4

I'm considering the Euclidean Klein-Gordon theory, with action $$S_{0}[\phi] = \frac{1}{2}\int~\mathrm d^{4}x ~\phi(x) \left[ - \partial_{x}^{2} + m^{2} \right] \phi(x).\tag{1} $$ My generating function is then given by:

$$\mathcal{Z}_{0}[J] = \int \mathcal{D}[\phi]\ \exp\left( - S_{0}[\phi] + \int ~\mathrm d^{4}x \ \phi(x) J(x)\right).\tag{2}$$

I'm supposed to derive the following equation of motion, involving the two-point function: $$ \left[ - \partial_{y}^{2} + m^{2} \right] \langle \phi(y) \phi(x) \rangle = \delta^{(4)}(y-x).\tag{3}$$

I've been told to do this by starting with the following definition of the one-point function:

$$ \langle \phi(x) \rangle = \frac{1}{\mathcal{Z}_{0}[0]} \frac{\delta \mathcal{Z}_{0}[J]}{\delta J(x)} \bigg|_{J~=~0} \tag{4}$$

I'm supposed to use the invariance of the functional integration under field re-defintions; IE. if we replace $\phi(x)$ with $\phi^{\prime} = \phi(x) + \epsilon(x)$.

$\ $

I've been looking online and the usual approach that I've seen involves looking at the equality $$\int \mathcal{D}[\phi] \exp( - S_{0}[\phi] ) \phi(x) = \int \mathcal{D}[\phi^{\prime}] \exp( - S_{0}[\phi^{\prime}] ) \phi^{\prime}(x) \tag{5}$$ and performing an expansion in $\epsilon$.

How would you do this starting with $\langle \phi(x)\rangle $?

Qmechanic
  • 220,844
QuantumEyedea
  • 3,794

2 Answers2

6

OP's sought-for formula (3) is a special case of Schwinger-Dyson (SD) equations

$$\left< \Omega \left| T_{\rm cov}\left\{ F[\phi]\frac{\delta S[\phi;J]}{\delta \phi(y)}\right\}\right| \Omega \right>_J~=~i\hbar\left< \Omega \left| T_{\rm cov}\left\{\frac{\delta F[\phi]}{\delta \phi(y)} \right\}\right| \Omega \right>_J\tag{A}$$

with $$F[\phi]~=~\phi(x),\tag{B}$$ and where $T_{\rm cov}$ denotes covariant time-ordering, i.e. time-differentiations inside its argument should be taken after/outside the usual time ordering $T$.

The SD eqs. (A) can be proven:

  1. either by formal integration by parts (by assuming no boundary contributions) $$0~=~\int\! {\cal D}\phi ~\frac{\delta }{\delta \phi(y)}\left( F[\phi]~e^{\frac{i}{\hbar}S[\phi;J]}\right)\tag{C}$$ inside the path integral $$\int\! {\cal D}\phi ~F[\phi]e^{\frac{i}{\hbar}S[\phi;J]}~=~Z[J]~\left< \Omega \left| T_{\rm cov}\{ F[\phi]\}\right| \Omega \right>_J~; \tag{D}$$

  2. or equivalently, by formal infinitesimal field redefinitions/reparametrizations of the integration variables in the path integral (D) (by assuming the path integral measure ${\cal D}\phi$ is translation invariant);

  3. or via the operator formalism, cf. e.g. Ref. 1.

OP is asked to use method 2.

References:

  1. M.D. Schwartz, QFT and the Standard Model, 2014; Section 7.1.
Qmechanic
  • 220,844
3

Here is a rough sketch of how it goes. Some factors of i or something is not taken care of.

Integral $\mathcal{Z}_{0}[J] = \int \mathcal{D}[\phi]\ \exp^{ i\left( - S_{0}[\phi] + \int ~\mathrm d^{4}x \ \phi(x) J(x)\right)}$ can be evaluated explicitly by discretizing space-time. After discretization and dividing the integral into intervals (the usual way of evaluation) you get,

$\int dq_1dq_2...dq_N\ exp^{(\frac{i}{2})q.(-\partial^2+m^2).q+iJ.q}$, where $q$ is a matrix form column element and this integral evaluates to $Ne^{-(i/2)J.D.J}$ (simple Gaussian integration with N a constant factor), where D is inverse of the differential operator $(-\partial^2+m^2)$ and using the $D.D^{-1}=1$ in continuum limit gives,

$(-\partial^2+m^2)D(x-y)=\delta^4(x-y)$ .

Definition of n-point function is, $\langle\phi(x_1)\phi(x_2).....\phi(x_n)\rangle=\left. \frac{1}{i^n}\frac{\delta^nZ_0}{\delta J(x_1)\delta J(x_2)....\delta J(x_n)}\right|_{J~=~0}$

Now take, $Z_0(J)=e^{-(i/2)J.D.J}$ with N factor omitted from the definition of n-point function and the definition of $Z_0$.

The two-point function is given as, $\langle\phi(x)\phi(y)\rangle=-\left.\frac{\delta^2Z_0}{\delta J(x)\delta J(y)}\right|_{J~=~0}$. Doing the calculation with the given $Z_0$ and putting $J=0$ yields, $\langle\phi(x)\phi(y)\rangle=iD(x,y)$ from which we see $(-\partial^2+m^2)\langle\phi(x)\phi(y)\rangle=\delta^4(x-y)$.

For reference, consult Ryder's qft book. One point function identically vanishes but the functional differentiation of the expression again gives you the two-point function.

ved
  • 797