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I can take the following functional derivative

$$ C(p)=\frac{\delta}{\delta \phi(p')} \frac{\delta}{\delta \phi(-p')} \int_{-\infty}^{\infty} dp \phi(p)\phi(-p) = 2\delta(0). $$

where I am left with an extra delta function (on top of the one that killed the integral) because I took two functional derivatives. If I discretize the momentum I would expect this to be equivalent to:

$$ D(p_j)=\frac{\partial}{\partial \phi(p_j)} \frac{\partial}{\partial \phi(-p_j)} \sum_{i} \phi(p_i)\phi(-p_i) = 2. $$

As noted by loewe, the units do not match in the above. The only scale available to fix this is the momentum integration interval, let's call it $P$. If instead we take $\int \textrm{d}p \to P \sum_p$, $\frac{\delta}{\delta \phi(p)} \to P^{-1} \frac{\partial}{\partial \phi_p}$ and $\delta(0) \to P^{-1}$ the units stay the same under discretization. However, it is not completely clear why these replacements are entirely appropriate.

Moreover this doesn't really solve my confusion that in one case we seem to have $$ \int dp C(p) f(p) = 2f(0)$$ which only depends on the value of $f$ at $p=0$ while in the other we have $$ \sum_{p_j} D(p_j) f(p_j) = 2 \sum_{p_j} f(p_j)$$ which depends on the value of $f$ at all momenta.

How do I resolve this? What am I misunderstanding?

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    $\begingroup$ I think your third equation is incorrect, as you noted $C(p) = 2 \delta(0)$ and hence it is independent of $p$. $\endgroup$
    – loewe
    Feb 10 at 13:10
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    $\begingroup$ @loewe. You are right. I mentally inserted the derivatives into the sum. $\endgroup$
    – mike stone
    Feb 10 at 13:30
  • $\begingroup$ @loewe, I think this is still my main confusion. $\delta(0)$ is not really momentum independent. It treats $p=0$ as a preferred momentum value. We could have also considered the case $\frac{\delta}{p'} \frac{\delta}{p''} \int dp \phi(p) \phi(-p)$ and the correct result we want in that case is a delta function $\delta(p'-p'')$. This is just a specific case where the delta function has 0 as it's argument meaning that if this appears in an integral it should pick out the 0 term. $\endgroup$
    – Kvothe
    Feb 10 at 14:52
  • $\begingroup$ Oops, I am just mistaken. I am just misinterpreting the meaning of $\delta(0)$. $\endgroup$
    – Kvothe
    Feb 10 at 15:35

2 Answers 2

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The units of the functional derivative are \begin{equation} \left[ \frac{\delta}{\delta \phi(p)} \right] = \frac{[\textrm{Length}]}{[\phi]} \end{equation} since the delta function $\delta(p)$ has units of $[\textrm{Length}]$. In contrast, the units of the regular derivative are \begin{equation} \left[ \frac{\partial}{\partial \phi_p} \right] = \frac{1}{[\phi]} \end{equation} Hence, your two expressions do not even have the same units. With $\int \textrm{d}p \to L^{-1} \sum_p$, $\frac{\delta}{\delta \phi(p)} \to L \frac{\partial}{\partial \phi_p}$ and $\delta(0) \to L$, the discretization works out perfectly.

Edit to answer comment: $L$ is the size of the system and thus serves as infrared cutoff for all momenta, hence the smallest unit of momentum is $\Delta p = L^{-1}$. This explains the (standard) replacement $\int \textrm{d}p \to \Delta p \sum_p = L^{-1} \sum_p$. To see why the replacement for the delta function is natural, note that it has the property that its integral over all $p$ is 1 but it vanishes for any $p$ further from $p=0$ than $\Delta p$. Hence, we can think of it as a box function of width $\Delta p = L^{-1}$ and height $\Delta p^{-1} = L$. This explains $\delta(0) \to L$. This also fixes the prefactor of the discretized version of the functional derivative if we require that $\frac{\delta \phi(p)}{\delta \phi(p')} = \delta(p-p')$.

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  • $\begingroup$ I agree that the units before and after discretizing don't match. And that the only way to make them match is to use the integration interval (the only scale in the problem) when making the replacements. The delta function replacement seems ad-hoc and surprising. To me it looks like the delta function really changes the behaviour of the expression and not in a way that the introduction of an overall factor of a scale can explain. See my edits. I would love to hear your thoughts. $\endgroup$
    – Kvothe
    Feb 10 at 11:40
  • $\begingroup$ See the edits in my answer. $\endgroup$
    – loewe
    Feb 10 at 13:28
  • $\begingroup$ @Kvothe While it's true that the integration scale is the only scale in the problem, so your argument in the comments is right, you can also give a more specific argument. The Riemann integral is defined as $\int dx f(x) = \lim_{N\rightarrow \infty} \sum_{i=1}^N f(x_i) \Delta x/N$, where (assuming for the sake of argument a specific way of discretizing where each lattice point is equally spaced) $\Delta x=x_{N}-x_1$, and $N$ is the number of grid points. The factor of $\Delta/N$ is the factor of $L^{-1}$ that lowe is referring to. $\endgroup$
    – Andrew
    Feb 10 at 13:44
  • $\begingroup$ You can also repeat the above exercise by treating the delta function as a limit to derive the appropriate factor of $L$. $\endgroup$
    – Andrew
    Feb 10 at 13:45
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You second equation with the "2" on the RHS is wrong. It should be $2N$ where $N$ is the number of possible "$i$"s. In the continuum limit this is infinite, just as is the $\delta(0)$.

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  • $\begingroup$ I don't think so, one derivative gives a Kronecker Delta which removes the summation over $i$, so no factor of $N$. $\endgroup$
    – loewe
    Feb 10 at 12:48
  • $\begingroup$ I agree with loewe that this answer is wrong. The derivative is with respect to a specific $p_j$ (there is no sum over this index!). The derivatives will match for two terms only out of the sum over $i$. (Either $p_j=p_i or$ $p_j=-p_i$). $\endgroup$
    – Kvothe
    Feb 10 at 14:48

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