form $\cos(\frac{\bm{Q}}{2e})$ [120, 121]. We believe that understanding the
Hamiltonian description of such unusual circuits may also lead to suggestions
for novel physical realizations of such circuits which can then exhibit new
physics.

[title=Ideal gyrator,label=exc:gyr1]

Figure 6.10: Transforming a capacitor into an inductor using a gyrator.

Consider the circuit in Fig. 6.9 with the particular case in which the
Lagrangians $\mathcal{L}_{1}$ and $\mathcal{L}_{2}$ are the Lagrangians of LC
oscillators with inductance $L$ and capacitance $C$ in parallel. Assume the
same inductance and the same capacitance on the two sides. Obtain the
Hamiltonian in this case. What is a mechanical analog of this system? Use the
Lagrangian formalism to show that it is possible to transform a capacitor into
an inductor using a gyrator as shown in Fig. 6.10. We can assume that $g1$ and
$g2$ are reference nodes in each disconnected graph and set
$\Phi_{g1}=\Phi_{g2}=0$. Obtain the value of the equivalent inductance $L$.

[ref=exc:gyr1] In the case of two LC oscillators the Lagrangian of Fig. 6.9
reads

$\mathcal{L}=\frac{C}{2}\dot{\Phi}_{1}^{2}-\frac{\Phi_{1}^{2}}{2L}+\frac{C}{2}\dot{\Phi}_{2}^{2}-\frac{\Phi_{2}^{2}}{2L}+\frac{G}{2}(\Phi_{1}\dot{\Phi}_{2}-\dot{\Phi}_{1}\Phi_{2}).$
(6.23)

To obtain the Hamiltonian we define the conjugate variables

$Q_{1}=\frac{\partial\mathcal{L}}{\partial\dot{\Phi}_{1}}=C\dot{\Phi}_{1}-\frac{G}{2}\Phi_{2},$
(6.24a)
$Q_{2}=\frac{\partial\mathcal{L}}{\partial\dot{\Phi}_{2}}=C\dot{\Phi}_{2}+\frac{G}{2}\Phi_{1}.$
(6.24b)

By taking the Legendre transform we get

$H=Q_{1}\dot{\Phi}_{1}+Q_{2}\dot{\Phi}_{2}-\mathcal{L}=\frac{(Q_{1}+\frac{G}{2}\Phi_{2})^{2}}{2C}+\frac{(Q_{2}-\frac{G}{2}\Phi_{1})^{2}}{2C}+\frac{\Phi_{1}^{2}}{2L}+\frac{\Phi_{2}^{2}}{2L}.$
(6.25)

We notice that this Hamiltonian has the same structure as the Hamiltonian of a
charged particle moving in two dimensions in a constant magnetic field
perpendicular to the plane, with, in addition, a paraboloid potential. This
can be seen even more clearly if we define the fictitious vector potential

$\bm{A}^{T}=(A_{1},A_{2})=\frac{G}{2}(-\Phi_{2},\Phi_{1}),$ (6.26)

and rewrite the Hamiltonian as

$H=\frac{\lvert\bm{Q}-\bm{A}\rvert^{2}}{2C}+\frac{\Phi_{1}^{2}}{2L}+\frac{\Phi_{2}^{2}}{2L},$
(6.27)

with $\bm{Q}=(Q_{1}\quad Q_{2})^{T}$. The Lagrangian of the circuit in Fig.
6.10 reads

$\mathcal{L}=\frac{G}{2}\bigl{(}\Phi_{1}\dot{\Phi}_{2}-\dot{\Phi}_{1}\Phi_{2}\bigr{)}+\frac{C}{2}\dot{\Phi}_{2}^{2}.$
(6.28)

The Euler-Lagrange equation associated with $\Phi_{2}$ reads

$\frac{d}{dt}\biggl{(}\frac{\partial\mathcal{L}}{\partial\dot{\Phi}_{2}}\biggr{)}-\frac{\partial\mathcal{L}}{\partial\Phi_{2}}=G\dot{\Phi}_{1}+C\ddot{\Phi}_{2}=0\implies\dot{\Phi}_{2}=-\frac{G}{C}\Phi_{1}+\alpha,$
(6.29)

with $\alpha\in\mathbb{R}$. Substituting this into the Lagrangian Eq. (6.28),
we obtain

$\mathcal{L}=\frac{G}{2}\Phi_{1}\biggl{(}-\frac{G}{C}\Phi_{1}+\alpha\biggr{)}-\frac{G}{2}\dot{\Phi}_{1}\Phi_{2}+\frac{C}{2}\dot{\Phi}_{2}\biggl{(}-\frac{G}{C}\Phi_{1}+\alpha\biggr{)}=\\\
-\frac{1}{2C/G^{2}}\Phi_{1}^{2}-\frac{G}{2}\frac{d(\Phi_{1}\Phi_{2})}{dt}+\frac{\alpha
G}{2}\Phi_{1}+\frac{\alpha
C}{2}\dot{\Phi}_{2}=-\frac{1}{2C/G^{2}}\Phi_{1}^{2}-\frac{G}{2}\frac{d(\Phi_{1}\Phi_{2})}{dt}+\frac{C}{2}\alpha^{2},$
(6.30)

which is the Lagrangian of an inductor with $L=C/G^{2}$ plus a total time
derivative and a constant. The total time derivative does not change the
Euler-Lagrange equations, see Exercise LABEL:exc:gauge, and can thus be
omitted.

[title=A circulator from a gyrator, label=exc:circ-gyr]

Figure 6.11: Carlin’s construction of a circulator. Here the three terminals
are labelled and are supposed to identify with the terminals in Fig. 6.2.

Let us consider the circuit depicted in Fig. 6.11. It represents a way to
construct a circulator, the three-terminal element in Fig. 6.2, from a
gyrator, known as Carlin’s construction, connected to an external circuit with
some lagrangians $\mathcal{L}_{i}$. It is thus sort of the opposite of Fig.
6.6 in which one obtains a gyrator from the element $\bm{Y}_{c}$. Starting
from the Lagrangian of a gyrator introduced in Eq. (6.12), and setting
$\Phi_{g}=0$ as reference node flux, show that the Lagrangian of the
circulator can be written as

$\mathcal{L}_{\mathrm{cir}}=G(\Phi_{1}\dot{\Phi}_{2}+\Phi_{2}\dot{\Phi}_{3}+\Phi_{3}\dot{\Phi}_{1}).$
(6.31)

_Hint: remember that summing a total time derivative to the Lagrangian does
not change the equations of motion, see Exercise LABEL:exc:gauge._ Building on
the intuition developed in Exercise LABEL:exc:gyr1, derive the admittance
matrix $\bm{Y}_{c}(s)$ of a circulator (as in Eq. (6.3)) from the equations of
motion. From the admittance matrix obtain the scattering matrix assuming that
all the transmission lines at the ports of the circulator have characteristic
impedance $Z_{0}$. Obtain the gyration constant $G$ for which the scattering
matrix becomes the scattering matrix of an ideal impedance-matched circulator
$\bm{S}_{c}$ in Eq. (6.4). What happens if the transmission line is not
impedance-matched?

[ref=exc:circ-gyr] The Lagrangian of the circulator part of the circuit in
Fig. 6.11 follows immediately from the Lagrangian $\mathcal{L}_{\mathrm{gyr}}$
in Eq. (6.11)

$\mathcal{L^{\prime}}_{\mathrm{cir}}=\frac{G}{2}\left(\Phi_{1}\dot{\Phi}_{2}-\dot{\Phi}_{1}\Phi_{2}+\Phi_{2}\dot{\Phi}_{3}-\dot{\Phi}_{2}\Phi_{3}+\Phi_{3}\dot{\Phi}_{1}-\dot{\Phi}_{3}\Phi_{1}\right).$
(6.32)

We see that it is not of the form of Eq. 6.31, but it can be brought into that
form by adding to the Lagrangian a total time derivative, i.e., an equivalent
circulator Lagrangian is

$\mathcal{L}_{\mathrm{cir}}=\mathcal{L^{\prime}}_{\mathrm{cir}}+\frac{G}{2}\frac{d}{dt}(\Phi_{1}\Phi_{2}+\Phi_{2}\Phi_{3}+\Phi_{3}\Phi_{1}).$
(6.33)

We can now proceed to get the admittance matrix. Consider the total Lagrangian

$\mathcal{L}=\mathcal{L}_{1}+\mathcal{L}_{2}+\mathcal{L}_{3}+\mathcal{L}_{\mathrm{cir}},$
(6.34)

with
$\mathcal{L}_{k}=\mathcal{L}_{k}(\\{\Phi_{k},\dot{\Phi}_{k}\\}),\,k=\\{1,2,3\\}$.
The currents $\mathrm{i}_{\mathcal{L}_{k}}$ passing through the element with
Lagrangian $\mathcal{L}_{k}$ are defined as

$\mathrm{i}_{\mathcal{L}_{k}}=\frac{d}{dt}\biggl{(}\frac{\partial\mathcal{L}_{k}}{\partial\dot{\Phi}_{k}}\biggr{)}-\frac{\partial\mathcal{L}_{k}}{\partial\Phi_{k}}.$
(6.35)

At the same time we have

$0=\frac{d}{dt}\biggl{(}\frac{\partial\mathcal{L}}{\partial\dot{\Phi}_{k}}\biggr{)}-\frac{\partial\mathcal{L}}{\partial\Phi_{k}}=\mathrm{i}_{\mathcal{L}_{k}}+\frac{d}{dt}\biggl{(}\frac{\partial\mathcal{L}_{\rm
cir}}{\partial\dot{\Phi}_{k}}\biggr{)}-\frac{\partial\mathcal{L}_{\rm
cir}}{\partial\Phi_{k}}.$ (6.36)

Taking the convention that $\mathrm{i}_{k}=\mathrm{i}_{\mathcal{L}_{k}}$ as
current going into the terminal, the Euler-Lagrange equations associated to
the variables $\Phi_{1},\Phi_{2}$ and $\Phi_{3}$ then read respectively

$\mathrm{i}_{1}+G\dot{\Phi}_{3}-G\dot{\Phi}_{2}=0,$ (6.37a)
$\mathrm{i}_{2}+G\dot{\Phi}_{1}-G\dot{\Phi}_{3}=0,$ (6.37b)
$\mathrm{i}_{3}+G\dot{\Phi}_{2}-G\dot{\Phi}_{1}=0,$ (6.37c)

which we can write compactly as

$\left(\begin{array}[]{c}\mathrm{i}_{1}(t)\\\ \mathrm{i}_{2}(t)\\\
\mathrm{i}_{3}(t)\end{array}\right)=\left(\begin{array}[]{cccc}0&G&-G\\\
-G&0&G\\\ G&-G&0\end{array}\right)\left(\begin{array}[]{c}\mathrm{v}_{1}(t)\\\
\mathrm{v}_{2}(t)\\\ \mathrm{v}_{3}(t)\end{array}\right).$ (6.38)

We can thus identify the admittance matrix of the circulator (in the Laplace
domain) as

$\mathbf{Y}_{\mathrm{c}}(s)=G\left(\begin{array}[]{cccc}0&1&-1\\\ -1&0&1\\\
1&-1&0\end{array}\right).$ (6.39)

To get the scattering matrix from the admittance matrix we use Eq. (5.15)
where $Z_{0}$ is the characteristic impedance of the transmission lines. We
get

$\bm{S}_{\rm
c}=\frac{1}{3G^{2}Z_{0}^{2}+1}\left(\begin{array}[]{ccc}1-G^{2}Z_{0}^{2}&2GZ_{0}(GZ_{0}-1)&2GZ_{0}(GZ_{0}+1)\\\
2GZ_{0}(GZ_{0}+1)&1-G^{2}Z_{0}^{2}&2GZ_{0}(GZ_{0}-1)\\\
2GZ_{0}(GZ_{0}-1)&2GZ_{0}(GZ_{0}+1)&1-G^{2}Z_{0}^{2}\\\ \end{array}\right),$
(6.40)

which only for $G=Z_{0}^{-1}$ becomes an impedance-matched circulator without
any unwanted reflection as in Eq. (6.4). If there is no perfect impedance
matching, we see that there is transmission from, say, port 1 to both port 2
and 3 as well as reflection.

##  Chapter 7 Noise, or all that can go wrong

The physical sources of noise, dissipation and limited control of
superconducting devices are manifold and a subject beyond these Lecture Notes.
In this chapter we first discuss the modeling of noise in Sections 7.1 and
7.2, and then, in Section 7.3, we discuss some aspects of noise sensitivity
and noise protection in superconducting qubits.

###  7.1 Lindblad Master Equation Model

As we have seen in Section 5.3 and Appendix B.3, it is possible to obtain the
Hamiltonian $H$ and the (nonnegative) decay rates $\kappa_{m}$associated with
the modes ($\hat{a}_{m}$) of a $N$-port linear reciprocal network to which $N$
Josephson junctions are attached. A simple dynamical model in which to use
these parameters is the Lindblad master equation:

$\dot{\rho}=-\frac{i}{\hbar}[H,\rho]+\sum_{m=1}^{M}\kappa_{m}\mathcal{D}[\hat{a}_{m}](\rho),$
(7.1)

using the definition ${\cal D}[A](\rho)=A\rho
A^{\dagger}-\frac{1}{2}\\{A^{\dagger}A,\rho\\}$ for any operator $A$. Here $H$
is the Hamiltonian such as in Eq. (5.33) and $\hat{a}_{m}$ is the annihilation
operator of mode $m=1,\ldots,M$ of the network.

Such master equation implicitly assumes that the noise is Markovian: the
environment which introduces the noise is sufficiently large to be unchanged
by the weak interaction with the system whose internal dynamics is governed by
$H$. If this approximation is not warranted, one has to use dynamical
equations which include both system and environment generally. In some cases
one can build models in which the environment is entirely classical, but non-
Markovian, characterized by long temporal correlations, such as in random
telegraph noise. In the Fourier domain, noise with long temporal correlations
has strong low frequency components. Generically, such noise is called $1/f$
noise as its amplitude is often seen to scale as $1/f$. Examples are the noise
due to an environment of so-called two-level defects, noise due to quasi-
particle dynamics, or other temporally-correlated flux or charge noise. For
temporally-correlated noise, one does not typically get an exponential decay
of qubit coherence, but decay of the form $\exp(-\gamma t^{\alpha})$ with
$\alpha>1$, or a power law decay is possible, see e.g. [122, 123].

In the master equation above, only energy loss is modeled, but if the
frequency of the mode is relatively low as compared to the dilution fridge
temperature $T\approx 20\,\mathrm{mK}$, one should include both loss and gain,
which leads to

$\dot{\rho}=-\frac{i}{\hbar}[H,\rho]+\sum_{m=1}^{M}\kappa_{m,-}\mathcal{D}[\hat{a}_{m}](\rho)+\sum_{m=1}^{M}\kappa_{m,+}\mathcal{D}[\hat{a}_{m}^{\dagger}](\rho).$
(7.2)

with $\kappa_{m,-}=\kappa_{m}(1+\overline{n}(\omega_{m}))$ and
$\kappa_{m,+}=\kappa_{m}\overline{n}(\omega_{m})$ with thermal occupation
number

$\displaystyle\overline{n}(\omega_{m})=\frac{1}{e^{\beta\hbar\omega_{m}}-1},$
(7.3)

and thermodynamic beta $\beta=\frac{1}{k_{B}T}$. Even though for a transmon
qubit at, say, $5\,\mathrm{GHz}$, $\overline{n}$ is tiny from a thermal
perspective, additional sources of excitations are known to exist, see
Exercise LABEL:exc:temp.

Now imagine that there is a single Josephson junction $N=1$ and hence we want
to consider the decoherence of the qubit associated with that Josephson
junction. Then Eq. (7.2) does not look like a very direct and efficient way of
doing this, as the noise comes in through the $M$ modes to which the qubit is
coupled. A more direct model starts from a qubit circuit connected to a lossy
circuit branch and will be considered in Section 7.2 and Appendix B.4.

In the remainder of this section we discuss some aspects of a generic qubit
Lindblad equation —assuming it can be derived to model noise— through two
exercises.

A qubit version of such Lindblad equation in Eq. (7.2) with
$H=-\frac{\hbar\Omega}{2}Z$, reads

$\dot{\rho}=-\frac{i}{\hbar}[H,\rho]+\kappa_{-}\mathcal{D}[\sigma^{-}](\rho)+\kappa_{+}\mathcal{D}[\sigma^{+}](\rho),$
(7.4)

with $\kappa_{+}=\kappa_{-}e^{-\beta\hbar\Omega}$. Solving the dynamics of the
qubit Lindblad equation is straightforward, see Exercise LABEL:exc:lind. The
qubit state $\rho(t)$ will equilibrate to the Gibbs state
$\rho_{\beta}=e^{-\beta H}/{\rm Tr}(e^{-\beta H})$ at sufficiently long time
$t$. The relaxation time $T_{1}$ sets the rate of decay of the diagonal
elements $\bra{0}(\rho(t)-\rho_{\beta})\ket{0}\propto e^{-t/T_{1}}$ where
$T_{1}$ can be related to $\kappa_{\pm}$ as $1/T_{1}=\kappa_{+}+\kappa_{-}$.
The off-diagonal elements of $\rho(t)$ will decay as
$\bra{0}\rho(t)\ket{1}\propto e^{-t/T_{2}}$ with $T_{2}=2T_{1}$. To model
other sources of (fast, Markovian) dephasing than just dephasing due to
relaxation, one can introduce a pure dephasing time $T_{\phi}$ defined through

$\displaystyle\frac{1}{T_{2}}=\frac{1}{2T_{1}}+\frac{1}{T_{\phi}},$ (7.5)

where $T_{2}$ can be taken as the ‘experimentally measured’ dephasing time. In
the qubit Lindblad equation, one can then model such additional dephasing by
including an additional pure dephasing term $\gamma\mathcal{D}[Z]$ with
$\gamma=\frac{1}{2T_{\phi}}$. In practice, the measured $T_{1}$ and $T_{2}$
times of superconducting qubits —$T_{2}$ is often measured with and without
applied echo or dynamical decoupling pulses which remove slow-varying noise
components— fluctuate considerably over time, and when a transmon is flux-
tuned to different frequencies, see e.g. [124], and studies on the impact of
cosmic rays [125] and references therein. Clearly, much more detailed physical
models are necessary to capture the non-equilibrium effects at varying time
scales [126].

[title=Dephasing and relaxation in the qubit Lindblad equation,label=exc:lind]
Observe that $\sigma^{-}=\ket{0}\bra{1}=\frac{1}{2}(X+iY)$. Determine
$\frac{d}{dt}\left({\rm Tr}\,\sigma^{-}\rho(t)\right)={\rm
Tr}\,\sigma^{-}\dot{\rho}(t)$ with $\rho(t)$ the solution of the Lindblad
equation in Eq. (7.4). Then solve for ${\rm Tr}\,\sigma^{-}(\rho(t)$ given
${\rm Tr}\,X\rho(t=0)=1$ and ${\rm Tr}\,Y\rho(t=0)=0$, i.e.
$\rho(t=0)=\ket{+}\bra{+}$ . Verify that
$\frac{2}{T_{2}}=\kappa_{+}+\kappa_{-}$. Include the additional dephasing term
$\gamma\mathcal{D}[Z]$ in the Lindblad equation and verify how $T_{\phi}$
relates to $\gamma$ by again examining $\frac{d}{dt}\left({\rm
Tr}\,\sigma^{-}\rho(t)\right)$. Solve for ${\rm Tr}\,\ket{0}\bra{0}\rho(t)$ as
well, confirming that $T_{2}=2T_{1}$.

