over Dirac and coordinate indices in
(294) one can write

$\displaystyle\rho$ $\displaystyle=$
$\displaystyle\,\int\\!\frac{d^{4}K}{(2\pi)^{4}}\mbox{tr}\left[iS[\mu](K)\
(-i)\gamma_{0}[\mu](K,-K;0)\right]$ (296) $\displaystyle=$
$\displaystyle\,\int\\!\frac{d^{4}K}{(2\pi)^{4}}\mbox{tr}\left[S[\mu](K)\
\gamma_{0}[\mu](K,-K;0)\right]\,,$

where ’tr’ indicates the trace over the Dirac indices.

Now if we consider free fermions, then the full propagator $S[\mu]$ can be
replaced by the free fermion propagator $S_{0}[\mu]$ and the number density in
(296) can be written as

$\displaystyle\rho_{0}$ $\displaystyle=$ $\displaystyle\
\,\int\\!\frac{d^{4}K}{(2\pi)^{4}}\mbox{tr}\left[S_{0}[\mu](K)\
\gamma_{0}[\mu](K,-K;0)\right],$ (297)

The expression for number density in (297) corresponds to Feynman diagram in
Fig. 22. The free fermionic propagator for momentum $K$ in helicity
representation is given in (327) as

$S_{0}(K)=\frac{\gamma_{0}-\vec{\gamma}\cdot\hat{k}}{2d_{+}(k_{0},k)}\
+\frac{\gamma_{0}+\vec{\gamma}\cdot\hat{k}}{2d_{-}(k_{0},k)},$ (298)
$\mbox{with}\,\,\,d_{\pm}=k_{0}\mp k\ \ .$ (299)

Using (298) in (297) and performing the trace over Dirac matrices, we get

$\displaystyle\rho_{0}(T,\mu)=2\int\frac{d^{3}k}{(2\pi)^{3}}\,\frac{1}{\beta}\sum_{k_{0}=(2n+1)\pi
iT+\mu}\left[\frac{1}{k_{0}-k}+\frac{1}{k_{0}+k}\right].$ (300)

We note here that the chemical potential is not considered in the propagator
but considered in the discrete frequency as $k_{0}=i\omega_{n}+\mu$. This
shifts the pole of $\tanh$ by an amount $\mu$ in (144). This will lead to same
result as will see below.

For computing the frequency sum in (300), we use the standard technique of
contour integration as given in (146) in presence of $\mu$ as

$\frac{1}{2\pi
i}\oint\limits_{C}\left[\frac{1}{k_{0}-k}+\frac{1}{k_{0}+k}\right]\frac{\beta}{2}\mbox{tanh}\left(\frac{\beta(k_{0}-\mu)}{2}\right)dk_{0}=\frac{\beta}{2}\
\frac{1}{2\pi i}\times(-2\pi i)\sum{\mbox{Residues}}\ .$ (301)

It is noted that the first term of (301) has a simple pole at $k_{0}=k$. On
the other hand the second term also has a simple pole at $k_{0}=-k$. After
calculating the residues, one obtains the number density HM ; HMT

$\displaystyle\rho_{0}(T,\mu)$ $\displaystyle=$
$\displaystyle-\int\frac{d^{3}k}{(2\pi)^{3}}\left[\tanh\frac{\beta(k-\mu)}{2}-\tanh\frac{\beta(k+\mu)}{2}\right]$
(302) $\displaystyle=$ $\displaystyle
2\int\frac{d^{3}k}{(2\pi)^{3}}\left[n(k-\mu)-n(k+\mu)\right],$

where $n(x)=1/(e^{\beta x}+1)$, is the Fermi-Dirac distribution function.

Now, by integrating the first line of (302) w.r.t. $\mu$, one obtains the
pressure for non-interacting fermion gas as

$\displaystyle{\mathcal{P}}_{0}(T,\mu)=2T\int\frac{d^{3}k}{(2\pi)^{3}}\left[\beta
k+\ln\left(1+e^{-\beta(k-\mu)}\right)+\ln\left(1+e^{-\beta(k+\mu)}\right)\right],$
(303)

where the first term is the zero-point energy that produces usual vacuum
divergence. It also agrees with that obtained in (290). The entropy density
for non-interacting can be obtained from pressure as

$\displaystyle{\cal S}_{0}(T,\mu)$ $\displaystyle=$
$\displaystyle\frac{\partial{\cal P}_{0}}{\partial
T}=2\int\frac{d^{3}k}{(2\pi)^{3}}\Big{[}\ln\left(1+e^{-\beta(k-\mu)}\right)+\ln\left(1+e^{-\beta(k+\mu)}\right)$
(304)
$\displaystyle\left.+\frac{\beta(k-\mu)}{e^{\beta(k-\mu)}+1}+\frac{\beta(k+\mu)}{e^{\beta(k+\mu)}+1}\right].$

##  7 General Structure of Fermionic Two-point Functions at $T\neq 0$

###  7.1 Fermion Self-Energy

A theory possessing only fermions and gauge bosons with no bare masses for
fermions is chirally invariant for all orders. It is to be noted that the
theory is also parity invariant. At $T=0$, chiral invariance has two
implications: i) there are no ${\bar{\psi}}\psi$ coupling in any finite order
of perturbation theory, ii) the general form of the fermion self-energy can be
written as HAW

$\Sigma(P)=-{\cal A}P\\!\\!\\!\\!/,$ (305)

for particle momentum $P\equiv(p_{0}=\omega,\bm{\vec{p}})$, $p=|\bm{\vec{p}}|$
and ${\cal A}$ is Lorentz invariant structure function which is function of
$P^{2}$.

The effective fermion propagator can be written from (316) as

$S^{\star}(P)=\frac{1}{P\\!\\!\\!\\!/-\Sigma(P)}=\frac{P\\!\\!\\!\\!/}{(1+{\cal
A})P^{2}}.$ (306)

The poles, $P^{2}=0$ are on the light cone, $\omega=p$ and $(1+{\cal A})$
modifies the residues.

At $T\neq 0$, the above item i) still holds whereas item ii) does not. At
$T\neq 0$, the system will not be in a vacuum because at such high temperature
there will be antiparticles present in equal numbers as the particles. This
constitutes a heat bath which introduces a special Lorentz frame. So, the heat
bath has four velocity $u^{\mu}=(1,0,0,0)$ with $u^{\mu}u_{\mu}=1$. The
presence of four velocity means that the most general ansatz for fermion self-
energy HAW  will be of the form

$\displaystyle\Sigma(P)$ $\displaystyle=$ $\displaystyle-{\cal
A}P\\!\\!\\!\\!/-{\cal B}u\\!\\!\\!/\,,$ (307)

where ${\cal B}$ is another Lorentz invariant structure function in addition
to ${\cal A}$. Since $P^{2}=\omega^{2}-p^{2}$, one can interpret
$\omega=p_{0}=P^{\mu}u_{\mu}$ and $p=\left(P^{\mu}u_{\mu}-P^{2}\right)^{1/2}$
as Lorentz invariant energy and momentum, respectively. The Lorentz invariant
structure functions are obtained as follows:

We now write from (307) as

$\displaystyle\Sigma P\\!\\!\\!\\!/$ $\displaystyle=-{\cal A}P^{2}-{\cal
B}P\\!\\!\\!\\!/u\\!\\!\\!/,$ (308a) $\displaystyle\Sigma u\\!\\!\\!/$
$\displaystyle=-{\cal A}P\\!\\!\\!\\!/u\\!\\!\\!/-{\cal B}.$ (308b)

Taking trace of (308a) and (308b), we get

$\displaystyle{\rm{Tr}}\left[\Sigma P\\!\\!\\!\\!/\,\right]$
$\displaystyle=-4{\cal A}P^{2}-4{\cal B}\left(P\cdot u\right),$ (309a)
$\displaystyle\left(P\cdot u\right){\rm{Tr}}\left[\Sigma u\\!\\!\\!/\right]$
$\displaystyle=-4{\cal A}\left(P\cdot u\right)^{2}-4{\cal B}\left(P\cdot
u\right).$ (309b)

Solving (309a) and (309b), one obtains

${\cal A}(\omega,p)=\frac{1}{4}\frac{{\rm{Tr}}\left[\Sigma
P\\!\\!\\!\\!/\,\right]-\left(P\cdot u\right){\rm{Tr}}\left[\Sigma
u\\!\\!\\!/\right]}{\left(P\cdot u\right)^{2}-P^{2}}.$ (310)

Further one can write

$\displaystyle P^{2}{\rm{Tr}}\left[\Sigma u\\!\\!\\!/\right]$
$\displaystyle=-4{\cal A}P^{2}\left(P\cdot u\right)-4{\cal B}P^{2},$ (311a)
$\displaystyle\left(P\cdot u\right){\rm{Tr}}\left[\Sigma
P\\!\\!\\!\\!/\,\right]$ $\displaystyle=-4{\cal A}P^{2}\left(P\cdot
u\right)-4{\cal B}\left(P\cdot u\right)^{2}.$ (311b)

Solving (311a) and (311b), one obtains

${\cal B}(\omega,p)=\frac{1}{4}\frac{P^{2}{\rm{Tr}}\left[\Sigma
u\\!\\!\\!/\right]-\left(P\cdot u\right){\rm{Tr}}\left[\Sigma
P\\!\\!\\!\\!/\,\right]}{\left(P\cdot u\right)^{2}-P^{2}}.$ (312)

Now in the rest frame of the heat bath, $u^{\mu}=(1,0,0,0)$, the most general
ansatz for fermionic self-energy reads HAW  as

$\displaystyle\Sigma(P)$ $\displaystyle=$ $\displaystyle-{\cal
A}(\omega,p)P\\!\\!\\!\\!/-{\cal B}(\omega,p)\gamma_{0},$ (313)

with structure functions

$\displaystyle{\cal A}(\omega,p)$
$\displaystyle=\frac{1}{4p^{2}}\left({\rm{Tr}}\left[\Sigma
P\\!\\!\\!\\!/\,\right]-\omega{\rm{Tr}}\left[\Sigma\gamma_{0}\right]\right),$
(314a) $\displaystyle{\cal B}(\omega,p)$
$\displaystyle=\frac{1}{4p^{2}}\left(P^{2}{\rm{Tr}}\left[\Sigma\gamma_{0}\right]-\omega{\rm{Tr}}\left[\Sigma
P\\!\\!\\!\\!/\,\right]\right).$ (314b)

###  7.2 Fermion Propagator

Figure 23: Pictorial representation of Dyson-Schwinger equation for effective
fermion propagator.

In Fig.23 we represent the full propagator by $S^{*}(P)$ and the bare
propagator by $S_{0}(P)$ and self-energy as $\Sigma(P)$, then the full
propagator is written as,

$\displaystyle S^{*}(P)$ $\displaystyle=$ $\displaystyle
S_{0}(P)+S_{0}(P)\Sigma(P)S^{*}(P)$ $\displaystyle S^{*}(P)S^{*-1}(P)$
$\displaystyle=$ $\displaystyle
S_{0}(P)S^{*-1}(P)+S_{0}(P)\Sigma(P)S^{*}(P)S^{*-1}(P)$ $\displaystyle 1$
$\displaystyle=$ $\displaystyle S_{0}(P)S^{*-1}(P)+S_{0}(P)\Sigma(P)$
$\displaystyle S_{0}^{-1}(P)$ $\displaystyle=$ $\displaystyle
S_{0}^{-1}(P)S_{0}(P)S^{*-1}(P)+S_{0}^{-1}(P)S_{0}(P)\Sigma(P)$ $\displaystyle
S^{*-1}(P)$ $\displaystyle=$ $\displaystyle S_{0}^{-1}(P)-\Sigma(P)$
$\displaystyle S^{*-1}(P)$ $\displaystyle=$
$\displaystyle\not{P}-\Sigma(P)\,,$ (315)

which is known as fermionic Dyson-Schwinger equation. The effective fermion
propagator can be obtained as

$S^{\star}(P)=\frac{1}{P\\!\\!\\!\\!/-\Sigma(P)}.$ (316)

Using (307) one can write the effective propagator as

$S^{\star}(P)=\frac{1}{(1+{\cal A})P\\!\\!\\!\\!/+{\cal
B}u\\!\\!\\!/}=\frac{(1+{\cal A})P\\!\\!\\!\\!/+{\cal B}u\\!\\!\\!/}{[(1+{\cal
A})P\\!\\!\\!\\!/+{\cal
B}u\\!\\!\\!/]^{2}}=\frac{P\\!\\!\\!\\!/-\Sigma(P)}{\cal
D}=\frac{S^{\star-1}(P)}{\cal D},$ (317)

where the Lorentz invariant quantity ${\cal D}$ is given as

$\displaystyle{\cal D}{(p,u)}$ $\displaystyle=$ $\displaystyle\left[(1+{\cal
A})P\\!\\!\\!\\!/+{\cal B}u\\!\\!\\!/\right]^{2}=(1+{\cal
A})^{2}{P\\!\\!\\!\\!/}^{2}+2(1+{\cal A}){\cal B}P\cdot u+{\cal B}^{2}.$ (318)

In the rest frame of heat bath, Eq.(318) reads as

$\displaystyle{\cal D}{(p,\omega)}$ $\displaystyle=$ $\displaystyle(1+{\cal
A})^{2}(\omega^{2}-p^{2})+2(1+{\cal A}){\cal B}\omega+{\cal
B}^{2}=\left[(1+{\cal A})\omega+{\cal B}\right]^{2}-(1+{\cal A})^{2}p^{2}$
(319) $\displaystyle=$ $\displaystyle\left[(1+{\cal A})(\omega-p)+{\cal
B}\right]\left[(1+{\cal A})(\omega+p)+{\cal B}\right]={\cal D}_{+}{\cal
D}_{-},$

where

${\cal D}_{\pm}(p,\omega)=(1+{\cal A})(\omega\mp p)+{\cal B}.$ (320)

In free case ${\cal A}={\cal B}=0$ and Eq.(320) becomes

$d_{\pm}(p,\omega)=\omega\mp p.$ (321)

Combining (319) and (317), one can write the effective propagator as

$S^{\star}(P)=\frac{S^{\star-1}(P)}{{\cal D}_{+}{\cal D}_{-}}.$ (322)

We can write the self energy in (313) as

$\displaystyle\Sigma(P)$ $\displaystyle=$ $\displaystyle-{\cal
A}(\omega,p)P\\!\\!\\!\\!/-{\cal B}(\omega,p)\gamma_{0}$ (323)
$\displaystyle=$ $\displaystyle-({\cal A}\omega+{\cal B})\gamma_{0}+{\cal A}p\
{\vec{\gamma}}\cdot\bm{\hat{p}}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left[-({\cal A}\omega+{\cal B})\gamma_{0}-({\cal
A}\omega+{\cal B})\gamma_{0}-{\cal A}p\gamma_{0}+{\cal A}p\gamma_{0}+{\cal
A}p\ {\vec{\gamma}}\cdot\bm{\hat{p}}+{\cal A}p\
{\vec{\gamma}}\cdot\bm{\hat{p}}\right.$ $\displaystyle\ \ \ \ \left.-({\cal
A}\omega+{\cal B}){\vec{\gamma}}\cdot\bm{\hat{p}}+({\cal A}\omega+{\cal
B}){\vec{\gamma}}\cdot\bm{\hat{p}}\right]$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left[-({\cal A}\omega+{\cal
B})(\gamma_{0}-{\vec{\gamma}}\cdot\bm{\hat{p}})-({\cal A}\omega+{\cal
B})(\gamma_{0}+{\vec{\gamma}}\cdot\bm{\hat{p}})-{\cal
A}p(\gamma_{0}-{\vec{\gamma}}\cdot\bm{\hat{p}})+{\cal
A}p(\gamma_{0}+{\vec{\gamma}}\cdot\bm{\hat{p}})\right]$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\left[\left({\cal A}(\omega+p)+{\cal
B}\right)(\gamma_{0}-{\vec{\gamma}}\cdot\bm{\hat{p}})+\left({\cal
A}(\omega-p)+{\cal
B}\right)(\gamma_{0}+{\vec{\gamma}}\cdot\bm{\hat{p}})\right].$

Now we can write

$\displaystyle P\\!\\!\\!\\!/$ $\displaystyle=$
$\displaystyle\gamma_{0}\omega-p{\vec{\gamma}}\cdot\bm{\hat{p}}=\frac{1}{2}\left[\gamma_{0}\omega+\gamma_{0}\omega-\gamma_{0}p+\gamma_{0}p-p{\vec{\gamma}}\cdot\bm{\hat{p}}-p{\vec{\gamma}}\cdot\bm{\hat{p}}+\omega{\vec{\gamma}}\cdot\bm{\hat{p}}-\omega{\vec{\gamma}}\cdot\bm{\hat{p}}\right]$
(324) $\displaystyle=$
$\displaystyle\frac{1}{2}\left[(\omega-p)(\gamma_{0}+{\vec{\gamma}}\cdot\bm{\hat{p}})+(\omega+p)(\gamma_{0}-{\vec{\gamma}}\cdot\bm{\hat{p}})\right].$

In the rest frame of heat bath the inverse of the effective propagator in
(315) can now be written as

$\displaystyle S^{\star-1}(P)$ $\displaystyle=$
$\displaystyle{P\\!\\!\\!\\!/-\Sigma(P)}$ (325) $\displaystyle=$
$\displaystyle\frac{1}{2}\left[\left\\{(\omega-p)+{\cal A}(\omega-p)+{\cal
B}\right\\}(\gamma_{0}+{\vec{\gamma}}\cdot\bm{\hat{p}})+\left\\{(\omega+p)+{\cal
A}(\omega+p)+{\cal
B}\right\\}(\gamma_{0}-{\vec{\gamma}}\cdot\bm{\hat{p}})\right]$
$\displaystyle=$ $\displaystyle\frac{1}{2}\left[(1+{\cal A})(\omega-p)+{\cal
B}\right](\gamma_{0}+{\vec{\gamma}}\cdot\bm{\hat{p}})+\frac{1}{2}\left[(1+{\cal
A})(\omega+p)+{\cal B}\right](\gamma_{0}-{\vec{\gamma}}\cdot\bm{\hat{p}})$
$\displaystyle=$
$\displaystyle\frac{1}{2}(\gamma_{0}+{\vec{\gamma}}\cdot\bm{\hat{p}}){\cal
D}_{+}+\frac{1}{2}(\gamma_{0}-{\vec{\gamma}}\cdot\bm{\hat{p}}){\cal D}_{-}$

Using (325) in (322), one finally obtains the effective fermion propagator as

$S^{\star}(P)=\frac{1}{2}\frac{(\gamma_{0}-{\vec{\gamma}}\cdot\bm{\hat{p}})}{{\cal
D}_{+}(\omega,p)}+\frac{1}{2}\frac{(\gamma_{0}+{\vec{\gamma}}\cdot\bm{\hat{p}})}{{\cal
D}_{-}(\omega,p)},$ (326)

which is decomposed in helicity eigenstates.

