Four-point correlation function path integral for free scalars

Question

In An Introduction to Quantum Field Theory by Peskin and Schroeder, section 9.2, they calculate the four-point correlation function for a free real scalar field $\phi(x)$ using the path integral formulation: $$ \langle \Omega |T\{\phi(x_1) \phi(x_2) \phi(x_3) \phi(x_4)\}|\Omega\rangle = \lim_{T\to\infty(1-i\epsilon)} \frac{\int \mathcal{D}\phi\ \phi(x_1) \phi(x_2) \phi(x_3) \phi(x_4) \exp \left[i\int d^4x \mathcal{L}\right]}{\int \mathcal{D}\phi\ \exp\left[i\int d^4x \mathcal{L}\right]} $$

Setting the fields to be defined on the points $x_i$ in a discrete lattice of volume V and performing a Fourier series $\phi(x_i)\to \phi_n(k_n)$, he arrives at the following:

$$ \text{Numerator} = \frac{1}{V^4}\sum_{m,l,p,q}e^{-i(k_m\cdot x_1 + k_l\cdot x_2+ k_p\cdot x_3+ k_q\cdot x_4)}\left( \prod_{k_n^0 >0} \int d\text{Re} \phi_n \, d\text{Im} \phi_n \right)\times \\ \times (\text{Re}\phi_m + i\text{Im}\phi_m) (\text{Re}\phi_l + i\text{Im}\phi_l) (\text{Re}\phi_p + i\text{Im}\phi_p) (\text{Re}\phi_q + i\text{Im}\phi_q)\times \\ \times \exp\left[ -\frac{i}{V}\sum_{k_n^0>0} (m^2-k_n^2)\left[(\text{Re}\phi_n)^2 + (\text{Im}\phi_n)^2\right] \right] $$

Now, most of the terms vanish, since the integrand would be odd, but there are some values of $m,l,p,q$ for which the integral is non-zero. For example, if $k_l = -k_m$ and $k_q = -k_p$, then the fact that $φ (x_i)$ is real, i.e. $\phi(k_n) = (\phi(-k_n))^*$, means that we get:

$$ \frac{1}{V^4}\sum_{m,p}e^{-ik_m\cdot (x_1 - x_2)}e^{i k_p\cdot (x_3+ x_4)}\left( \prod_{k_n^0 >0} \int d\text{Re} \phi_n \, d\text{Im} \phi_n \right)\times \\ \times \left[(\text{Re}\phi_m)^2 + (\text{Im}\phi_m)^2\right] \left[(\text{Re}\phi_p)^2 + (\text{Im}\phi_p)^2\right]\times \\ \times \exp\left[ -\frac{i}{V}\sum_{k_n^0>0} (m^2-k_n^2)\left[(\text{Re}\phi_n)^2 + (\text{Im}\phi_n)^2\right] \right] = \\ = \frac{1}{V^4}\sum_{m,p}e^{-ik_m\cdot (x_1 - x_2)}e^{i k_p\cdot (x_3+ x_4)}\left( \prod_{k_n^0>0}\frac{-i\pi V}{m^2-k^2_n} \right)\frac{-iV}{m^2-k_m^2 - i\epsilon} \frac{-iV}{m^2-k_p^2 - i\epsilon}, $$

which, when returning to the continuum, becomes:

$$ \left( \prod_{k_n^0>0}\frac{-i\pi V}{m^2-k^2_n} \right)D_F(x_1-x_2)D_F(x_3-x_4), $$

where $D_F(x-y)$ is the Feynman propagator. However, I have a problem with this argument. For the values corresponding to $k_m = k_p$, we would have a quartic term $\left[(\text{Re}\phi_m)^2 + (\text{Im}\phi_m)^2\right]^2 = (\text{Re}\phi_m)^4 + (\text{Im}\phi_m)^4 + 2 (\text{Re}\phi_m)^2 (\text{Im}\phi_m)^2 $ in the integral which, using the results from Gaussian integration:

$$ \int dx\ x^2 e^{-ax^2} = \frac{1}{2a}\sqrt{\frac{\pi}{a}}, \quad \int dx\ x^4 e^{-ax^2} = \frac{3}{4a^2}\sqrt{\frac{\pi}{a}}, $$

would mean that we get an additional factor of 2 only for the case when $k_m = k_p$. This would mess up the result with the Feynman propagators, right? Can anyone see where my mistake lies?

Answer 1

It should read for $k_m=k_p=-k_l=-k_q$ $$(|\Re\phi_m + i\Im\phi_m|^2)^2 = [\Re\phi_m^2 + \Im\phi_m^2]^2$$ where $\Re = \text{Re}$ and $\Im = \text{Im}$. Since the integration is symmetric in $\Re\phi_n$ and $\Im\phi_n$ this case is actually of the form $$\int x^4 e^{-ax^2} = \frac{3}{4\color{red}{a^2}}\sqrt{\frac{\pi}{a}}$$ so it does not cancel but seems to give, as the OP mentions, a non-zero contribution to the four point function. However it seems to be a special case of the more general case $k_l=-k_m$ and $k_q=-k_p$. Since the formula above does not restrict $k_l$ in relation to $k_q$ or $k_p$, they result should still hold for the more restrictive case.

It can be seen that the result for this sub-case agrees with the more general case in the question an corresponds to diagonal terms in the sum over $m,p$ \begin{align} \frac{1}{V^4}\sum_{\color{red}{m}}e^{-ik_m\cdot (x_1 - x_2 - x_3 - x_4)}\left( \prod_{k_n^0>0}\frac{-i\pi V}{m^2-k^2_n} \right)\left(\frac{-iV}{m^2-k_m^2 - i\epsilon}\right)^2\\ = \left(\prod_{k_n^0>0}\frac{-i\pi V}{m^2-k^2_n} \right)\int d^4 y\, D_F(x_1-y)D_F(x_2+x_3+x_4+y) \end{align} Where the property of the Fourier transform and convolution was used (in the continuum). So this seems to represent some sort of tadpole diagram which might explain why it is set to zero (my guess).