[ref=exc:lind] We have

${\rm Tr}\,X\dot{\rho}(t)=\frac{i\Omega}{2}{\rm
Tr}\,X[Z,\rho(t)]-\frac{1}{2}(\kappa_{-}+\kappa_{+}){\rm
Tr}\,X\rho(t)=\Omega{\rm Tr}\,Y\rho(t)-\frac{1}{2}(\kappa_{-}+\kappa_{+}){\rm
Tr}\,X\rho(t)$

and

${\rm Tr}\,Y\dot{\rho}(t)=\frac{i\Omega}{2}{\rm
Tr}\,Y[Z,\rho(t)]-\frac{1}{2}(\kappa_{-}+\kappa_{+}){\rm
Tr}Y\rho(t)=-\Omega{\rm Tr}X\rho(t)-\frac{1}{2}(\kappa_{-}+\kappa_{+}){\rm
Tr}Y\rho(t)$

or

$\frac{d}{dt}\left({\rm
Tr}\,\sigma^{-}\rho(t)\right)=(-i\Omega-\frac{1}{2}(\kappa_{-}+\kappa_{+})){\rm
Tr}\,(\sigma^{-}\rho(t)),$

which gives the solution ${\rm Tr}\,\sigma^{-}\rho(t)=\frac{1}{2}e^{-i\Omega
t-(\kappa_{+}+\kappa_{-})t/2}$ with the given initial condition, showing the
exponential decay with $T_{2}=\frac{2}{\kappa_{+}+\kappa_{-}}$. We have
$\mathcal{D}[Z](\rho(t))=Z\rho Z-\rho$ and thus $\gamma{\rm
Tr}X\mathcal{D}[Z](\rho)=-2\gamma{\rm Tr}X\rho(t)$ and the same for
$\gamma{\rm Tr}\,Y\mathcal{D}[Z](\rho)=-2\gamma{\rm Tr}Y\rho(t)$. This gives
an additional decay $e^{-2\gamma t}\equiv e^{-t/T_{\phi}}$ so
$T_{\phi}=\frac{1}{2\gamma}$. We have

$\frac{d}{dt}({\rm Tr}\ket{0}\bra{0}\rho(t))=\kappa_{-}{\rm
Tr}\ket{1}\bra{1}\rho(t)-\kappa_{+}{\rm
Tr}\ket{0}\bra{0}\rho(t)=\kappa_{-}-(\kappa_{+}+\kappa_{-}){\rm
Tr}\ket{0}\bra{0}\rho(t).$

or ${\rm
Tr}\ket{0}\bra{0}\rho(t)=Ce^{-(\kappa_{+}+\kappa_{-})t}+\frac{\kappa_{-}}{\kappa_{+}+\kappa_{-}}$
for some constant $C$, confirming that
$\frac{1}{T_{1}}=\kappa_{+}+\kappa_{-}$. Note that ${\rm
Tr}\ket{0}\bra{0}\rho_{\beta}=\frac{\kappa_{-}}{\kappa_{+}+\kappa_{-}}$.

[title=Thermal Equilibration?,label=exc:temp] A transmon qubit at
$f=\Omega/2\pi=5$GHz is left alone to idle and we imagine that it equilibrates
to the temperature of the dilution fridge which is kept at 20mK. What is the
probability to find the transmon qubit in $\ket{0}$ after it equilibrates? Do
the analysis both by treating the transmon as a qubit as well as a purely
harmonic system (what is the difference in answers?). You can use that $50{\rm
mK}\equiv 1{\rm GHz}$. When you compare your answers with an experimentally
estimated probability to find the transmon qubit in $\ket{0}$, see e.g. Fig. 1
in [127] where the ground-state occupancy is estimated as $80\%$, you see that
this thermal model is far too simple: there are other sources of noise which
make that the qubit is not measured as $\ket{0}$. Same question for a
fluxonium qubit, treating it as a qubit at $f=\Omega/2\pi=300$MHz. What does
this tell you about fluxonium qubit initialization?

[ref=exc:temp] We imagine that the qubit equilibrates to the Gibbs state
$\rho=\frac{e^{-\beta H}}{{\rm Tr}(e^{-\beta H})}$ with
$\beta=\frac{1}{k_{B}T}$. If we treat the transmon as a qubit we have
$H=-\frac{\hbar\Omega}{2}Z$, $\hbar\Omega=hf$ and
$\mathbb{P}(0)=\bra{0}\rho\ket{0}=\frac{1}{1+e^{-\beta\hbar\Omega}}=\frac{1}{1+e^{-5/(4/5)}}\approx
0.998$. If we treat the transmon as a harmonic system with
$H=\hbar\Omega(\hat{n}+\frac{1}{2})$, we have
$\mathbb{P}(0)=1-e^{-\beta\hbar\Omega}=1-e^{-5/(4/5)}\approx 0.998$ which is
identical within this precision. For the fluxonium qubit
$\mathbb{P}(0)=\bra{0}\rho\ket{0}=\frac{1}{1+e^{-(3/10)/(4/5)}}=\frac{1}{1+e^{-3/8})}\approx
0.593$. Due to its low frequency, a fluxonium qubit needs to be actively
measured (reset) to initialize the qubit to $\ket{0}$, as its thermal
population in $\ket{1}$ is rather high.

###  7.2 A qubit circuit with a lossy circuit branch

We imagine that in one of the qubit circuits we have been discussing in these
Lecture Notes, there is also some lossy element, which can be described as an
additional circuit branch between the qubit nodes with admittance
$\mathcal{Y}(\omega)=1/\mathcal{Z}(\omega)$. Since it is lossy, $\mathcal{Y}$,
and $\mathcal{Z}$, will have non-zero real parts as we have seen in the
single-port discussion in Section 5.3. This branch may be just a simple
resistor $R$, but it can also be composed of several circuit elements. We will
confine ourselves to the case where ${\rm Re}(\mathcal{Y})$ is small (so that
$\mathcal{Z}$, or $R$, is large); this makes sense because we will only
attempt to make a qubit in cases where the loss is small, so that the lossy
branch is only a small perturbation on the circuit.

We are thus considering a scenario such as in Fig. 7.1 where a LC oscillator,
or any other type of qubit like a transmon or flux qubit, is weakly
capacitively coupled to such resistive element. This resistive element can
also physically be an (unobserved) transmission line.

We observe that a ‘lossy Foster’ analysis in Section 5.3 in which each LC
oscillator has a resistance in parallel, does not fit the set-up in Figure 7.1
very well. The claim of the lossy Foster circuit is that one can write any
$Z(s)$, for example $Z(s)$ in parallel with the LC oscillator in Fig. 7.1, as
a series of RLC circuits (see Fig. 5.5 with $N=1$). However, in this case this
is a somewhat singular representation as we are missing some capacitors and
inductors. A more direct approach, explained in Appendix B.4 for an LC
oscillator in parallel with an arbitrary, but small admittance, is more
suitable. The physics approach to model the decoherence of the qubit is to to
treat the resistive element as a continuum bath of oscillators, with a certain
spectral density, in a thermal state at a given temperature, and consider the
system-bath Hamiltonian. This is a so-called Caldeira-Leggett model originally
studied in Ref. [128] (see also Refs. [7, 129] for the application to
electrical circuits). Then to such full system + bath Hamiltonian, one can
apply a Born-Markov analysis [71] through which one obtains a Lindblad master
equation for the qubit system only, treating the bath as a large unperturbed
weakly-coupled environment. The dissipative terms in this Lindblad master
equation which drive the qubit to a thermal state then set the relaxation time
$T_{1}$ as in Eq. (7.4). For example, without going through the derivation, we
quote a result from [50], Appendix 3 (see its derivation as Eq. (131) of [4]
and Section $3$ of [130]) for the decay rate:

$T_{1}^{-1}=\frac{1}{\hbar}|\langle
0|\hat{\phi}_{i}-\hat{\phi}_{j}|1\rangle|^{2}\left(\frac{\Phi_{0}}{2\pi}\right)^{2}\Omega\mbox{Re}(\mathcal{Y}(\Omega))\left(\coth\frac{\beta\hbar\Omega}{2}+1\right)$
(7.6)

where $\hbar\Omega$ is the energy splitting between the state $\ket{0}$ and
$\ket{1}$ of the qubit. The nodes $i$ and $j$ are those between which the
lossy impedance (with admittance $\mathcal{Y}(\omega)$) is attached as a
branch. Note that $\coth{\beta\hbar\Omega/2}=1+2\overline{n}(\Omega)$ with
$\overline{n}(\Omega))$ with thermal occupation number as in Eq. (7.3). We see
appearing here the operators for the dimensionless node fluxes of nodes $i$
and $j$; if one node is ground, the corresponding operator is simply omitted.

Eq. (7.6) also holds in the case when several qubit circuit in series are
inserted between these nodes $\phi_{i}$ and $\phi_{j}$ (and it could be that
the qubit wave functions are largely independent of $\phi_{i}$ and
$\phi_{j}$). But it is worthwhile to consider the application of this equation
to the very simplest example, that of a single LC circuit shunted by a
resistance $R$. The evaluation of the matrix element proceeds using Eqs.
(2.28b),(2.18a), yielding

$\langle
0|\hat{\phi}|1\rangle=\frac{2\pi}{\Phi_{0}}\sqrt{\frac{\hbar}{2}\sqrt{\frac{L}{C}}}.$
(7.7)

Plugging this into Eq. (7.6), with the additional replacements
$\Omega\mapsto\omega_{r}=1/\sqrt{LC}$ and
$\mbox{Re}(\mathcal{Y}(\omega_{r}))=1/R$, and in the low-temperature limit
($\coth(\cdot)+1\rightarrow 2$), we get the simple result

$T_{1}^{-1}=\frac{1}{RC}.$ (7.8)

This straightforward identification of $T_{1}$ with “RC decay” is discussed
further in [131]. In Appendix B.4 we show that this identification holds more
generally for LC circuits in parallel with a weakly perturbing admittance (see
Eq. (B.60)).

It is illuminating to collect the constants differently in Eq. (7.6) so that
it is rewritten as

$T_{1}^{-1}=|\langle
0|\hat{\phi}_{i}-\hat{\phi}_{j}|1\rangle|^{2}\frac{\Omega}{2\pi}R_{Q}\mbox{Re}(\mathcal{Y}(\Omega))\left(\coth\frac{\beta\hbar\Omega}{2}+1\right).$
(7.9)

where $R_{Q}$ is the quantum of resistance in Eq. (3.22). This rewrite makes
the units of the expression very transparent, and it makes clear that one
aspect of $T_{1}^{-1}$ being “small” is that
$\mbox{Re}\,{\mathcal{Y}}(\Omega)\ll R_{Q}^{-1}$.

Figure 7.1: An unobserved transmission line (not driven) with a characteristic
impedance $Z_{0}$ in (a) can be replaced by a resistor with resistance $Z_{0}$
in (b).

[title=Qubit relaxation caused by a long transmission
line,label=exc:transloss] In Fig. 2.11 we showed a simple resonant circuit
driven by a voltage source. The real situation is usually more like what it
shown in Fig. 7.1(a). The voltage source is actually many meters away,
connected to the experiment by a long length of transmission line (cf. Fig.
5.3). As seen by the resonant circuit, it is well known that an infinite
transmission line with wave impedance $Z_{0}$ has the same action as an actual
resistor of that same resistance $Z_{0}$ (see e.g. [76]). Of course, the
transmission line is not infinite, but there is in reality, after about a
meter of transmission line, a filter with a resistor with resistance $Z_{0}$
in it. So the situation is well represented as in Fig. 7.1(b). The actual
resistor is still in the coldest part of the cryostat, so we can say
$T=20\,\mathrm{mK}$. Many aspects of this problem resemble the analysis given
in Ref. [132], where an experiment is also presented.

Using the temperature given and the following parameters:
$Z_{0}=50\,\mathrm{\Omega}$, $E_{C}/h=250\,\mathrm{MHz}$ (see Eq. (2.21)),
$E_{L}/h=12.5\,\mathrm{GHz}$ (see Eq. (2.22)), and $C_{g}/C=0.005$
(reasonable, following [133]), compute $T_{1}$, and compare it with the period
of oscillation of the resonator. Treat the RC series combination on the left
of Fig. 7.1(b) as a single composite branch.

[ref=exc:transloss] Note that $\omega_{r}=1/\sqrt{LC}$. The relevant
quantities for our lossy branch are

$\displaystyle\mathcal{Z}(\omega_{r})$ $\displaystyle=$ $\displaystyle
Z_{0}+\frac{1}{i\omega_{r}C_{g}}$ $\displaystyle\mathcal{Y}(\omega_{r})$
$\displaystyle=$
$\displaystyle\frac{i\omega_{r}C_{g}}{1+i\omega_{r}C_{g}Z_{0}}$
$\displaystyle\mbox{Re}(\mathcal{Y}(\omega_{r}))$ $\displaystyle=$
$\displaystyle\frac{\omega_{r}^{2}C_{g}^{2}Z_{0}}{1+\omega_{r}^{2}C_{g}^{2}Z_{0}^{2}}$
$\displaystyle\mbox{Re}(\mathcal{Y}(\omega_{r}))$ $\displaystyle=$
$\displaystyle\frac{C_{g}^{2}Z_{0}/CL}{1+C_{g}^{2}Z_{0}^{2}/CL}$ (7.10)

Note that the factor in the denominator of the last equation can be expressed

$\frac{Z_{0}^{2}C_{g}^{2}}{LC}=Z_{0}^{2}\frac{C}{L}\left(\frac{C_{g}}{L}\right)^{2}=\frac{(2\pi)^{2}}{8}\left(\frac{Z_{0}}{R_{Q}}\right)^{2}\left(\frac{C_{g}}{C}\right)^{2}\frac{E_{L}}{E_{C}}\ll
1.$ (7.11)

This is true even though $E_{L}>E_{C}$, because the other factors are very
small. Thus, also noting the magnitude of the temperature, we can approximate

$\mbox{Re}({\mathcal{Y}}(\omega_{r}))\approx\frac{C_{g}^{2}Z_{0}}{CL}=\frac{(2\pi)^{2}}{8}\frac{1}{R_{Q}}\frac{Z_{0}}{R_{Q}}\left(\frac{C_{g}}{C}\right)^{2}\frac{E_{L}}{E_{C}},\,\,\,\,\mbox{
and }\,\,\,\,\coth{\beta\hbar\omega_{r}/2}\approx 1$ (7.12)

To evaluate the matrix element, we use the harmonic oscillator result from Eq.
(2.30a), so that

$|\langle 0|\hat{\phi}_{2}|1\rangle|^{2}=\sqrt{\frac{2E_{C}}{E_{L}}}.$ (7.13)

Collecting everything together, using Eq. (7.9) gives

$\displaystyle T_{1}^{-1}$ $\displaystyle=$
$\displaystyle\sqrt{\frac{2E_{C}}{E_{L}}}\cdot\frac{\omega_{r}}{2\pi}\cdot
R_{Q}\cdot\frac{(2\pi)^{2}}{8}\frac{1}{R_{Q}}\frac{Z_{0}}{R_{Q}}\left(\frac{C_{g}}{C}\right)^{2}\frac{E_{L}}{E_{C}}\cdot
2$ (7.14) $\displaystyle=$
$\displaystyle\frac{(2\pi)^{2}}{2\sqrt{2}}\sqrt{\frac{E_{L}}{E_{C}}}\frac{Z_{0}}{R_{Q}}\left(\frac{C_{g}}{C}\right)^{2}\frac{\omega_{r}}{2\pi}$
$\displaystyle=$ $\displaystyle 0.00019\cdot 6{\rm GHz}\approx 1.1{\rm MHz}.$

This gives a reasonable, but not state of the art, decay rate of about
$1\,\mathrm{\mu s}$. Longer lifetimes are achieved by using smaller values of
the coupling capacitance $C_{g}$, and sometimes by the use of a “Purcell
filter”, which strongly reduces ${\rm Re}(\mathcal{Y})$ at the frequency
$\omega_{r}$.

###  7.3 Noise sensitivity and protection

Imagine we wish to engineer a Hamiltonian $H_{\rm ideal}$ of a single
superconducting qubit, using the conjugate operators $\hat{\phi}$ and
$\hat{q}$. In practice, this system is coupled to other systems, quantum or
classical, whose dynamics affects the system introducing noise, and let the
Hamiltonian including the coupling to the environment be $\tilde{H}$. One can
roughly classify noise as either charge noise or flux noise, meaning that it
couples to either the charge variable $\hat{q}$ or the flux variable
$\hat{\phi}$ in $\tilde{H}$.

Both flux noise and charge noise can contribute to both qubit dephasing and
qubit relaxation: what noise is dominant depends on coupling strengths and the
character and symmetry of the qubit states. A simple model for the rate
$1/T_{1}$ is given by Fermi’s golden rule which says that

$\frac{1}{T_{1}}=\frac{2\pi}{\hbar}|\bra{0}\tilde{H}\ket{1}|^{2}\rho(\Omega),$
(7.15)

where $\rho(\Omega)$ is the density of states of the environmental degrees of
freedom at the qubit frequency $\Omega$. One can compare this expression with
Eq. (7.6) where the coupling is expressed in terms of the branch flux
operator, the admittance $\mathcal{Y}(\Omega)$ and thermal properties of the
environment at frequency $\Omega$. Clearly, whether flux and charge noise
contribute to relaxation via Eq. (7.15), depends on whether terms which
involve the charge $\hat{q}$ and/or flux operator $\hat{\phi}$ in $\tilde{H}$
induce transitions between $\ket{0}$ and $\ket{1}$.

##### Noise sensitivity

As we know, the transmon qubit is designed such that it is less sensitive to
charge noise. This is expressed in the plots in Fig. 4.2 where one sees a
reduced sensitivity towards noise in the offset charge $n_{g}$ of the energies
of all eigenstates $\ket{\psi_{k}}$. Let $\tilde{H}$ be the Hamiltonian when
the offset charge $n_{g}$ is slightly different than some ideal value
$n_{g}^{\rm ideal}$, i.e., $\tilde{H}(n_{g}=n_{g}^{\rm ideal})=H_{\rm ideal}$
and so the insensitivity is expressed as

$\bra{\psi_{m}}\tilde{H}\ket{\psi_{m}}\approx\bra{\psi_{m}}H_{\rm
ideal}\ket{\psi_{m}},$ (7.16)

where $\ket{\psi_{m}}$ are eigenstates of $H_{\rm ideal}$. For the transmon
this comes about due to the weak dependence on $n_{g}$ in $E_{m}(n_{g})$ in
Eq. (4.30) for large $E_{J}/E_{C}$. This insensitivity lengthens the dephasing
time —limiting the $Z$ error strength— of the qubit due to charge noise, since
it implies that

$\bra{0}\tilde{H}\ket{0}-\bra{1}\tilde{H}\ket{1}\approx\bra{0}H_{\rm
ideal}\ket{0}-\bra{1}H_{\rm ideal}\ket{1}.$ (7.17)

If a qubit is insensitive to dephasing errors one expects that the dephasing
$T_{2}$ is long, and dominated by relaxation-induced dephasing, i.e.,
$\frac{1}{T_{2}}\gtrapprox\frac{1}{2T_{1}}$.

