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Let $\langle{0|}T\phi(x)\phi(y)|0\rangle$ be the vacuum expectation value of the 2 point correlator for the free scalar field. Page six in these notes say that we can calculate this correlator by taking the derivatives of the generating functional $Z[J]$

$$\tag{1}\langle{0|}T\phi(x)\phi(y)|0\rangle = (-i)^2\frac{1}{Z_{0}}\frac{\delta}{\delta J(x)}\frac{\delta}{\delta J(y)}Z[J]$$

where $Z_{0}$ is the generating functional in the absence of sources

However, when I explicitly calculate this using $$Z[J] = \exp\left(-\frac{1}{2}\int{dxdyJ(x)D(x,y)J(y)}\right)$$ I get $$\frac{\delta}{\delta J(x)} \exp\left(-\frac{1}{2}\int{dxdyJ(x)D(x,y)J(y)}\right) = \frac{-1}{2}\left(\frac{\delta}{\delta J(x)}\int{dxdyJ(x)D(x,y)J(y)}\right)\exp(…)$$

Which is equal to $$\frac{-1}{2}\left(\int{dxdy\delta(x-y)D(x,y) + \int{dxdyJ(y)D(x,y)}\exp(…)}\right)$$

Then taking the derivative with respect to $J(y)$ gives $$\frac{-1}{2}\int{dxdyD(x,y)}\exp(…)$$

Then plugging this into the first equation gives

$$\langle{0|}T\phi(x)\phi(y)|0\rangle = \frac{1}{4}\int{dxdyD(x,y)}$$

Which isn’t the Feynman propagator $D(x,y)$

So what’s wrong?

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    $\begingroup$ Minor comment to the post (v3): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. $\endgroup$
    – Qmechanic
    Jun 14, 2023 at 23:08
  • $\begingroup$ This is exactly the question I was looking for. I was also confused in similar points $\endgroup$
    – Plantation
    Oct 3, 2023 at 6:52

1 Answer 1

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It's best not to confuse the dummy variables of integration with the variable appearing in the functional derivative

\begin{align} &\frac{\delta}{\delta J(x)} \exp\left(-\frac{1}{2}\int{J(x')D(x',y')J(y')}\,\mathrm{d}x'\,\mathrm{d}y'\right) \\ &= \frac{-1}{2}\left(\frac{\delta}{\delta J(x)}\int{J(x')D(x',y')J(y')}\,\mathrm{d}x'\,\mathrm{d}y'\right)Z[J] \\ &= \frac{-1}{2}\left(\int{\frac{\delta J(x')}{\delta J(x)}D(x',y')J(y')} + {J(x')D(x',y')\frac{\delta J(y')}{\delta J(x)}}\,\mathrm{d}x'\,\mathrm{d}y'\right)Z[J] \\ &= \frac{-1}{2}\left(\int{\delta(x' - x)D(x',y')J(y')} + {J(x')D(x',y')\delta(y' - x)}\,\mathrm{d}x'\,\mathrm{d}y'\right)Z[J] \\ &= \frac{-1}{2}\left(\int D(x,y')J(y')\,\mathrm{d}y' + \int J(x')D(x',x)\,\mathrm{d}x'\right)Z[J] \end{align} We just use the product rule, then simple delta function identities.

This is, of course, just one of the functional derivatives. The other functional derivative can be computed similarly.

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    $\begingroup$ Brilliant! I thought that might’ve been the reason, so thanks for confirming it. $\endgroup$
    – user310742
    Jun 15, 2023 at 1:47

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