In free fermion case, the propagator becomes

$S(P)=\frac{1}{2}\frac{(\gamma_{0}-{\vec{\gamma}}\cdot\bm{\hat{p}})}{d_{+}(\omega,p)}+\frac{1}{2}\frac{(\gamma_{0}+{\vec{\gamma}}\cdot\bm{\hat{p}})}{d_{-}(\omega,p)},$
(327)

where $d_{\pm}(\omega,p)$ are given in (321) and has already been used in
Sec.6.3.3 in (298).

##  8 General structure of Gauge Boson Two-point Functions at $T\neq 0$

###  8.1 Covariant Description

. The general structure of the gauge boson self-energy in vacuum Peskin  is
given as

$\displaystyle\Pi^{\mu\nu}(P^{2})=V^{\mu\nu}\Pi(P^{2}),$ (328)

where the form factor $\Pi(P^{2})$ is Lorentz invariant and depends only on
the four scalar $P^{2}$. The vacuum projection operator is given by

$\displaystyle V^{\mu\nu}=\eta^{\mu\nu}-\frac{P^{\mu}P^{\nu}}{P^{2}},$ (329)

which satisfies the gauge invariance through the transversality condition

$\displaystyle P_{\mu}\Pi^{\mu\nu}=0,$ (330)

with $\eta^{\mu\nu}\equiv(1,-1,-1,-1)$ and $P\equiv(\omega,\bm{\vec{p}})$. It
is also symmetric under the exchange of $\mu\leftrightarrow\nu$ as

$\Pi_{\mu\nu}(P^{2})=\Pi_{\nu\mu}(P^{2}).$ (331)

The presence of the heat bath or the finite temperature ($\beta=1/T$) breaks
the Lorentz invariance of the system. One collects all the four vectors and
tensors in order to construct a general covariant structure of the gauge boson
self-energy at finite temperature. These are $P^{\mu}$ and $\eta^{\mu\nu}$
from vacuum, and the four-velocity $u^{\mu}$ of the heat bath. With these one
can form four types of tensors, namely
$P^{\mu}P^{\nu},P^{\mu}u^{\nu}+u^{\mu}P^{\nu},u^{\mu}u^{\nu}$ and
$\eta^{\mu\nu}$ Ashok_Das ; Weldon:1982aq . These four tensors can form two
independent tensors by virtue of two constraints provided by the
transversality condition in (330). One can form two mutually orthogonal
projection tensors from these two independent tensors in order to construct
Lorentz-invariant structure of the gauge boson two point functions at finite
temperature.

Now, we define the Lorentz scalars, vectors and tensors that characterise the
heat bath:

$\displaystyle u^{\mu}$ $\displaystyle=$ $\displaystyle(1,0,0,0),$
$\displaystyle P^{\mu}u_{\mu}$ $\displaystyle=$ $\displaystyle P\cdot
u=\omega,$ (332)

###  8.2 Tensor Decomposition

Similar to vaccum, we can define $\tilde{\eta}^{\mu\nu}$ transverse to
$u^{\mu}$ as

$\displaystyle\tilde{\eta}^{\mu\nu}$
$\displaystyle=\eta^{\mu\nu}-u^{\mu}u^{\nu}\,$ (333a) $\displaystyle
u_{\mu}\tilde{\eta}^{\mu\nu}$
$\displaystyle=u_{\mu}\eta^{\mu\nu}-u_{\mu}u^{\mu}u^{\nu}=u^{\nu}-u^{\nu}=0\,.$
(333b)

So $u^{\mu}$ and $\tilde{\eta}^{\mu\nu}$ are transverse.

Any four vector can be decomposed parallel and orthogonal component with
respect to $u^{\mu}$:

$\displaystyle P^{\mu}_{\shortparallel}$ $\displaystyle=(P\cdot
u)u^{\mu}=\omega u^{\mu},$ (334a) $\displaystyle P^{\mu}_{\perp}$
$\displaystyle=\tilde{P}^{\mu}=P^{\mu}-P^{\mu}_{\shortparallel}=P^{\mu}-\omega
u^{\mu}\,.$ (334b)

Now,

$\displaystyle\tilde{P}^{2}$ $\displaystyle=$
$\displaystyle\left(P^{\mu}-\omega u^{\mu}\right)\left(P_{\mu}-\omega
u_{\mu}\right)=P^{2}-\omega^{2}-\omega^{2}+\omega^{2}=P^{2}-\omega^{2}=-p^{2}\,.$
(335)

We can also define any four vector parallel and perpendicular to $P^{\mu}$

$\displaystyle u^{\mu}_{\shortparallel}$ $\displaystyle=\frac{(P\cdot
u)P^{\mu}}{P^{2}}=\frac{\omega P^{\mu}}{P^{2}}\,,$ (336a)
$\displaystyle\bar{u}^{\mu}$ $\displaystyle\equiv
u^{\mu}_{\perp}=u^{\mu}-u^{\mu}_{\shortparallel}=u^{\mu}-\frac{\omega
P^{\mu}}{P^{2}}\,.$ (336b)

So, $P^{\mu}\bar{u}_{\mu}=0$.

Again,

$\displaystyle V^{\mu\nu}$
$\displaystyle=\eta^{\mu\nu}-\frac{P^{\mu}P^{\nu}}{P^{2}}\,,$ (337a)
$\displaystyle P_{\mu}V^{\mu\nu}$ $\displaystyle=0\,.$ (337b)

Given these, it is possible to construct only two independent second rank
symmetric tensors at finite temperature from
$\eta^{\mu\nu},P^{\mu}P^{\nu},u^{\mu}u^{\nu},P^{\mu}u^{\nu}+u^{\mu}P^{\nu}$
which are orthogonal to $P^{\mu}$. These two tensors are Ashok_Das

$A^{\mu\nu}=\tilde{\eta}^{\mu\nu}-\frac{\tilde{P}^{\mu}\tilde{P}^{\nu}}{\tilde{P}^{2}}\,,$
(338)

and

$B^{\mu\nu}=\frac{P^{2}}{\tilde{P}^{2}}\bar{u}^{\mu}\bar{u}^{\nu}=\frac{\bar{u}^{\mu}\bar{u}^{\nu}}{\bar{u}^{2}}\,,$
(339)

where

$A^{\mu\nu}+B^{\mu\nu}=V^{\mu\nu}=\eta^{\mu\nu}-\frac{P^{\mu}P^{\nu}}{P^{2}}\,.$
(340)

We show that $A^{\mu\nu}$ and $B^{\mu\nu}$ are orthogonal to $P^{\mu}$:

$\displaystyle P_{\mu}A^{\mu\nu}$ $\displaystyle=$ $\displaystyle
P_{\mu}\tilde{\eta}^{\mu\nu}-P_{\mu}\frac{\tilde{P}^{\mu}\tilde{P}^{\nu}}{\tilde{P}^{2}}$
(341) $\displaystyle=$ $\displaystyle
P_{\mu}\left(\eta^{\mu\nu}-u^{\mu}u^{\nu}\right)-\frac{P_{\mu}\left(P^{\mu}-\omega
u^{\mu}\right)\tilde{P}^{\nu}}{\tilde{P}^{2}}$ $\displaystyle=$
$\displaystyle\left(P^{\nu}-\omega
u^{\nu}\right)-\left(P^{2}-\omega^{2}\right)\frac{\tilde{P}^{\nu}}{\tilde{P}^{2}}$
$\displaystyle=$
$\displaystyle\tilde{P}^{\nu}-\frac{\tilde{P}^{2}\tilde{P}^{\nu}}{\tilde{P}^{2}}$
$\displaystyle=$ $\displaystyle 0\,.$

$\displaystyle P_{\mu}B^{\mu\nu}$ $\displaystyle=$
$\displaystyle\frac{P^{2}}{\tilde{P}^{2}}P_{\mu}\left(u^{\mu}-\frac{\omega
P^{\mu}}{P^{2}}\right)\left(u^{\nu}-\frac{\omega P^{\nu}}{P^{2}}\right)$ (342)
$\displaystyle=$
$\displaystyle\frac{P^{2}}{\tilde{P}^{2}}\left[\left(\omega-\frac{\omega
P^{2}}{P^{2}}\right)\left(u^{\nu}-\frac{\omega P^{\nu}}{P^{2}}\right)\right]$
$\displaystyle=$ $\displaystyle 0\,.$

$A^{\mu\nu}$ and $B^{\mu\nu}$ also satisfy following relations:

$\displaystyle A^{\mu\nu}B_{\mu\nu}$
$\displaystyle=\left(\tilde{\eta}_{\mu\nu}-\frac{\tilde{P}_{\mu}\tilde{P}_{\nu}}{\tilde{P}^{2}}\right)\frac{P^{2}}{\tilde{P}^{2}}\bar{u}^{\mu}\bar{u}^{\nu}=0\,$
(343a) $\displaystyle A_{\mu\nu}A^{\nu\rho}$
$\displaystyle=\left(\tilde{\eta}_{\mu\rho}-\frac{\tilde{P}_{\mu}\tilde{P}_{\rho}}{\tilde{P}^{2}}\right)\left(\tilde{\eta}^{\rho\nu}-\frac{\tilde{P}^{\rho}\tilde{P}^{\nu}}{\tilde{P}^{2}}\right)=A_{\mu}^{\rho}\,,$
(343b) $\displaystyle B_{\mu\nu}B^{\nu\rho}$
$\displaystyle=\left(V_{\mu\rho}-A_{\mu\rho}\right)\left(V^{\rho\nu}-A^{\rho\nu}\right)$
$\displaystyle=V_{\mu\rho}V^{\rho\nu}-2A_{\mu\rho}V^{\rho\nu}+A_{\mu\rho}A^{\rho\nu}$
$\displaystyle=V_{\mu}^{\nu}-2A_{\mu}^{\nu}+A_{\mu}^{\nu}=B_{\mu}^{\rho}\,,$
(343c) $\displaystyle A^{\mu\nu}A_{\mu\nu}$ $\displaystyle=2\,,$ (343d)
$\displaystyle B^{\mu\nu}B_{\mu\nu}$ $\displaystyle=1\,,$ (343e)
$\displaystyle A_{\mu\nu}(P)$ $\displaystyle=A_{\nu\mu}(P)=A_{\mu\nu}(-P),$
(343f) $\displaystyle B_{\mu\nu}(P)$
$\displaystyle=B_{\nu\mu}(P)=B_{\mu\nu}(-P)\,.$ (343g)

Finally, one can obtain

$\displaystyle A^{\mu\nu}$
$\displaystyle=\eta^{\mu\nu}-u^{\mu}u^{\nu}-\frac{\left(P^{\mu}-\omega
u^{\mu}\right)\left(P^{\nu}-\omega u^{\nu}\right)}{P^{2}-\omega^{2}}$
$\displaystyle=\frac{1}{P^{2}-\omega^{2}}\left[(P^{2}-\omega^{2})(\eta^{\mu\nu}-u^{\mu}u^{\nu})-P^{\mu}P^{\nu}-\omega^{2}u^{\mu}u^{\nu}+\omega(P^{\mu}u^{\nu}+u^{\mu}P^{\nu})\right]\,,$
(344a) $\displaystyle B^{\mu\nu}$
$\displaystyle=\frac{P^{2}}{\tilde{P}^{2}}\bar{u}^{\mu}\bar{u}^{\nu}=\frac{P^{2}}{P^{2}-\omega^{2}}\left[(u^{\mu}-\frac{\omega
P^{\mu}}{P^{2}})(u^{\nu}-\frac{\omega P^{\nu}}{P^{2}})\right]$
$\displaystyle=\frac{P^{2}}{P^{2}-\omega^{2}}\left[u^{\mu}u^{\nu}-\frac{\omega
P^{\mu}u^{\nu}}{P^{2}}-\frac{\omega
u^{\mu}P^{\nu}}{P^{2}}+\frac{\omega^{2}P^{\mu}P^{\nu}}{P^{4}}\right]$
$\displaystyle=\frac{1}{P^{2}\left(P^{2}-\omega^{2}\right)}\left[P^{4}u^{\mu}u^{\nu}+\omega^{2}P^{\mu}P^{\nu}-\omega
P^{2}(P^{\mu}u^{\nu}+u^{\mu}P^{\nu})\right]\,.$ (344b)

###  8.3 General Structure of Self-energy of a Vector Particle in a Thermal
Medium

The self-energy of a vector particle in a medium (finite temperature/density)
can be written as

$\Pi_{\mu\nu}=\Pi_{T}(\omega,p)A_{\mu\nu}+\Pi_{L}(\omega,p)B_{\mu\nu},$ (345)

It obeys the current conservation or transversality condition as

$\displaystyle P^{\mu}\Pi_{\mu\nu}$ $\displaystyle=$
$\displaystyle\Pi_{T}P^{\mu}A_{\mu\nu}+\Pi_{L}P^{\mu}B_{\mu\nu}$ (346)
$\displaystyle=$ $\displaystyle 0\,.$

Also note that at zero temperature

$\Pi^{(0)}_{\mu\nu}(\omega,p)=\Pi^{(0)}(P^{2})\left(\eta_{\mu\nu}-\frac{P_{\mu}P_{\nu}}{P^{2}}\right)\,,$
(347)

where

$\Pi^{(0)}_{L}(\omega,p)=\Pi^{(0)}_{T}(\omega,p)=\Pi^{(0)}(P^{2})\,.$ (348)

Using (345) we can write

$\displaystyle\Pi_{00}(\omega,p)$ $\displaystyle=$
$\displaystyle\Pi_{T}(\omega,p)A_{00}+\Pi_{L}(\omega,p)B_{00}\,.$ (349)

We obtain from (344a) and (344b), respectively, as

$\displaystyle A_{00}$ $\displaystyle=0\,,$ (350a) $\displaystyle B_{00}$
$\displaystyle=-\frac{p^{2}}{P^{2}}\,.$ (350b)

Using (350a) and (350b) in (349), one obtains Ashok_Das

$\displaystyle\Pi_{L}(\omega,p)$ $\displaystyle=$
$\displaystyle\left(-\frac{P^{2}}{p^{2}}\right)\Pi_{00}(\omega,p)\,.$ (351)

Again, we can write

$\eta^{\mu\nu}\Pi_{\mu\nu}=\Pi_{T}\eta^{\mu\nu}A_{\mu\nu}+\Pi_{L}\eta^{\mu\nu}B_{\mu\nu}\,.$
(352)

One obtains

$\displaystyle\eta^{\mu\nu}A_{\mu\nu}$ $\displaystyle=$
$\displaystyle\eta^{\mu\nu}A_{\mu\nu}-u^{\mu}u_{\mu}-\frac{\tilde{P}^{\mu}\tilde{P}_{\nu}}{\tilde{P}^{2}}$
$\displaystyle=D-1-1=D-2,$ (353)

where $D=4$ is the dimension of the system. Also, one can obtain

$\displaystyle\eta^{\mu\nu}B_{\mu\nu}$ $\displaystyle=$
$\displaystyle\eta^{\mu\nu}\frac{P^{2}}{\tilde{P}^{2}}\bar{u}_{\mu}\bar{u}_{\nu}=\frac{P^{2}}{\tilde{P}^{2}}\bar{u}_{\mu}\bar{u}^{\mu}=\frac{P^{2}}{\tilde{P}^{2}}\left(u_{\mu}-\frac{\omega
P_{\mu}}{P^{2}}\right)\left(u^{\mu}-\frac{\omega P^{\mu}}{P^{2}}\right)$ (354)
$\displaystyle=$
$\displaystyle\frac{P^{2}}{\tilde{P}^{2}}\left[u_{\mu}u^{\mu}-\frac{\omega}{P^{2}}\left(P_{\mu}u^{\mu}+P^{\mu}u_{\mu}\right)+\frac{\omega^{2}}{P^{4}}P_{\mu}P^{\mu}\right]$
$\displaystyle=$
$\displaystyle\frac{P^{2}}{\tilde{P}^{2}}\left(1-\frac{2\omega^{2}}{P^{2}}+\frac{\omega^{2}}{P^{2}}\right)=\frac{P^{2}}{P^{2}-\omega^{2}}\left(\frac{P^{2}-\omega^{2}}{P^{2}}\right)=1\,.$

Equation (353) denotes the presence of two transverse modes whereas (354)
corresponds to a longitudinal mode of the gauge boson.