Observe that a different question is whether the matrix element
$|\bra{0}\tilde{H}\ket{1}|$ is small. Note that by definition one has
$|\bra{0}H_{\rm ideal}\ket{1}|=0$ since $\ket{0},\ket{1}$ are the ideal
eigenstates, and one thus considers the effect of the (time-varying,
stochastic) perturbation in $\tilde{H}$ which couples the qubit to the
environment. Just from the fact that the transmon is in the regime $E_{C}\ll
E_{J}$ and any charge noise comes with energy scale $E_{C}$, charge noise
causing transitions between $\ket{0}$ and $\ket{1}$ — bit flips $X$ — will be
suppressed. For these general reasons, a flux-type qubit where the
contributions for capacitance terms scaling with $E_{C}$ are relatively small,
is naturally protected against charge noise.

####  7.3.1 Flux sweetspots

Any loop in a superconducting lumped-circuit representation of a device
through which a current can run, is affected by (stray, fluctuating) magnetic
fields and can lead to flux noise.

A common form of protection against classical flux noise, is obtained when the
ideal Hamiltonian $H_{\rm ideal}$ operates at a so-called external flux-sweet
spot. Imagine setting the potential $U(\phi,\phi_{\rm ext})$ to the point
$\phi_{\rm ext}=\phi_{\rm ext}^{\rm ideal}$. Given a sufficiently small amount
of noise on the external flux $\phi_{\rm ext}$, we can then write

$\tilde{H}(\phi_{\rm ext})=H_{\rm ideal}+(\phi_{\rm ext}-\phi_{\rm ext}^{\rm
ideal})\frac{\partial U}{\partial\phi_{\rm ext}}\biggl{|}_{\phi_{\rm ext}^{\rm
ideal}}+\frac{1}{2}(\phi_{\rm ext}-\phi_{\rm ext}^{\rm
ideal})^{2}\frac{\partial^{2}U}{\partial\phi_{\rm
ext}^{2}}\biggl{|}_{\phi_{\rm ext}^{\rm ideal}}+\ldots$

Now we can consider how fluctuations in $\phi_{\rm ext}$ affect the qubit
frequency, i.e.

$\bra{0}\tilde{H}(\phi_{\rm ext})\ket{0}-\bra{1}\tilde{H}(\phi_{\rm
ext})\ket{1}=\hbar\Omega^{\rm ideal}\\\ +(\phi_{\rm ext}-\phi_{\rm ext}^{\rm
ideal})\left(\bra{0}\frac{\partial U}{\partial\phi_{\rm
ext}}\biggl{|}_{\phi_{\rm ext}^{\rm ideal}}\ket{0}-\bra{1}\frac{\partial
U}{\partial\phi_{\rm ext}}\biggl{|}_{\phi_{\rm ext}^{\rm
ideal}}\ket{1}\right)+\mathcal{O}((\phi_{\rm ext}-\phi_{\rm ext}^{\rm
ideal})^{2}),$

where $\Omega^{\rm ideal}$ is the targeted transition frequency. We say that
$\phi_{\rm ext}^{\rm ideal}$ is a flux sweet-spot when

$\bra{0}\frac{\partial U}{\partial\phi_{\rm ext}}\biggl{|}_{\phi_{\rm
ext}^{\rm ideal}}\ket{0}-\bra{1}\frac{\partial U}{\partial\phi_{\rm
ext}}\biggl{|}_{\phi_{\rm ext}^{\rm ideal}}\ket{1}=0,$ (7.18)

A common case is when $U$ depends on $\cos(\phi-\phi_{\rm ext}^{\rm ideal})$
with $\phi_{\rm ext}^{\rm ideal}=k\pi$ for $k\in\mathbb{Z}$, such that
$\bra{\psi}\frac{\partial U}{\partial\phi_{\rm ext}}\big{|}_{\phi_{\rm
ext}^{\rm ideal}}\ket{\psi}\propto\bra{\psi}\sin(\phi)\ket{\psi}=0$ for any
eigenstate $\ket{\psi}$ due to the parity symmetry of the ideal Hamiltonian,
as discussed in Section 3.4. Thus $\phi_{\rm ext}^{\rm ideal}=k\pi$ are then
flux sweet-spots where one expects the longest $T_{2}$ time.

[label=exc:ss] Show that small fluctuations in $\phi_{\rm ext}$ at a flux
sweetspot for which Eq. (7.18) holds, can still generate bit-flip ($X$) errors
and thus lower $T_{1}$, i.e.,

$|\bra{0}\tilde{H}(\phi_{\rm ext})\ket{1}|,$ (7.19)

is not necessarily small when $U$ is parity-symmetric at $\phi_{\rm ext}^{\rm
ideal}$, $U(\phi,\phi_{\rm ext}=\phi_{\rm ext}^{\rm ideal})=U(-\phi,\phi_{\rm
ext}=\phi_{\rm ext}^{\rm ideal})$ and we wish to be able to drive the $0-1$
transition of the qubit.

[ref=exc:ss] To first order we have

$|\bra{0}\tilde{H}(\phi_{\rm ext})\ket{1}|=(\phi_{\rm ext}-\phi_{\rm ext}^{\rm
ideal})|\bra{0}\frac{\partial U}{\partial\phi_{\rm ext}}\biggl{|}_{\phi_{\rm
ext}^{\rm ideal}}\ket{1}|.$ (7.20)

Since $H_{\rm ideal}$ is parity-symmetric, $\frac{\partial
U}{\partial\phi_{\rm ext}}$ is an odd function $f_{\rm odd}(-\phi)=-f_{\rm
odd}(\phi)$. The states $\ket{0}$ and $\ket{1}$ should have different parity
symmetry to be able to drive the qubit (due to Eqs. (3.65) and (3.66)), hence
there is no symmetry or sweet-spot reason why Eq. (7.19) should be small.

####  7.3.2 Built-in protection

Here we discuss forms of noise protection which are based on shaping the form
of the wave function as in a quantum error correcting code. For general
perspective on protected qubits we refer to Ref. [134].

One can generally write

$\tilde{H}=H_{\rm ideal}+f(\hat{\phi})+g(\hat{q}),$

where $f(.)$ and $g(.)$ are some functions which can also depend on additional
quantum or classical environment degrees of freedom, or be time-dependent. For
the purpose of the next arguments, the form of this dependence is not relevant
so we don’t specify it exactly.

Imagine we define a qubit with $\psi_{0}(\phi)$ and $\psi_{1}(\phi)$ the
wavefunctions of the state $\ket{0}$ resp. $\ket{1}$, which are eigenstates of
$H_{\rm ideal}$. If the wavefunctions $\psi_{0}(\phi)$ and $\psi_{1}(\phi)$
have approximately disjoint support, then

$\bra{0}f(\hat{\phi})\ket{1}=\int_{-\infty}^{\infty}d\phi\,\psi_{0}^{*}(\phi)f(\phi)\psi_{1}(\phi)\approx
0,$ (7.21)

Similarly, we have for the function $g(\hat{q})=\hat{q}^{k}$ for any finite
$k$:

$\bra{0}g(\hat{q})\ket{1}=\int_{-\infty}^{\infty}d\phi\,\psi_{0}^{*}(\phi)\left(-i\frac{\partial}{\partial\phi}\right)^{k}\psi_{1}(\phi)\approx
0.$ (7.22)

Note, however, that if $g(\hat{q})=e^{i\epsilon\hat{q}}$, that is, a
displacement of sufficient magnitude $\epsilon$, then the support of
$\psi_{0}(\phi)$ could be displaced to that of $\psi_{1}(\phi)$ and
$\bra{0}g(\hat{q})\ket{1}\neq 0$. However, we don’t naturally expect such
displacement terms in the Hamiltonian since $\hat{q}$ represents the charge
operator.

Eqs. (7.21) and (7.22) imply that the noisy Hamiltonian $\tilde{H}$ cannot
cause any transition from $\ket{0}$ to $\ket{1}$ and vice-versa, leading to
the qubit being protected against bit-flip $X$ errors induced by charge or
flux noise. Said differently, energy exchange with the environment inducing
$\ket{0}\leftrightarrow\ket{1}$ is suppressed as we cannot couple into these
transitions and one expects that the qubit relaxation time $T_{1}$ is very
long for this reason.

An example of a qubit with such ‘built-in’ $T_{1}$-protection 111Of course, if
we happen to call the states $\ket{0}$ and $\ket{1}$, $\ket{+}$ and $\ket{-}$
instead, then one would call this built-in protection against phase-flip $Z$
errors, leading to a long dephasing time. is the $0$-$\pi$ qubit discussed in
Section 3.3.5 as the wave functions for $\ket{0}$ and $\ket{1}$ are localized
in the well around $\phi=0$ and the well around $\phi=\pi$. It is clear that
the requirement of having disjoint support requires the potential-well
structure to be sufficiently broad. For a qubit like the transmon, the wave
functions definitely do not have disjoint supports (think approximately two
lowest-energy wave functions of the harmonic oscillator which involve Hermite
functions with index 0 and 1).

####  7.3.3 Double protection?

We have seen that disjoint support of wavefunctions induces a protection
against $X$ errors, but is it possible to shield a qubit from both $X$ as well
as $Z$ errors in this manner?

At first sight, this seems difficult as it would require that

$\displaystyle\bra{+}f(\hat{\phi})\ket{-}=\int_{-\infty}^{\infty}d\phi\,\psi_{+}^{*}(\phi)f(\phi)\psi_{-}(\phi)\overset{?}{\approx}0,$
(7.23)
$\displaystyle\bra{+}g(\hat{q})\ket{-}=\int_{-\infty}^{\infty}d\phi\,\psi_{+}^{*}(\phi)g\left(-i\frac{\partial}{\partial\phi}\right)\psi_{-}(\phi)\overset{?}{\approx}0,$
(7.24)

where $\psi_{\pm}(\phi)\approx(\psi_{0}(\phi)\pm\psi_{1}(\phi))/\sqrt{2}$ are
the wavefunctions of the states $\ket{\pm}=(\ket{0}\pm\ket{1})/\sqrt{2}$. It
is clear that the states $\psi_{\pm}(\phi)$ generally do not have disjoint
support, as they are superpositions. But still Eqs. (7.23) and (7.24) can be
zero for another reason, as follows. We can examine the $\ket{\pm}$ states
using the wavefunctions $\tilde{\psi}_{\pm}(q)$ in the conjugate
$\hat{q}$-basis. In fact, it is possible to design wavefunctions
$\psi_{0,1}(\phi)$ such that the wavefunctions $\tilde{\psi}_{\rm\pm}(q)$ also
have nonoverlapping support: this type of qubit is called a GKP qubit after
its inventors Gottesman, Kitaev, Preskill in 2001 [135]. For the GKP qubit,
$\psi_{0}(\phi)$ is a wavefunction sharply peaked at $\phi=2k\alpha\sqrt{\pi}$
for $k\in\mathbb{Z}$, while $\psi_{1}(\phi)$ is supported near
$\phi=(2k+1)\alpha\sqrt{\pi}$ for some chosen bias $\alpha$. Fourier
transforming implies that $\tilde{\psi}_{+}(q)$ has support on
$q=2k\sqrt{\pi}/\alpha$ while $\tilde{\psi}_{-}(q)$ has disjoint support on
$q=(2k+1)\sqrt{\pi}/\alpha$, hence with a gap of $\Delta q=\sqrt{\pi}/\alpha$
between the supports. Such disjoint support then suffices to claim that Eq.
(7.24) holds, i.e. we will have

$\bra{+}g(\hat{q})\ket{-}=\int_{-\infty}^{\infty}dq\,\tilde{\psi}_{+}^{*}(q)g(q)\tilde{\psi}_{-}(q)\approx
0.$ (7.25)

However, even given disjoint support in the $\hat{q}$-basis, when, say,
$f(\hat{\phi})\propto\cos(\hat{\phi})=(e^{i\hat{\phi}}+e^{-i\hat{\phi}})/2$,
the support of $\tilde{\psi}_{+}(q)$ can get shifted by $\pm 1$ since

$e^{i\hat{\phi}}\ket{q}=\ket{q+1}.$ (7.26)

This implies that one should choose the bias $\alpha$ such that $\Delta
q=\sqrt{\pi}/\alpha>1$ in the presence of some additional $\cos\phi$ noise
term. However, the crafting of a potential which has the GKP wave functions as
ground states could however already lead to terms which effectively act as,
say, $\cos(2\phi)$ etc., generating larger charge fluctuations, which require
a small $\alpha$. Hence, it is not clearly possible to fulfill these
conditions with the appropriate squeezing bias $\alpha$: what is possible also
depends on the inductive and capacitive energy scales which play into the
relative strength of the zero point fluctuation bias $q_{\rm zpf}$ versus
$\phi_{\rm zpf}=\frac{1}{2q_{\rm zpf}}$ in Eq. (2.30b). We refer to [120] for
the construction of a Hamiltonian with approximate GKP qubit eigenstates using
a gyrator. The active (driven) realization of GKP qubit states in a
superconducting 3D cavity resonator was first realized in [136].

In general, we note that a built-in protection of a qubit against noise also
makes the qubit harder to couple to. Ideally a Hamiltonian can be tuned from a
protected to a less protected regime to allow for gates to take place [54].

When realizing a protected superconducting qubit in a system of many degrees
of freedom, one can ask about how many degrees of freedom should be involved
to make logical transitions from $\ket{0}\leftrightarrow\ket{1}$ and
$\ket{+}\leftrightarrow\ket{-}$. For example, the Möbius strip qubit [57],
partially discussed in Section 3.3.6, can have a protection which grows with
$N$ as it involves $N$ degrees of freedom to make some logical transitions.
This quantum error correcting code perspective is further developed in [59].

##  Chapter 8 Models of superconducting amplifiers

In this chapter we give three ‘stand-alone’ exercises on the topic of
amplifiers. Amplifiers built from circuit-QED components can be close to
achieving the quantum limit in terms of adding noise. These amplifiers play an
important role in the dispersive measurement of qubits since the microwave
signal which is used for the measurement is of low-intensity and requires low-
noise amplification before it can read-off as a classical voltage at room
temperature. We refer to [137, 138] for more background.

###  8.1 Exercises on amplifiers

[title=Haus-Caves limit of a bosonic amplifier,label=exc:HC] In this exercise
we will derive the so-called quantum limit of a linear, phase-insensitive
bosonic amplifier. A very simple model of such amplifier is the relation
between the operators of an output mode $\hat{b}$ and an input mode $\hat{a}$
as

$\hat{b}=\sqrt{G}\hat{a}+\sqrt{G-1}\hat{c}^{\dagger},$ (8.1)

with gain $G$ and $\hat{c}^{\dagger}$ is the creation operator of a bath mode
at some relevant frequency $\omega$ obeying
$[\hat{c},\hat{c}^{\dagger}]=\mathds{1}$; $\hat{a}$ similarly obeys
$[\hat{a},\hat{a}^{\dagger}]=\mathds{1}$. Such relation between input and
output fields could have been obtained using an actual circuit and the input-
output formalism in Section 4.5, see Exercise LABEL:exc:JRM. We assume that
the bath state is a thermal bosonic state with $\langle\hat{c}\rangle_{T}=0$
at all temperatures $T$. Explain why quantum mechanics forbids the existence
of an amplifier with $\hat{b}=\sqrt{G}\hat{a}$, where $G>1$, i.e., we need to
have a contribution from another bath mode as in Eq. (8.1). Verify that
$[\hat{b},\hat{b}^{\dagger}]=\mathds{1}$ from Eq. (8.1). The reason that the
amplifier is called phase-insensitive is that both input quadratures
$X_{1}=(\hat{a}+\hat{a}^{\dagger})/2$ and
$X_{2}=i(\hat{a}^{\dagger}-\hat{a})/2$ (dropping hats on these new operators
to simplify notation) get amplified by the same amount. Verify indeed that
$\langle Y_{1}\rangle^{2}=G\langle X_{1}\rangle^{2}$ and similarly $\langle
Y_{2}\rangle^{2}=G\langle X_{2}\rangle^{2}$ with output quadratures
$Y_{1}=(\hat{b}+\hat{b}^{\dagger})/2$ and
$Y_{2}=i(\hat{b}^{\dagger}-\hat{b})/2$. Show that the variance of the $Y_{1}$
quadrature satisfies the inequality

$\langle(Y_{1}-\langle Y_{1}\rangle)^{2}\rangle\equiv(\Delta Y_{1})^{2}\geq
G(\Delta X_{1})^{2}+\frac{1}{4}\lvert G-1\rvert.$ (8.2)

An analogous formula holds for $(\Delta Y_{2})^{2}$. Show that the inequality
in Eq. (8.2) becomes an equality when the bath mode is in the vacuum state at
$T=0$. Comment: A different way of expressing the Haus-Caves limit is obtained
by dividing both sides of Eq. (8.2) by $G$ such that one

$\frac{(\Delta Y_{1})^{2}}{G}\geq(\Delta X_{1})^{2}+A_{N},$

with $A_{N}\geq\frac{1}{4}(1-G^{-1})$ the added noise number which is at least
$1/4$. For a high temperature bath in which thermally added noise dominates,
one can use the high-temperature approximation $kT\gg\hbar\omega$ to
approximate
$A_{N}(\omega)\approx\bar{n}(\omega)\approx\frac{kT}{\hbar\omega}$.

[ref=exc:HC]

We know that the input and output modes are bosonic modes and so they must
satisfy the bosonic commutation relation
$[\hat{a},\hat{a}^{\dagger}]=[\hat{b},\hat{b}^{\dagger}]=\mathds{1}$. However
if $\hat{b}=\sqrt{G}\hat{a}$, then $[\hat{b},\hat{b}^{\dagger}]=G\mathds{1}$,
which forces $G=1$, i.e., there is no amplification nor de-amplification. To
verify $[\hat{b},\hat{b}^{\dagger}]=\mathds{1}$, we use the bosonic
commutation relations of $\hat{a}$ and $\hat{c}$. We have $\langle
Y_{i}\rangle=\sqrt{G}\langle X_{i}\rangle+\sqrt{G-1}\langle X_{i,{\rm
bath}}\rangle_{T}=\sqrt{G}\langle X_{i}\rangle$ due to
$\langle\hat{c}\rangle_{T}=0$ (using $X_{1,{\rm
bath}}=(\hat{c}+\hat{c}^{\dagger})/2$ and $X_{2,{\rm
bath}}=i(\hat{c}^{\dagger}+\hat{c})/2$). We consider $(\Delta
Y_{1})^{2}=\langle Y_{1}^{2}\rangle-\langle Y_{1}\rangle^{2}$ and $\langle
Y_{i}^{2}\rangle=G\langle X_{i}^{2}\rangle+2\sqrt{G(G-1)}\langle
X_{i}\rangle\langle X_{i,{\rm bath}}\rangle_{T}+(G-1)\langle X_{i,{\rm
bath}}^{2}\rangle_{T}=G\langle X_{i}^{2}\rangle+(G-1)\langle X_{i,{\rm
bath}}^{2}\rangle_{T}$ while $\langle Y_{i}\rangle$ is given in the previous
question. We have

$\langle X_{i,{\rm
bath}}^{2}\rangle_{T}=\frac{1}{4}\langle\hat{c}\hat{c}^{\dagger}+\hat{c}^{\dagger}\hat{c}\rangle_{T}=\frac{1}{4}[1+2\langle\hat{c}^{\dagger}\hat{c}\rangle_{T}]=\frac{1}{4}[1+2\bar{n}(\omega)],$

with thermal occupation number
$\bar{n}(\omega)=\frac{1}{e^{\frac{\hbar\omega}{kT}}-1}\geq 0$. This leads to
the lower bound in Eq. (8.2). In the vacuum $\bar{n}(\omega)=0$ for any
$\omega$.

[title=Degenerate parametric amplifier,label=exc:deg_par_amp]

Figure 8.1: LC oscillator with a SQUID loop threaded by two time-dependent
fluxes.

