4
$\begingroup$

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 $?

$\endgroup$
6
$\begingroup$

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.
$\endgroup$
  • 1
    $\begingroup$ That's what I would usually post, can't believe I missed the chance :) +1. Just one correction: you don't really need the source term $J$. $\endgroup$ – Prof. Legolasov Dec 4 '16 at 12:50
  • 1
    $\begingroup$ Yeah, $J$ is just a passive spectator here. $\endgroup$ – Qmechanic Dec 4 '16 at 13:19
3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ I see the logic. You start out with the above definition of $< \phi(x) \phi(y) >$, and you get $<\phi(x)\phi(y)> = i D(x,y)$ where $D(x,y)$ is defined by $( -\partial^{2} + m^{2} ) D(x,y) = \delta(x-y)$. This all makes sense...but then why am I being told to start out with $<\phi(x)>$? $\endgroup$ – Greg.Paul Dec 4 '16 at 2:36
  • 2
    $\begingroup$ Two point function is obtained from the functional differentiation of one point function, so you do start with the one point function. You may take a look in Ryder's qft, chapter 6 for details. It is much better to do the functional differentiation in place of field redefinition and expansion in power of some infinitesimal parameter. $\endgroup$ – ved Dec 5 '16 at 5:13

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.