Using (353) and (354) in (352) we obtain Ashok_Das

$\Pi_{T}(\omega,p)=\frac{1}{D-2}\left[{\Pi}^{\mu}_{\mu}(\omega,p)-\Pi_{L}(\omega,p)\right]=\frac{1}{2}\left[{\Pi}^{\mu}_{\mu}(\omega,p)-\Pi_{L}(\omega,p)\right].$
(355)

###  8.4 Massless Vector Gauge Boson Propagator in Covariant Gauge

Figure 24: Effective gauge boson propagator.

We represent the full propagator by $D_{mn}$ and the bare propagator by
$D^{0}_{\mu\nu}$ and each blob that appears in the summation as
$\Pi_{\mu\nu}$. One can work out this summation in Fig. 24 in tensorial form
for the effective gauge boson propagator in Fig. 24 as

$\displaystyle D^{\mu\rho}$ $\displaystyle=$
$\displaystyle{D^{0}}^{\mu\rho}+{D^{0}}^{\mu\alpha}\Pi_{\alpha\beta}{D}^{\beta\rho}$
$\displaystyle D^{\mu\rho}D^{-1}_{\rho\nu}$ $\displaystyle=$
$\displaystyle{D^{0}}^{\mu\rho}D^{-1}_{\rho\nu}+{D^{0}}^{\mu\alpha}\Pi_{\alpha\beta}D^{\beta\rho}D^{-1}_{\rho\nu}$
$\displaystyle\delta^{\mu}_{\nu}$ $\displaystyle=$
$\displaystyle{D^{0}}^{\mu\rho}D^{-1}_{\rho\nu}+{D^{0}}^{\mu\alpha}\Pi_{\alpha\beta}\delta^{\beta}_{\nu}$
$\displaystyle\delta^{\mu}_{\nu}$ $\displaystyle=$
$\displaystyle{D^{0}}^{\mu\rho}D^{-1}_{\rho\nu}+{D^{0}}^{\mu\alpha}\Pi_{\alpha\nu}$
$\displaystyle\delta^{\mu}_{\nu}(D^{0}_{\mu\gamma})^{-1}$ $\displaystyle=$
$\displaystyle{D^{0}}^{\mu\rho}(D^{0}_{\mu\gamma})^{-1}D^{-1}_{\rho\nu}+{D^{0}}^{\mu\alpha}(D^{0}_{\mu\gamma})^{-1}\Pi_{\alpha\nu}$
$\displaystyle(D^{0}_{\nu\gamma})^{-1}$ $\displaystyle=$
$\displaystyle\delta^{\rho}_{\gamma}D^{-1}_{\rho\nu}+\delta^{\alpha}_{\gamma}\Pi_{\alpha\nu}$
$\displaystyle D^{-1}_{\nu\gamma}$ $\displaystyle=$
$\displaystyle(D^{0}_{\nu\gamma})^{-1}-\Pi_{\nu\gamma}.$ (356)

Effective photon propagator is given by

${D^{-1}_{\mu\nu}=(D^{0}_{\mu\nu})^{-1}-\Pi_{\mu\nu}}\,,$ (357)

which is known as Dyson-Schwinger equation.

For massless gauge boson propagator is given in (414) as

$D^{0}_{\mu\nu}=-\frac{\eta_{\mu\nu}}{P^{2}}+{(1-\xi)}\frac{P_{\mu}P_{\nu}}{P^{4}}\,$
(358)

where $\xi$ is the gauge fixing parameter with $\xi=1$ in Feynman gauge and
$\xi=0$ in Landau gauge. We will discuss the gauge fixing and gauge boson
propagator in subsec 9.4.

Dyson-Schwinger equation is given in (357) as

$D^{-1}_{\mu\nu}=(D^{0}_{\mu\nu})^{-1}-\Pi_{\mu\nu}\,.$ (359)

So we have to calculate the inverse of $D^{0}_{\mu\nu}$.

Lets define

$(D^{0}_{\mu\rho})^{-1}=a\eta_{\mu\rho}+bP_{\mu}P_{\rho}\,.$ (360)

We know

$\displaystyle D^{0}_{\mu\rho}({D^{0}}^{\rho\nu})^{-1}$ $\displaystyle=$
$\displaystyle\delta_{\mu}^{\nu}$
$\displaystyle\frac{1}{P^{2}}\left(-\eta_{\mu\rho}+(1-\xi)\frac{P_{\mu}P_{\rho}}{P^{2}}\right)\left(a\eta^{\rho\nu}+bP^{\rho}P^{\nu}\right)$
$\displaystyle=$ $\displaystyle\delta_{\mu}^{\nu}$
$\displaystyle\frac{1}{P^{2}}\left(-a\delta_{\mu}^{\nu}+\frac{a(1-\xi)}{P^{2}}P_{\mu}P^{\nu}-bP_{\mu}P^{\nu}+b(1-\xi){P_{\mu}P^{\nu}}\right)$
$\displaystyle=$ $\displaystyle\delta_{\mu}^{\nu}\,.$ (361)

Equating coefficients of both side, we get

${a=-P^{2}}\,\,,\,\,{b=-\frac{1-\xi}{\xi}}\,.$ (362)

Using (362) in (360), one obtains

$(D^{0}_{\mu\nu})^{-1}=-P^{2}\eta_{\mu\nu}-\frac{1-\xi}{\xi}P_{\mu}P_{\nu}.$
(363)

We know from (340)

$\displaystyle P_{\mu}P_{\nu}$ $\displaystyle=$ $\displaystyle
P^{2}(\eta_{\mu\nu}-A_{\mu\nu}-B_{\mu\nu})\,.$ (364)

Substituting this in $(D^{0}_{\mu\nu})^{-1}$, one can have

$\displaystyle(D^{0}_{\mu\nu})^{-1}$ $\displaystyle=$
$\displaystyle-P^{2}\eta_{\mu\nu}-\frac{1-\xi}{\xi}P^{2}(\eta_{\mu\nu}-A_{\mu\nu}-B_{\mu\nu})$
(365) $\displaystyle=$
$\displaystyle-\frac{P^{2}}{\xi}\eta_{\mu\nu}+\frac{1-\xi}{\xi}P^{2}(A_{\mu\nu}+B_{\mu\nu})\,,$

which is the inverse of $D^{0}_{\mu\nu}$.

Using (365) and (345) in (357) one can find

$\displaystyle D_{\mu\nu}^{-1}$ $\displaystyle=$
$\displaystyle-\frac{P^{2}}{\xi}\eta_{\mu\nu}+\frac{1-\xi}{\xi}P^{2}(A_{\mu\nu}+B_{\mu\nu})-\Pi_{T}A_{\mu\nu}-\Pi_{L}B_{\mu\nu}$
(366) $\displaystyle=$
$\displaystyle-\frac{P^{2}}{\xi}\eta_{\mu\nu}+(P_{m}^{2}-\Pi_{T})A_{\mu\nu}+(P_{m}^{2}-\Pi_{L})B_{\mu\nu}\,,$

where $P_{m}^{2}=\frac{(1-\xi)}{\xi}P^{2}$. Eq.(366) is inverse of
$D_{\mu\nu}$. Now we have to find out $D_{\mu\nu}$ from $D_{\mu\nu}^{-1}$.

Lets define

$D_{\mu\rho}=cP_{\mu}P_{\rho}+dA_{\mu\rho}+eB_{\mu\rho}\,.$ (367)

We know

$D_{\mu\rho}(D^{\rho\nu})^{-1}=\delta_{\mu}^{\nu}\,$
$\frac{cP^{2}}{\xi}P_{\mu}P^{\nu}+d\left(P^{2}+\Pi_{T}\right)A_{\mu}^{\nu}+e\left(P^{2}+\Pi_{L}\right)B_{\mu}^{\nu}=-\delta_{\mu}^{\nu}\,.$
(368)

Substituting $A_{\mu}^{\nu}$ and $B_{\mu}^{\nu}$ from (344a) and (344b), the
above equation becomes

$\displaystyle\frac{cP^{2}}{\xi}P_{\mu}P^{\nu}+d\left(P^{2}+\Pi_{T}\
\right)\left[\delta_{\mu}^{\nu}-\frac{P^{2}}{P^{2}-\omega^{2}}u_{\mu}u^{\nu}-\frac{P_{\mu}P^{\nu}}{P^{2}-\omega^{2}}+\frac{\omega}{P^{2}-\omega^{2}}\left(u_{\mu}P^{\nu}+P_{\mu}u^{\nu}\right)\right]$
(369) $\displaystyle+$ $\displaystyle
e\left(P^{2}+\Pi_{L}\right)\frac{P^{2}}{P^{2}-\omega^{2}}\left[u_{\mu}u^{\nu}-\frac{\omega}{P^{2}}\left(u_{\mu}P^{\nu}+P_{\mu}u^{\nu}\right)+\frac{\omega^{2}}{P^{4}}P_{\mu}P^{\nu}\right]=-\delta_{\mu}^{\nu}\,.$

Now equating coefficients on both side, we get

$\displaystyle{\mbox{Coefficients of}}\,\,\delta_{\mu}^{\nu}:\hskip
56.9055ptd$ $\displaystyle=-\frac{1}{P^{2}+\Pi_{T}}\,,$ (370a)
$\displaystyle{\mbox{Coefficients of}}\,\,u_{\mu}P^{\nu}+P_{\mu}u^{\nu}:\hskip
56.9055pte$ $\displaystyle=-\frac{1}{P^{2}+\Pi_{L}}\,,$ (370b)
$\displaystyle{\mbox{Coefficients of}}\,\,P_{\mu}P^{\nu}:\hskip 56.9055ptc$
$\displaystyle=-\frac{\xi}{P^{4}}\,.$ (370c)

Using (370a), (370b) and (370c) in (367), one obtains the effective propagator
of interacting photon Ashok_Das  in presence of thermal medium as

${D_{\mu\nu}=-\frac{\xi}{P^{4}}P_{\mu}P_{\nu}-\frac{1}{P^{2}+\Pi_{T}}A_{\mu\nu}-\frac{1}{P^{2}+\Pi_{L}}B_{\mu\nu}}\,.$
(371)

###  8.5 Massive Vector Boson Propagator

The free propagator for massive vector boson is given as

$\displaystyle D^{0}_{\mu\nu}$ $\displaystyle=$
$\displaystyle\frac{-\eta_{\mu\nu}+\frac{P_{\mu}P_{\nu}}{m_{v}^{2}}}{P^{2}-m_{v}^{2}}$
(372) $\displaystyle\equiv$
$\displaystyle\frac{-\eta_{\mu\nu}+\frac{P_{\mu}P_{\nu}}{m_{v}^{2}}}{X}\,,$

where $X=P^{2}-m_{v}^{2}$ and $m_{v}$ is the mass of the vector boson.

The inverse of $D^{0}_{\mu\nu}$ can be written as

$(D^{0}_{\rho\nu})^{-1}=a\eta_{\rho\nu}+bP_{\rho}P_{\nu}\,.$ (373)

We have

$\displaystyle D^{0}_{\mu\rho}{(D^{0\rho\nu}})^{-1}$ $\displaystyle=$
$\displaystyle\delta_{\mu}^{\nu}$
$\displaystyle\left(\frac{-\eta_{\mu\rho}+\frac{P_{\mu}P_{\rho}}{m_{v}^{2}}}{X}\right)\left(a\eta^{\rho\nu}+bP^{\rho}P^{\nu}\right)$
$\displaystyle=$ $\displaystyle\delta_{\mu}^{\nu}$
$\displaystyle-a\delta_{\mu}^{\nu}-bP_{\mu}P^{\nu}+a\frac{P_{\mu}P^{\nu}}{m_{v}^{2}}+b\frac{P^{2}P_{\mu}P^{\nu}}{m_{v}^{2}}$
$\displaystyle=$ $\displaystyle X\delta_{\mu}^{\nu}\,.$ (374)

Equating coefficients of $\delta_{\mu}^{\nu},$ and $P_{\mu}P^{\nu}$ yields,

$a=-X=-(P^{2}-m_{v}^{2})\,,$ (375)

and

$\displaystyle\frac{a}{m_{v}^{2}}-b+\frac{bP^{2}}{m_{v}^{2}}$ $\displaystyle=$
$\displaystyle 0$ $\displaystyle b$ $\displaystyle=$ $\displaystyle 1\,.$
(376)

Now the inverse of the free propagator in (373) becomes

$(D^{0}_{\mu\nu})^{-1}=-(P^{2}-m_{v}^{2})\eta_{\mu\nu}+P_{\mu}P_{\nu}\,.$
(377)

Using (340) we get

$P_{\mu}P_{\nu}=P^{2}\left(\eta_{\mu\nu}-A_{\mu\nu}-B_{\mu\nu}\right)\,,$
(378)

one gets

$\displaystyle(D^{0}_{\mu\nu})^{-1}$ $\displaystyle=$
$\displaystyle-(P^{2}-m_{v}^{2})\eta_{\mu\nu}+P^{2}\left(\eta_{\mu\nu}-A_{\mu\nu}-B_{\mu\nu}\right)$
(379) $\displaystyle=$ $\displaystyle
m_{v}^{2}\eta_{\mu\nu}-P^{2}\left(A_{\mu\nu}+B_{\mu\nu}\right)\,.$

Putting back this value in Dyson equation in (357)

$D_{\mu\nu}^{-1}=m_{v}^{2}\eta_{\mu\nu}-P^{2}\left(A_{\mu\nu}+B_{\mu\nu}\right)-\Pi_{\mu\nu}\,.$
(380)

Using Eq.(345) we get

$\displaystyle D_{\mu\nu}^{-1}$ $\displaystyle=$ $\displaystyle
m_{v}^{2}\eta_{\mu\nu}-P^{2}\left(A_{\mu\nu}+B_{\mu\nu}\right)-\Pi_{T}A_{\mu\nu}-\Pi_{L}B_{\mu\nu}$
(381) $\displaystyle=$ $\displaystyle
m_{v}^{2}\eta_{\mu\nu}-\left(P^{2}+\Pi_{T}\right)A_{\mu\nu}-\left(P^{2}+\Pi_{L}\right)B_{\mu\nu}$

Now we have to find $D_{\mu\nu}$. Lets define

$D^{\rho\nu}=\alpha P^{\rho}P^{\nu}+\beta A^{\rho\nu}+\gamma B^{\rho\nu}\,.$
(382)