In this exercise we consider an amplifier which realizes phase-sensitive
amplification or single-mode squeezing. Squeezing and other quantum optical
transformations are well described in [69]. Consider the circuit in Fig. 8.1.
Obtain the Lagrangian and the Hamiltonian of the system. Consider a time-
dependent flux $\Phi_{\mathrm{ext}}(t)=\mathcal{E}_{P}\cos\omega_{P}t$ and
write the potential term in the Hamiltonian, using the first order Jacobi-
Anger expansion (see https://en.wikipedia.org/wiki/Jacobi–Anger_expansion for
the full series)

$\cos(z\cos\theta)\approx J_{0}(z)-2J_{2}(z)\cos(2\theta),$ (8.3)

where $J_{0}(z)$ (resp. $J_{2}(z)$) are the 0th (resp. 2nd) Bessel function of
the first kind. Also, expand the potential to second order. Then, introduce
annihilation and creation operators for the mode. Show that for a specific
value of $\omega_{P}$ (what value?), one gets the Hamiltonian of a degenerate
parametric amplifier

$H_{\rm DPA}=\frac{\hbar\varepsilon}{2}(\hat{a}^{2}+\hat{a}^{\dagger 2}),$
(8.4)

in the interaction (rotating frame) picture of the resonant frequency of the
mode. Determine $\varepsilon$. The time evolution operator associated to
$H_{\rm DPA}$ has the form of a unitary squeezing operator

$S(\xi)=\exp\biggl{[}\frac{1}{2}\biggl{(}\xi^{*}\hat{a}^{2}-\xi\hat{a}^{\dagger
2}\biggr{)}\biggr{]},$ (8.5)

where generically $\xi=re^{i\theta}$ is a complex number. It can be shown, see
e.g. the proof at https://en.wikipedia.org/wiki/Squeeze_operator or [69] that
for such operator one has

$S^{\dagger}(\xi)\hat{a}S(\xi)=\hat{a}\cosh
r-\hat{a}^{\dagger}e^{i\theta}\sinh r.$ (8.6)

Using Eq. (8.6), compute the average and the variance of the two quadratures
$X_{\theta/2}$ and $X_{\theta/2+\pi/2}$ (using the definition
$X_{\varphi}=(\hat{a}e^{-i\varphi}+\hat{a}^{\dagger}e^{i\varphi})/2$) in a
vacuum squeezed state $\ket{\xi}=S(\xi)\ket{0}$. Why does this model realize a
phase-sensitive amplifier? According to these results what happens when the
system starts in the vacuum and undergoes a time evolution with the
Hamiltonian in Eq. (8.4)? Consider now attaching a transmission line to the
system. Write down the Heisenberg-Langevin equation for the annihilation and
creation operators $\hat{a}(t)$ and $\hat{a}^{\dagger}(t)$ (as in Section
4.5), and solve the system in the steady-state, i.e., express $\hat{b}_{\rm
out}(t)$ in terms of the input fields. What parameters determine the amount of
squeezing in $\hat{b}_{\rm out}$, i.e., give $\cosh(r)$. Comment: This
analysis can be improved by Fourier transforming the Langevin equations and
considering $\hat{b}_{\rm out}[\omega]$ as a function of $\hat{b}_{\rm
in}[\omega]$ and $\hat{b}_{\rm in}^{\dagger}[\omega]$ allowing one to
understand the frequency-dependence of the squeezing gain.

[ref=exc:deg_par_amp] The Lagrangian reads

$\mathcal{L}=\frac{C}{2}\dot{\Phi}-\frac{\Phi^{2}}{2L}+E_{J}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}\biggl{(}\Phi+\frac{\Phi_{\mathrm{ext}}(t)}{2}\biggr{)}\biggr{]}+E_{J}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}\biggl{(}\Phi-\frac{\Phi_{\mathrm{ext}}(t)}{2}\biggr{)}\biggr{]}.$
(8.7)

Identifying the conjugate variable $Q=\partial\mathcal{L}/\partial\dot{\Phi}$
we obtain the Hamiltonian

$H=\frac{Q^{2}}{2C}+\frac{\Phi^{2}}{2L}-2E_{J}\cos\biggl{[}\frac{\pi}{\Phi_{0}}\Phi_{\mathrm{ext}}(t)\biggr{]}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}\Phi\biggr{]}.$
(8.8)

We now consider $\Phi_{\mathrm{ext}}(t)=\mathcal{E}_{P}\cos(\omega_{P}t)$. We
have the potential

$U=\frac{\Phi^{2}}{2L}-2E_{J}\cos\biggl{[}\frac{\pi}{\Phi_{0}}\mathcal{E}_{P}\cos(\omega_{P}t)\biggr{]}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}\Phi\biggr{]},$
(8.9)

which, using the Jacobi-Anger expansion of $\cos(z\cos\theta)$, expanding the
second cosine to second order in $\Phi$ and neglecting terms that do not alter
the Euler-Lagrange equations, becomes

$U=\frac{\Phi^{2}}{2L}+2E_{J}\biggl{[}J_{0}\biggl{(}\frac{\pi\mathcal{E}_{P}}{\Phi_{0}}\biggr{)}-2J_{2}\biggl{(}\frac{\pi\mathcal{E}_{P}}{\Phi_{0}}\biggr{)}\cos(2\omega_{P}t)\biggr{]}\biggl{(}\frac{2\pi}{\Phi_{0}}\biggr{)}^{2}\Phi^{2}.$
(8.10)

In order to keep a compact notation we define an equivalent inductance $L_{H}$
via the relation

$\frac{1}{L_{H}}=\frac{1}{L}+4E_{J}J_{0}\biggl{(}\frac{\pi\mathcal{E}_{P}}{\Phi_{0}}\biggr{)}\biggl{(}\frac{2\pi}{\Phi_{0}}\biggr{)}^{2},$
(8.11)

and a parameter

$\lambda=-4E_{J}J_{2}\biggl{(}\frac{\pi\mathcal{E}_{P}}{\Phi_{0}}\biggr{)}\biggl{(}\frac{2\pi}{\Phi_{0}}\biggr{)}^{2},$
(8.12)

to rewrite the Hamiltonian as

$H=\frac{Q^{2}}{2C}+\frac{\Phi^{2}}{2L_{H}}+\lambda\cos(2\omega_{P}t)\Phi^{2}.$
(8.13)

We now introduce annihilation and creation operators ($Z=\sqrt{L_{H}/C}$,
$\omega_{r}=1/\sqrt{L_{H}C}$)

$\Phi=\sqrt{\frac{\hbar Z}{2}}(\hat{a}+\hat{a}^{\dagger}),$ (8.14a)
$Q=i\sqrt{\frac{\hbar}{2Z}}(\hat{a}^{\dagger}-\hat{a}),$ (8.14b)

in terms of which the Hamiltonian reads

$H=\hbar\omega_{r}\hat{a}^{\dagger}\hat{a}+\frac{\hbar\varepsilon}{2}\biggl{(}e^{i2\omega_{P}t}+e^{-i2\omega_{P}t}\biggr{)}(\hat{a}^{2}+\hat{a}^{\dagger
2}+1+2\hat{a}^{\dagger}\hat{a}),$ (8.15)

where we defined the parameter $\varepsilon=\lambda Z/2$. We now go to an
interaction picture with $H_{0}=\hbar\omega_{r}\hat{a}^{\dagger}\hat{a}$. The
Hamiltonian in the interaction picture becomes

$\displaystyle H_{I}(t)=H_{\rm
DPA}=e^{iH_{0}t/\hbar}He^{-iH_{0}t/\hbar}-H_{0}=\frac{\hbar\varepsilon}{2}\biggl{(}e^{i2\omega_{P}t}+e^{-i2\omega_{P}t}\biggr{)}$
$\displaystyle\times\biggl{(}a^{2}e^{-2i\omega_{r}t}+a^{\dagger
2}e^{-2i\omega_{r}t}+1+2\hat{a}^{\dagger}\hat{a}\biggr{)}.$ (8.16)

In order to obtain the desired Hamiltonian we need to choose a particular
frequency $\omega_{P}$ so that only the desired term is time-independent while
the effect of the others averages out in a RWA. This frequency is clearly
$\omega_{P}=\omega_{r}$. We finally get the Hamiltonian of a degenerate
parametric amplifier (within the RWA approximation):

$H\overset{\mathrm{RWA}}{=}\frac{\hbar\varepsilon}{2}(\hat{a}^{2}+\hat{a}^{\dagger
2}).$ (8.17)

Using Eqs. (8.6) and its Hermitian conjugate, it is immediate to see that the
average of a generic quadrature in a vacuum squeezed state
$\ket{\xi}=S(\xi)\ket{0}$ is zero. In fact, it is sufficient to show that the
average of $\hat{a}$ is zero:

$\bra{\xi}\hat{a}\ket{\xi}=\bra{0}S^{\dagger}(\xi)\hat{a}S(\xi)\ket{0}=\cosh
r\bra{0}\hat{a}\ket{0}-e^{i\theta}\sinh r\bra{0}\hat{a}^{\dagger}\ket{0}=0,$
(8.18)

and similarly $\bra{\xi}\hat{a}^{\dagger}\ket{\xi}=0$. This implies that the
variance of a generic quadrature $(\Delta X_{\varphi})^{2}$ is equal to the
average of $X_{\varphi}^{2}$ in a vacuum squeezed state. In order to compute
this variance, let us first derive how a generic quadrature $X_{\varphi}$
transforms under the action of the squeezing operator. Using again Eqs. (8.6)
and its Hermitian conjugate we get

$S^{\dagger}(\xi)X_{\theta/2}S(\xi)=[\cosh(r)-\sinh(r)]X_{\theta/2},$ (8.19)

and

$S^{\dagger}(\xi)X_{\theta/2+\pi/2}S(\xi)=[\cosh(r)+\sinh(r)]X_{\theta/2+\pi/2}.$
(8.20)

We can now compute the variance of $X_{\theta/2}$ in a vacuum-squeezed state:

$(\Delta
X_{\theta/2})^{2}=\bra{\xi}X_{\theta/2}^{2}\ket{\xi}=[\cosh^{2}(r)+\sinh^{2}(r)-2\sinh(r)\cosh(r)]\bra{0}X_{\theta/2}^{2}\ket{0}=\frac{1}{4}e^{-r}.$
(8.21)

Similarly, we can deduce the variance of the quadrature $X_{\theta/2+\pi/2}$

$(\Delta X_{\theta/2+\pi/2})^{2}=\frac{1}{4}e^{r}.$ (8.22)

We see that the variance of the quadrature $X_{\theta/2}$ is reduced
(squeezed) compared to the vacuum, while the variance of the conjugate
quadrature $X_{\theta/2+\pi/2}$ is increased (anti-squeezed). These two
quadratures are special because they are the only quadratures for which the
uncertainty relation $(\Delta X_{\theta/2})^{2}(\Delta
X_{\theta/2+\pi/2})^{2}\geq\frac{1}{16}$ is satisfied with equality sign for
the squeezed vacuum as you can check. The amplification is phase-sensitive
since different quadratures (with different phases) are squeezed to a
different extent.

Let us consider the time evolution operator associated to the Hamiltonian Eq.
(8.4),

$U_{I}(t)=\exp\biggl{[}-\frac{i\varepsilon t}{2}(\hat{a}^{2}+\hat{a}^{\dagger
2})\biggr{]},$ (8.23)

which is a squeezing operator with $\xi(t)=i\varepsilon t=\epsilon
te^{i\pi/2}$, hence $\theta=\pi/2$. We conclude that if we start from the
vacuum, the time evolution operator of Eq. (8.23) generates a vacuum squeezed
state in which the quadrature $X_{\pi/4}$ is squeezed, while the quadrature
$X_{3\pi/4}$ is anti-squeezed. The modulus $r$ of the squeezing parameter
$\xi$ increases linearly with time. The Langevin equations associated to
$\hat{a}(t)$ and $\hat{a}^{\dagger}(t)$ (dropping time-dependence for
simplicity) are

$\frac{d\hat{a}}{dt}=\frac{1}{i\hbar}[\hat{a},H_{\rm
DPA}]-\frac{\kappa}{2}\hat{a}+\sqrt{\kappa}\hat{b}_{\rm in},$ (8.24a)
$\frac{d\hat{a}^{\dagger}}{dt}=\frac{1}{i\hbar}[\hat{a}^{\dagger},H_{\rm
DPA}]-\frac{\kappa}{2}\hat{a}^{\dagger}+\sqrt{\kappa}\hat{b}_{\rm
in}^{\dagger},$ (8.24b)

which give

$\frac{d\hat{a}}{dt}=-i\varepsilon\hat{a}^{\dagger}-\frac{\kappa}{2}\hat{a}+\sqrt{\kappa}\hat{b}_{\rm
in},$ (8.25a)
$\frac{d\hat{a}^{\dagger}}{dt}=i\varepsilon\hat{a}-\frac{\kappa}{2}\hat{a}^{\dagger}+\sqrt{\kappa}\hat{b}_{\rm
in}^{\dagger},$ (8.25b)

with input-output relation

$\hat{b}_{\rm out}=\sqrt{\kappa}\hat{a}-\hat{b}_{\rm in}.$ (8.26)

In the steady state they become algebraic equations

$-i\varepsilon\hat{a}^{\dagger}-\frac{\kappa}{2}\hat{a}+\sqrt{\kappa}\hat{b}_{\rm
in}=0,$ (8.27a)
$i\varepsilon\hat{a}-\frac{\kappa}{2}\hat{a}^{\dagger}+\sqrt{\kappa}\hat{b}_{\rm
in}^{\dagger}=0.$ (8.27b)

Solving for $\hat{a}$ we get

$\hat{a}=-\frac{\kappa}{\kappa^{2}-4\varepsilon^{2}}\frac{4i\varepsilon}{\sqrt{\kappa}}\hat{b}_{\rm
in}^{\dagger}+\frac{2\kappa^{2}}{\kappa^{2}-4\varepsilon^{2}}\frac{1}{\sqrt{\kappa}}\hat{b}_{\rm
in},$ (8.28)

and using the input-output relations Eq. (8.26) we obtain the output field
annihilation operator

$\hat{b}_{\rm
out}(t)=\frac{\kappa^{2}+4\varepsilon^{2}}{\kappa^{2}-4\varepsilon^{2}}\hat{b}_{\rm
in}(t)-i\frac{4\varepsilon\kappa}{\kappa^{2}-4\varepsilon^{2}}\hat{b}_{\rm
in}^{\dagger}(t),$ (8.29)

You can verify that this equation is of the form $\hat{b}_{\rm
out}=\hat{b}_{\rm in}\cosh(r)-\hat{b}_{\rm in}^{\dagger}e^{i\theta}\sinh(r)$
and $\cosh(r)=\frac{\kappa^{2}+4\epsilon^{2}}{\kappa^{2}-4\epsilon^{2}}$,
depending on the strength of the squeezing term $\epsilon$ and the coupling
(decay) strength into the transmission line $\kappa$. Note that the amount of
squeezing $r$ does not increase linearly in time, but is limited by the
strength of $\kappa$ versus $\epsilon$.

[title=Nondegenerate parametric amplifier: Josephson ring
modulator,label=exc:JRM]

Figure 8.2: Circuit of a Josephson ring modulator.

Consider the circuit of a Josephson Ring Modulator (JRM) in Fig. 8.2 with
equal junctions and a loop threaded by a constant external magnetic flux
$\Phi_{\mathrm{ext}}$. The goal of this exercise is to show how it can be used
to realize the Hamiltonian of a non-degenerate parametric amplifier in Eq.
(8.33). While the degenerate amplifier with Hamiltonian in Eq. (8.4) in the
previous Exercise enacts single-mode squeezing, the non-degenerate amplifier
in Eq. (8.33) enacts two-mode squeezing, that is, the two modes on which it
acts are different/non-degenerate. Obtain the Lagrangian of this system taking
the tree shown in orange in Fig. 8.2, assuming $\Phi_{3}=0$ as reference
ground node. Then, introduce two consecutive changes of variables. First

$\Phi_{1}-\Phi_{2}=\tilde{\Phi}_{1}-\tilde{\Phi}_{2}+\frac{\Phi_{\rm
ext}}{4},$ (8.30a) $\Phi_{2}=\tilde{\Phi}_{2}+\frac{\Phi_{\rm ext}}{4},$
(8.30b) $\Phi_{4}=\tilde{\Phi}_{4}-\frac{\Phi_{\rm ext}}{4},$ (8.30c)

and afterwards

$\Phi_{X}=\tilde{\Phi}_{1},$ (8.31a)
$\Phi_{Y}=\tilde{\Phi}_{4}-\tilde{\Phi}_{2},$ (8.31b)
$\Phi_{Z}=\tilde{\Phi}_{1}-\tilde{\Phi}_{2}-\tilde{\Phi}_{4}.$ (8.31c)

Obtain the Hamiltonian in terms of $\Phi_{X},\Phi_{Y},\Phi_{Z}$ and their
conjugate variables. By setting $\Phi_{\rm ext}=\Phi_{0}/2$ (half a flux
quantum), show that the system approximately realizes the Hamiltonian of three
harmonic oscillators coupled by the coupling term
$H_{c}=\nu\Phi_{X}\Phi_{Y}\Phi_{Z}$: give the parameter $\nu$. _Hint: You can
expand the potential up to second order._ Finally, introduce annihilation and
creation operators and show that by a careful choice of the characteristic
frequencies $\omega_{X}$, $\omega_{Y}$, $\omega_{Z}$ (what choice?) one
obtains the three-wave mixing Hamiltonian

$H_{3W}/\hbar\approx\omega_{X}\hat{a}_{X}^{\dagger}\hat{a}_{X}+\omega_{Y}\hat{a}_{Y}^{\dagger}\hat{a}_{Y}+\omega_{Z}\hat{a}_{Z}^{\dagger}\hat{a}_{Z}+\eta(\hat{a}_{X}^{\dagger}\hat{a}_{Y}^{\dagger}\hat{a}_{Z}+\mathrm{h.c.}),$
(8.32)

with $\eta$ the three-wave mixing parameter. Consider the Hamiltonian in Eq.
(8.32). In what follows we call mode $X$ the signal mode, i.e.,
$\hat{a}_{X}=\hat{a}_{S}$, mode $Y$ the idler mode, i.e.,
$\hat{a}_{Y}=\hat{a}_{I}$ and mode $Z$ the pump mode, i.e.,
$\hat{a}_{Z}=\hat{a}_{P}$. We drive the pump mode with a coherent drive at
frequency $\omega_{P}$ with large amplitude so that we can substitute
$\hat{a}_{P}$ with some time-dependent (approximate) expectation value
$\langle\hat{a}_{P}\rangle=i\alpha_{P}e^{-i\omega_{P}t}$ with some real
amplitude $\alpha_{P}$. Convert the Hamiltonian to the interaction picture
(rotating frame) at the frequencies of the modes $X$ and $Y$ to get the two-
mode squeezing Hamiltonian

$\tilde{H}_{I3W}/\hbar=i\lambda\biggl{(}\hat{a}_{S}^{\dagger}\hat{a}_{I}^{\dagger}-\hat{a}_{S}\hat{a}_{I}\biggr{)}.$
(8.33)

We now assume the signal mode to be coupled to a transmission line with input
field $\hat{b}_{S,\rm in}(t)$ (coupling set by decay rate $\kappa_{S}$), while
the idler mode is coupled to another transmission line with $\hat{b}_{I,\rm
in}(t)$ (coupling set by decay rate $\kappa_{I}$). Write down the Heisenberg-
Langevin equations of motion for $\hat{a}_{S}(t)$ and
$\hat{a}_{I}^{\dagger}(t)$ as in Section 4.5, and obtain the steady-state
solution $\hat{b}_{S,\rm out}(t)$ as a function of input fields of signal and
idler. Discuss the effect of the idler mode when it starts in the vacuum state
(see Exercise LABEL:exc:HC). Comment: This analysis can be improved by Fourier
transforming the Langevin equations and considering $\hat{b}_{S,\rm
out}[\omega]$ as a function of $\hat{b}_{S,\rm in}[\omega]$ and
$\hat{b}_{I,\rm in}^{\dagger}[\omega]$ allowing one to understand the
frequency-dependence, see [138].

