I have an exercise to calculate the following one dimensional integral explicitly using the Wick theorem:

$$<q(t_1)q(t_2)q(t_3)q(t_4)q(t_5)q(t_6)>=\frac{\int Dq q(t_1)q(t_2)q(t_3)q(t_4)q(t_5)q(t_6)e^{-S_E} }{\int Dqe^{-S_E}}$$ where $$S_E=\int_{-\infty}^{+\infty}\Big(\frac{1}{2}\dot{q}^2+\frac{m^2q^2}{2} \Big)dt. $$ I know what the answer is going to be as it is just using Wick theorem. But as the problem is to explicitly calculate, I assume that I actually need to calculate the integral myself.

This is what I have done: $$\int Dq q(t_1)q(t_2)q(t_3)q(t_4)q(t_5)q(t_6)e^{-S_E} =\left. \frac{\delta}{\delta f(t_1)}\dots\frac{\delta}{\delta f(t_6)}\int e^{-S_E+\int\! dt~ fq}\right|_{f=0} . $$

I need some help in how to proceed form here.

  • $\begingroup$ From there is a gaussian integral. The expoent is quadratic in q. Do you know how to do it? $\endgroup$ – OkThen Nov 11 '16 at 15:07
  • $\begingroup$ I think so, I get $\sqrt{2\pi}$? $\endgroup$ – Fredovich Nov 11 '16 at 15:42
  • $\begingroup$ It's wrong. I'm sorry that I'm not able to write a better answer now. I'm traveling and without complete access to internet. Please, take a look at harmonic oscillator propagator. Wikipedia may suffice. In time, if you haven't solved it or someone hasn't posted something, I will write a complete answer. $\endgroup$ – OkThen Nov 12 '16 at 1:08
  • $\begingroup$ I look it up and I can solve the integral now, thanks. Do you think I interpreted the exercise correctly? I am still a bit unsure about that. $\endgroup$ – Fredovich Nov 16 '16 at 10:25

Sorry for the delay.

I will work with integrals defined on $\mathbf{R}^{n}$. A path integral can be thought as the limit $n \to \infty$. Generalizations to complex numbers are also straightforward.


I will use the summation convention. Every repeated index is summed over. For example,

$$ \sum_{i,j=1}^{n} x_{i} A_{ij} x_{j} \text{ is written as } x_{i} A_{ij} x_{j}. $$

Gaussian Integrals.

Consider the integral

$$ Z(A) = \int d^{n} x \, \exp\left( - \frac{1}{2} x_{i} A_{ij} x_{j} \right). $$

It converges if the eigenvalues of A are non-negative and non-vanishing. One can also prove that

$$ Z(A) = ( 2 \pi )^{n/2} \left( \det A \right)^{-1/2}. $$

by several methods. A possible way is to consider an orthogonal transformation which diagonalizes $A$, such that

$$ A = O D O^{T} , \quad O^{T} O = 1 \quad \text{and} \quad D_{ij} = a_{i} \delta_{ij} $$

where $a_{i}$ are the eigenvalues of $A$. The integral factorizes because

$$ x_{i} A_{ij} x_{j} = x_{i} O_{ik} a_{k} O_{jk} x_{j} = a_{i} y_{i}^{2} $$

for $y_{i} = O_{ij} x_{j}$. And it converges to

$$ Z(A) = \Pi_{i=1}^{n} \int dy_{i} e^{-a_{i} y_{i}^{2}/2} = (2 \pi)^{n/2} (a_{1} a_{2} \cdots a_{n})^{-1/2} = (2 \pi)^{n/2} (\det A)^{-1/2}. $$

Of special interest is the case

$$ Z(A,b) = \int d^{n}x \exp \left( - \frac{1}{2} x_{i} A_{ij} x_{j} + b_{i} x_{i} \right) $$

which gives

$$ Z(A, b) = (2 \pi)^{n/2} (\det A)^{-1/2} \exp \left[ \frac{1}{2} b_{i} \Delta_{ij} b_{j} \right] $$

where $\Delta$ is the inverse of $A$ (prove this result yourself. Hint: Consider the change of variables

$$ x_{i} = \Delta_{ij} b_{j} + y_i $$

from $x_{i}$ to $y_{i}$).

Wick Theorem.

A Gaussian integrand can be considered a probability distribution. We can use it to calculate expectation values:

$$ \langle F(x) \rangle = N \int d^{n} x \, F(x) \, \exp \left( - \frac{1}{2} x_{i} A_{ij} x_{j} \right) $$

and the constant $N$ is determined from the condition $\langle 1 \rangle = 1$. It is

$$ N = Z^{-1}(A, 0) = (2 \pi)^{-n/2} (\det A)^{1/2}. $$

The function

$$ Z(A,b)/Z(A,0) = \langle e^{b_{i} x_{i}} \rangle $$

is then the generating function of the moments of distribution. Expectation values are obtained by differentiating with respect to $b_{i}$:

$$ \langle x_{k_{1}} \cdots x_{k_{l}} \rangle = \left( \frac{\partial}{\partial b_{k_{1}}} \cdots \frac{\partial}{\partial b_{k_{l}}} \right) \exp\left[ \frac{1}{2} b_{i} \Delta_{ij} b_{j} \right] |_{b=0}. $$

This can be inspected by expanding both sides of the generating function in powers of $b_{i}$.

If $F(x)$ is polynomial in $x$, then

$$ \langle F(x) \rangle = F\left( \frac{\partial}{\partial b} \right) \exp \left[ \frac{1}{2} b_{i} \Delta_{ij} b_{j} \right]|_{b=0}. $$

This is known as Wick's theorem. Each time one differentiates the exponential, one gets a factor of $b$. Since $b$ is set to zero at the end, one must differentiate this factor of $b$ later, otherwise the corresponding contribution vanishes.

Thus, the expectation value, in a Gaussian theory, is given by all the possible ways of pairing the indices $k_{1}$ to $k_{l}$ in $\Delta_{k_{i} k_{j}}$. One finds, successively,

$$ \langle x_{i_1} x_{i_2} \rangle = \Delta_{i_{1} i_{2}} , $$

$$ \langle x_{i_{1}} x_{i_{2}} x_{i_{3}} x_{i_{4}} \rangle = \Delta_{ i_{1} i_{2} } \Delta_{i_{3} i_{4}} + \Delta_{i_{1} i_{3} } \Delta_{i_{2} i_{4}} + \Delta_{i_{1} i_{4}} \Delta_{i_{2} i_{3}}. $$


Quantum Mechanics.

You can now make the substitution $x_{i} \mapsto x(t)$. So instead of using vectors and partial derivatives, you will need functions and functional differentiation.

Of ultimate importance is $\Delta$, which is now called propagator. It was defined via

$$ A_{ij} \Delta_{jk} = \delta_{ik} $$

and now is interpreted in the continuum limit:

$$ A \Delta(t-t') = \delta (t - t'), $$

where $A$ is the quadratic piece in the Lagrangian. In your case

$$ A = \frac{d^{2}}{dt^{2}} + \omega^{2}. $$

In the literature, this $\Delta$ is known as the Green function.

I have given you all ingredients so that you can do this exercise. Hope this gives you some insight.


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