We can write

$\displaystyle D_{\mu\rho}(D^{\rho\nu})^{-1}$ $\displaystyle=$
$\displaystyle\delta_{\mu}^{\nu}$ $\displaystyle\left(\alpha
P_{\mu}P_{\rho}+\beta A_{\mu\rho}+\gamma
B_{\mu\rho}\right)\left(m_{v}^{2}\eta^{\rho\nu}-P^{2}\left(A^{\rho\nu}+B^{\rho\nu}\right)-\Pi_{T}A^{\rho\nu}-\Pi_{L}B^{\rho\nu}\right)$
$\displaystyle=$ $\displaystyle\delta_{\mu}^{\nu}$ $\displaystyle\alpha
m_{v}^{2}P_{\mu}P^{\nu}+\beta
m_{v}^{2}A_{\mu}^{\nu}-\beta\left(P^{2}+\Pi_{T}\right)A_{\mu}^{\nu}+\gamma
m_{v}^{2}B_{\mu}^{\nu}-\gamma\left(P^{2}+\Pi_{L}\right)B_{\mu}^{\nu}$
$\displaystyle=$ $\displaystyle\delta_{\mu}^{\nu}$ $\displaystyle\alpha
m_{v}^{2}P_{\mu}P^{\nu}+\beta\left(m_{v}^{2}-P^{2}-\Pi_{T}\right)A_{\mu}^{\nu}+\gamma\left(m_{v}^{2}-P^{2}-\Pi_{L}\right)$
$\displaystyle=$ $\displaystyle\delta_{\mu}^{\nu}\,.$ (383)

Substituting $A_{\mu}^{\nu},B_{\mu}^{\nu}$ from Eq.(344a),Eq.(344b) and
equating coefficients we get,

$\displaystyle{\mbox{Coefficients of}\,\,\,\delta_{\mu}^{\nu}}:\hskip
56.9055pt\beta$ $\displaystyle=\frac{1}{m_{v}^{2}-P^{2}-\Pi_{T}}\,,$ (384a)
$\displaystyle{\mbox{Coefficients of}\,\,\,u_{\mu}u^{\nu}}:\hskip
56.9055pt\gamma$ $\displaystyle=\frac{1}{m_{v}^{2}-P^{2}-\Pi_{L}}\,,$ (384b)
$\displaystyle{\mbox{Coefficients of}\,\,\,P_{\mu}P^{\nu}}:\hskip
56.9055pt\alpha$ $\displaystyle=\frac{1}{P^{2}m_{v}^{2}}\,.$ (384c)

Using (384a) to (384c) in (382) one gets the propagator for massive vector
boson in a thermal medium

$D_{\mu\nu}=\frac{P_{\mu}P_{\nu}}{P^{2}m_{v}^{2}}-\frac{A_{\mu\nu}}{{P^{2}-m_{v}^{2}+\Pi_{T}}}-\frac{B_{\mu\nu}}{P^{2}-m_{v}^{2}+\Pi_{L}}\,.$
(385)

##  9 Quantum Electrodynamics (QED)

Quantum electrodynamics (QED) is an abelian gauge theory. The symmetry group
is $U(1)$ abelian group which are also commutative group. In QED, the
interaction between two spin $1/2$ fermionic fields is mediated by
electromagnetic field photon, $A_{\mu}$, which is a gauge field. Before doing
anything else it is essential to introduce gauge and gauge fixing first.

###  9.1 Dirac Field

The Dirac Lagrangian density in (245) describes the free fermion and given as

$\displaystyle{\cal L}_{D}$ $\displaystyle=$
$\displaystyle{\bar{\psi}}\left(i\partial\\!\\!\\!/-m\right)\psi.$ (386)

As we have seen in subsec. 6.3.1 that it is invariant under a global phase
transformation, $e^{-ie\alpha}\psi(X)$, with a fixed phase parameter which
does not depend upon space and time. This is a global transformation. If the
phase factor $\alpha$ is any differentiable function of space-time,
$\alpha(X)$, i.e., at each space-time point it is different, then the
transformation,

$\psi(X)\rightarrow e^{-ie\alpha(X)}\psi(X),$ (387)

is called local transformation. The Lagrangian density in (386) is no longer
invariant under such local transformation as seen in (263):

$\displaystyle{\cal L}_{D}\rightarrow{\cal L}^{\prime}_{D}$ $\displaystyle=$
$\displaystyle\bar{\psi}e^{ie\alpha(X)}\left(i\partial\\!\\!\\!/-m\right)\psi
e^{-ie\alpha(X)}$ (388) $\displaystyle=$
$\displaystyle\bar{\psi}\left(i\partial\\!\\!\\!/-m\right)\psi\,+\,e\bar{\psi}\partial\\!\\!\\!/\alpha(X)\psi$
$\displaystyle=$ $\displaystyle{\cal
L}_{D}\,+\,e\bar{\psi}\partial\\!\\!\\!/\alpha(X)\psi.$

One needs to include a gauge potential (field) $A_{\mu}$ in the theory. As we
will see below that this gauge field $A_{\mu}$ together with the original
fermionic fields make the Lagrangian invariant under such local phase
transformation. This is also called local gauge transformation.

Under the local gauge transformation, the modified Dirac invariant Lagrangian
in (386) now reads as

$\displaystyle{\cal L}_{D}$ $\displaystyle=$
$\displaystyle{\bar{\psi}}\left(i\partial\\!\\!\\!/-m\right)\psi-e{\bar{\psi}}\gamma^{\mu}A_{\mu}\psi={\bar{\psi}}\left(iD\\!\\!\\!\\!/-m\right)\psi,$
(389)

where the original partial differential operator is replaced by the covariant
differential operator

$\displaystyle\partial_{\mu}\rightarrow D_{\mu}$ $\displaystyle=$
$\displaystyle\partial_{\mu}+ieA_{\mu},$ (390)

along with the transformation of the gauge field as

$\displaystyle A_{\mu}\rightarrow A_{\mu}+\partial_{\mu}\alpha(X).$ (391)

[Now lets check the invariance of (389) under local gauge transformation:
$\psi(X)\rightarrow\psi^{\prime}=e^{-ie\alpha(X)}\psi$

$\displaystyle\bar{\psi}^{\prime}D_{\mu}\psi^{\prime}$ $\displaystyle=$
$\displaystyle\bar{\psi}^{\prime}[\partial_{\mu}+ieA_{\mu}]e^{-ie\alpha(X)}\psi$
(392) $\displaystyle=$
$\displaystyle\bar{\psi}^{\prime}e^{-ie\alpha(X)}\partial_{\mu}\psi-
ie\bar{\psi}^{\prime}(\partial_{\mu}\alpha(X))e^{-ie\alpha(X)}\psi+ie\bar{\psi}^{\prime}A_{\mu}\psi
e^{-ie\alpha(X)}.$

If (392) vis-a-vis (389) is to be invariant under local transformation of the
fermionic field $\psi(X)\rightarrow\psi^{\prime}=e^{-ie\alpha(X)}\psi$, the
gauge field has also to be transformed as $A_{\mu}\rightarrow
A_{\mu}+\partial_{\mu}\alpha(X)$ as given in (391). So

$\displaystyle\bar{\psi}^{\prime}D_{\mu}\psi^{\prime}$ $\displaystyle=$
$\displaystyle\bar{\psi}^{\prime}e^{-ie\alpha(X)}\partial_{\mu}\psi-
ie\bar{\psi}^{\prime}(\partial_{\mu}\alpha(X))e^{-ie\alpha(X)}\psi+ie\bar{\psi}^{\prime}(A_{\mu}+\partial_{\mu}\alpha(X))e^{-ie\alpha(X)}\psi$
(393) $\displaystyle=$
$\displaystyle\bar{\psi}(\partial_{\mu}+ieA_{\mu})=\bar{\psi}D_{\mu}\psi.$

This suggests that (389) is invariant under local gauge transformation.]

###  9.2 Pure Gauge Field

Following Maxwell’s equations both electric field ${\mathbf{E}}$ and the
magnetic field ${\mathbf{B}}$ can be written from pure gauge field $A_{\mu}$
in a manifestly covariant form as

$\displaystyle F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu},$ (394)

which is known as electromagnetic field tensor. This is a gauge invariant
quantity and also antisymmetric, $F_{\mu\nu}=-F_{\nu\mu}$, under the exchange
of the Lorentz indices $\mu\leftrightarrow\nu$. One can now construct a
Lorentz scalar out of $F_{\mu\nu}$, which can be included in the Lagrangian
density for pure gauge field as

$\displaystyle{\cal L}_{\gamma}$ $\displaystyle=$
$\displaystyle-\frac{1}{4}F_{\mu\nu}F^{\mu\nu},$ (395)

where the normalisation factor $1/4$ is chosen in such a way that it gives the
correct equation of motion for electromagnetic field.

###  9.3 Electromagnetic Lagrangian

Now one can write the total electromagnetic Lagrangian density pbpal ; Peskin
describing fermions, electromagnetic field and interaction between them as

$\displaystyle{\cal L}_{em}$ $\displaystyle=$ $\displaystyle{\cal
L}_{\gamma}+{\cal L}_{D}$ (396) $\displaystyle=$
$\displaystyle-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+{\bar{\psi}}\left(iD\\!\\!\\!\\!/-m\right)\psi$
$\displaystyle=$
$\displaystyle-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+{\bar{\psi}}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi-e\bar{\psi}\gamma^{\mu}\psi
A_{\mu}.$

Figure 25: Electron-photon interaction vertex.

We note the following:

  1. $\bullet$

The first term corresponds to the free Lagrangian density of a gauge field
(photon)

  2. $\bullet$

The second term is the Lagrangian density that describes the free fermions
with mass $m$. Now if one wants to include chemical potential $\mu$, one
should follow as done in (268) in Subsec. 6.3.1 . The presence of the chemical
potential is like shifting the temporal component of the gauge field by
$\partial_{0}-i\mu$.

  3. $\bullet$

The third term originates from the local $U(1)$ gauge symmetry and corresponds
to the interaction Lagrangian density of fermionic and gauge field in which
gauge field interact with fermionic field through the dimensionless coupling
parameter $e$. This interaction is represented in Fig. 25. Using this
interaction in perturbative techniques, one can compute Feynman diagrams for
the theory.

  4. $\bullet$

The second and third term together will lead to the equation of motion for
gauge field $\partial_{\mu}F^{\mu\nu}=-j^{\nu}$ .

  5. $\bullet$

If one checks the gauge invariance of second and third term together, then it
will lead to a conserved current density and the Lagrangian density will then
differ by a total derivative, but leaves the equation of motion unchanged .

###  9.4 Gauge Fixing

One of the problem of (396) is that it can not be quantised. One way to see
this is that the propagator for photon does not exist. As we have already
experienced from scalar and fermionic fields that the propagator is obtained
from free theory, we start with the free Lagrangian density for gauge field
(photon)

$\displaystyle{\cal L}_{\gamma}$ $\displaystyle=$
$\displaystyle-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}=-\frac{1}{4}\left[\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}\right]\left[\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}\right]$
(397) $\displaystyle=$
$\displaystyle-\frac{1}{2}\partial_{\mu}A_{\nu}\left[\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}\right]+\partial_{\mu}(\cdots)$
$\displaystyle=$
$\displaystyle\frac{1}{2}A_{\nu}\left[\eta^{\mu\nu}\Box-\partial^{\mu}\partial^{\nu}\right]A_{\mu},$

where we have interchanged $\mu\rightarrow\nu$ in the second term inside the
first square braces in the second line. We have done the integration by parts
to arrive at third line. Now using the Fourier decomposition of the gauge
field one can get

$\displaystyle{\cal L}_{\gamma}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}A_{\nu}\left[-\eta^{\mu\nu}P^{2}+P^{\mu}P^{\nu}\right]A^{\mu},$
(398)

where the quantity inside the square braces should in principle be the inverse
of photon propagator say $(D_{0}^{\mu\nu})^{-1}$ as

$\displaystyle(D_{0}^{\mu\nu})^{-1}$ $\displaystyle=$
$\displaystyle-\eta^{\mu\nu}P^{2}+P^{\mu}P^{\nu}.$ (399)

The inverse of $(D_{0}^{\mu\nu})^{-1}$ should give the photon propagator as

$\displaystyle D_{0}^{\mu\nu}=a\eta^{\mu\nu}+bP^{\mu}P^{\nu}.$ (400)

Now

$\displaystyle D_{0}^{\mu\lambda}(D_{0})^{-1}_{\lambda\nu}$ $\displaystyle=$
$\displaystyle\delta^{\mu}_{\nu}$
$\displaystyle\left(a\eta^{\mu\lambda}+bP^{\mu}P^{\lambda}\right)\left(-\eta_{\lambda\nu}P^{2}+P_{\lambda}P_{\nu}\right)$
$\displaystyle=$ $\displaystyle\delta^{\mu}_{\nu}$ $\displaystyle-
aP^{2}\delta^{\mu}_{\nu}+aP^{\mu}P_{\nu}-bP^{2}P^{\mu}P_{\nu}+bP^{2}P^{\mu}P^{\nu}$
$\displaystyle=$ $\displaystyle\delta^{\mu}_{\nu}$ $\displaystyle-
aP^{2}\delta^{\mu}_{\nu}+aP^{\mu}P_{\nu}$ $\displaystyle=$
$\displaystyle\delta^{\mu}_{\nu}.$ (401)

Comparing the coefficients in both sides, one gets $a=-1/P^{2}$ and $a=0$ but
$b$ remains undetermined. The matrix $(D^{\mu\nu}_{0})^{-1}$ is not
invertible, so the propagator $D^{\mu\nu}_{0}$ does not exist.

[Another way: In general any second rank tensor can be written as

${\cal M}^{\mu\nu}=c{\cal P}^{\mu\nu}_{L}+d{\cal P}^{\mu\nu}_{T},$ (402)

where the longitudinal and transverse projection operator in orthogonal
subspace are

$\displaystyle{\cal P}_{L}^{\mu\nu}$ $\displaystyle=$ $\displaystyle
P^{\mu}P^{\nu}/P^{2},$ $\displaystyle{\cal P}_{T}^{\mu\nu}$ $\displaystyle=$
$\displaystyle\left(\eta^{\mu\nu}-{P^{\mu}P^{\nu}}/{P^{2}}\right),$ (403)

which satisfy the properties of the projection operator as

${\cal P}_{L}^{\mu\nu}+{\cal P}_{T}^{\mu\nu}=\eta^{\mu\nu},\,\,\,{\cal
P}_{L}^{2}={\cal P}_{L},\,\,\,{\cal P}_{T}^{2}={\cal P}_{T},\,\,\,{\cal
P}_{L}{\cal P}_{T}=0$ (404)

The inverse of (402) can be written as

$({\cal M}^{\mu\nu})^{-1}=c^{-1}{\cal P}^{\mu\nu}_{L}+d^{-1}{\cal
P}^{\mu\nu}_{T}.$ (405)

Now (399) can be written in terms of projection operator as

$\displaystyle(D^{\mu\nu})^{-1}$ $\displaystyle=$
$\displaystyle-\eta^{\mu\nu}P^{2}+P^{\mu}P^{\nu}=(0){\cal
P}_{L}^{\mu\nu}+(-P^{2}){\cal P}_{T}^{\mu\nu}.$ (406)

Now comparing (406) with (402) one gets $c=0$ and $d=-P^{2}$. Therefore, (405)
suggests that $d^{-1}=-1/P^{2}$ and $c^{-1}$ does not exist as $c^{-1}=1/0$.
Thus the inverse of (406) also does not exist. ]

Some of the reasons are noted below:

  1. $\bullet$

The momenta conjugate to the $A^{\mu}$ are given by

$\displaystyle\pi^{\mu}$ $\displaystyle=$ $\displaystyle\frac{\delta{\cal
L}_{em}}{\delta{\dot{A}}_{\mu}}=\frac{\delta{\cal
L}_{em}}{\delta(\partial_{0}A_{\mu})}=F^{\mu 0}\,$ (407)

which give $\pi^{0}=0$, as the diagonal components of $F^{\mu\nu}$ are zero.
This clearly indicates that one of the canonical momenta does not exist. The
equations defining canonical momenta cannot therefore be inverted to express
quantities $\dot{A}_{\mu}$ in terms of the momenta. So, the Hamiltonian
formalism also does not exist.

  2. $\bullet$

In Maxwell’s equations there are six field variables from both $\mathbf{E}$
and $\mathbf{B}$ but the two homogeneous equations imply four constraints on
the electromagnetic field components so that there only are two independent
component of electromagnetic fields. Therefore, all components of the gauge
field $A^{\mu}$ are not independent but they are connected by gauge
transformation, even though only two components are independent.

