Quantum Field Theory – Deriving Two-Point Correlation Function for Scalar Fields

correlation-functionsfourier transformpath-integralquantum-field-theory

I am reading Peskin and Schroeder section 9.2. I don't understand the step when they evaluate equation (9.18)

$$<\Omega| T \phi_H(x_1)\phi_H(x_2)|\Omega>= {\rm lim}_{T\to \infty (1-i\epsilon)}\frac{\int D\phi\, \phi(x_1)\,\phi(x_2)\exp\left[i\int_{-T}^T d^4x \mathcal{L}\right]}{\int D\phi \,\exp\left[i\int_{-T}^T d^4x \mathcal{L}\right]}\tag{9.18}$$

for the scalar field case ($\mathcal{L}=\frac{1}{2}\partial\mu\phi\partial^\mu\phi-\frac{1}{2}m^2\phi^2$). In particular I cannot reproduce the step from eq (9.26) to the next equation in the book.

Best Answer

Eq. (9.26) is the numerator of eq. (9.18). Using a Fourier decomposition of $\phi(x_i)=\frac{1}{V}\sum_ne^{-ik_n\cdot x_i}\phi(k_n)$, with $k_n^\mu=2\pi n^\mu/L$, we get:

$$\frac{1}{V^2}\sum_{m\, l} e^{-i(k_m\cdot x_1+k_l\cdot x_2)}\left(\prod_{k_n^0>0}\int dRe\phi_n\, dIm\phi_n\right)\left(Re\phi_m+iIm\phi_m\right)\left(Re\phi_l+iIm\phi_l\right)\times\,\times{\rm Exp}\left[-i\frac{1}{V}\sum_{k_n^0>0}(m^2-k_n^2)\left((Re\phi_n)^2+(Im\phi_n)^2\right)\right].\tag{9.26}$$

Note that the parentheses around $\left(\prod \int...\right)$ are essential!

The $\phi$'s that appear above are in momentum space and we use the shortcut $\phi(k_m)=\phi_m$.

I understand now that only the $m=-l$ term in the sum survives. And I also understand that the cross terms ($Im\phi_n Re\phi_l$ and $Im\phi_l\, Re\phi_n$) vanish. Then the expression reduces to:

$$\frac{1}{V^2}\sum_{m} e^{-ik_m(x_1- x_2)}\left(\prod_{k_n^0>0}\int dRe\phi_n\, dIm\phi_n {\rm Exp}\left[-i\frac{1}{V}(m^2-k_n^2)\left((Re\phi_n)^2+(Im\phi_n)^2\right)\right]\right)\times\\\times\left((Re\phi_m)^2+(Im\phi_m)^2\right)~~~(*)$$

where I have used $k_{-l}=-k_l$ and $\phi(-k_m)=\phi^*(k_m)$ and hence $Re\phi_{-l}=\phi_l$ and $Im\phi_{-l}=-Im\phi_l$.

Next I evaluate the integrals that appea in the product in the equation above. I distinguish the case $n=m$ and $n\neq m$.

For $n=m$ I get:

$$\int dRe\phi_m\, dIm\phi_m\,(Re\phi_m)^2{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)\left((Re\phi_m)^2+(Im\phi_m)^2\right)\right]\\ + \int dRe\phi_m\, dIm\phi_m\,(Im\phi_m)^2{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)\left((Re\phi_m)^2+(Im\phi_m)^2\right)\right]\\ =\int dRe\phi_m\,(Re\phi_m)^2{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)(Re\phi_m)^2\right]\int dIm\phi_m\,{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)(Im\phi_m)^2\right]\\ + \int dIm\phi_m\,(Im\phi_m)^2{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)(Im\phi_m)^2\right]\int dRe\phi_m\,{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)(Re\phi_m)^2\right]\\ = 2 \int dx\,x^2{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)x^2\right]\int dy\,{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)y^2\right] $$