[ref=exc:JRM] The Lagrangian in terms of node variables reads

$\mathcal{L}=\frac{C_{J}}{2}\dot{\Phi}_{2}^{2}+\frac{C_{J}}{2}\dot{\Phi}_{4}^{2}+\frac{C_{J}}{2}(\dot{\Phi}_{1}-\dot{\Phi}_{2})^{2}+\frac{C_{J}}{2}(\dot{\Phi}_{1}-\dot{\Phi}_{4})^{2}+\frac{C_{X}}{2}\dot{\Phi}_{1}^{2}+\frac{C_{Y}}{2}(\dot{\Phi}_{4}-\dot{\Phi}_{2})^{2}\\\
+E_{J}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}\Phi_{2}\biggr{]}+E_{J}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}\Phi_{4}\biggr{]}+E_{J}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}(\Phi_{1}-\Phi_{2})\biggr{]}+E_{J}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}(\Phi_{1}-\Phi_{4}-\Phi_{\mathrm{ext}})\biggr{]}.$
(8.34)

The first suggested change of variables results in

$\mathcal{L}=\frac{C_{J}}{2}\dot{\tilde{\Phi}}_{2}^{2}+\frac{C_{J}}{2}\dot{\tilde{\Phi}}_{4}^{2}+\frac{C_{J}}{2}(\dot{\tilde{\Phi}}_{1}-\dot{\tilde{\Phi}}_{2})^{2}+\frac{C_{J}}{2}(\dot{\tilde{\Phi}}_{1}-\dot{\tilde{\Phi}}_{4})^{2}+\frac{C_{X}}{2}\dot{\tilde{\Phi}}_{1}^{2}+\frac{C_{Y}}{2}(\dot{\tilde{\Phi}}_{4}-\dot{\tilde{\Phi}}_{2})^{2}\\\
+E_{J}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}\biggl{(}\tilde{\Phi}_{2}+\frac{\Phi_{\mathrm{ext}}}{4}\biggr{)}\biggr{]}+E_{J}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}\biggl{(}\tilde{\Phi}_{4}-\frac{\Phi_{\mathrm{ext}}}{4}\biggr{)}\biggr{]}+E_{J}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}\biggl{(}\tilde{\Phi}_{1}-\tilde{\Phi}_{2}+\frac{\Phi_{\mathrm{ext}}}{4}\biggr{)}\biggr{]}\\\
+E_{J}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}\biggl{(}\tilde{\Phi}_{1}-\tilde{\Phi}_{4}-\frac{\Phi_{\mathrm{ext}}}{4}\biggr{)}\biggr{]},$
(8.35)

while the second change of variables diagonalizes the kinetic part of the
Lagrangian and brings it into the simple form

$\mathcal{L}=\frac{C_{J}+C_{X}}{2}\dot{\Phi}_{X}^{2}+\frac{C_{J}+C_{Y}}{2}\dot{\Phi}_{Y}^{2}+\frac{C_{J}}{2}\dot{\Phi}_{Z}^{2}\\\
+4E_{J}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}\frac{\Phi_{\mathrm{ext}}}{4}\biggr{]}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}\frac{\Phi_{X}}{2}\biggr{]}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}\frac{\Phi_{Y}}{2}\biggr{]}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}\frac{\Phi_{Z}}{2}\biggr{]}\\\
+4E_{J}\sin\biggl{[}\frac{2\pi}{\Phi_{0}}\frac{\Phi_{\mathrm{ext}}}{4}\biggr{]}\sin\biggl{[}\frac{2\pi}{\Phi_{0}}\frac{\Phi_{X}}{2}\biggr{]}\sin\biggl{[}\frac{2\pi}{\Phi_{0}}\frac{\Phi_{Y}}{2}\biggr{]}\sin\biggl{[}\frac{2\pi}{\Phi_{0}}\frac{\Phi_{Z}}{2}\biggr{]}.$
(8.36)

From this Lagrangian it is easy to obtain the Hamiltonian

$H=\frac{Q_{X}^{2}}{2C_{\Sigma X}}+\frac{Q_{Y}^{2}}{2C_{\Sigma
Y}}+\frac{Q_{Z}^{2}}{2C_{J}}-4E_{J}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}\frac{\Phi_{\rm
ext}}{4}\biggr{]}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}\frac{\Phi_{X}}{2}\biggr{]}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}\frac{\Phi_{Y}}{2}\biggr{]}\cos\biggl{[}\frac{2\pi}{\Phi_{0}}\frac{\Phi_{Z}}{2}\biggr{]}\\\
-4E_{J}\sin\biggl{[}\frac{2\pi}{\Phi_{0}}\frac{\Phi_{\rm
ext}}{4}\biggr{]}\sin\biggl{[}\frac{2\pi}{\Phi_{0}}\frac{\Phi_{X}}{2}\biggr{]}\sin\biggl{[}\frac{2\pi}{\Phi_{0}}\frac{\Phi_{Y}}{2}\biggr{]}\sin\biggl{[}\frac{2\pi}{\Phi_{0}}\frac{\Phi_{Z}}{2}\biggr{]},$
(8.37)

with the conjugate variables $Q_{X}=C_{\Sigma X}\dot{\Phi}_{X}$,
$Q_{Y}=C_{\Sigma Y}\dot{\Phi}_{Y}$, $Q_{Z}=C_{J}\dot{\Phi}_{Z}$ and the
abbreviations $C_{\Sigma X}=C_{X}+C_{J}$, $C_{\Sigma Y}=C_{Y}+C_{J}$. Setting
$\Phi_{\mathrm{ext}}=\Phi_{0}/2$ and expanding the potential up to second
order as suggested, one obtains

$H=\frac{Q_{X}^{2}}{2C_{\Sigma X}}+\frac{Q_{Y}^{2}}{2C_{\Sigma
Y}}+\frac{Q_{Z}^{2}}{2C_{J}}+\frac{1}{2L_{JRM}}(\Phi_{X}^{2}+\Phi_{Y}^{2}+\Phi_{Z}^{2})+\nu\Phi_{X}\Phi_{Y}\Phi_{Z},$
(8.38)

where we defined a characteristic inductance of the JRM,
$L_{JRM}=\Phi_{0}^{2}/(4\sqrt{2}E_{J}\pi^{2})$, and the coupling parameter
$\nu=-2\sqrt{2}E_{J}\pi^{3}/\Phi_{0}^{3}$.
In order to pass to a quantum description, we introduce annihilation and
creation operators for the modes $X,Y$ and $Z$ as

$\Phi_{K}=\sqrt{\frac{\hbar Z_{K}}{2}}(\hat{a}_{K}+\hat{a}_{K}^{\dagger}),$
(8.39a)
$Q_{K}=i\sqrt{\frac{\hbar}{2Z_{K}}}(\hat{a}_{K}^{\dagger}-\hat{a}_{K}),$
(8.39b)

with $K=\\{X,Y,Z\\}$ and $Z_{K}$ the characteristic impedance of mode $K$ and
we rewrite the Hamiltonian as

$H=\hbar\omega_{X}\hat{a}_{X}^{\dagger}\hat{a}_{X}+\hbar\omega_{Y}\hat{a}_{Y}^{\dagger}\hat{a}_{Y}+\hbar\omega_{Z}\hat{a}_{Z}^{\dagger}\hat{a}_{Z}+\hbar\eta(\hat{a}_{X}+\hat{a}_{X}^{\dagger})(\hat{a}_{Y}+\hat{a}_{Y}^{\dagger})(\hat{a}_{Z}+\hat{a}_{Z}^{\dagger}),$
(8.40)

where we introduced the characteristic frequencies of the resonators and the
parameter

$\hbar\eta=\nu\biggl{(}\frac{\hbar}{2}\biggr{)}^{3/2}\sqrt{Z_{X}Z_{Y}Z_{Z}}.$
(8.41)

Let us take a closer look at Eq. (8.40). We would like to have an argument for
keeping only the term $a_{X}^{\dagger}a_{Y}^{\dagger}a_{Z}$ and their
Hermitian conjugate. Of course, what we need to impose is that this term is
the only energy preserving (resonant) term among all the terms that appear in
the expansion of the coupling Hamiltonian. This is the case if the frequency
of the mode $Z$ (the pump) is equal to the sum of the other two. So we set the
condition $\omega_{Z}=\omega_{X}+\omega_{Y}$. Let us rewrite the Hamiltonian
in the new notation, keeping only the three wave mixing terms

$H_{3W}/\hbar\approx\omega_{S}\hat{a}_{S}^{\dagger}\hat{a}_{S}+\omega_{I}\hat{a}_{I}^{\dagger}\hat{a}_{I}+\omega_{P}\hat{a}_{P}^{\dagger}\hat{a}_{P}+\eta(\hat{a}_{S}^{\dagger}\hat{a}_{I}^{\dagger}\hat{a}_{P}+\mathrm{h.c.}),$
(8.42)

with the three-wave mixing condition $\omega_{P}=\omega_{I}+\omega_{S}$. By
assuming that we strongly drive the pump we substitute the operator
$\hat{a}_{P}$ with its average $i\alpha_{P}e^{-i\omega_{P}t}$ to get

$H_{3W}/\hbar\approx\omega_{S}\hat{a}_{S}^{\dagger}\hat{a}_{S}+\omega_{I}\hat{a}_{I}^{\dagger}\hat{a}_{I}+i\lambda\biggl{(}\hat{a}_{S}^{\dagger}\hat{a}_{I}^{\dagger}e^{-i\omega_{P}t}-\hat{a}_{S}\hat{a}_{I}e^{+i\omega_{P}t}\biggr{)},$
(8.43)

where we called $\lambda=\eta\alpha_{P}$. Going to an interaction picture at
the resonators’ frequencies, i.e., with
$H_{0}/hbar=\omega_{S}\hat{a}_{S}^{\dagger}\hat{a}_{S}+\omega_{I}\hat{a}_{I}^{\dagger}\hat{a}_{I}$,
we have $\tilde{H}_{3W}=e^{iH_{0}t}H_{3W}e^{-iH_{0}t}-H_{0}$, obtaining Eq.
(8.33). We have the following Heisenberg-Langevin equations for
$\hat{a}_{I}^{\dagger}$ and $\hat{a}_{S}$:

$\frac{d\hat{a}_{I}^{\dagger}}{dt}=\frac{1}{i\hbar}[\hat{a}_{I}^{\dagger},\tilde{H}_{I3W}]-\frac{\kappa_{I}}{2}\hat{a}_{I}^{\dagger}+\sqrt{\kappa_{I}}\hat{b}_{I,\rm
in}^{\dagger},$ (8.44a)
$\frac{d\hat{a}_{S}}{dt}=\frac{1}{i\hbar}[\hat{a}_{S},\tilde{H}_{I3W}]-\frac{\kappa_{S}}{2}\hat{a}_{S}+\sqrt{\kappa_{S}}\hat{b}_{S,\rm
in},$ (8.44b)

with input-output relations $\hat{b}_{I,\rm
out}=\sqrt{\kappa_{I}}\hat{a}_{I}-\hat{b}_{I,\rm in}$ and $\hat{b}_{S,\rm
out}=\sqrt{\kappa_{S}}\hat{a}_{S}-\hat{b}_{S,\rm in}$. Computing the
commutators we get

$\frac{d\hat{a}_{I}^{\dagger}}{dt}=\lambda\hat{a}_{S}-\frac{\kappa_{I}}{2}\hat{a}_{I}^{\dagger}+\sqrt{\kappa_{I}}\hat{b}_{I,\rm
in}^{\dagger},$ (8.45a)
$\frac{d\hat{a}_{S}}{dt}=\lambda\hat{a}_{I}^{\dagger}-\frac{\kappa_{S}}{2}\hat{a}_{S}+\sqrt{\kappa_{S}}\hat{b}_{S,\rm
in}.$ (8.45b)

The steady state solutions are readily obtained as

$\hat{a}_{S}(t)=\frac{2\lambda}{\kappa_{S}}\hat{a}_{I}^{\dagger}+\frac{2}{\sqrt{\kappa_{S}}}\hat{b}_{S,\rm
in},$ (8.46a)
$\hat{a}_{I}^{\dagger}(t)=\frac{2\lambda}{\kappa_{I}}\hat{a}_{S}+\frac{2}{\sqrt{\kappa_{I}}}\hat{b}_{I,\rm
in}^{\dagger},$ (8.46b)

which imply

$\hat{a}_{S}(t)=\frac{1}{1-Q^{2}}\frac{4\lambda}{\kappa_{S}\sqrt{\kappa_{I}}}\hat{b}_{I,\rm
in}^{\dagger}(t)+\frac{1}{1-Q^{2}}\frac{2}{\sqrt{\kappa}_{S}}\hat{b}_{S,\rm
in}(t),$ (8.47)

with $Q=2\lambda/(\sqrt{\kappa_{1}\kappa_{2}})$. Using $\hat{b}_{S,\rm
out}=\sqrt{\kappa_{S}}\hat{a}_{S}-\hat{b}_{S,\rm in}$ we can write the output
field of the signal as

$\hat{b}_{S,\rm out}(t)=\frac{1+Q^{2}}{1-Q^{2}}\hat{b}_{S,\rm
in}(t)+\frac{2Q}{1-Q^{2}}\hat{b}_{I,\rm in}^{\dagger}(t).$ (8.48)

We see that the idler mode will extra noise even when it is in the vacuum
state, as discussed in the analysis of Eq. (8.1) in Exercise LABEL:exc:HC.

##  Appendix A A review of canonical quantization

In this appendix, we review the construction of a Hamiltonian from a
Lagrangian and the formalism of canonical quantization of classical
conservative (energy-conserving) dynamics. We refer the reader to specialized
texts on classical mechanics such as Ref. [139] for more details.

###  A.1 Principle of minimal action and gauge invariance

We start with a phase space of $n$ independent variables $x_{1},\ldots,x_{n}$
with $\dot{x_{i}}=dx_{i}/dt$ and each $x_{i}\in\mathbb{R}$. In a mechanical
scenario these could be the positions of $n$ particles. In the electrical
circuit case these could be a set of independent node fluxes, or independent
node charges. For an electrical circuit, we have freedom in this choice of
variables as long as we have verified that they form an independent set of
variables.

Given is the Lagrangian
$\mathcal{L}(x_{1},\ldots,x_{n},\dot{x}_{1},\ldots,\dot{x}_{n})$. For
simplicity, we assume that $\mathcal{L}$ has no explicit time-dependence. For
a general mechanical system, in the absence of a magnetic field, the
Lagrangian will be a sum of the kinetic energy $T$ minus the potential energy
$U$, i.e.,

$\mathcal{L}(x_{1},\ldots,x_{n},\dot{x}_{1},\ldots,\dot{x}_{n})=T(\dot{x}_{1},\ldots,\dot{x}_{n})-U(x_{1},\ldots,x_{n}).$
(A.1)

With the Lagrangian we can define the classical action as

$\mathcal{S}=\int_{t_{1}}^{t_{2}}dt\,\mathcal{L}(x_{1},\ldots,x_{n},\dot{x}_{1},\ldots,\dot{x}_{n}).$
(A.2)

The principle of minimal action, i.e., $\delta S=0$, implies the classical
equations of motion, called the Euler-Lagrange equations:

$\frac{d}{dt}\left(\frac{\partial\mathcal{L}}{\partial\dot{x}_{i}}\right)-\frac{\partial\mathcal{L}}{\partial
x_{i}}=0,\quad\forall i=1,\dots,n.$ (A.3)

Thus, the construction of the Lagrangian is determined by the classical
dynamical equations and vice versa. However, the Lagrangian is not unique and
it can be changed by a gauge transformation, namely

$\mathcal{L}\rightarrow\mathcal{L}+\frac{df(x_{1},\ldots,x_{n})}{dt}$ (A.4)

with $f(x_{1},\ldots,x_{n})$ an arbitrary function, without this affecting the
equations of motion. Clearly, such change maps the action
$\mathcal{S}\rightarrow\mathcal{S}+f(t_{2})-f(t_{1})$ which indeed does not
change which trajectory minimizes the action.

[label=exc:gauge] Verify that Eq. (A.3) follows from
$\frac{d}{dt}\left(\frac{\partial\mathcal{L}^{\prime}}{\partial\dot{x}_{i}}\right)-\frac{\partial\mathcal{L}^{\prime}}{\partial
x_{i}}=0$ with
$\mathcal{L}^{\prime}=\mathcal{L}+\frac{df(x_{1},\ldots,x_{n})}{dt}$, using
$\frac{df}{dt}=\sum_{i}\frac{\partial f}{\partial x_{i}}\dot{x}_{i}$ as $f(.)$
only depends on $x_{1},\ldots,x_{n}$.

[ref=exc:gauge] Consider
$\frac{d}{dt}\frac{\partial}{\partial\dot{x}_{i}}\frac{df}{dt}=\frac{d}{dt}\frac{\partial
f}{\partial x_{i}}=\sum_{j}\frac{\partial^{2}f}{\partial x_{j}\partial
x_{i}}\dot{x}_{j}=\frac{\partial}{\partial x_{i}}\frac{df}{dt}$.

###  A.2 Legendre transformation: the Hamiltonian

To define a Hamiltonian, one introduces, for each independent variable
$x_{i}$, a conjugate variable $p_{i}$ defined as

$p_{i}=\frac{\partial\mathcal{L}}{\partial\dot{x}_{i}}.$ (A.5)

For a simple mechanical system with kinetic energy given by
$T=\sum_{i}m_{i}\dot{x}_{i}^{2}/2$, the conjugate variables are momenta given
by $p_{i}=m_{i}\dot{x}_{i}$.

One can then define a new function, namely the Hamiltonian, which depends on
the variables $\\{x_{i}\\}$ and their conjugates $\\{p_{i}\\}$, but no longer
on $\\{\dot{x}_{i}\\}$. More precisely, the Hamiltonian
$\mathcal{H}(x_{1},\dots,x_{n},p_{1},\dots,p_{n})$ is defined as the Legendre
transformation of the Lagrangian, i.e.,

$\mathcal{H}(x_{1},\ldots,x_{n},p_{1},\ldots,p_{n})=\sup_{\dot{x}_{1},\ldots,\dot{x}_{n}}\left[\sum_{i=1}^{n}p_{i}\dot{x_{i}}-\mathcal{L}(x_{1},\ldots,x_{n},\dot{x}_{1},\ldots,\dot{x}_{n})\right].$
(A.6)

When $\mathcal{L}(x_{1},\ldots,x_{n},\dot{x}_{1},\ldots,\dot{x}_{n})$ is a
convex function of the variables $\dot{x}_{1},\ldots,\dot{x}_{n}$, the
supremum in Eq. (A.6) occurs at a unique point $\dot{x}_{1}^{\rm
sup},\ldots,\dot{x}_{n}^{\rm sup}$ since
$\sum_{i}p_{i}\dot{x}_{i}-\mathcal{L}$ is concave in $\dot{x}_{i}$. At this
supremum

$\frac{\partial}{\partial\dot{x}_{i}}\left[\sum_{i=1}^{n}p_{i}\dot{x}_{i}-\mathcal{L}(x_{1},\ldots,x_{n},\dot{x}_{1},\ldots,\dot{x}_{n})\right]=0\implies
p_{i}=\frac{\partial\mathcal{L}}{\partial\dot{x}_{i}},\quad\forall i.$ (A.7)

Then we can derive the Hamiltonian in Eq. (A.6) by determining $p_{i}$ and
solving for $\dot{x}_{i}$ to eliminate the $\dot{x}_{i}$ variables in
$\mathcal{H}$.

