1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

Notation.

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,

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

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

EDIT.

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.