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I am studying Fradkin's lecture notes on advanced field theory, and in the section on Wilson RG he has a list of integrals of the propagators of the fast fields. One of such integrals is: $$ \int d^D l \left[G_{0,>}(l)\right]^3 = \frac{\Lambda^{2D-6}}{(1+t)^3}\frac{S_D}{(2\pi)^D}\delta s $$

I don't see how we get the right hand side. Here $G_{0,>}(l)$ is the free propagator of the fast fields (those with momenta in the shell that we integrate out): $$ G_{0,>}(l) = \int_{e^{-\delta s}\Lambda < |p|<\Lambda}\frac{d^D p}{(2\pi)^D}\frac{e^{ip\cdot l}}{p^2+t\Lambda^2} $$

Here are my thoughts so far. I understand that after Fourier transforming we get something like: $$ \int\frac{d^D p_1}{(2\pi)^D}\frac{1}{p_1^2+t\Lambda^2}\int\frac{d^D p_2}{(2\pi)^D}\frac{1}{p_2^2+t\Lambda^2}\int d^D p_3\frac{\delta^D(p_1+p_2+p_3)}{p_3^2+t\Lambda^2} $$

Here we need to make sure that when we integrate $p_3$ out, we have to restrict the other two momenta such that $e^{-\delta s}\Lambda<|p_1+p_2|<\Lambda$. That is where I get stuck. I am not sure if this is correct, but I set the $p_1$ direction as the "z" axis for the sake of $p_2$ integration, such that $p_1 \cdot p_2 = |p_1||p_2|\cos(\theta)$, and we should get something like: $$ \begin{split} &\int\frac{d^D p_1}{(2\pi)^D}\frac{1}{p_1^2+t\Lambda^2}\int\frac{d^D p_2}{(2\pi)^D}\frac{1}{p_2^2+t\Lambda^2}\frac{\delta(\cos\theta + \frac{1}{2})}{p_1^2+p_2^2+2|p_1||p_2|\cos(\theta)+t\Lambda^2} \\&= \frac{S_D^2}{(2\pi)^{2D}}\frac{\Lambda^{2D-6}\delta s}{(1+t)^2}\frac{\int_0^\pi d\theta \frac{\sin^{D-2}\theta}{(2+t+2\cos\theta)}\delta(\cos\theta+ \frac{1}{2})}{\int_0^\pi d\theta \sin^{D-2}\theta} \end{split} $$

where the delta function for $\cos\theta$ is there to make sure the sum of momenta are on the shell and for the second equality I multiply and divide by the integral of $\sin^{D-2}\theta$ to get an additional factor of $S_D$, since I think only that angular integral for $p_2$ will be modified. Here I get the right power counting and the factor of $(1+t)^{-3}$, but the numerical factors don't match Fradkin's value. Any help would be appreciated!

Edit: Found a similar question this one is related to.

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  • $\begingroup$ Did you evaluate the integration of the first equation in 15.61 of his notes? $\endgroup$
    – Mass
    Nov 13, 2023 at 21:32
  • $\begingroup$ @Mass Yes, I evaluated the three integrals before the one I mention in the question and got the same results as the text. $\endgroup$
    – SSh2402
    Nov 16, 2023 at 23:50
  • $\begingroup$ I think you should use the differentiation with respect to $t$ under the integration. First, take the case of $\ell = 0$, which will immediately give you the rest of the integrals. Now, do just generalize the result for $\ell \neq 0$ case. $\endgroup$
    – Mass
    Nov 17, 2023 at 6:41
  • $\begingroup$ @Mass I am not really sure what that will achieve. Which integration should I take the derivative w.r.t. t under? And how does that lead to the other propagators? $\endgroup$
    – SSh2402
    Nov 20, 2023 at 16:53
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    $\begingroup$ @NandagopalManoj Ah, sorry, I see what you mean. All of the angular integrals aside from $\theta$ that we have to do for $p_2$ are unaffected by the constraint we have. So what I do is simply multiply and divide by $\int_0^\pi d\theta' \sin^{D-2}\theta'$ to recover a factor of $S_D$ with the rest of the unaffected integrals. That leads to the expression I have after renaming the dummy variable in the denominator. $\endgroup$
    – SSh2402
    Jan 9 at 19:24

