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I have the following integral

$$I=\int D\varphi \; e^{-\int d^4p d^4p' \left[ -\frac{1}{2}\varphi(p) g(p) \delta(p+p') \varphi(p') \right]}.\tag{1}$$

This is the continuum limit of a gaussian matrix integral: $$\int d\phi_1 \cdots d\phi_N \exp^{-\sum_{i,j}^N \phi^i M_{ij}\phi^j} = \sqrt{\frac{\pi}{\det(M)}}$$

Thus I find

$$ I = C \det( g(p) \delta(p+p') )^{-1/2} = C \prod_i - (g(p_i) g(-p_i))^{-1/2} \tag{2}$$

and

$$ \frac{\delta I}{\delta g(p)} = -\frac{1}{2}g(p)^{-1} I. \tag{3}$$

Here $C$ is some (possibly diverging) constant factor. The RHS is not invariant under $p\to-p$.

On the other hand I can compute

$$ \frac{\delta I}{\delta g(p)} = \frac{1}{2}\int D\varphi \; \varphi(p)\varphi(-p) \; e^{-\int d^4p' d^4p'' \left[ -\frac{1}{2}\varphi(p') g(p') \delta(p'+p'') \varphi(p'') \right]} \tag{4}$$

which is clearly invariant under $p \to -p$. What mistake am I making?

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    $\begingroup$ Strictly speaking the Gaussian matrix integral formula has a $\det [ \frac{1}{2} ( M + M^T )]$. Obviously this is really a trivial statement since the antisymmetric part of $M$ doesn't really show up in the Gaussian integrand. The same is true in your case. The antisymmetric part of $g$, namely $g(p)-g(-p)$ does not show up in your integrand. We can therefore assume, without loss of generality that $g(p) = g(-p)$. $\endgroup$
    – Prahar
    Commented Feb 4, 2022 at 13:07
  • $\begingroup$ @Prahar, could you expand on this. It was good to point out that I should have $M+M^T$ but as you already pointed out that gives the exact same result (the determinant of M is already invariant under $p\to -p$). Are you saying that equation (3) is invalid? Or equation (4)? I don't agree with your statement that we can assume $g(p)=g(-p)$ without loss of generality. I think you mean that ONLY in a certain equation where the integral kills any odd part we can replace the full $g(p)$ by its even part $g_{\text{even}}(p)$. $\endgroup$
    – Kvothe
    Commented Feb 4, 2022 at 13:48
  • $\begingroup$ Equation 3 is invalid. Use the correct version of the Gaussian integral and then take the derivative carefully. $\endgroup$
    – Prahar
    Commented Feb 4, 2022 at 14:02
  • $\begingroup$ @Prahar, but $\det(M)=\det(M^T)$ as follows from the eigenvalues of an anti-diagonal matrix. So the correct gaussian matches the wrong one I used, right? Are you saying equation (2) is already wrong? Or that I failed in taking its derivative in (3). (The derivative seems straightforward since $g(p_i)$ is just a number.) $\endgroup$
    – Kvothe
    Commented Feb 4, 2022 at 14:10
  • $\begingroup$ Yes. Clearly, (2) is also wrong since you are not using the correct Guassian integral formula. You are insisting on keeping the antisymmetric part of $g$ explicit so in that case you should replace $M \to (M+M^T)/2$ in the Gaussian formula. The determinant in (2) will then similarly change from $g(p) \to [g(p)+g(-p)]/2$. Alternatively, as I mentioned before since the antisymmetric part of $g(p)$ never appears anywhere, you might as well assume that $g(p)$ is symmetric. You can pick your favorite way to resolve the issue. $\endgroup$
    – Prahar
    Commented Feb 4, 2022 at 14:12

1 Answer 1

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Let's look at the term in the exponent $$ S = \int d^4 p d^4 p' \phi(p) \phi(p') g(p) \delta^4(p+p') \tag{1} $$ Interchange $p \leftrightarrow p'$ and we have \begin{align} S &= \int d^4 p' d^4 p \phi(p') \phi(p) g(p') \delta^4(p'+p) \\ &= \int d^4 p d^4 p' \phi(p) \phi(p') g(-p) \delta^4(p+p') \tag{2} \end{align} Adding (1) and (2), we find $$ S = \int d^4 p d^4 p' \phi(p) \phi(p') [ \frac{g(p) + g(-p)}{2} ] \delta^4(p+p') $$ Consequently, only the symmetric part of $g$ enters the action. Defining $g_s(p) = \frac{g(p) + g(-p)}{2}$ so that $g_s(p) = g_s(-p)$, we find $$ S = \int d^4 p d^4 p' \phi(p) \phi(p') g_s(p) \delta^4(p+p') $$ You can now go through your derivation and replace $g\to g_s$ everywhere. Your problem will disappear.

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  • $\begingroup$ Thank you! I agree with the manipulations that you do. I'm still confused about which of the equations that I wrote were wrong. What if I really have a derivative with respect to a $g(p)$ that might not be even. Is my equation (4) wrong? Can the $g(p)$ in the exponent not act as source in this case because only the even part contributes to the integral? $\endgroup$
    – Kvothe
    Commented Feb 4, 2022 at 13:58

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