  3. $\bullet$

The matrix in the transverse projection operator in (399) has zero eigenvalue
as can be seen

$\displaystyle\left(-\eta^{\mu\nu}P^{2}+P^{\mu}P^{\nu}\right)P_{\nu}$
$\displaystyle=$ $\displaystyle 0$
$\displaystyle\left(\eta^{\mu\nu}-\frac{P^{\mu}P^{\nu}}{P^{2}}\right)P_{\nu}$
$\displaystyle=$ $\displaystyle 0$ $\displaystyle{\cal P}_{T}^{\mu\nu}P_{\nu}$
$\displaystyle=$ $\displaystyle 0.$ (408)

The (408) is the transversality condition that projects on to orthogonal
subspace. This also indicates that the presence of zero eigenvalues of the
${\cal P}_{T}^{\mu\nu}$ is a direct consequence of the gauge invariance of the
theory. Gauge invariance implies that the theory contains fewer degrees of
freedom (only transverse degrees of freedom), which reflects itself the
presence of zero eigenvalues in the quadratic part of the Lagrangian density
in (395) vis-a-vis (397). However, gauge field $A^{\mu}$ is represented by
four components $A^{\mu}$, all of which are not independent but connected by
gauge transformation. There are only two independent components and one needs
to eliminate or restrict the additional degrees of gauge freedom. It is also
to be noted that the degrees of gauge freedom absent in (395) should not
reappear through the interaction. That is ensured through the interaction of
the gauge field with a conserved fermionic current as noted in the last item
after (396) in previous page.

Before quantisation this redundancy is dealt by fixing the gauge that restrict
the gauge degrees of freedom in the theory. The Lagrangian density for the
gauge field (photon) in $U(1)$ gauge theory can be rewritten along with an
addition term (referred as gauge fixing term with a gauge parameter $\xi$) as

$\displaystyle{\cal L}_{\gamma}$ $\displaystyle=$
$\displaystyle-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\frac{1}{2\xi}(\partial_{\nu}A^{\nu})(\partial_{\mu}A^{\mu})$
(409) $\displaystyle=$
$\displaystyle-\frac{1}{4}\left[\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}\right]\left[\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}\right]-\frac{1}{2\xi}(\partial_{\nu}A^{\nu})(\partial_{\mu}A^{\mu})$
$\displaystyle=$
$\displaystyle-\frac{1}{2}\partial_{\mu}A_{\nu}\left[\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}\right]-\frac{1}{2\xi}(\partial_{\nu}A^{\nu})(\partial_{\mu}A^{\mu})+\partial_{\mu}(\cdots)$
$\displaystyle=$
$\displaystyle\frac{1}{2}A_{\nu}\left[\eta^{\mu\nu}\Box-\left(1-\frac{1}{\xi}\right)\partial^{\mu}\partial^{\nu}\right]A_{\mu},$

where the inverse of the propagator can be written as

$\displaystyle(D_{0}^{\mu\nu})^{-1}=\left[-\eta^{\mu\nu}P^{2}+\left(1-\frac{1}{\xi}\right)P^{\mu}P^{\nu}\right],$
(410)

and the inverse of which should give us the correct form the propagator as

$\displaystyle
D_{0}^{\mu\nu}=\left[a\eta^{\mu\nu}+b\left(1-\frac{1}{\xi}\right)P^{\mu}P^{\nu}\right]\,.$
(411)

As before solving for coefficients $a$ and $b$:

$\displaystyle D_{0}^{\mu\lambda}(D_{0})^{-1}_{\lambda\nu}$ $\displaystyle=$
$\displaystyle\delta^{\mu}_{\nu}$
$\displaystyle\left[a\eta^{\mu\lambda}+b\left(1-\frac{1}{\xi}\right)P^{\mu}P^{\lambda}\right]\left[-\eta_{\lambda\nu}P^{2}+\left(1-\frac{1}{\xi}\right)P_{\lambda}P_{\nu}\right]$
$\displaystyle=$ $\displaystyle\delta^{\mu}_{\nu}$ $\displaystyle-
aP^{2}\delta^{\mu}_{\nu}+\left(1-\frac{1}{\xi}\right)\left[a-bP^{2}+b\left(1-\frac{1}{\xi}\right)P^{2}\right]P^{\mu}P_{\nu}$
$\displaystyle=$ $\displaystyle\delta^{\mu}_{\nu},$ (412)

where comparing the coefficients, one gets

$\displaystyle a$ $\displaystyle=$ $\displaystyle-\frac{1}{P^{2}},$
$\displaystyle\left[a-bP^{2}+b\left(1-\frac{1}{\xi}\right)P^{2}\right]$
$\displaystyle=$ $\displaystyle 0$ $\displaystyle\Rightarrow\,\,\,b$
$\displaystyle=$ $\displaystyle-\frac{\xi}{P^{4}}.$ (413)

With these the photon propagator from (411) reads as

$\displaystyle
D_{0}^{\mu\nu}=-\frac{1}{P^{2}}\left[\eta^{\mu\nu}-\left(1-\xi\right)\frac{P^{\mu}P^{\nu}}{P^{2}}\right].$
(414)

The Feynman propagator will read as

$\displaystyle{\cal
D}_{0}^{\mu\nu}=iD_{0}^{\mu\nu}=-\frac{i}{P^{2}}\left[\eta^{\mu\nu}-\left(1-\xi\right)\frac{P^{\mu}P^{\nu}}{P^{2}}\right].$
(415)

It is worth noting at this point if a gauge field has mass then the gauge
fixing is not required, as we will see later when the massive vector boson
discussed in subsec 8.5. In different gauges $\xi$ takes different values as
$\xi=1$ (Feynman gauge) and $\xi=0$ (Landau gauge). However, the final result
is independent of gauge choice, so one can choose it conveniently for the
purpose.

After gauge fixing the QED Lagrangian reads as

$\displaystyle{\cal L}_{em}$ $\displaystyle=$
$\displaystyle-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\frac{1}{2\xi}(\partial_{\mu}A^{\mu})^{2}+{\bar{\psi}}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi-e\bar{\psi}\gamma^{\mu}\psi
A_{\mu}.$ (416)

Now one can show that this gauge fixing term does not change the Lagrangian or
the Maxwell’s equations as long as the current is conserved: The new equation
of motion becomes

$\displaystyle\Box
A^{\nu}-\left(1-\frac{1}{\xi}\right)\partial^{\nu}\left(\partial_{\mu}A^{\mu}\right)$
$\displaystyle=$ $\displaystyle-j^{\nu}.$ (417)

Operating 4-divergence and using the current conservation
$\partial_{\mu}j^{\mu}=0$ one gets

$\displaystyle\Box\left(\partial_{\mu}A^{\mu}\right)$ $\displaystyle=$
$\displaystyle-\partial_{\mu}j^{\mu}=0$
$\displaystyle\rightarrow\,\,\,\partial_{\mu}A^{\mu}$ $\displaystyle=$
$\displaystyle 0,$ (418)

with appropriate boundary condition on $A^{\mu}$ so that (418) is satisfied.
This implies that one can always transform $A^{\mu}$ according to (391) so
that it satisfies (418). This is called gauge fixing condition in covariant
gauge. Now one can compute canonical momenta and do the Hamiltonian
formulation.

###  9.5 Free Photon Partition Function

The free photon Lagrangian density can be written from (416) as

$\displaystyle{\cal L}_{\gamma}$ $\displaystyle=$
$\displaystyle-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}.$ (419)

In vacuum, photon partition function can be written as,

$\displaystyle\mathcal{Z}$ $\displaystyle=$
$\displaystyle\int\mathcal{D}A_{\mu}\exp{[iS]}$ (420) $\displaystyle=$
$\displaystyle\int\mathcal{D}A_{\mu}\exp{\left[{i\int
d^{4}X\left(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\right)}\right]}\,,$

where ${\cal D}A_{\mu}={\cal D}A_{0}{\cal D}A_{1}{\cal D}A_{2}{\cal D}A_{3}$.
Gauge transformations should not change anything physically.

We now choose a covariant gauge condition as

$G(A)=\partial_{\mu}A^{\mu}=w(X),$ (421)

which can be imposed to (420) by inserting a identity Yang  given as

$\int\mathcal{D}\alpha(x)\delta\left(G(A,\alpha)-w(X)\right)\Big{|}{\frac{\delta\left(G(A,\alpha)-w(X)\right)}{\delta\alpha}}\Big{|}=1\,.$
(422)

Lorentz gauge condition can be recovered by taking $w=0$. There is still a
residual gauge freedom as one shifts the gauge field

$A_{\mu}\rightarrow A_{\mu}+\partial_{\mu}\alpha(X)\,,$ (423)

where the phase factor, $\alpha(X)$ is differentiable function of space-time.
Now we can write the gauge condition in (421) as

$G(A,\alpha)=\partial_{\mu}A^{\mu}+\partial_{\mu}\partial^{\mu}\alpha\,.$
(424)

The determinant term can be obtained as

$\frac{\delta\left(G(A,\alpha(X))-w(X)\right)}{\delta\alpha(Y)}=\partial_{\mu}\partial^{\mu}\delta^{(4)}(X-Y)\,.$
(425)

So (420) becomes,

$\mathcal{Z}=\int\mathcal{D}A_{\mu}\mathcal{D}\alpha(X)\delta\left(G(A,\alpha)-w(X)\right)\Big{|}{\frac{\delta\left(G(A,\alpha)-w(X)\right)}{\delta\alpha}}\Big{|}\exp{\left[{i\int
d^{4}X\left(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\right)}\right]}\,.$ (426)

Since there is a residual gauge freedom, we shift the gauge field as (423) and
write (426) as

$\mathcal{Z}=\int\mathcal{D}A_{\mu}\,\,\mathcal{D}\alpha(X)\,\,\delta\left(G(A)-w(X)\right)\,\det{\partial^{2}}\exp{\left[{i\int
d^{4}X\left(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\right)}\right]}\,.$ (427)

Now the integrand does not contain $\alpha$ and the $\alpha$ integration gives
diverging result. This is due to the redundancy of the residual gauge
transformation. Now, $w(X)$ is an arbitrary function of $X$ and the behaviour
of $w(X)$ is not known, so the integral involved in the partition function can
not be solved. One can avoid this problem by averaging over $w(X)$ around zero
with a Gaussian width $\xi$

$\int\mathcal{D}w\frac{1}{\sqrt{2\pi\xi}}\exp{\frac{w^{2}(X)}{2\xi}}\,,$ (428)

where $\xi$ is a gauge fixing parameter that one chooses for the convenience
of calculations and $\frac{1}{\sqrt{2\pi\xi}}$ is normalisation factor. Now
the integration over $w$ is performed in (427) and one gets

$\mathcal{Z}=\int\mathcal{D}A_{\mu}\,\,\det{\partial^{2}}\exp{\left[i\int
d^{4}X\left(-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\frac{(\partial_{\mu}A^{\mu})^{2}}{2\xi}\right)\right]}\,.$
(429)

Now we can write $\det{\partial^{2}}$ term using Grassmann property given in
(81) as

$\det{\partial^{2}}=\int\mathcal{D}\bar{C}\,\,\mathcal{D}C\,\exp({-\bar{C}\partial^{2}C})\,,$
(430)

where $C$ is Grassmann field which is also known as ghost field. This plays a
crucial role to cancel the redundant gauge degrees of freedom. Now,

$\displaystyle\mathcal{Z}$ $\displaystyle=$
$\displaystyle\int\mathcal{D}A\,\,\mathcal{D}\bar{C}\,\,\mathcal{D}C\,\exp{\left[i\int
d^{4}X(\mathcal{L}_{\gamma}+\mathcal{L}_{\textrm{gf}}+\mathcal{L}_{\textrm{gh})}\right]}$
(431) $\displaystyle=$
$\displaystyle\mathcal{Z}_{\gamma+\textrm{gf}}\mathcal{Z}_{\textrm{gh}}\,,$

where gauge fixing and ghost terms, respectively, are

$\displaystyle\mathcal{L}_{\textrm{gf}}$ $\displaystyle=$
$\displaystyle-\frac{(\partial_{\mu}A^{\mu})^{2}}{2\xi}\,,$
$\displaystyle\mathcal{L}_{\textrm{gh}}$ $\displaystyle=$
$\displaystyle-\bar{C}\,\partial^{2}C\,.$

Let us calculate the contributions of gauge and gauge fixing part of (431) to
the partition function as

$\displaystyle\mathcal{Z}_{\gamma+\textrm{gf}}$
$\displaystyle{=\atop{t\rightarrow-i\tau}}$
$\displaystyle\int\mathcal{D}A\exp{\left(\frac{i}{2}\int
d^{4}XA_{\nu}\left[\eta^{\mu\nu}\Box-(1-\frac{1}{\xi})\partial^{\mu}\partial^{\nu}\right]A_{\mu}\right)}\,$
(432) $\displaystyle=$
$\displaystyle\int\mathcal{D}A\exp{\left(-\frac{1}{2}\int_{0}^{\beta}d\tau\,d^{3}\bm{\vec{x}}\,A_{i}\left[\delta^{ij}\Box_{\tau}+(1-\frac{1}{\xi})\partial^{i}\partial^{j}\right]A_{j}\right)}\,,$

where we have used $\eta^{\mu\nu}\leftrightarrow-\delta^{ij}$ .

The Fourier transform of the gauge field is given as

$\displaystyle A_{i}(X)=A_{i}(\bm{\vec{x}},\tau)$
$\displaystyle{=\atop{t\rightarrow-i\tau}}\atop{p_{0}\rightarrow i\omega_{n}}$
$\displaystyle\frac{1}{\sqrt{V\beta}}\sum_{P}\,e^{-iP\cdot X}\,A_{i}(P)$ (433)
$\displaystyle=$
$\displaystyle\frac{1}{\sqrt{V\beta}}\sum_{n,\bm{\vec{p}}}\,e^{i\bm{\vec{p}\cdot\vec{x}}}\,\,e^{-i\omega_{n}\tau}\,A_{i}(\omega_{n},\bm{\vec{p}})\,.$

One can write the partition function in (432) in frequency momentum space as

$\displaystyle\mathcal{Z}_{\gamma+\textrm{gf}}$ $\displaystyle=$
$\displaystyle\int\mathcal{D}A(\omega_{n},\bm{\vec{p}})\exp{\left(-\sum_{n,\bm{\vec{p}}}\frac{1}{2}A^{*}_{i}(\omega_{n},\bm{\vec{p}})\left[\delta^{ij}(\omega_{n}^{2}+p^{2})-(1-\frac{1}{\xi})p^{i}p^{j}\right]A_{j}(\omega_{n},\bm{\vec{p}})\right)}$
(434) $\displaystyle=$
$\displaystyle\int\mathcal{D}A(\omega_{n},\bm{\vec{p}})\exp{\left(-\sum_{n,\bm{\vec{p}}}\frac{1}{2}A^{*}_{i}(\omega_{n},\bm{\vec{p}})D^{-1}_{ij}A_{j}(\omega_{n},\bm{\vec{p}})\right)}$
$\displaystyle=$
$\displaystyle\prod_{n,\bm{\vec{p}}}\sqrt{\frac{\pi^{D}}{\det{D^{-1}_{ij}(\xi)}}}$

where $D^{-1}_{ij}$ is a $4\times 4$ matrix and inverse of the gauge boson
propagator. Let us set $p=(0,0,p)$ as three space part of $p$ are in equal
footing now Yang . In Feynman gauge we have $\xi=1$.

$D^{-1}_{ij}=\left({\begin{array}[]{cccc}p^{2}+\omega_{n}^{2}&0&0&0\\\
0&p^{2}+\omega_{n}^{2}&0&0\\\ 0&0&p^{2}+\omega_{n}^{2}&0\\\
0&0&0&p^{2}+\omega_{n}^{2}\\\ \end{array}}\right)$
$\therefore\det{D^{-1}}_{ij}=(p^{2}+\omega_{n}^{2})^{4}.$ (435)
$\displaystyle\mbox{So,}\,\,\,\ln{\mathcal{Z}_{\gamma+\textrm{gf}}}$
$\displaystyle=$
$\displaystyle-\frac{1}{2}\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{n,\bm{\vec{p}}}\ln\textrm{det}[D_{ij}^{-1}]$
(436) $\displaystyle=$
$\displaystyle-4\times\frac{1}{2}\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{P_{E}}\ln{[P_{E}^{2}]}\,,$

where $P_{E}$ is Euclidean momentum and
$\sum\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{P_{E}}$ is a bosonic sum-integral.