As an example, we consider the general case in which the initial Lagrangian
has a quadratic, but potentially ‘off-diagonal’ kinetic energy. Let
$\bm{x}=\left(x_{1},\dots,x_{n}\right)^{T}$. The Lagrangian reads

$\mathcal{L}(\bm{x},\dot{\bm{x}})=\frac{1}{2}\dot{\bm{x}}^{T}\bm{C}\dot{\bm{x}}-U(\bm{x})=\frac{1}{2}\sum_{i,j=1}^{n}\bm{C}_{ij}\dot{x}_{i}\dot{x}_{j}-U(x_{1},\ldots,x_{n}),$
(A.8)

where the matrix $\bm{C}$ is symmetric and we assume $\bm{C}>0$, i.e.,
$\bm{C}$ has eigenvalues $d_{i}>0$, $i=1,\ldots,n$. In this case, we can
determine that $\bm{p}=\bm{C}\dot{\bm{x}}$ so that
$\dot{\bm{x}}=\bm{C}^{-1}\bm{p}$ at the supremum. Inserting this equality in
Eq. (A.6) gives compactly

$\mathcal{H}(\bm{x},\bm{p})=\frac{1}{2}\bm{p}^{T}\bm{C}^{-1}\bm{p}+U(\bm{x}).$
(A.9)

In a different analysis of the same problem we could have first switched to
new independent variables $\bm{y}=\bm{S}\bm{x}$, with $\bm{S}$ the matrix that
diagonalizes $\bm{C}$, in order to obtain

$\mathcal{L}(\bm{y},\dot{\bm{y}})=\frac{1}{2}\dot{\bm{y}}^{T}\bm{D}\dot{\bm{y}}-\underbrace{U(\bm{S}^{-1}\bm{y})}_{\tilde{U}(\bm{y})}=\frac{1}{2}\sum_{i=1}^{n}d_{i}\dot{y}_{i}^{2}-\tilde{U}(y_{1},\dots,y_{n}),$
(A.10)

with the diagonal matrix $\bm{D}=\bm{S}^{T}\bm{C}\bm{S}$ with diagonal entries
$d_{i}$. Only then we could have introduced the conjugate variable as
$\bm{p}=\partial\mathcal{L}/\partial\dot{\bm{y}}=\bm{D}\dot{\bm{y}}$ so that

$\mathcal{H}(\bm{y},\bm{p})=\frac{1}{2}\bm{p}^{T}\bm{D}^{-1}\bm{p}+\tilde{U}(\bm{y})=\sum_{i=1}^{n}\frac{p_{i}^{2}}{2d_{i}}+\tilde{U}(y_{1},\ldots,y_{n}).$
(A.11)

It is instructive to discuss the case when $\bm{C}\geq 0$, meaning that some
of the eigenvalues $d_{i}$ of $\bm{C}$, are zero, thus preventing $\bm{C}$
from being invertible. In this case, the associated variable $y_{i}$ (with
$d_{i}=0$) only occurs in the potential $\tilde{U}$. Such cases can
classically be solved by realizing that all coordinates $y_{i}$ without
kinetic energy contribution are simply to be put at a value for which
$\partial\tilde{U}(y_{1},\ldots,y_{n})/\partial y_{i}=0$ as this is the Euler-
Lagrange equation for those coordinates. When the potential is convex, this
configuration is one which minimizes the potential
$\tilde{U}(y_{1},\ldots,y_{n})$ energy, more generally one sets these values
to a local minimum 111Local maxima are unstable when $d_{i}$ is only slightly
nonzero..

[label=exc:relat]The Lagrangian of a relativistic particle with mass $m$ and
spatial coordinate ${\bf x}=(x_{1},x_{2},x_{3})$, is

$\mathcal{L}=-mc^{2}\sqrt{1-\frac{\sum_{i=1}^{3}\dot{x}_{i}^{2}}{c^{2}}}-U(x_{1},x_{2},x_{3}),$

where $c$ is the speed of light and $U(x_{1},x_{2},x_{3})$ is some potential.
Verify that $\mathcal{L}$ is convex in $\dot{x_{i}}$ and derive the
Hamiltonian.

[ref=exc:relat] By plotting $f(x)=-\sqrt{1-x^{2}}$ for $-1\leq x\leq 1$ one
can verify $f(x)$ is convex and so is $\mathcal{L}$. We have relativistic
momenta
$p_{i}=\frac{\partial\mathcal{L}}{\partial\dot{x}_{i}}=\frac{m\dot{x}_{i}}{\sqrt{1-v^{2}/c^{2}}}$
or $\sum_{i=1}^{3}\dot{x}_{i}^{2}=v^{2}=\frac{p^{2}}{m^{2}+p^{2}/c^{2}}$, so
that
$\mathcal{H}=\frac{mv^{2}}{\sqrt{1-v^{2}/c^{2}}}+mc^{2}\sqrt{1-v^{2}/c^{2}}+U(x_{1},x_{2},x_{3})=\sqrt{p^{2}c^{2}+m^{2}c^{4}}+U(x_{1},x_{2},x_{3})$.

In various physical situations, the Lagrangian is explicitly time-dependent,
i.e., $\mathcal{L}({\bf x},{\bf\dot{x}},t)$. For example, the potential is one
that is induced by adding a time-dependent driving force $F(t)=A\cos(\omega
t)$ to a mass-spring system with position $x$. The potential will be
$U(x)=\frac{1}{2}Kx^{2}-F(t)x$ corresponding to the total force $F_{\rm
tot}(t)=-\partial U/\partial x=-Kx+F(t)$. In such case, when the explicit
time-dependence occurs in the potential energy $U$, it is simply carried on in
the Hamiltonian $\mathcal{H}(t)$, i.e. for the driven mass-spring system we
have $\mathcal{H}=\frac{p^{2}}{2m}+\frac{1}{2}Kx^{2}-F(t)x$.

On the other hand, when the explicit time-dependence affects the kinetic
energy, it enters the Euler-Lagrange equation, Eq. (A.3), in its total time-
derivative $d/dt$. For example, a mechanical system with a time-dependent mass
$m(t)$ and a potential $U(x)$ obeys the Euler-Lagrange equation
$m(t)\ddot{x}+\dot{x}\dot{m}=dU/dx$. In addition, when we switch from
Lagrangian to Hamiltonian, such time-dependence can enter the definition of
the conjugate variables and through this makes the definition of the
Hamiltonian time-dependent. For example, when treating time-dependent fluxes,
the conjugate charge in Eq. (2.89) in Section 2.6 is explicitly time-dependent
through $\Phi_{\rm ext}(t)$. For conjugate variables which are time-dependent
(time is simply a classical parameter), one can still write down the Poisson
bracket and apply quantization.

###  A.3 Poisson bracket and quantization

Using conjugate variables one can define a Poisson bracket for functions
$f(\bm{x},\bm{p})$ and $g(\bm{x},\bm{p})$ as follows

$\\{f,g\\}=\sum_{i=1}^{n}\left[\frac{\partial f}{\partial x_{i}}\frac{\partial
g}{\partial p_{i}}-\frac{\partial f}{\partial p_{i}}\frac{\partial g}{\partial
x_{i}}\right].$ (A.12)

It follows from this definition that

$\\{x_{i},p_{j}\\}=\delta_{ij},\\{x_{i},x_{j}\\}=\\{p_{i},p_{j}\\}=0.$ (A.13)

One can also apply the Poisson bracket to the Hamiltonian function defined in
Eq. (A.6) and obtain

$\\{x_{i},\mathcal{H}\\}=\frac{\partial\mathcal{H}}{\partial
p_{i}}=\dot{x}_{i},$ (A.14)

i.e., the Poisson bracket with the Hamiltonian determines the time dynamics.
The last equality follows from how $\mathcal{H}$ is defined as a function of
$p_{i}$. Similarly, we have

$\\{p_{i},\mathcal{H}\\}=-\frac{\partial\mathcal{H}}{\partial
x_{i}}=\dot{p}_{i},$ (A.15)

where we have used that $-\partial\mathcal{H}/\partial
x_{i}=\partial\mathcal{L}/\partial x_{i}$. In the last equality we used the
Euler-Lagrange equation and the definition of $p_{i}$. Eqs. (A.14) and (A.15)
are known as Hamilton’s equations.

Quantization means that we promote the conjugate variables
$x_{i}\in\mathbb{R}$ and $p_{j}\in\mathbb{R}$ to operators
$x_{i}\rightarrow\hat{x}_{i}$, $p_{j}\rightarrow\hat{p}_{j}$,
$\mathcal{H}\rightarrow H$. Paul Dirac came up with the idea that the Poisson
bracket should be replaced by the commutator due to ‘their great similarity’,
see roughly minutes 15.30-23.00 in Dirac Lecture 1 of 4 on Quantum Mechanics.
The mapping is

$\\{f,g\\}\rightarrow\frac{1}{i\hbar}[\hat{f},\hat{g}].$ (A.16)

This directly gives the canonical commutation relation

$[\hat{x}_{i},\hat{p}_{j}]=i\hbar\delta_{ij}\mathds{1}.$ (A.17)

through Eq. (A.13). In addition, the quantization of Eq. (A.14) gives
$[\hat{x}_{i}(t),H]=i\hbar\frac{d\hat{x}_{i}(t)}{dt}$ where $\hat{x}_{i}(t)$
is the time-evolved Heisenberg operator $\hat{x}_{i}(t)\equiv
e^{itH/\hbar}\hat{x}_{i}e^{-itH/\hbar}$.

[label=exc:field]For a scalar (continuum) field of variables $\xi(x_{\mu})$
depending on space-time coordinates $x_{\mu}=(t,x,y,z)$, the (non-
relativistic) action equals $\mathcal{S}=\int d^{3}x\int
dt\,\mathcal{L}(\xi(x_{\mu}),\frac{\partial\xi}{\partial x_{\mu}})$. Show that
the principle of minimal action,
$\frac{\delta\mathcal{S}}{\delta\xi(x_{\mu})}=0$ for the field configuration
$\xi(x_{\mu})$, implies that $\xi(x_{\mu})$ obeys the equation of motion

$\frac{\partial\mathcal{L}}{\partial\xi}-\frac{\partial}{\partial
x_{\mu}}\frac{\partial\mathcal{L}}{\partial(\frac{\partial\xi}{\partial
x_{\mu}})}=0.$ (A.18)

(Comment: non-relativistic means that we make no distinction between $x_{\mu}$
and $x^{\mu}$ while these are usually related as $x_{\mu}=g_{\mu\nu}x^{\nu}$
with summation implied and $g_{\mu\nu}$ is the metric, nor is the action is
here manifestly Lorentz-invariant.)

[ref=exc:field] If we vary the field locally, i.e.,
$\xi^{\prime}(x_{\mu})=\xi(x_{\mu})+\delta\xi(x_{\mu})$, it leads to

$\mathcal{S}^{\prime}=\int d^{3}x\int
dt\,\mathcal{L}\left(\xi(x_{\mu})+\delta\xi,\frac{\partial(\xi+\delta\xi)}{\partial
x_{\mu}}\right)=\mathcal{S}+\int d^{3}x\int
dt\,\left[\frac{\partial\mathcal{L}}{\partial\xi}\delta\xi+\sum_{\mu}\frac{\partial\mathcal{L}}{\partial(\frac{\partial\xi}{\partial
x_{\mu}})}\frac{\partial(\delta\xi)}{\partial x_{\mu}}\right],$

and partial integration of the last term, assuming that $\delta\xi$ (and
$\xi(x_{\mu})$) vanish at the boundaries of the space-time coordinates, we get

$\delta\mathcal{S}=\int d^{3}x\int
dt\,\delta\xi(x_{\mu})\left[\frac{\partial\mathcal{L}}{\partial\xi}-\frac{\partial}{\partial
x_{\mu}}\frac{\partial\mathcal{L}}{\partial(\frac{\partial\xi}{\partial
x_{\mu}})}\right],$

so that $\delta\mathcal{S}=0$ for any variation $\delta\xi(x_{\mu})$ leads to
Eq. (A.18).

##  Appendix B Harmonic systems and beyond: elimination of high-energy
variables

In this appendix, we discuss how to extract the uncoupled normal modes from a
classical Lagrangian of coupled modes in a quadratic potential. Such
Lagrangians corresponding to ‘linear’ systems — the equations of motion are
linear — are ubiquitous in circuit-QED. If the potential energy admits an
expansion as a polynomial in the degrees of freedom and the terms beyond the
quadratic potential are weak, one can first first solve the harmonic system
and express the remaining nonlinear terms of the potential in terms of the
normal modes of the harmonic system.

In Section B.2 we go beyond this and discuss the Born-Oppenheimer
approximation which allows one to eliminate ‘high-frequency’ or ‘fast’
variables to focus on obtaining low-energy effective dynamics with fewer
degrees of freedom. This method can also be used to eliminate degrees of
freedom in the Lagrangian which have very large kinetic energy (capacitance or
’mass’ going to zero).

###  B.1 Normal modes of a harmonic system

We first argue, through one simple Exercise, that any linear dependence in the
potential energy on the variables $x_{i}$ can be absorbed in a quadratic
dependence by a change of variables.

Given a time-independent Lagrangian $\mathcal{L}(\bm{x},\bm{\dot{x}})$ with
potential

$U(\bm{x})=\bm{a}^{T}\bm{x}+\frac{1}{2}\bm{x}^{T}\bm{K}\bm{x},$ (B.1)

with symmetric matrix $\bm{K}>0$ and any vector $\bm{a}$, show how to write
this as

$U^{\prime}(\bm{x}^{\prime})=\frac{1}{2}\bm{x}^{\prime
T}\bm{K}\bm{x}^{\prime},$ (B.2)

up to a constant, using $\bm{x}^{\prime}=\bm{x}+\bm{b}$ for some $\bm{b}$.

[ref=ex:lin] Transform $\bm{x}^{\prime}=\bm{x}+\bm{b}$ so that
$\bm{\dot{x}}^{\prime}=\bm{\dot{x}}$ with $\bm{b}=\bm{K^{-1}}\bm{a}$ using
that $\bm{K}$ is invertible and symmetric and dropping the constant term
$\frac{1}{2}\bm{b}^{T}\bm{b}$.

Now consider some linear dependence on $\bm{\dot{x}}$ in the Lagrangian
$\mathcal{L}$, i.e.,

$\mathcal{L}=\frac{1}{2}\bm{\dot{x}^{\prime}}^{T}\bm{C}\bm{\dot{x}^{\prime}}+\bm{a}^{T}\bm{\dot{x}}-U(\bm{x}).$
(B.3)

where $U(\bm{x})$ is arbitrary. By choosing
$\mathcal{L}^{\prime}=\mathcal{L}-\bm{a}^{T}\bm{\dot{x}}$, or
$f(x_{1},\ldots,x_{n})=-\sum_{i}a_{i}x_{i}$ in Eq. (A.4), such dependence can
be gauged away entirely.

Hence besides special cases, e.g. when $\bm{a}$ in Eq. (B.3) is explicitly-
time-dependent or there are magnetic fields or gyrators at play, we can assume
that the Lagrangian is of the following form

$\mathcal{L}(\bm{x},\bm{\dot{x}})=\frac{1}{2}\bm{\dot{x}}^{T}\bm{C}\bm{\dot{x}}-U(\bm{x}).$
(B.4)

The potential is assumed to be of the form

$U(\bm{x})=\frac{1}{2}\sum_{i,j}K_{ij}x_{i}x_{j}+U_{\rm
nl}(x_{1},\ldots,x_{n}),$ (B.5)

where $\bm{K}\geq 0$ and $U_{\rm nl}$ is a higher-order (cubic or more)
polynomial in the $x_{i}$. We thus assume that the potential $U(\bm{x})$ can
be expanded as a polynomial in the variables $x_{i}$ and that we have
transformed away any linear part.

####  B.1.1 Diagonalization

If $U_{\rm nl}$ in Eq. (B.5) is zero, the system is harmonic. We then show how
to transform the variables $\bm{\dot{x}}$ and $\bm{x}$ such that they
represent a set of uncoupled harmonic oscillators, each with resonant
frequency $\omega_{i}$, see Eq. (B.10). We will perform the appropriate change
of variables at the Lagrangian level and so we do not have to worry about the
preservation of canonical commutation relations, which will be automatically
imposed when we define the conjugate variables.

We can write the quadratic part of the Lagrangian compactly as

$\mathcal{L}_{\rm
lin}(\bm{x},\bm{\dot{x}})=\frac{1}{2}\bm{\dot{x}}^{T}\bm{C}\bm{\dot{x}}-\frac{1}{2}\bm{x}^{T}\bm{K}\bm{x}.$
(B.6)

The Euler-Lagrange equation, Eq. (A.3), associated with the Lagrangian
$\mathcal{L}_{\rm lin}$ in Eq. (B.6) can be written as

$\bm{C}\frac{d^{2}\bm{x}}{dt^{2}}=-\bm{K}\bm{x}\implies\frac{d^{2}\bm{x}}{dt^{2}}=-\bm{C}^{-1}\bm{K}\bm{x},$
(B.7)

and is manifestly linear in the variables (hence the nomenclature linear or
harmonic systems).

Solving Eq. (B.7) is equivalent to diagonalizing the matrix
$\bm{C}^{-1}\bm{K}$. We remark that $\bm{C}^{-1}\bm{K}$ is not necessarily
symmetric. In the following we will make use of the following theorem (adapted
from Ref. [140], page 485).

######  Theorem B.1.1.

Let $\bm{A},\bm{B}$ be symmetric, real $n\times n$ matrices.

  1. 1.

If $\bm{A}>0$, then there is a non-singular, real $n\times n$ matrix $\bm{S}$
such that $\bm{A}=\bm{S}\bm{S}^{T}$ and
$\bm{B}=(\bm{S}^{-1})^{T}\bm{\Lambda}\bm{S}^{-1}$, where $\Lambda$ is a real,
diagonal matrix.

  2. 2.

If $\bm{A}>0$, then $\bm{AB}$ is diagonalizable and its diagonal matrix is
$\bm{\Lambda}$.

