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From the standard text book about quantum field theory, we know that if we consider $$\mathcal{L}=\frac{1}{2}(\partial_{\mu} \phi)^2-\frac{m^2}{2}\phi^2,$$ the partition function of this Gaussian theory can be worked out $$\frac{Z[J]}{Z[0]}=\frac{\int D\phi\,e^{\frac{i}{2}\int d^4x\,\phi[-(\partial_\mu^2+m^2)]\phi+i\int\,d^4x\,J\phi}}{\int D\phi\,e^{\frac{i}{2}\int d^4x\mathcal{L}}}=e^{-\frac{1}{2}\int d^4xd^4y J(x)G(x-y)J(y)},$$ where we use integration by parts $\int d^4x(\partial_\mu\phi)^2=-\int d^4x\phi\partial^2_{\mu}\phi$, and $iG(x,y)=(\partial_\mu^2+m^2)^{-1}$. On the other hand, $$\langle T\phi(x)\phi(y)\rangle=-\frac{1}{Z[0]}\frac{\delta^2Z[J]}{\delta J(x)\delta J(y)}|_{J=0}=G(x-y).$$

But how about this kind of correlation function $\langle \partial_i \phi(x) \partial_j\phi(0) \rangle$ ? May be we can use some similar strategy, $$\frac{Z'[J]}{Z[0]}=\frac{\int D\phi\,e^{\frac{i}{2}\int d^4x\,\phi[-(\partial_\mu^2+m^2)]\phi+i\int\,d^4x\,J^\mu\partial_\mu\phi}}{\int D\phi\,e^{\frac{i}{2}\int d^4x\mathcal{L}}}=e^{-\frac{1}{2}\int d^4xd^4y (\partial_\mu J^\mu(x))G(x-y)(\partial_\nu J^\nu(y))},$$ and $$\langle T\partial_i\phi(x)\partial_j\phi(y)\rangle=-\frac{1}{Z[0]}\frac{\delta^2Z'[J]}{\delta J^i(x)\delta J^j(y)}|_{J=0},$$ then we will meet some terms like $$\int d^4x_1d^4y_1\frac{\delta \partial_\mu J^\mu(x_1)}{\delta J^i(x)}G(x_1-y_1)(\partial_\nu J^\nu(y)),$$ can we exchange $\delta$ and $\partial$ here? Then finally we may find $$\langle \partial_i \phi(x) \partial_j\phi(0) \rangle\sim \partial_i\partial_j\,(\partial_\mu^2+m^2)^{-1},$$ and after Fourier transformation, we may have $\frac{k_ik_j}{\omega^2-k^2+m^2}$.

Actually, I find this correlation in some book like Altland and Simons' book, Condensed matter field theory, 2nd, page 393, exercise and Xiao-Gang Wen's book, quantum field theory of many body systems, problem 3.3.3. I want to know if there is any standard method to calculate this kind of correlation, since it seems not rigorous in above "derivation".

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    $\begingroup$ Why don't you just compute $\langle \phi(x) \phi(y) \rangle$ using the above method, and then just hit that with derivatives? In practice this is 100% what I would do. $\endgroup$ Commented Nov 23, 2022 at 1:08
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    $\begingroup$ For time ordering and differentiation, see e.g. physics.stackexchange.com/q/359961/2451 $\endgroup$
    – Qmechanic
    Commented Nov 23, 2022 at 5:38
  • $\begingroup$ very good question and thanks for including the books; when you say (...)"Simon's book"(...) you don't mean George.F.Simmons? (I think someone else) but asking to be sure; $\endgroup$ Commented Nov 23, 2022 at 10:24
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    $\begingroup$ @WilliamMartens No, that book “condensed matter theory “ has two authors. $\endgroup$
    – ZJX
    Commented Nov 23, 2022 at 10:32
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    $\begingroup$ I also find something called covariant time ordering which is involved in the path integral. But it seems that not much textbooks talk about this. $\endgroup$
    – ZJX
    Commented Nov 23, 2022 at 10:34

1 Answer 1

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Your idea of introducing the generating functional $$ Z[J] := \int \mathrm{D}\phi \exp\!\left(\mathrm{i}\,S[\phi]+\mathrm{i}\int\mathrm{d}^d{x} \ J^\mu\partial_\mu\phi\right)$$ is correct. Indeed you see that1 $$-\frac{\delta^2\ Z[J]}{\delta J^\mu(x)\ \delta J^\nu(y)}\Bigg|_{J=0} = \Big\langle\partial_\mu\phi(x)\partial_\nu\phi(y)\Big\rangle.$$

In your Gaussian theory, you were worried about exchanging $\partial_\mu$ and $\delta$. You don't have to do that though. Note that in this case an integration by parts gives \begin{align} Z[J] &= \exp\!\left(-\frac12\int\mathrm{d}^d x\,\mathrm{d}^d y\ \partial_\mu J^\mu(x)\;G(x-y)\;\partial_\nu J^\nu(y)\right)=\\ &=\exp\!\left(\frac12\int\mathrm{d}^d x\,\mathrm{d}^d y\ J^\mu(x)\;\partial_\mu \partial_\nu G(x-y)\; J^\nu(y)\right), \end{align} where both derivatives are wrt $x$ in the last line, since $\partial_y f(x-y) = - \partial_x f(x-y)$. From here on you are free to take your $J(x)$ variational derivatives as you usually would and find indeed $$\Big\langle\partial_\mu\phi(x)\partial_\nu\phi(y)\Big\rangle = - \partial_\mu \partial_\nu\ G(x-y).$$

Finally, @QuantumEyedea's comment about first computing $\big\langle\phi(x)\phi(y)\big\rangle$ and then taking derivatives is absolutely correct. It does come with limitations, however. If you want to compute a correlation function of polynomials of the field and its derivatives: $$ \Big\langle \mathcal{O}_1(x)\cdots \mathcal{O}_n(x)\Big\rangle,$$ where $\mathcal{O}_i(x) := \mathrm{P}_i[\phi,\partial\phi,\partial^2\phi,\cdots](x)$ are polynomials, with @QuantumEyedea's method you would have to compute each $m$-point function of $\phi$ and then take appropriate derivatives, which will be chaotic and daunting. With the generating functional method, the computation is much more organised and straightforward. This is exactly what happens in practice when people compute string theory amplitudes.


1 setting $Z[0]=1$ for simplicity

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