Now we have to calculate ghost contribution to the partition function from
(431) as

${\mathcal{Z}}_{\textrm{
gh}}=\int\mathcal{D}{\bar{C}}\,\,\mathcal{D}C\,\exp{\left(-i\int
d^{4}X\,\bar{C}\,\partial^{2}\,C\right)}=\int\mathcal{D}{\bar{C}}\,\,\mathcal{D}C\,\,\exp{\left(\int_{0}^{\beta}d\tau\,d^{3}x\,\bar{C}\,\Box_{\tau}\,C\right)}\,.$
(437)

The Fourier transform of the ghost field is given as

$\displaystyle
C(\tau,\bm{\vec{x}})=\frac{1}{\sqrt{V\beta}}\sum_{n,\bm{\vec{p}}}\,e^{i\bm{\vec{p}\cdot\vec{x}}}\,e^{-i\omega_{n}\tau}\,C(\omega_{n},\bm{\vec{p}})\,.$
(438)

Combining (437) and (438), one can get the ghost partition function in
frequency momentum space as

$\displaystyle{\mathcal{Z}}_{\textrm{ gh}}$ $\displaystyle=$
$\displaystyle\int\mathcal{D}{\bar{C}(\omega_{n},\bm{\vec{p}})}\mathcal{D}C(\omega_{n},\bm{\vec{p}})\exp{\left(-\sum_{n,\bm{\vec{p}}}\bar{C}(\omega_{n},\bm{\vec{p}})(\omega_{n}^{2}+p^{2})C(\omega_{n},\bm{\vec{p}})\right)}$
(439) $\displaystyle=$
$\displaystyle\prod_{n,\bm{\vec{p}}}(\omega_{n}^{2}+p^{2}).$
$\mbox{So,}\,\,\,\ln{\mathcal{Z}_{\textrm{gh}}}=2\times\frac{1}{2}\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{P_{E}}\ln{[P_{E}^{2}]}\,.$
(440)

The logarithm of the photon partition function can be written as

$\displaystyle\ln{\mathcal{Z}}$ $\displaystyle=$
$\displaystyle\ln{\mathcal{Z}}_{\gamma+\textrm{gf}}+\ln{\mathcal{Z}}_{\textrm{gh}}$
(441) $\displaystyle=$
$\displaystyle-4\times\frac{1}{2}\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{P_{E}}\ln{[P_{E}^{2}]}+2\times\frac{1}{2}\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{P_{E}}\ln{[P_{E}^{2}]}$
$\displaystyle=$
$\displaystyle-\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{P_{E}}\ln{[P_{E}^{2}]}\,.$

So, ghost contribution cancels two unphysical degrees of freedom of photon.
Now there are two physical transverse degrees of freedom of photon.

By calculating the frequency sum one gets,

$\ln{\mathcal{Z}}=-2V\int\frac{d^{3}p}{(2\pi)^{3}}\left[\frac{\beta\omega_{p}}{2}+\ln{(1-\exp{(-\beta\omega_{p})})}\right]\,.$
(442)

which agrees with one bosonic degree of freedom in (16).

So free energy density of non-interacting photon at finite temperature is
given by (ignoring vacuum part)

$\displaystyle F_{0}$ $\displaystyle=$
$\displaystyle-\frac{T}{V}\ln{\mathcal{Z}}=2\int\frac{d^{3}p}{(2\pi)^{3}}\ln{(1-\exp{(-\beta\omega_{p})})}$
(443) $\displaystyle=$ $\displaystyle-\frac{\pi^{2}T^{4}}{45}\,.$

The pressure for non-interacting photon

${\cal P}_{0}=-F_{0}=\frac{\pi^{2}T^{4}}{45}\,.$ (444)

###  9.6 One-loop Fermion Self-energy $\Sigma$ and Structure Functions ${\cal
A}$ and ${\cal B}$ in HTL approximation

We have electron-photon interaction Lagrangian from (416) as

$\mathcal{L}_{int}=-e\bar{\psi}\gamma_{\mu}\psi A^{\mu}.$ (445)

Figure 26: One loop fermion self-energy diagram.

The one loop fermion self-energy in Fig. 26 can be written in Feynman gauge as

$\displaystyle\Sigma(P)$ $\displaystyle=$ $\displaystyle
x^{2}C_{F}T\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{\\{K\\}}\gamma_{\mu}\frac{K\\!\\!\\!\\!/}{K^{2}}\gamma^{\mu}\frac{1}{(P-K)^{2}}$
(446) $\displaystyle=$
$\displaystyle-2x^{2}C_{F}T\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{\\{K\\}}\frac{K\\!\\!\\!\\!/}{K^{2}\
Q^{2}}\,,$

where $Q=(P-K)$ and $\sum\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{\\{K\\}}$ is a
fermionic sum-integral. Also $x$ and $C_{F}$ are, respectively, coupling and
Casimir invariant for a given group. For $U(1)$ group $x=e$ and $C_{F}=1$ and
Eq.(446) corresponds to electron self-energy. For $SU(3)$ group $x=g$ and
$C_{F}=4/3$ and Eq.(446) corresponds to quark self-energy with internal photon
line should be replaced by gluon line.

We now would like to compute the structure functions as given in the rest
frame of the heat bath, respectively, in (310) and (312) as

$\displaystyle{\cal A}(\omega,p)$
$\displaystyle=\frac{1}{4p^{2}}\left({\rm{Tr}}\left[\Sigma
P\\!\\!\\!\\!/\,\right]-\omega{\rm{Tr}}\left[\Sigma
u\\!\\!\\!/\right]\right),$ (447a) $\displaystyle{\cal B}(\omega,p)$
$\displaystyle=\frac{1}{4p^{2}}\left(P^{2}{\rm{Tr}}\left[\Sigma
u\\!\\!\\!/\right]-(P\cdot u){\rm{Tr}}\left[\Sigma
P\\!\\!\\!\\!/\,\right]\right).$ (447b)

Now, we can write

$\displaystyle{\textrm{Tr}}\left[\Sigma P\\!\\!\\!\\!/\,\right]$
$\displaystyle\,=\,-8x^{2}C_{F}T\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{\\{K\\}}\frac{k_{0}\omega-\bm{{\vec{k}}\cdot{\vec{p}}}}{K^{2}Q^{2}}\,,$
(448a) $\displaystyle{\textrm{Tr}}\left[\Sigma u\\!\\!\\!/\right]$
$\displaystyle\,=\
-8x^{2}C_{F}T\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{\\{K\\}}\frac{k_{0}}{K^{2}Q^{2}}\,.$
(448b)

Using (448a) and (448b) in (447a), we get

${\cal
A}(\omega,p)=\frac{2x^{2}C_{F}}{p^{2}}T\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{\\{K\\}}\frac{\bm{{\vec{k}}\cdot{\vec{p}}}}{K^{2}Q^{2}}\,.$
(449)

We use the following frequency sum in mixed representation

$\displaystyle T\sum_{k_{0}}\frac{1}{K^{2}Q^{2}}$ $\displaystyle=$
$\displaystyle\frac{1}{4kq}\left[\left(1-n_{F}(k)+n_{B}(q)\right)\left(\frac{1}{\omega+k+q}-\frac{1}{\omega-
k-q}\right)\right.$ (450)
$\displaystyle\left.+\left(n_{B}(q)+n_{F}(k)\right)\left(\frac{1}{\omega+k-q}-\frac{1}{\omega-k+q}\right)\right]\,,$

where $n_{F\atop B}(y)=(e^{y/T}\pm 1)^{-1}$. When loop momentum is hard,
$K\sim T$ , compare to the external momentum $P$ is called hard thermal loop
(HTL) approximation BP . Under this HTL approximation666The details of HTL
approximation will be discussed in section 12. one can make following
simplifications as

$\displaystyle q\,=\,\left|\bm{\vec{p}}-\bm{\vec{k}}\right|$
$\displaystyle\,=\,\sqrt{p^{2}+k^{2}-2pk\cos\theta}=\sqrt{p^{2}+k^{2}-2pkc}\approx
k-pc=k-{\bm{\vec{p}\cdot\hat{k}}}\,,$ (451a) $\displaystyle n_{B}(q)$
$\displaystyle\,=n_{B}(k-{\bm{\vec{p}\cdot\hat{k}}})\approx
n_{B}(k)-{\bm{\vec{p}\cdot\hat{k}}}\frac{dn_{B}(k)}{dk}\,,$ (451b)
$\displaystyle\omega\pm k\pm q$ $\displaystyle\,=\omega\pm k\pm
k\mp\,{\bm{\vec{p}\cdot\hat{k}}}\approx\pm 2k\,,$ (451c)
$\displaystyle\omega\pm k\mp q$ $\displaystyle\,=\omega\pm k\mp
k\pm\,{\bm{\vec{p}\cdot\hat{k}}}\approx\omega\pm{\bm{\vec{p}\cdot\hat{k}}}\ .$
(451d)

Using (451a) to (451d) in (450), one can write

$\displaystyle T\sum_{k_{0}}\frac{1}{K^{2}Q^{2}}$ $\displaystyle=$
$\displaystyle\frac{1}{4k^{2}}\left[\left(1-n_{F}(k)+n_{B}(k)-{\bm{\vec{p}\cdot\hat{k}}}\frac{dn_{B}(k)}{dk}\right)\left(\frac{1}{k}\right)\right.$
(452)
$\displaystyle\left.+\left(n_{B}(k)+n_{F}(k)-{\bm{\vec{p}\cdot\hat{k}}}\frac{dn_{B}(k)}{dk}\right)\left(\frac{1}{\omega+{\bm{\vec{p}\cdot\hat{k}}}}-\frac{1}{\omega-{\bm{\vec{p}\cdot\hat{k}}}}\right)\right]\,,$

Combining (452) and (449), we get

$\displaystyle{\cal A}(\omega,p)$ $\displaystyle=$
$\displaystyle\frac{2x^{2}C_{F}}{p^{2}}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\bm{{\vec{p}}\cdot{\vec{k}}}}{4k^{2}}\left[\left(1-n_{F}(k)+n_{B}(k)-{\bm{\vec{p}\cdot\hat{k}}}\frac{dn_{B}(k)}{dk}\right)\left(\frac{1}{k}\right)\right.$
(453)
$\displaystyle\left.+\left(n_{B}(k)+n_{F}(k)-{\bm{\vec{p}\cdot\hat{k}}}\frac{dn_{B}(k)}{dk}\right)\left(\frac{1}{\omega+{\bm{\vec{p}\cdot\hat{k}}}}-\frac{1}{\omega-{\bm{\vec{p}\cdot\hat{k}}}}\right)\right]\,.$

We note that the term $1/k$ inside the square bracket contributes as
proportional to $T$, which is sub-leading in $T$. This term can be neglected.
We now evaluate the second term only that contributes as $T^{2}$, which is
leading order in $T$. We further note that the term
${\bm{\vec{p}\cdot\hat{k}}}\frac{dn_{B}(k)}{dk}$ has soft momentum and is
neglected. With this the (453) reduces as

$\displaystyle{\cal A}(\omega,p)$ $\displaystyle=$
$\displaystyle\frac{2x^{2}C_{F}}{4p^{2}}\int\frac{k\,dk\,d\Omega}{(2\pi)^{3}}\left(n_{B}(k)+n_{F}(k)\right)\left[\frac{{\bm{{\vec{p}}\cdot{\hat{k}}}}}{\omega+{\bm{\vec{p}\cdot\hat{k}}}}-\frac{{\bm{{\vec{p}}\cdot{\hat{k}}}}}{\omega-{\bm{\vec{p}\cdot\hat{k}}}}\right]\,.$
(454)

Now putting $\cos\theta\rightarrow-\cos\theta$ in the first term inside the
square bracket, the above equation then becomes

$\displaystyle{\cal A}(\omega,p)$ $\displaystyle=$
$\displaystyle-\frac{x^{2}C_{F}}{p^{2}}\int_{0}^{\infty}k\,dk\left(n_{B}(k)+n_{F}(k)\right)\
\int\frac{d\Omega}{(2\pi)^{3}}\frac{{\bm{{\vec{p}}\cdot{\hat{k}}}}}{\omega-{\bm{\vec{p}\cdot\hat{k}}}}\,.$
(455)

We can perform the $k$-integration as

$\int_{0}^{\infty}k\,dk\left(n_{B}(k)+n_{F}(k)\right)=\frac{\pi^{2}T^{2}}{4}.$
(456)

Using (456) we can obtain

$\displaystyle{\cal A}(\omega,p)$ $\displaystyle=$
$\displaystyle-\frac{m^{2}_{\textrm{th}}}{p^{2}}\int\frac{d\Omega}{4\pi}\frac{{\bm{{\vec{p}}\cdot{\hat{k}}}}}{\omega-{\bm{\vec{p}\cdot\hat{k}}}}$
(457) $\displaystyle=$
$\displaystyle-\frac{m^{2}_{\textrm{th}}}{p^{2}}\int\frac{d\Omega}{4\pi}\frac{{\bm{{\vec{p}}\cdot{\hat{k}}}}}{P\cdot\hat{K}}\,,$

where ${\hat{K}}=(1,\bm{\hat{k}})$ is a light like vector and the thermal mass
of the fermion is given as

$\displaystyle m_{\textrm{th}}^{2}=\frac{x^{2}C_{F}T^{2}}{8}\,.$ (458)

For electron the thermal mass becomes
$m_{\textrm{th}}^{2}=\frac{e^{2}T^{2}}{8}$ whereas for quark
$m_{\textrm{th}}^{2}=\frac{g^{2}T^{2}}{6}$.

The angular integration can be performed using the following integrals

$\displaystyle\int_{-1}^{1}\frac{dy}{a-by}$
$\displaystyle=\frac{1}{b}\ln\frac{a+b}{a-b}\,,$ (459a)
$\displaystyle\int_{-1}^{1}\frac{y\ dy}{a-by}$
$\displaystyle=-\frac{2}{b}+\frac{a}{b^{2}}\ln\frac{a+b}{a-b}.$ (459b)

After performing the angular integration in (457) we finally get HAW ; VVK

$\mathcal{A}(\omega,p)\,=\,\frac{m_{\textrm{th}}^{2}}{p^{2}}\left[1-\frac{\omega}{2p}\ln\left(\frac{\omega+p}{\omega-p}\right)\right]\,,\\\
$ (460)

Using (448a) and (448b) in (447b) we write the structure function ${\cal
B}(\omega,p)$ as

$\displaystyle{\cal B}(\omega,p)$ $\displaystyle=$
$\displaystyle\frac{2x^{2}C_{F}}{p^{2}}\left[p^{2}T\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{\\{K\\}}\frac{k_{0}}{K^{2}Q^{2}}-(P\cdot
u)T\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{\\{K\\}}\frac{\bm{\vec{p}\cdot\vec{k}}}{K^{2}Q^{2}}\right]\,.$
(461)

The frequency sum in the second term inside square bracket is already done in
(450) . The frequency sum needed for the first term is given in mixed
representation as

$\displaystyle T\sum_{k_{0}}\frac{k_{0}}{K^{2}Q^{2}}$ $\displaystyle=$
$\displaystyle-\frac{1}{4q}\left[\left(1-n_{F}(k)+n_{B}(q)\right)\left(\frac{1}{\omega+k+q}+\frac{1}{\omega-
k-q}\right)\right.$ (462)
$\displaystyle\left.+\left(n_{B}(q)+n_{F}(k)\right)\left(\frac{1}{\omega+k-q}+\frac{1}{\omega-k+q}\right)\right]\,,$

Using HTL approximations in (451a) to (451d), the frequency sum becomes

$\displaystyle T\sum_{k_{0}}\frac{k_{0}}{K^{2}Q^{2}}$ $\displaystyle=$
$\displaystyle-\frac{1}{4k}\left[\left(n_{B}(k)+n_{F}(k)-{\bm{\vec{p}\cdot\hat{k}}}\frac{dn_{B}(k)}{dk}\right)\left(\frac{1}{\omega+\bm{\vec{p}\cdot\hat{k}}}+\frac{1}{\omega-\bm{\vec{p}\cdot\hat{k}}}\right)\right]\,,$
(463)

Substituting (452) and (463) in (461), we get

$\displaystyle{\cal B}(\omega,p)$ $\displaystyle=$
$\displaystyle\frac{2x^{2}C_{F}}{p^{2}}\left[-p^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{4k}\left(n_{B}(k)+n_{F}(k)-{\bm{\vec{p}\cdot\hat{k}}}\frac{dn_{B}(k)}{dk}\right)\left(\frac{1}{\omega+\bm{\vec{p}\cdot\hat{k}}}+\frac{1}{\omega-\bm{\vec{p}\cdot\hat{k}}}\right)\right.$
(464) $\displaystyle\left.-(P\cdot
u)\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{4k^{2}}\left(1-n_{F}(k)+n_{B}(k)-{\bm{\vec{p}\cdot\hat{k}}}\frac{dn_{B}(k)}{dk}\right)\frac{1}{k}\right.$
$\displaystyle\left.-(P\cdot
u)\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{4k^{2}}\left(n_{B}(k)+n_{F}(k)-{\bm{\vec{p}\cdot\hat{k}}}\frac{dn_{B}(k)}{dk}\right)\left(\frac{\bm{\vec{p}\cdot\vec{k}}}{\omega+\bm{\vec{p}\cdot\hat{k}}}-\frac{\bm{\vec{p}\cdot\vec{k}}}{\omega-\bm{\vec{p}\cdot\hat{k}}}\right)\right]$