###### Proof.

The matrix $\bm{A}$ is positive definite and thus there exists an orthogonal
matrix $\bm{O}_{A}$ that diagonalizes $\bm{A}$, i.e.,
$\bm{O}_{A}^{T}\bm{A}\bm{O}_{A}=\bm{D}$ where $\bm{D}$ is a diagonal matrix
with positive diagonal entries. Defining the matrix
$\bm{T}=\bm{O}_{A}\bm{D}^{1/2}$, we have
$\bm{T}^{-1}\bm{A}(\bm{T}^{-1})^{T}=\mathds{1}$. The matrix
$\bm{T}^{T}\bm{B}\bm{T}$ is real and symmetric and thus can also be
diagonalized by an orthogonal matrix $\bm{O}$ such that
$\bm{O}^{T}\bm{T}^{T}\bm{B}\bm{T}\bm{O}=\bm{\Lambda}$. Defining the
nonsingular matrix $\bm{S}=\bm{T}\bm{O}$, we see that
$\bm{S}^{-1}\bm{A}(\bm{S}^{-1})^{T}=\mathds{1}$ and
$\bm{S}^{T}\bm{B}\bm{S}=\bm{\Lambda}$. The inversion of the previous formulas
completes the proof of point 1.
Point 2 simply follows from point 1. In fact, we can write
$\bm{A}=\bm{S}\bm{S}^{T}$ and
$\bm{B}=(\bm{S}^{-1})^{T}\bm{\Lambda}\bm{S}^{-1}$ from which we get
$\bm{A}\bm{B}=\bm{S}\bm{\Lambda}\bm{S}^{-1}$. ∎

In addition, it can be shown that if $\bm{B}\geq 0$, then $\bm{\Lambda}$ has
nonnegative eigenvalues (if $\bm{B}>0$, $\bm{\Lambda}$ has positive
eigenvalues). We omit the proof of this fact here, but we refer the interested
reader to Theorems 4.5.8 and 7.6.1 in Ref. [140]. Our matrices $\bm{C}^{-1}$
($=\bm{A}$) and $\bm{K}$ ($=\bm{B}$) satisfy the assumptions of Theorem B.1.1
and let’s call $\bm{\Lambda}=\bm{\omega^{2}}$ with nonnegative eigenvalues.
Let $\bm{S}$ be the matrix that diagonalizes $\bm{C}^{-1}\bm{K}$, that is

$\bm{S}^{-1}\bm{C}^{-1}\bm{K}\bm{S}=\bm{\Lambda},$ (B.8)

and such that $\bm{C}^{-1}=\bm{S}\bm{S}^{T}/C_{0}$ where $C_{0}>0$ is an
arbitrary parameter with the same dimensions as the elements of $\bm{C}$ (we
introduce the parameter $C_{0}$ to ensure that $\bm{S}$ is dimensionless).
Notice that $\bm{C}^{-1}\bm{K}$ is not necessarily symmetric and thus,
$\bm{S}$ is not an orthogonal matrix, although it must be invertible. Note
that in case $\bm{K}$ does not have full rank, i.e., the quadratic potential
is flat ‘in some directions’, there will be corresponding zero eigenvalues in
$\bm{\Lambda}$.

We now define new normal-mode variables as

$\bm{X}=\bm{S}^{-1}\bm{x}.$ (B.9)

Using the properties of Theorem B.1.1 the quadratic part of the Lagrangian Eq.
(B.6) can be written in terms of the normal-mode variables $\bm{X}$ as

$\mathcal{L}_{\rm
lin}(\bm{X},\bm{\dot{X}})=\frac{C_{0}}{2}\bm{\dot{X}}^{T}\bm{\dot{X}}-\frac{C_{0}}{2}\bm{X}^{T}\bm{\Lambda}\bm{X}=\sum_{i=1}^{n}\left[\frac{C_{0}}{2}\dot{X}_{i}^{2}-\frac{C_{0}}{2}\omega_{i}^{2}X_{i}^{2}\right],$
(B.10)

which is the Lagrangian of a collection of uncoupled harmonic oscillators with
resonant frequencies $\omega_{i}$.

Defining the conjugates to these normal modes as

$\bm{P}=\frac{\partial\mathcal{L}}{\partial\bm{\dot{X}}}=C_{0}\bm{\dot{X}},$
(B.11)

we also immediately obtain the Hamiltonian in terms of the normal modes

$\mathcal{H}(\bm{X},\bm{P})=\frac{1}{2C_{0}}\bm{P}^{T}\bm{P}+\frac{1}{2}C_{0}\bm{X}^{T}\bm{\Lambda}\bm{X}.$
(B.12)

Quantization implies that $[\hat{X}_{i},\hat{P}_{j}]=i\hbar\delta_{ij}$. By
introducing annihilation operators

$\hat{a}_{i}=\sqrt{\frac{C_{0}\omega_{i}}{2\hbar}}\left(\hat{X}_{i}+\frac{i}{C_{0}\omega_{i}}\hat{P}_{i}\right)$

with $[\hat{a}_{i},\hat{a}_{j}^{\dagger}]=\delta_{ij}\mathds{1}$ we can write
the quantized Hamiltonian as

$H=\sum_{i=1}^{n}\hbar\omega_{i}\left(\hat{a}_{i}^{\dagger}\hat{a}_{i}+\frac{1}{2}\right).$
(B.13)

Now we can include the non-linear term directly into the classical
Hamiltonian. Using the normal modes and their conjugate momenta, we have

$\mathcal{H}(\bm{X},\bm{P})=\frac{1}{2C_{0}}\bm{P}^{T}\bm{P}+\frac{1}{2}C_{0}\bm{X}^{T}\bm{\Lambda}\bm{X}+U_{\rm
nl}(\bm{S}\bm{X}).$ (B.14)

Using
$\hat{X}_{i}=\sqrt{\frac{\hbar}{2C_{0}\omega_{i}}}(a_{i}+a_{i}^{\dagger})$ we
can write the nonlinearity $U_{\rm nl}(\sum_{j}S_{ij}X_{j})$ in terms of
creation and annihilation operators, see also Section 5.2 and Appendix B.3 for
the use of this treatment in black-box quantization in circuit-QED.

###  B.2 Eliminating high-frequency modes

Having determined the normal modes of a system, we imagine that for some
subset of modes, the frequencies $\omega_{i}$ are much larger than for the
other modes. We can call these the high-energy modes $X_{i,{\rm high}}$
(versus low-energy modes $X_{i,{\rm low}}$). Then, as a very simple
approximation, one may choose to put these high-energy modes directly in their
ground vacuum state, and replace the operators $X_{i,{\rm high}}$ and
$P_{i,{\rm high}}$ by taking expectation values with respect to the vacuum
state. We then work in a reduced Hilbert space where one considers only the
remaining low-energy modes. Such substitution will of course also affect the
nonlinearity $U_{\rm nl}(\bm{S}\bm{X})$ in Eq. (B.5) which will reduce to some
nonlinearity on the remaining ‘active’ modes. In this simple approach the
effect of the nonlinearity is not included in determining the frequencies of
the high-energy or the low-energy modes.

A more accurate approximation which goes by the name of Born-Oppenheimer
approximation is as follows 111See Ref. [141] for the original work by Born
and Oppenheimer (in German).. This approximation is used in the molecular
physics literature where the ‘fast’ high-energy (light mass) variables are the
motional degrees of freedom of the electrons and the slow (heavy) variables
are those of the nuclei. In what follows, we adapt it to our case of harmonic
oscillators coupled by a nonlinear potential. We first identify a set of low
frequency modes with positions $\\{X_{i,{\rm low}}\\}$ and high frequency
modes to be eliminated with positions $\\{X_{i,{\rm high}}\\}$. Then we return
to the Lagrangian, obtained by applying Eq. (B.9) to Eq. (B.5) and collecting
the linear term in Eq. (B.10)

$\mathcal{L}(\bm{X},\bm{\dot{X}})=\frac{C_{0}}{2}\bm{\dot{X}}^{T}\bm{\dot{X}}-\frac{C_{0}}{2}\bm{X}^{T}\bm{\omega}^{2}\bm{X}-U_{\rm
nl}(\bm{S}\bm{X}).$ (B.15)

We expand this Lagrangian to second order in the variables $X_{i,{\rm high}}$,
keeping all dependence on $X_{j,{\rm low}}$:

$U_{\rm nl}(\bm{S}\bm{X})=\sum_{j}c_{j}(\\{X_{i,{\rm low}}\\})X_{j,{\rm
high}}+\frac{1}{2}\bm{X}_{\rm high}^{T}\bm{N}(\\{X_{i,{\rm
low}}\\})\bm{X}_{\rm high}+\tilde{U}_{\rm nl}(\\{X_{i,\rm
low}\\})+O(|\bm{X}_{\rm high}|^{3}),$ (B.16)

where $c_{j}(\\{X_{i,{\rm low}\\}}\\})$ are some coefficients with depend on
$X_{i,{\rm low}}$. Here we have collected all terms which solely depend on
$X_{i,{\rm low}}$ in $\tilde{U}_{\rm nl}(\\{X_{i,\rm low}\\})$. Thus the
‘positions’ of the low-frequency (slow) modes create a potential for the high-
frequency (fast) modes 222If one expands around a point where $\bm{N}$ has
negative eigenvalues, one has to include higher-order nonlinear terms to have
a confining potential.. With this expansion, the form of the Lagrangian for
the vector of variables $\bm{X}_{\rm high}$ is that of a harmonic system as
the linear term in $\bm{X}_{\rm high}$ can be transformed away into a
quadratic term (see Exercise B.2) assuming that the symmetric matrix $\bm{N}$
is invertible. More generally, in the Born-Oppenheimer approximation one
solves for the ground-state problem of the high-energy variables (a possibly
non-harmonic system).

Neglecting the higher-order dependence on $\bm{X}_{\rm high}$, we can
diagonalize the Lagrangian for the high-energy modes and obtain a form as in
Eq. (B.10), where the frequencies $\tilde{\omega}_{k}$ of these new uncoupled
high-energy modes will depend on $\bm{X}_{\rm low}$. Then when we put these
high-frequency modes in their ground state: they will contribute their vacuum
energy $\frac{1}{2}\sum_{k,{\rm
high}}\hbar\tilde{\omega}_{k}(\\{\hat{X}_{i,{\rm low}}\\})$. The final
Hamiltonian in this approximation reads

$H\approx\sum_{i,{\rm low}}\hbar\omega_{i}\left(\hat{a}_{i,{\rm
low}}^{\dagger}\hat{a}_{i,{\rm
low}}+\frac{1}{2}\right)+\frac{1}{2}\sum_{k,{\rm
high}}\hbar\tilde{\omega}_{k}(\\{\hat{X}_{i,{\rm low}}\\})+\tilde{U}_{\rm
nl}(\\{\hat{X}_{i,{\rm low}}\\}).$ (B.17)

Here the first term comes about through the first two terms in Eq. (B.15) only
for the variables $X_{i,{\rm low}}$ with their associated operators
$\hat{a}_{i,{\rm low}}$. Thus the high-energy modes are eliminated inducing an
effective (nonlinear) coupling between the previously decoupled low-energy
modes. This method was used for handling electrical circuits in Ref. [142].

Figure B.1: Series combination of a linear inductance $L$ and a nonlinear
inductor (red) in parallel to a shunting capacitance $C$. The blue branch
highlights the intrinsic capacitance $C^{\prime}$ of the nonlinear inductor,
which we consider to be either vanishingly small (regular case) or absent
(singular case).

####  B.2.1 Illustration of the Born-Oppenheimer method

We now go on to a simple example that illustrates the application of the Born-
Oppenheimer approach. We consider the series combination of a linear
inductance $L$ and a generic nonlinear inductor with intrinsic capacitance
$C^{\prime}$, all in parallel with a total shunting capacitance $C$; see Fig.
B.1. If $C^{\prime}=0$ we have a singular circuit for which a Hamiltonian
cannot be straightforwardly derived as one degree of freedom only enters the
potential energy (see the discussion in Section 3.3.1).

Here we imagine $C^{\prime}$ is nonzero but very small. Then, we can
straightforwardly transform the Lagragian

$\mathcal{L}=\frac{C\dot{\Phi}^{2}}{2}+\frac{C^{\prime}\dot{\Phi}_{c}^{2}}{2}-\frac{(\Phi-\Phi_{c})^{2}}{2L}-U_{\rm
nl}(\Phi_{c}),$ (B.18)

to obtain the quantized, circuit Hamiltonian:

$H_{r}=\frac{\hat{Q}^{2}}{2C}+\frac{\hat{Q}_{c}^{2}}{2C^{\prime}}+\frac{(\hat{\Phi}-\hat{\Phi}_{c})^{2}}{2L}+U_{\rm
nl}(\hat{\Phi}_{c}).$ (B.19)

The divergence of the second term of Eq. (B.19) for $C^{\prime}\rightarrow 0$
indicates that $\Phi_{c}$ will be the “fast” high-energy variable.

The Born-Oppenheimer approximation [142] will allow us to derive an effective
low-energy Hamiltonian as a function of $\Phi$ and $Q$ only. To this end, we
first solve the stationary Schrödinger equation associated with the fast
degree of freedom, $\Phi_{c}$, for fixed values of $\Phi$ and $Q$. Thus, we
identify the high-energy part of $H_{r}$, exhibiting fast dynamics
classically, as

$H_{\text{fast}}=\frac{\hat{Q}_{c}^{2}}{2C^{\prime}}+\frac{(\hat{\Phi}-\hat{\Phi}_{c})^{2}}{2L}+U_{\rm
nl}(\hat{\Phi}_{c}),$ (B.20)

and we solve

$H_{\text{fast}}\psi_{\Phi,n}(\Phi_{c})=E_{\Phi,n}\psi_{\Phi,n}(\Phi_{c})$
(B.21)

for the eigenfunctions $\psi_{\Phi,n}(\Phi_{c})$ and the associated
eigenenergies $E_{\Phi,n}$, which both are labeled by $n\in\mathbb{N}_{0}$ and
parametrized by $\Phi$. The ground state energy ($n=0$) is then considered as
an effective low-energy potential for the low-energy (or “slow”) variable,
$\Phi$, whose dynamics is captured by the effective Hamiltonian

$H_{r,\text{eff}}=\frac{\hat{Q}^{2}}{2C}+U_{\text{BO}}(\hat{\Phi}),$ (B.22)

with the Born-Oppenheimer potential given by the fast-variable ground state
energy:

$U_{\text{BO}}(\Phi)=E_{\Phi,n=0}-E_{\Phi=0,n=0}.$ (B.23)

It is handy to introduce here an energy offset of $U_{\text{BO}}(\Phi)$ such
that $U_{\text{BO}}(0)=0$ in order to avoid additive constants that are
divergent as $C^{\prime}\rightarrow 0$.

We will shortly make a particular choice for $U_{\rm nl}$. Of course, the
nonlinear potential that we know is $-E_{J}\cos(\Phi_{c})$. Since we are only
here trying to illustrate the mathematical treatment of Born-Oppenheimer
theory, we will make another choice that makes the analysis easier. For many
situations, including when we have the Josephson potential, Born-Oppenheimer
becomes a purely numerical exercise, which we want to avoid here. We will
explain in the end the relation of our simplified $U_{\rm nl}$ to the
realistic case of the Josephson potential in Fig. 3.1.

In what follows, it is convenient to think about the $\Phi$ degree of freedom
as a classical variable, while only $\Phi_{c}$ is quantized. Thus, we will not
use hats on the $\Phi$ variable, but we will quantize it only at the end of
the procedure. Before fixing a particular $U_{\rm nl}$, it is informative to
rescale the fast Hamiltonian. We introduce the ${\rm LC^{\prime}}$-resonator
frequency and a flux zero point fluctuation pareameter, defined as

$\omega^{\prime}_{r}=\frac{1}{\sqrt{LC^{\prime}}},\qquad\delta\Phi=\sqrt{\hbar}\biggl{(}\frac{L}{C^{\prime}}\biggr{)}^{1/4},$
(B.24)

respectively, and we express $H_{\text{fast}}$ as

$H_{\text{fast}}=\hbar\omega^{\prime}_{r}\left[\frac{\hat{p}^{2}+\hat{z}^{2}}{2}+\frac{U_{\rm
nl}(\delta\Phi\hat{z}+\Phi)}{\hbar\omega^{\prime}_{r}}\right].$ (B.25)

Note that the parameter $\delta\Phi$ is essentially $\Phi_{\rm zpf}$ of Eq.
(2.29), differing by factor $\sqrt{2}$. The dimensionless conjugate variables
$\hat{z}$ and $\hat{p}$ are defined as
$\hat{z}=(\hat{\Phi}_{c}-\Phi)/\delta\Phi$ and
$\hat{p}=\hat{Q}_{c}\delta\Phi/\hbar$, and they satisfy the canonical
commutation relation $[\hat{z},\hat{p}]=i\mathds{1}$.

We define the parameter
$\epsilon=1/\sqrt{\hbar\omega^{\prime}_{r}}=\sqrt[4]{LC^{\prime}}/\sqrt{\hbar}=\sqrt{L}/\delta\Phi$,
and we divide out the prefactor in Eq. (B.25), obtaining

$\epsilon^{2}H_{\text{fast}}=H_{0}+\epsilon^{2}U_{\rm
nl}\left(\frac{\sqrt{L}}{\epsilon}\hat{z}+\Phi\right),$ (B.26)

with the dimensionless harmonic-oscillator Hamiltonian
$H_{0}=(\hat{p}^{2}+\hat{z}^{2})/2$. Note that we are looking to get the
$\Phi$ dependence of the ground state of this Hamiltonian, which appears only
in the final term. Furthermore, we note that since $\epsilon$ can be viewed as
a small parameter when we consider $C^{\prime}\rightarrow 0$, we may be able
to progress by treating the final term in Eq. (B.26) as a perturbation on the
ground state of $H_{0}$ (as in the expansion in Eq. (B.16)). The ground state
energy of $H_{0}$ is just $1/2$, so we will seek small corrections in
$\epsilon$ to this. Note that we will have to divide out the $\epsilon^{2}$
from the left hand side of Eq. (B.26) in the end, so that the ground state
energy of $H_{\rm fast}$ will diverge; this is correct, and due to zero-point
fluctuations. But the $\Phi$ dependence of this ground state energy will be in
the perturbative corrections, and will not diverge for $U_{\rm nl}$ functions
of interest.

Of course, the case of most interest is the Josephson potential, but the
perturbative calculation is not feasible to work out in closed form in this
case, so we take another model $U_{\rm nl}$ potential, namely

$U_{\rm nl}(\Phi)=E_{\rm
nl}\left|\frac{\Phi}{\Phi_{0}}\right|^{\nicefrac{{3}}{{2}}}.$ (B.27)

We will explain at the end why this mathematical curiosity is informative
about the case of the Josephson potential.