For $m\neq n$ I get:

$$\int dRe\phi_n\, dIm\phi_n{\rm Exp}\left[-i\frac{1}{V}(m^2-k_n^2)\left((Re\phi_n)^2+(Im\phi_n)^2\right)\right]\\ + \int dRe\phi_n\, dIm\phi_n{\rm Exp}\left[-i\frac{1}{V}(m^2-k_n^2)\left((Re\phi_n)^2+(Im\phi_n)^2\right)\right]\\ = \left( \int dx {\rm Exp}\left[-i\frac{1}{V}(m^2-k_n^2)x^2\right]\right)^2 $$ Note that no $Re\phi_m$ and $Im\phi_m$ factors appear, since the $m=n$ case includes these factors already.

Putting everything together eq (*) becomes:

$$(*)=\frac{1}{V^2}\sum_{m} e^{-ik_m(x_1- x_2)}\left(2 \int dx\,x^2{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)x^2\right]\int dy\,{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)y^2\right]\right)\times\\ \times\left(\prod_{k_n^0>0, n\neq m}\left( \int dx {\rm Exp}\left[-i\frac{1}{V}(m^2-k_n^2)x^2\right]\right)^2\right)~~(**)$$

Now it is easy to see that the denominator in (9.18) is:

$$\int D\phi e^{i S_0}=\prod_{k_n^0>0}\int Re\phi_n\, dIm\phi_n\, {\rm Exp}\left[-\frac{i}{V}(m^2-k_n^2\left((Re\phi_n)^2+(Im\phi_n)^2\right))\right]\\ =\prod_{k_n^0>0} \left(\int dx {\rm Exp}\left[-\frac{i}{V}(m^2-k_n^2)x^2\right]\right)$$

This is almost what appears in $(**)$. We manipulate $(**)$ further:

$$(**)=\frac{1}{V^2}\sum_{m} e^{-ik_m(x_1- x_2)}2 \frac{\int dx\,x^2{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)x^2\right]}{\int dy\,{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)y^2\right]}\left(\prod_{k_n^0>0}\left( \int dx {\rm Exp}\left[-i\frac{1}{V}(m^2-k_n^2)x^2\right]\right)^2\right)\\ =\frac{1}{V^2}\sum_{m} e^{-ik_m(x_1- x_2)}2 \frac{\int dx\,x^2{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)x^2\right]}{\int dy\,{\rm Exp}\left[-i\frac{1}{V}(m^2-k_m^2)y^2\right]}\,e^{i S_0}~~~(***)$$

I evaluate the remaining integrals by substituting $k_m^2-m^2\to k_m^2-m^2+i\epsilon$. I get for the integral

$$\frac{\int dx\,x^2{\rm Exp}\left[i\frac{1}{V}(k_m^2-m^2+i\epsilon)x^2\right]}{\int dy\,{\rm Exp}\left[i\frac{1}{V}(k_m^2-m^2+i\epsilon)y^2\right]}=\frac{1}{2}\frac{-iV}{ -i\epsilon- k_m^2+ m^2}$$

Inserting this into (***) I arrive at:

$$(***)=\frac{1}{V}\sum_{m} e^{-ik_m(x_1- x_2)}\frac{-i}{ m^2-k_m^2-i\epsilon}e^{i S_0}~~~(****)$$

Now inserting everything into (9.18):

$$(9.18)=\frac{****}{e^{iS_0}}=\frac{1}{V}\sum_{m} e^{-ik_m(x_1- x_2)}\frac{-i}{ m^2-k_m^2-i\epsilon}$$

Finally I use $\frac{1}{V}\sum_{k_n}\to \int \frac{d^4k}{(2\pi)^4}$:

$$<\Omega| T \phi_H(x_1)\phi_H(x_2)|\Omega>= \int \frac{d^4k}{(2\pi)^4}e^{-ik(x_1- x_2)}\frac{i}{ k^2-m^2+i\epsilon}$$

and this is eq (9.27) in Peskin Schoeder.