As before we neglect the second term inside the square bracket that gives a
contribution proportional to $T$, which is sub-leading in $T$. Also
${\bm{\vec{p}\cdot\hat{k}}}\frac{dn_{B}(k)}{dk}$ term is neglected as the soft
momentum is associated with it. Then, the above equation can be written as

$\displaystyle{\cal B}(\omega,p)$ $\displaystyle=$
$\displaystyle\frac{x^{2}C_{F}}{p^{2}}\left[\int_{0}^{\infty}k\
dk\left(n_{B}(k)+n_{F}(k)\right)\int\frac{d\Omega}{(2\pi)^{3}}\frac{(P\cdot
u)(\bm{\vec{p}\cdot\hat{k}})-p^{2}}{\omega-\bm{\vec{p}\cdot\hat{k}}}\right]\,.$
(465)

After performing the $k$-integration using (456), we get

$\displaystyle\mathcal{B}(\omega,p)$ $\displaystyle\,=\,$
$\displaystyle\frac{m_{\textrm{th}}^{2}}{p^{2}}\int\frac{d\Omega}{4\pi}\frac{(P\cdot
u)(\bm{\vec{p}\cdot\hat{k}})-p^{2}}{P\cdot\hat{K}}\,,$ (466)

Now performing the angular integrations using (459a) and (459b), we finally
get HAW ; VVK

$\displaystyle\mathcal{B}(\omega,p)$ $\displaystyle\,=\,$
$\displaystyle\frac{m_{\textrm{th}}^{2}}{p}\left[-\frac{\omega}{p}+\left(\frac{\omega^{2}}{p^{2}}-1\right)\frac{1}{2}\ln\left(\frac{\omega+p}{\omega-p}\right)\right]\,.$
(467)

The fermion self-energy can be written from (446) as

$\displaystyle\Sigma(P)$ $\displaystyle=$
$\displaystyle-2x^{2}C_{F}T\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{K}\left[\frac{k_{0}\gamma_{0}}{K^{2}Q^{2}}-\frac{\vec{\gamma}\cdot\bm{\vec{k}}}{K^{2}Q^{2}}\right]\,.$
(468)

Following the same procedures as those structure functions, the fermion self-
energy contribution within HTL approximation can be written as

$\displaystyle\Sigma(P)$ $\displaystyle=$ $\displaystyle
x^{2}C_{F}\int_{0}^{\infty}k\
dk\left(n_{B}(k)+n_{F}(k)\right)\int\frac{d\Omega}{(2\pi)^{3}}\left[\frac{\gamma_{0}}{\omega-\bm{\vec{p}\cdot\hat{k}}}-\frac{\vec{\gamma}\cdot\bm{\hat{k}}}{\omega-\bm{\vec{p}\cdot\hat{k}}}\right]\,,$
(469)

which after $k$-integration becomes

$\displaystyle\Sigma(P)$ $\displaystyle=$ $\displaystyle
m_{\textrm{th}}^{2}\int\frac{d\Omega}{4\pi}\frac{\hat{\not{K}}}{P\cdot\hat{K}}.$
(470)

After explicit calculations, we obtain

$\displaystyle\Sigma(P)$ $\displaystyle=$
$\displaystyle\frac{m_{\textrm{th}}^{2}}{2p}\ln\left(\frac{\omega+p}{\omega-p}\right)\gamma_{0}+\frac{m_{\textrm{th}}^{2}}{p}\left[1-\frac{\omega}{2p}\ln\left(\frac{\omega+p}{\omega-p}\right)\right]\left({\vec{\gamma}}\cdot\bm{\hat{p}}\right)\,.$
(471)

This expression can also be obtained directly by combining (313), (460) and
(467).

###  9.7 Dispersion of Fermionic Quasiparticles and Collective Excitations in
HTL Approximation

Figure 27: Plot of quasiparticles dispersion in HTL approximation.

We can obtain ${\cal D}_{\pm}(\omega,p)$ by combining (320), (460) and (467)
as

$\displaystyle{\cal D}_{\pm}(\omega,p)$ $\displaystyle=$
$\displaystyle\left[1+\frac{m_{\textrm{th}}^{2}}{p^{2}}\left(1-\frac{\omega}{2p}\ln\left(\frac{\omega+p}{\omega-p}\right)\right)\right](\omega\mp
p)+\frac{m_{\textrm{th}}^{2}}{p}\left[-\frac{\omega}{p}+\left(\frac{\omega^{2}}{p^{2}}-1\right)\frac{1}{2}\ln\left(\frac{\omega+p}{\omega-p}\right)\right]$
(472) $\displaystyle=$ $\displaystyle(\omega\mp
p)-\frac{m_{\textrm{th}}^{2}}{p}\left[\frac{1}{2}\left(1\mp\frac{\omega}{p}\right)\ln\left(\frac{\omega+p}{\omega-p}\right)\pm
1\right]\,.$

One can obtain the in-medium fermion propagator reads from (326)

$S^{\star}(P)=\frac{1}{2}\frac{(\gamma_{0}-{\vec{\gamma}}\cdot\bm{\hat{p}})}{{\cal
D}_{+}(\omega,p)}+\frac{1}{2}\frac{(\gamma_{0}+{\vec{\gamma}}\cdot\bm{\hat{p}})}{{\cal
D}_{-}(\omega,p)},$ (473)

The charge invariance demands that ${\cal D}_{\pm}(-\omega,p)=-{\cal
D}_{\mp}(\omega,p)$ which implies that ${\cal A}(-\omega,p)={\cal
A}(\omega,p)$ and ${\cal B}(-\omega,p)=-{\cal B}(\omega,p)$. ${\cal
D}(\omega,p)$ has imaginary part for space like momenta
$P\,(p_{0}^{2}<p^{2})$, it is also useful to define the parity properties for
both real and imaginary parts of ${\cal D}(\omega,p)$ as ${\mathrm{Re}}{\cal
D}_{+}(-\omega,p)=-{\mathrm{Re}}{\cal D}_{-}(\omega,p)$ and
${\mathrm{Im}}{\cal D}_{+}(-\omega,p)={\mathrm{Im}}{\cal D}_{-}(\omega,p)$.

Although the effective propagator in (473) manifests the chiral symmetry, the
poles of ${\cal D}_{\pm}(\omega,p)$ do not occur at light cone, $\omega=\pm
p$. This means that the poles of the effective fermion propagator will be away
from the light cone in the time like domain. This is because the extra term
${\cal B}\gamma_{0}$ appears in self-energy in (313) due to the breaking of
Lorentz invariance at finite temperature HAW . The zeros of ${\cal
D}_{\pm}(\omega,p)$ define dispersion property of a quark in the thermal bath.
${\cal D}_{+}(\omega,p)=0$ has two ploes at $\omega=\omega_{+}(p)$ and
$\omega=-\omega_{-}(p)$ whereas ${\cal D}_{-}(\omega,p)=0$ has two ploes at
$\omega=\omega_{-}(p)$ and $\omega=-\omega_{+}(p)$. Only the positive energy
solutions are displayed in Fig. 27. A mode with energy $\omega_{+}$ represents
the in-medium propagation of a particle excitation This is a Dirac spinors and
eigenstate of $(\gamma_{0}-{\vec{\gamma}}\cdot\bm{\hat{p}})$ with chirality to
helicity ratio $+1$. On the other hand there is a new long wavelength mode
known as plasmino with energy $\omega_{-}$ and eigenstate of
$(\gamma_{0}+{\vec{\gamma}}\cdot\bm{\hat{p}})$ with chirality to helicity
ratio $-1$. The $\omega_{-}$ branch has a minimum at low momentum and then
approaches free dispersion curve at large momentum. It is important to note
that the minimum leads to Van Hove singularities in soft dilepton rate BPY
and meson spectral function KMT .

Below we present the approximate analytic solutions of $\omega_{\pm}(p)$ for
small and large values of momentum $p$. For small values of momentum
($p<<m_{\textrm{th}}$), the dispersion relations are

$\displaystyle\omega_{+}(p)$
$\displaystyle\,\approx\,m_{\textrm{th}}+\frac{1}{3}p+\frac{1}{3}\frac{p^{2}}{m_{\textrm{th}}}-\frac{16}{135}\frac{p^{3}}{m^{2}_{\textrm{th}}}\,,$
(474a) $\displaystyle\omega_{-}(p)$
$\displaystyle\,\approx\,m_{\textrm{th}}-\frac{1}{3}p+\frac{1}{3}\frac{p^{2}}{m_{\textrm{th}}}+\frac{16}{135}\frac{p^{3}}{m^{2}_{\textrm{th}}}\,,$
(474b)

whereas for large values of momentum ($m_{\textrm{th}}<<p<<T$), one obtains

$\displaystyle\omega_{+}(p)$
$\displaystyle\,\approx\,p+\frac{m^{2}_{\textrm{th}}}{p}-\frac{m^{4}_{\textrm{th}}}{2p^{3}}\ln\frac{m^{2}_{\textrm{th}}}{2p^{2}}+\frac{m^{6}_{\textrm{th}}}{4p^{5}}\left[\ln^{2}\frac{m^{2}_{\textrm{th}}}{2p^{2}}+\ln\frac{m^{2}_{\textrm{th}}}{2p^{2}}-1\right]\,,$
(475a) $\displaystyle\omega_{-}(p)$
$\displaystyle\,\approx\,p+2p\exp\left(-\frac{2p^{2}+m^{2}_{\textrm{th}}}{m^{2}_{\textrm{th}}}\right)\,.$
(475b)

We note that at large (hard) momentum the collective mode with chirality to
helicity ratio $+1$ resembles the free particle in vacuum whereas the long
wave length mode, plasmino with chirality to helicity ratio $-1$ decouples
from the plasma. These are clearly evident from (475a) and (475b). At small
momenta both collective modes are equally important which could be seen from
(474a) and (474b). In addition to the pole contributions coming from time like
domain ($\omega^{2}>p^{2}$), ${\cal D}_{\pm}(\omega,p)$ contains a
discontinuous part corresponding to Landau damping coming from space like
domain ($\omega^{2}<p^{2}$) due to the presence of logarithmic term in (472).

###  9.8 Spectral Representation of Fermion Propagator

From (326) the in-medium fermion propagator reads as

$S^{\star}(P)=\frac{1}{2}\frac{(\gamma_{0}-{\vec{\gamma}}\cdot\bm{\hat{p}})}{{\cal
D}_{+}(\omega,p)}+\frac{1}{2}\frac{(\gamma_{0}+{\vec{\gamma}}\cdot\bm{\hat{p}})}{{\cal
D}_{-}(\omega,p)},$ (476)

where ${\cal D}_{\pm}(\omega,p)$ are given in (472) as

$\displaystyle{\cal D}_{\pm}(\omega,p)$ $\displaystyle=$ $\displaystyle
p(\frac{\omega}{p}\mp
1)-\frac{m_{\textrm{th}}^{2}}{p}\left[\frac{1}{2}\left(1\mp\frac{\omega}{p}\right)\ln\left(\frac{\frac{\omega}{p}+1}{\frac{\omega}{p}-1}\right)\pm
1\right]\,.$ (477)

According to (321), ${\cal D}_{\pm}(\omega,p)=d_{\pm}(\omega,p)=\omega\pm p$,
for free massless case. The corresponding free spectral function can be
obtained following (696) as

$\displaystyle\rho^{f}_{\pm}(\omega,k)$ $\displaystyle=$
$\displaystyle\lim_{\epsilon\rightarrow
0}\frac{1}{\pi}{\textrm{Im}}\left.\frac{1}{d_{\pm}(\omega,p)}\right|_{\omega\rightarrow\omega+i\epsilon}$
(478) $\displaystyle=$ $\displaystyle\lim_{\epsilon\rightarrow
0}\frac{1}{\pi}{\textrm{Im}}\frac{1}{\omega\mp
p+i\epsilon}=\frac{\delta(\omega\mp p)}{\left|\frac{\ d(\omega\mp
p)}{d\omega}\right|}=\delta(\omega\mp p)\,.$

As discussed in subsec9.7 that ${\cal D}_{\pm}(\omega,p)$ has solutions at
$\omega_{\pm}(k)$ and $-\omega_{\mp}(p)$ and a cut part due to space like
momentum , $\omega^{2}<p^{2}$. In-medium spectral function corresponding to
the effective fermion propagator in (476) will have both pole and cut
contribution as

$\rho_{\pm}(\omega,p)=\rho^{\textrm{pole}}_{\pm}(\omega,p)+\rho^{\textrm{cut}}_{\pm}(\omega,p)\,.$
(479)

The pole part of the spectral function can be obtained following (696) as

$\displaystyle\rho^{\textrm{pole}}_{\pm}(\omega,p)$ $\displaystyle=$
$\displaystyle\lim_{\epsilon\rightarrow
0}\frac{1}{\pi}{\textrm{Im}}\left.\frac{1}{{\cal
D}_{\pm}(\omega,p)}\right|_{\omega\rightarrow\omega+i\epsilon}=\frac{\delta(\omega-\omega_{\pm})}{\left|\frac{d{\cal
D_{\pm}}}{d\omega}\right|_{\omega=\omega_{\pm}}}+\frac{\delta(\omega+\omega_{\mp})}{\left|\frac{d{\cal
D_{\mp}}}{d\omega}\right|_{\omega=-\omega_{\mp}}}\,$ (480) $\displaystyle=$
$\displaystyle\frac{(\omega^{2}-p^{2})}{2m^{2}_{\textrm{th}}}\delta(\omega-\omega_{\pm})+\frac{(\omega^{2}-p^{2})}{2m^{2}_{\textrm{th}}}\delta(\omega+\omega_{\mp})\,.$

For $\omega^{2}<p^{2}$, there is a discontinuity in
$\ln\frac{\omega+p}{\omega-p}$ as $\ln{y}=\ln\left|y\right|-i\pi\ ,$ which
leads to the spectral function, $\rho^{\textrm{cut}}_{\pm}(\omega,p)$,
corresponding to the discontinuity in ${\cal D}_{\pm}(\omega,p)$ can be
obtained from (694) as

$\displaystyle\rho^{\textrm{cut}}_{\pm}(\omega,p)$ $\displaystyle=$
$\displaystyle\frac{1}{2\pi i}\textrm{Disc}\frac{1}{{\cal
D}_{\pm}(\omega,p)}=\frac{1}{\pi}\lim_{\epsilon\rightarrow
0}\textrm{Im}\left.\frac{1}{{\cal
D}_{\pm}(\omega,p)}\right|_{{\omega\rightarrow\omega+i\epsilon}\atop{\omega<p}}$
(481) $\displaystyle=$
$\displaystyle\\!\\!\frac{\frac{m^{2}_{\textrm{th}}}{2p}\left(\pm\frac{\omega}{p}-1\right)\Theta(p^{2}-\omega^{2})}{\left[\omega\mp
p-\frac{m^{2}_{\textrm{th}}}{p}\left(\pm
1-\frac{p\mp\omega}{2p}\ln\frac{p+\omega}{p-\omega}\right)\right]^{2}+\left[\pi\frac{m^{2}_{\textrm{th}}}{2p}\left(1\mp\frac{\omega}{p}\right)\right]^{2}}$
$\displaystyle=$ $\displaystyle\beta_{\pm}(\omega,p)\Theta(p^{2}-\omega^{2}).$

###  9.9 Calculation of $\Pi_{L}$ and $\Pi_{T}$ from One-loop Photon Self-
energy in HTL Approximation

Figure 28: One loop photon self-energy diagram.

We have QED interaction Lagrangian from (416)

$\mathcal{L}_{int}=-e\bar{\psi}\gamma_{\mu}\psi A^{\mu}.$ (482)

So using Fig. 28 photon self energy can be written as

$\displaystyle\Pi_{\mu\nu}$ $\displaystyle=$
$\displaystyle-\int\frac{d^{4}K}{(2\pi)^{4}}\mathsf{Tr}\left[(-ie\gamma_{\mu})\frac{i\not{K}}{K^{2}}(-ie\gamma_{\nu})\frac{i(\not{P}+\not{K})}{(K+P)^{2}}\right]$
(483) $\displaystyle=$
$\displaystyle-e^{2}\int\frac{d^{4}K}{(2\pi)^{4}}\mathsf{Tr}{\left[\frac{\gamma_{\mu}\not{K}\gamma_{\nu}(\not{P}+\not{K})}{K^{2}(P+K)^{2}}\right]}$
$\displaystyle=$
$\displaystyle-4e^{2}\int\frac{d^{4}K}{(2\pi)^{4}}\left[\frac{2K_{\mu}K_{\nu}-\eta_{\mu\nu}K^{2}-\eta_{\mu\nu}K\cdot
P+(K_{\mu}P_{\nu}+P_{\mu}K_{\nu})}{K^{2}(P+K)^{2}}\right]\,.$

Now in HTL approximation one can neglect external soft momentum BP ;
Weldon:1982aq ; Fradkin ; KK , i.e., one or higher power of $P$. So we get,

$\displaystyle\Pi_{\mu\nu}$ $\displaystyle\approx$
$\displaystyle-8e^{2}\int\frac{d^{4}K}{(2\pi)^{4}}\left[\frac{K_{\mu}K_{\nu}}{K^{2}(P+K)^{2}}\right]+4\eta_{\mu\nu}\int\frac{d^{4}K}{(2\pi)^{4}}\frac{1}{(P+K)^{2}}$
(484) $\displaystyle\approx$
$\displaystyle-8e^{2}\int\frac{d^{4}K}{(2\pi)^{4}}\left[K_{\mu}K_{\nu}\Delta_{F}(K)\Delta_{F}(Q)\right]+4\eta_{\mu\nu}\int\frac{d^{4}K}{(2\pi)^{4}}\Delta_{F}(Q)\,,$

where $\Delta_{F}(K)=\frac{1}{K^{2}}$ and
$\Delta_{F}(Q)=\frac{1}{(P+K)^{2}}=\frac{1}{Q^{2}}$ with $Q=P+K$. We also note
that $F$ stands for fermionic part of the propagator.