For our choice, the perturbation in Eq. (B.26) becomes explicitly

$\epsilon^{2}U_{\rm
nl}\left(\frac{\sqrt{L}}{\epsilon}\hat{z}+\Phi\right)=\epsilon^{\nicefrac{{1}}{{2}}}E_{\rm
nl}\frac{L^{\nicefrac{{3}}{{4}}}}{\Phi_{0}^{\nicefrac{{3}}{{2}}}}\left|\hat{z}+\frac{\epsilon}{\sqrt{L}}\Phi\right|^{\nicefrac{{3}}{{2}}}.$
(B.28)

To proceed with first order perturbation theory, we compute

$\bra{0}\epsilon^{\nicefrac{{1}}{{2}}}E_{\rm
nl}\frac{L^{\nicefrac{{3}}{{4}}}}{\Phi_{0}^{\nicefrac{{3}}{{2}}}}\left|\hat{z}+\frac{\epsilon}{\sqrt{L}}\Phi\right|^{\nicefrac{{3}}{{2}}}\ket{0}=\epsilon^{\nicefrac{{1}}{{2}}}E_{\rm
nl}\frac{L^{\nicefrac{{3}}{{4}}}}{\pi^{\nicefrac{{1}}{{2}}}\Phi_{0}^{\nicefrac{{3}}{{2}}}}\int_{-\infty}^{\infty}dze^{-z^{2}}\left|z+\frac{\epsilon}{\sqrt{L}}\Phi\right|^{\nicefrac{{3}}{{2}}}\\\
=\epsilon^{\nicefrac{{1}}{{2}}}E_{\rm
nl}\frac{L^{\nicefrac{{3}}{{4}}}}{\pi^{\nicefrac{{1}}{{2}}}\Phi_{0}^{\nicefrac{{3}}{{2}}}}\frac{\pi\sqrt{\epsilon\Phi}}{2\sqrt[4]{L}}e^{-\frac{\epsilon^{2}\Phi^{2}}{4L}}\left[\frac{\epsilon^{2}\Phi^{2}}{L}I_{\frac{3}{4}}\left(\frac{\epsilon^{2}\Phi^{2}}{4L}\right)+\left(\frac{\epsilon^{2}\Phi^{2}}{L}+1\right)I_{-\frac{1}{4}}\left(\frac{\epsilon^{2}\Phi^{2}}{4L}\right)\right].$
(B.29)

Here we see the appearance of modified Bessel functions $I_{n}$ (integrals
courtesy of Mathematica). The Bessel functions here are non-analytic (fourth
root) in their arguments, but because these arguments go like $\Phi^{2}$, and
because of the additional $\sqrt{\Phi}$ factor, Eq. (B.29) is analytic in
$\Phi^{2}$. Expanding in a Taylor series:

$\bra{0}\epsilon^{\nicefrac{{1}}{{2}}}E_{\rm
nl}\frac{L^{\nicefrac{{3}}{{4}}}}{\Phi_{0}^{\nicefrac{{3}}{{2}}}}\left|\hat{z}+\frac{\epsilon}{\sqrt{L}}\Phi\right|^{\nicefrac{{3}}{{2}}}\ket{0}=\epsilon^{\nicefrac{{1}}{{2}}}E_{\rm
nl}\frac{L^{\nicefrac{{3}}{{4}}}}{\pi^{\nicefrac{{1}}{{2}}}\Phi_{0}^{\nicefrac{{3}}{{2}}}}\left({\mbox{const.}}+\frac{3\pi}{4\sqrt[4]{2}\Gamma(3/4}\frac{\epsilon^{2}\Phi^{2}}{L}+O\left((\epsilon\Phi)^{4}\right)\right).$
(B.30)

So, dividing by $\epsilon^{2}$, we get the following form for the Born-
Oppenheimer potential:

$U_{\text{BO}}(\Phi)=E_{\Phi,0}-E_{\Phi=0,0}\approx\frac{3\pi^{\nicefrac{{1}}{{2}}}}{4\sqrt[4]{2}\Gamma({\nicefrac{{3}}{{4}}})}\frac{E_{\rm
nl}}{L^{\nicefrac{{1}}{{4}}}\Phi_{0}^{\nicefrac{{3}}{{2}}}}\epsilon^{\nicefrac{{1}}{{2}}}\Phi^{2}=\frac{3\pi^{\nicefrac{{1}}{{2}}}}{4\sqrt[4]{2}\Gamma({\nicefrac{{3}}{{4}}})}\frac{E_{\rm
nl}(C^{\prime})^{\nicefrac{{1}}{{8}}}}{\hbar^{\nicefrac{{1}}{{4}}}L^{\nicefrac{{1}}{{8}}}\Phi_{0}^{\nicefrac{{3}}{{2}}}}\Phi^{2}.$
(B.31)

By comparing with Eq. (2.28b), the reader can see that this expression for the
Born-Oppenheimer potential is dimensionally consistent. Examination of the
Taylor expansion shows that this quadratic expression is valid over a range
that diverges as $\epsilon\rightarrow 0$. Thus, one observes that the
anharmonic potential has, due to quantum fluctuations, been turned into a
harmonic effective potential for the slow coordinate. But perhaps more
significant is that, due to the final $C^{\prime}$ dependence, this whole
potential is vanishing (very slowly!) as $C^{\prime}\rightarrow 0$. Thus, the
whole circuit from node $\Phi$ to ground is going to an open circuit (i.e.,
effectively infinite inductance). This is the most notable consequence of
quantum fluctuations. In Ref. [42], it was proven that this open-circuit
behavior is universal, and applies also to the case where a Josepshon junction
is present as the nonlinear element in Fig. B.1. The only thing that is
required for this universal behavior to hold is that the nonlinear potential
increases more slowly than $\Phi_{c}^{2}$ for large $|\Phi_{c}|$.

###  B.3 Normal modes and Cauer’s construction

Building on the analysis in Appendix B.1.1, we now show that an arbitrary
lossless, reciprocal impedance matrix can always be expanded as in Eq. (5.34),
a result that goes under the name of Foster’s theorem when we only have a
single port [100, 143]. We provide a derivation of this result based on the
Lagrangian formalism for electrical circuits that we have given in these
lecture notes. A similar approach has been put forward also in Ref. [144]. The
derivation makes the equivalence between the normal modes of a linear circuit
and the Foster expansion of the impedance matrix clear. The Lagrangian
formulation also allows the straightforward derivation of the Hamiltonian and
consequently the quantization of the circuit. In addition, we will point out
how nonlinear effects, due for instance to Josephson junctions coupled at the
ports, can be straightforwardly introduced in the model leading to the general
Hamiltonian in Eq. (5.82).

We start our derivation by considering the general case of an $N$-port network
pictorially depicted in Fig. 5.1. As we assume the network to be LTI, it is
completely characterized by its impulse-response matrix, i.e., its impedance
$\bm{Z}(s)$ in the Laplace domain, and the network must consist only of
interconnections of capacitances, inductances and mutual inductances (or ideal
transformers). We follow the general procedure for obtaining the Lagrangian of
a circuit described in Section 2.1. We make the following observations:

  * •

We can always choose the port fluxes $\Phi_{p_{n}}$, $n=1,\dots,N$ defined in
Eq. (5.27) as the variables in the problem;

  * •

Additionally, there will be internal degrees of freedom that will generally be
defined as the integral of the voltages across some arbitrary branches. These
internal degrees of freedom depend on the topology of the specific circuit. We
denote them as $\Phi_{In^{\prime}}$, with $n^{\prime}=1,\dots,N_{I}$ and
$N_{I}$ the total number of independent internal degrees of freedom.

Given these facts the Lagrangian of a linear, reciprocal and lossless network
can always be written as

$\mathcal{L}_{\mathrm{lin}}(\bm{\dot{\theta}};\bm{\theta})=\frac{1}{2}\bm{\dot{\theta}}^{T}\bm{C}\bm{\dot{\theta}}-\frac{1}{2}\bm{\theta}^{T}\bm{K}\bm{\theta},$
(B.32)

where we defined the total vector of fluxes

$\bm{\theta}=\begin{bmatrix}\Phi_{p_{1}},\dots,\Phi_{p_{N}},\Phi_{I_{1}},\dots,\Phi_{I_{N_{I}}}\end{bmatrix}^{T},$
(B.33)

i.e., $\bm{\theta}$ is a vector of length $M=N+N_{I}$ with $M$ the total
number of independent variables in the problem.

We now note that Eq. (B.32) is of the form of the Lagrangian in Eq. (B.6), and
thus we can simply follow the derivation there. Defining the normal-mode
variables (cf. Theorem B.1.1)

$\bm{\Phi}=\bm{S}^{-1}\bm{\theta},$ (B.34)

we can readily write

$\mathcal{L}_{\mathrm{lin}}(\dot{\bm{\Phi}},\bm{\Phi})=\sum_{m=1}^{M}\left[\frac{C_{0}}{2}\dot{\Phi}_{m}^{2}-\frac{C_{0}}{2}\omega_{m}^{2}\Phi_{m}^{2}\right].$
(B.35)

As expected the Lagrangian expressed as a function of the normal mode fluxes
can be interpreted as a collection of $M$ uncoupled harmonic oscillators with
frequency $\omega_{m}$ and (arbitrary) characteristic impedance given in Eq.
(5.30), using the reference capacitance $C_{0}$.

The previous derivation shows that such a network can always be replaced by an
equivalent circuit shown in Fig. 5.5. In fact, the circuit has a simple
interpretation. We have several uncoupled parallel LC oscillators that are
associated with the normal modes. Eq. (B.34) implies that the port fluxes (or
voltages) can be expressed as a linear combination of those associated with
the normal modes, specifically

$\Phi_{p_{n}}=\sum_{m=1}^{M}t_{mn}\Phi_{m},\quad n=1,\dots,N,$ (B.36)

where the $t_{mn}=S_{mn}\in\mathbb{R}$ have the interpretation of (signed)
turn ratios of the transformers in the circuit in Fig. 5.5 and are the
elements of the vectors $\bm{t}_{m}$ in Eq. (5.32).

It is now a simple exercise to show that the impedance matrix associated with
the circuit in Fig. 5.5 can be expanded as in Eq. (5.34). The impedance of the
$m$-th LC-oscillator in the circuit is given by

$Z_{m}(s)=\frac{1}{2C_{0}}\biggl{(}\frac{1}{s+i\omega_{m}}+\frac{1}{s-i\omega_{m}}\biggr{)}.$
(B.37)

Let us now apply a generic vector of currents at the ports

$\bm{I}(s)=\begin{bmatrix}I_{p_{1}}(s)&I_{p_{2}}(s)&\dots&I_{p_{N}}(s)\end{bmatrix}^{T}.$
(B.38)

The total current passing through the $m$-th LC-oscillator is given by

$I_{m}(s)=\sum_{n=1}^{N}t_{mn}I_{P_{n}}(s)=\bm{t}_{m}^{T}\bm{I}(s).$ (B.39)

As a consequence we get the following voltage across the $m$-th LC-oscillator:

$V_{m}(s)=Z_{m}(s)I_{m}(s)=\frac{1}{2C_{0}}\biggl{(}\frac{1}{s+i\omega_{m}}+\frac{1}{s-i\omega_{m}}\biggr{)}\bm{t}_{m}^{T}\bm{I}(s).$
(B.40)

The voltage drop across the $n$-th port is then given as

$V_{p_{n}}(s)=\sum_{m=1}^{M}t_{mn}V_{m}(s)=\sum_{m=1}^{M}\frac{1}{2C_{0}}\biggl{(}\frac{1}{s+i\omega_{m}}+\frac{1}{s-i\omega_{m}}\biggr{)}t_{mn}\bm{t}_{m}^{T}\bm{I}(s),$
(B.41)

and so we write the vector of port voltages as

$\bm{V}(s)=\underbrace{\biggl{(}\sum_{m=1}^{M}\frac{1}{2C_{0}}\biggl{(}\frac{1}{s+i\omega_{m}}+\frac{1}{s-i\omega_{m}}\biggr{)}\bm{t}_{m}\bm{t}_{m}^{T}\biggr{)}}_{\bm{Z}(s)}\bm{I}(s),$
(B.42)

with matrix $\bm{R}_{m}=\bm{t}_{m}\bm{t}_{m}^{T}$ which shows the result.

After quantizing the circuit, annihilation and creation operators for the
normal modes can be introduced as usual, to write

$\hat{\Phi}_{m}=\sqrt{\frac{\hbar
Z_{m}}{2}}\bigl{(}\hat{a}_{m}+\hat{a}_{m}^{\dagger}\bigr{)},$ (B.43)

which gives Eq. (5.29) and from which Eq. (5.33) also follows.

Finally, as discussed in Section 5.3, we also mention the weakly dissipative
case which can be modelled by adding resistances $R_{m}$ with $m=1,\dots,M$ in
parallel with the LC oscillators associated with each mode, as shown in Fig.
5.5. Weak dissipation corresponds to the resistance of the $R_{m}$s being
large.

In this case, following a similar reasoning as that in Section 5.3.1 for the
single RLC oscillator, in the weakly dissipative limit, the impedance matrix
can be approximated as

$\bm{Z}(s)\approx\sum_{m=1}^{M}\frac{1}{2C_{0}}\biggl{(}\frac{1}{s+i(\omega_{m}+i\kappa_{m}/2)}+\frac{1}{s+i(\omega_{m}-i\kappa_{m}/2)}\biggr{)}\bm{R}_{m},$
(B.44)

Recall that the condition for having weak dissipation for each mode is

$\kappa_{m}=\frac{1}{R_{m}C_{0}}\ll\omega_{m}.$ (B.45)

In order to determine the decay rates $\kappa_{m}$, and thus the effective
resistances $R_{m}$, in the circuit in Fig. 5.5, the procedure is completely
equivalent to the one described in Section 5.3.1 for the single-port case. Let
us select an arbitrary port, say port $1$, and let us take all other ports to
be open-circuited. If we let a current $I_{1}(s)$ flow through port $1$, we
get

$\bm{V}(s)=\begin{pmatrix}V_{p1}(s)\\\ V_{p2}(s)\\\ \vdots\\\
V_{pn}(s)\end{pmatrix}=\begin{pmatrix}Z_{11}(s)\\\ Z_{21}(s)\\\ \vdots\\\
Z_{M1}(s)\end{pmatrix}I_{1}(s).$ (B.46)

Note that in this setup port $1$ does not see the effect of the other ports,
and thus the circuit can be treated as the single-port case in Fig. 5.6. Thus,
port $1$ sees the effective admittance

$Y(s)=\frac{I_{1}(s)}{V_{p1}(s)}=\frac{1}{Z_{11}(s)}.$ (B.47)

This shows that, in the multi-port case, we can obtain the approximate damping
rate $\kappa_{m}$ of mode $m$ using Eq. (5.64) with the admittance $Y(s)$ in
Eq. (B.47) in the frequency domain.

###  B.4 LC circuit shunted by a small admittance

Figure B.2: LC oscillator in parallel with a shunting admittance.

In this appendix, we consider the circuit in Fig. B.2 focusing on the case of
small shunting admittance $Y_{\mathrm{sh}}(s)$. Our goal is two-fold. First,
we want to show how the small admittance modifies the resonant frequency of
the circuit and induces a decay rate. Second, we highlight the fact that the
decay rate we obtain with this simple classical reasoning matches the quantum
one in Eq. (7.6). In fact, we have already seen in Section 7.2 that Eq. (7.6)
reproduces the decay rate of an parallel RLC oscillator and we wish to confirm
that the same holds for a weak shunting admittance. Let us start by
considering the total admittance in the Laplace domain of the circuit in Fig.
B.2 that reads

$Y(s)=Y_{\mathrm{LC}}(s)+Y_{\mathrm{sh}}(s),$ (B.48)

with the admittance of the LC oscillator given by

$Y_{\mathrm{LC}}(s)=\frac{1}{Z_{\mathrm{LC}}(s)}=\left[\frac{1}{2C}\left(\frac{1}{s+i\omega_{r}}+\frac{1}{s-i\omega_{r}}\right)\right]^{-1},$
(B.49)

with resonant frequency $\omega_{r}=1/\sqrt{LC}$. Close to the pole $s\approx
i\omega$ the admittance of the LC oscillator is approximately

$Y_{\mathrm{LC}}(s)\approx 2C(s-i\omega_{r}).$ (B.50)

Thus, within this approximation, the total admittance becomes

$Y(s)\approx 2C(s-i\omega_{r})+Y_{\mathrm{sh}}(s)\approx
2C\left\\{s-i\left[\omega-\frac{\mathrm{Im}(Y_{\mathrm{sh}}(i\omega_{r}))}{2C}\right]+\frac{\mathrm{Re}(Y_{\mathrm{sh}}(i\omega_{r}))}{2C}\right\\}.$
(B.51)

Thus, we obtain a new pole

$\tilde{p}=i\left[\omega_{r}-\frac{\mathrm{Im}(Y_{\mathrm{sh}}(i\omega_{r}))}{2C}\right]-\frac{\mathrm{Re}(Y_{\mathrm{sh}}(i\omega_{r}))}{2C}=i\tilde{\omega}_{r}-\frac{\kappa_{\mathrm{sh}}}{2},$
(B.52)

where we defined the shifted resonant frequency

$\tilde{\omega}_{r}=\omega_{r}-\frac{\mathrm{Im}(Y_{\mathrm{sh}}(i\omega_{r}))}{2C},$
(B.53)

and the decay rate caused by the shunting admittance

$\kappa_{\mathrm{sh}}=\frac{\mathrm{Re}(Y_{\mathrm{sh}}(i\omega_{r}))}{C}.$
(B.54)

Notice that this procedure is valid self-consistently as long as the new pole
$\tilde{p}$ does not deviate too much from the original one at $i\omega_{r}$,
a condition that is satisfied if

$\biggl{|}\frac{\mathrm{Im}(Y_{\mathrm{sh}}(i\omega_{r}))}{2C}\biggr{|}\ll\omega_{r},\quad\biggl{|}\frac{\mathrm{Re}(Y_{\mathrm{sh}}(i\omega_{r}))}{2C}\biggr{|}\ll\omega_{r}.$
(B.55)

Repeating the procedure close to $s\approx-i\omega_{r}$, and using the fact
that $\forall\omega\in\mathbb{R}$ $\mathrm{Re}(i\omega)=\mathrm{Re}(-i\omega)$
and $\mathrm{Im}(i\omega)=-\mathrm{Im}(-i\omega)$ (see Chapter $4$ in [76] for
these properties), we also get another pole $\tilde{p}^{*}$ that is simply the
complex conjugate of $\tilde{p}$. Thus, following a similar reasoning as in
Section 5.3.1, we get that the impedance of the circuit in Fig. B.2 can be
approximated as

$Z(s)=\frac{1}{Y(s)}\approx\frac{1}{2C}\left(\frac{1}{s-\tilde{p}}+\frac{1}{s-\tilde{p}^{*}}\right).$
(B.56)

Up to now it is not clear why the coefficient $\kappa_{\mathrm{sh}}$ in Eq.
(B.54) is called a “decay rate”. In order to understand this we need to
analyze the problem in the time domain. Let us suppose a Dirac delta current
$\mathrm{i}(t)=\mathrm{i}_{0}\delta(t)$ flows through the circuit. In the
Laplace domain $I(s)=\mathrm{i}_{0}$ and thus we get

$V(s)=Z(s)I(s)\approx\frac{\mathrm{i}_{0}}{2C}\biggl{(}\frac{1}{s-\tilde{p}}+\frac{1}{s-\tilde{p}^{*}}\biggr{)}$
(B.57)

Now, remember that the Laplace transform of $g(t)=e^{pt}u(t)$ with $u(t)$ the
Heaviside step function is given by

$G(s)=\int_{-\infty}^{+\infty}dt\,e^{(p-s)t}u(t)=\int_{0}^{+\infty}dt\,e^{(p-s)t}=\frac{1}{s-p},$
(B.58)

when $\mathrm{Re}(p)<\mathrm{Re}(s)$. Thus, in the time domain we get that for
$t\geq 0$

$\mathrm{v}(t)=\frac{\mathrm{i}_{0}}{2C}\left(e^{\tilde{p}t}+e^{\tilde{p}^{*}t}\right)=\frac{\mathrm{i}_{0}}{2C}\left(e^{i\tilde{\omega}_{r}t}+e^{-i\tilde{\omega}_{r}t}\right)e^{-\frac{\kappa_{\mathrm{sh}}}{2}t},$
(B.59)

which shows why $\kappa_{\mathrm{sh}}/2$ is interpreted as a decay rate.

Finally, like for the RLC circuit analyzed in Section 7.2 (see the derivation
of Eq. (7.8)) the quantum formula for the decay rate in Eq. (7.6) at zero
temperature matches the classical derivation in Eq. (B.54) for a harmonic
oscillator:

$\frac{1}{T_{1}}=\frac{\mathrm{Re}(Y_{\mathrm{sh}}(i\omega_{r}))}{C}=\frac{\mathrm{Re}(\mathcal{Y}_{\mathrm{sh}}(\omega_{r}))}{C}=\kappa_{\mathrm{sh}}.$
(B.60)

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