We take the time-time part of $\Pi_{\mu\nu}$ as Aritra ; Bithika

$\displaystyle\Pi_{00}$ $\displaystyle=$
$\displaystyle-8e^{2}T\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{\\{K\\}}k_{0}^{2}\Delta_{F}(K)\Delta_{F}(Q)+4e^{2}T\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{\\{K\\}}\Delta_{F}(Q)$
(485) $\displaystyle=$
$\displaystyle-8e^{2}T\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{\\{K\\}}(K^{2}+k^{2})\Delta_{F}(K)\Delta_{F}(Q)+4e^{2}T\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{\\{K\\}}\Delta_{F}(Q)$
$\displaystyle=$
$\displaystyle-4e^{2}T\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{\\{K\\}}\Delta_{F}(K)-8e^{2}T\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{\\{K\\}}k^{2}\Delta_{F}(K)\Delta_{F}(Q)\,,$

where $\sum\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{\\{K\\}}$ is a fermionic sum-
integral and we have used $\Delta_{F}(Q)=1/(P+K)^{2}\approx
1/K^{2}=\Delta_{F}(K)$ in the first term.

Also we obtain Aritra ; Bithika

$\displaystyle\Pi_{\mu}^{\mu}$ $\displaystyle=$
$\displaystyle-8e^{2}T\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{\\{K\\}}K^{2}\Delta_{F}(K)\Delta_{F}(Q)+4e^{2}\delta_{\mu}^{\mu}T\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{\\{K\\}}\Delta_{F}(Q)$
(486) $\displaystyle=$ $\displaystyle
8e^{2}T\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{\\{K\\}}\Delta_{F}(Q)\approx
8e^{2}T\sum\\!\\!\\!\\!\\!\\!\\!\\!\\!\int\limits_{\\{K\\}}\Delta_{F}(K).$

We know the results of frequency sums:

$\displaystyle T\sum_{k_{0}}\Delta_{F}(K)$
$\displaystyle=T\sum_{k_{0}}\frac{1}{K^{2}}=\frac{1}{2k}\left(1-2n_{F}(k)\right),$
(487a) $\displaystyle T\sum_{k_{0}}\Delta_{F}(K)\Delta_{F}(Q)$
$\displaystyle=\frac{1}{4kq}\left[\left(1-n_{F}(k)-n_{F}(q)\right)\left(\frac{1}{\omega-
k-q}-\frac{1}{\omega+k+q}\right)\right.$
$\displaystyle\left.-\left(n_{F}(k)-n_{F}(q)\right)\left(\frac{1}{\omega+k-q}-\frac{1}{\omega-k+q}\right)\right]\,,$
(487b)

where $k=|\bm{\vec{k}}|$, $q=|\bm{\vec{k}+\vec{p}}|$ and the Fermi
distribution is given as $n_{F}(x)=1/(\exp(\beta x)+1)$.

Now using (487a) in (486), one gets

$\Pi_{\mu}^{\mu}=8e^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{2k}\left(1-2n_{F}(k)\right)\,.$
(488)

Neglecting vacuum part, one can have

$\displaystyle\Pi_{\mu}^{\mu}$ $\displaystyle=$
$\displaystyle-\frac{4e^{2}}{\pi^{2}}\int_{0}^{\infty}k\ n_{F}(k)\ dk$ (489)
$\displaystyle=$
$\displaystyle-\frac{4e^{2}}{\pi^{2}}\times\frac{\pi^{2}T^{2}}{12}$
$\displaystyle=$ $\displaystyle-\frac{e^{2}T^{2}}{3}=-m_{D}^{2}\,,$

where the Debye mass in QED is given as $m_{D}^{2}=e^{2}T^{2}/3$.

Now using (487a) and (487b) in (485), one can write the time-time component as

$\displaystyle\Pi_{00}$ $\displaystyle=$
$\displaystyle-4e^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{2k}\left(1-2n_{F}(k)\right)-8e^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{k^{2}}{4kq}$
(490)
$\displaystyle\times\left[\left(1-n_{F}(k)-n_{F}(q)\right)\left(\frac{1}{\omega-
k-q}-\frac{1}{\omega+k+q}\right)\right.$
$\displaystyle\left.-\left(n_{F}(k)-n_{F}(q)\right)\left(\frac{1}{\omega+k-q}-\frac{1}{\omega-k+q}\right)\right]\,.$

Under the HTL approximation BP  one can make following simplifications as

$\displaystyle q\,=\,\left|\bm{\vec{p}}+\bm{\vec{k}}\right|$
$\displaystyle\,=\,\sqrt{p^{2}+k^{2}+2pk\cos\theta}=\sqrt{p^{2}+k^{2}+2pkc}\approx
k+pc=k+{\bm{\vec{p}\cdot\hat{k}}}\,,$ (491a) $\displaystyle n_{F}(q)$
$\displaystyle\,=n_{B}(k+{\bm{\vec{p}\cdot\hat{k}}})\approx
n_{F}(k)+{\bm{\vec{p}\cdot\hat{k}}}\frac{dn_{F}(k)}{dk}\,,$ (491b)
$\displaystyle\omega\pm k\pm q$ $\displaystyle\,=\omega\pm k\pm
k\pm\,{\bm{\vec{p}\cdot\hat{k}}}\approx\pm 2k\,,$ (491c)
$\displaystyle\omega\pm k\mp q$ $\displaystyle\,=\omega\pm k\mp
k\mp\,{\bm{\vec{p}\cdot\hat{k}}}\approx\omega\mp{\bm{\vec{p}\cdot\hat{k}}}\ .$
(491d)

Now using (491a) to (491d) in (487b), one can write the time-time component as

$\displaystyle\Pi_{00}$ $\displaystyle=$
$\displaystyle-4e^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{2k}\left(1-2n_{F}(k)\right)-2e^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\left[\left(1-2n_{F}(k)-{\bm{\vec{p}\cdot\hat{k}}}\frac{dn_{F}(k)}{dk}\right)\left(-\frac{1}{k}\right)\right.$
(492)
$\displaystyle\left.+{\bm{\vec{p}\cdot\hat{k}}}\frac{dn_{F}(k)}{dk}\left(\frac{1}{\omega-{\bm{\vec{p}\cdot\hat{k}}}}-\frac{1}{\omega+{\bm{\vec{p}\cdot\hat{k}}}}\right)\right]$
$\displaystyle=$
$\displaystyle-4e^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{2k}\left(1-2n_{F}(k)\right)+4e^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{2k}\left(1-2n_{F}(k)\right)$
$\displaystyle-2e^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{{\bm{\vec{p}\cdot\hat{k}}}}{k}\frac{dn_{F}(k)}{dk}-2e^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{dn_{F}(k)}{dk}\left(\frac{{\bm{\vec{p}\cdot\hat{k}}}}{\omega-{\bm{\vec{p}\cdot\hat{k}}}}-\frac{{\bm{\vec{p}\cdot\hat{k}}}}{\omega+{\bm{\vec{p}\cdot\hat{k}}}}\right)$
$\displaystyle\approx$
$\displaystyle-2e^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{dn_{F}(k)}{dk}\left(\frac{{\bm{\vec{p}\cdot\hat{k}}}}{\omega-{\bm{\vec{p}\cdot\hat{k}}}}-\frac{{\bm{\vec{p}\cdot\hat{k}}}}{\omega+{\bm{\vec{p}\cdot\hat{k}}}}\right)\,,$

where we have neglected the third term in second line for soft momentum. Now
considering $\cos\theta=-\cos\theta$ in the second term of the last line we
get

$\displaystyle\Pi_{00}(\omega,p)$ $\displaystyle=$
$\displaystyle-2e^{2}\int_{0}^{\infty}\frac{k^{2}\
dk}{2\pi^{2}}\frac{dn_{F}(k)}{dk}\int\frac{d\Omega}{4\pi}\frac{{2\bm{\vec{p}\cdot\hat{k}}}}{\omega-{\bm{\vec{p}\cdot\hat{k}}}}$
(493) $\displaystyle=$
$\displaystyle\frac{4e^{2}}{T}\int_{0}^{\infty}\frac{k^{2}\
n_{F}(k)(1-n_{F}(k))\
dk}{2\pi^{2}}\int\frac{d\Omega}{4\pi}\,\frac{\omega-P\cdot{\hat{K}}}{P\cdot{\hat{K}}}$
$\displaystyle=$ $\displaystyle\frac{4e^{2}}{2\pi^{2}T}\int_{0}^{\infty}k^{2}\
n_{F}(k)(1-n_{F}(k))\ dk\int_{-1}^{1}\frac{dc}{2}\,\left(\frac{\omega}{\omega-
pc}-1\right)$ $\displaystyle=$
$\displaystyle\frac{4e^{2}}{2\pi^{2}T}\times\frac{\pi^{2}T^{3}}{6}\times\left[\frac{\omega}{2p}\ln\frac{\omega+p}{\omega-p}-1\right]=\frac{e^{2}T^{2}}{3}\left[\frac{\omega}{2p}\ln\frac{\omega+p}{\omega-p}-1\right]$
$\displaystyle=$ $\displaystyle
m_{D}^{2}\left[\frac{\omega}{2p}\ln\frac{\omega+p}{\omega-p}-1\right]\,.$

Using (493) in (351) one obtains the longitudinal component of the photon
self-energy Aritra ; Bithika  as

$\displaystyle\Pi_{L}(\omega,p)$ $\displaystyle=$
$\displaystyle-\frac{P^{2}}{p^{2}}\Pi_{00}(\omega,p)$ (494) $\displaystyle=$
$\displaystyle\frac{m_{D}^{2}P^{2}}{p^{2}}\left[1-\frac{\omega}{2p}\ln\frac{\omega+p}{\omega-p}\right]\,.$

Now using (489) and (494) in (355), one obtains the transverse component of
the photon self-energy Aritra ; Bithika  as

$\displaystyle\Pi_{T}(\omega,p)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\left[\Pi_{\mu}^{\mu}-\Pi_{L}\right]\,$ (495)
$\displaystyle=$
$\displaystyle\frac{1}{2}\left[-m_{D}^{2}-\frac{m_{D}^{2}P^{2}}{p^{2}}\left(1-\frac{\omega}{2p}\ln\frac{\omega+p}{\omega-p}\right)\right]$
$\displaystyle=$
$\displaystyle-\frac{m_{D}^{2}\omega^{2}}{2p^{2}}\left[1+\frac{p^{2}-\omega^{2}}{2\omega
p}\ln\frac{\omega+p}{\omega-p}\right]\,.$

The photon self-energies in (494) and (495) in HTL approximation can also be
derived from the Vlasov equation, i.e., a transport equation without collision
term considering a mean field. The only non-classical quantity, entering
thereby, is Fermi-Dirac distribution for the electrons and positrons. The
photon self-energy is obtained via the dielectric constants Elze ; SM ; TG
which follow from the solution of the Vlasov equation for the distribution
function assuming a small derivation from equilibrium LP . The coincidence of
the self-energy, found in the HTL approximation with the one in Vlasov
equation is caused by the equivalence of high temperature limit,
$T\rightarrow\infty$, and the classical limit, $\hbar\rightarrow 0$. As a
matter of fact Silin Silin  had found in 1960 the results given in (494) and
(495) for studying the properties of a relativistic but classical
electromagnetic plasma.

We now note that in the IR limit ($\omega\rightarrow 0$), one gets can also be
derived from

$\displaystyle\lim_{\omega\rightarrow 0}\Pi_{L}(\omega,p)$
$\displaystyle\,=\lim_{\omega\rightarrow
0}-\frac{P^{2}}{p^{2}}\Pi_{00}(\omega,p)=-m_{D}^{2}\,,$ (496a)
$\displaystyle\lim_{\omega\rightarrow 0}\Pi_{T}(\omega,p)$
$\displaystyle\,=0\,.$ (496b)

In QED the (496a) is the Debye electric screening mass of photon acts as a IR
regulator at the static electric scale ($\sim eT$). On the other hand (496b)
indicates that there is no screening for magnetic fields in QED as the 1-loop
photon transverse self-energy in leading order HTL approximation vanishes in
the IR limit and provides no magnetic screening mass for photon as it is a
nonperturbative effect which can not be calculated perturbatively Linde1 ;
Linde2 .

###  9.10 Dispersion Relation and Collective Excitations of Photon in HTL
Approximation

Now we can find the dispersion relations of photon using the thermal
propagator given in (371). The poles of the propagator give the dispersion
relations of photon in thermal medium and one needs to solve the following
equations:

$\displaystyle P^{2}+\Pi_{L}$
$\displaystyle=0\,\,\,\implies\omega^{2}-p^{2}+\frac{P^{2}m_{D}^{2}}{p^{2}}\left[1-\frac{\omega}{2p}\ln{\frac{\omega+p}{\omega-p}}\right]=0\,,$
(497a) $\displaystyle P^{2}+\Pi_{T}$
$\displaystyle=0\,\,\,\implies\omega^{2}-p^{2}-\frac{m_{D}^{2}}{2}\frac{\omega^{2}}{p^{2}}\left[1+\frac{p^{2}-\omega^{2}}{2\omega
p}\ln{\frac{\omega+p}{\omega-p}}\right]=0\,.$ (497b)

Figure 29: Dispersion of photon in heat bath.

The first condition in (497a) gives the longitudinal mode of propagation with
energy $\omega_{L}$ known as plasmon and the second condition in (497b) gives
the transverse mode with energy $\omega_{T}$, which is doubly degenerate as
evident from (343d). The plasmon mode with energy $\omega_{L}$ is a long
wavelength mode that arises solely due to the presence of the thermal medium.
The dispersion relations for photon at one loop are plotted in Fig 29. Energy
of photon splits in two modes due to the presence of a thermal medium. The
longitudinal one with energy $\omega_{L}$ is a long wavelength mode which
reduces to free dispersion very fast than the transverse one with energy
$\omega_{T}$. The same behaviour was also found by Silin Silin  for a
relativistic but classical electromagnetic plasma using Vlasov equation. We
further note that at large (hard) momentum the collective mode with energy
$\omega_{T}$ resembles the transverse photon in vacuum whereas the long wave
length mode, the plasmon with energy $\omega_{L}$, decouples from the plasma.
This is clearly evident from (499a) and (499b). At small momenta both
collective modes are equally important which can be seen from (498a) and
(498b).

Below we obtain approximate analytic solutions of $\omega_{L,T}$ for small and
large values of momentum. For small value of momentum ($p<<m_{D}$),

$\displaystyle\omega_{L}$
$\displaystyle\approx\,\frac{m_{D}}{\sqrt{3}}\left[1+\frac{9}{10}\frac{p^{2}}{m_{D}^{2}}-\frac{27}{280}\frac{p^{4}}{m_{D}^{4}}+\frac{9}{2000}\frac{p^{6}}{m_{D}^{6}}\right],$
(498a) $\displaystyle\omega_{T}$
$\displaystyle\approx\,\frac{m_{D}}{\sqrt{3}}\left[1+\frac{9}{5}\frac{p^{2}}{m_{D}^{2}}-\frac{81}{35}\frac{p^{4}}{m_{D}^{4}}+\frac{792}{125}\frac{p^{6}}{m_{D}^{6}}\right]\,.$
(498b)

For large value of momentum ($p>>m_{D}$),

$\displaystyle\omega_{L}$ $\displaystyle\approx\,p+2p\
\exp\left(-\frac{2(p^{2}+m_{D}^{2})}{m_{D}^{2}}\right),$ (499a)
$\displaystyle\omega_{T}$