1 Answer 1

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In the thin shell limit, we can take $p_i^2 \approx \Lambda^2$. The only nontrivial part which you correctly identified comes from the "on-shell" constraint when we integrate out $p_3$. Defining $P_i \equiv p_i /\Lambda$, \begin{align} &\frac{1}{(2\pi)^{2D}}\int \bigg[\prod_i d^Dp_i\bigg] \frac{\delta^D(p_1+p_2+p_3)}{(p_1^2 + t\Lambda^2)(p_2^2 + t\Lambda^2)(p_3^2 + t\Lambda^2)} \\ &\approx \frac{1}{(2\pi)^{2D}}\int \bigg[\prod_i {d^DP_i}\bigg] \Lambda^{3D - D - 6}\frac{\delta^D(P_1 + P_2 + P_3)}{(1+t)^3} \\ &= \frac{\Lambda^{2D-6}}{(2\pi)^{2D}(1+t)^3}\int \bigg[\prod_i d^DP_i\bigg] \delta^D(P_1 + P_2 + P_3) \end{align} You want to calculate the last integral with the constraint $1-s<\vert P_i\vert \leq 1$, where $s \ll 1$ (we will call this the shell). Imagine a shell centered at the origin and place $P_1$ somewhere on this shell, say $P_1 = (1,0,0,0,\dots )$ (the exact position will not matter). Now, the allowed values for $P_2$ such that $P_3$ is "on-shell" can be visualized by putting another shell centered at $P_1$, and looking at the intersection of the two shells. This is the allowed volume for $P_2$. The volume of this region can be readily calculated. It is a small rhombus of $120^\circ$ and height $\delta s$, times a $(D-1)$-sphere (in Fradkin's language which is non-standard). On top of this, $P_1$ can take values anywhere on the shell. Therefore, we get, \begin{align} &=\frac{\Lambda^{2D-6}}{(2\pi)^{2D}(1+t)^3} \delta s^3 S_{D} S_{D-1} \frac{2}{\sqrt{3}} \end{align} This clearly does not match Fradkin's result (even the $\delta s$ scaling!), but I'm hoping someone can point out my error and it helps us reach the answer.

Update

I just saw the post that you linked, it seems to be consistent with my answer. The answer there claims that to get Fradkin's answer, you need to take only one of the internal momenta on shell and the rest to be anywhere inside the momentum cutoff. This is equivalent to replacing $t\Lambda^2 \to m^2$, evaluating the integral for all $\vert p_i\vert \leq 1$ (cutoff regularized integral), and taking the derivative of the result w.r.t. $\log \Lambda$.

I tried evaluating this integral analytically and failed. I found references where they evaluate a very similar integral (same diagram at non-zero external momentum) using dimensional regularization and that required numerical estimates. I could not find any resource which performs this integral analytically -- David Tong's notes on statistical field theory mentions this integral but does not evaluate it (eqn. 3.42).

Regardless of whether that integral is analytically tractable, I think it is reasonable to conclude that the integral for the fast modes that Fradkin wrote must be $\sim \delta s^3$, and it is not clear to me what he is evaluating there.

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    $\begingroup$ I asked Eduardo Fradkin (whose office is next to mine) who said that you need to use a gaussian cutoff rather than a thin shell, so the $p^2+t\Lambda^2$ by $\Lambda^2(1+t)$ is not correct. But I still cannot reproduce the $G^3$ integral. $\endgroup$
    – mike stone
    Jan 13 at 0:01
  • $\begingroup$ Could you clarify: Gaussian cutoff in the UV momentum cutoff sense or the fast mode momentum shell sense? I assume the former? $\endgroup$ Jan 13 at 0:16
  • $\begingroup$ As I undestand what Eduardo said is the sharp cutoffs $e^{-s}\Lambda< |p|< \Lambda$ defining the thin shell have to be replaced with some sort of gaussian centered on $|p|=\Lambda$. If you not do this you have to cope with the oscillations in $G_0(r) \propto J_{d/2-1}(\Lambda r)$ He also said that the computations are easiest done in configuration ($x$)space rather than in momentum space. I have not got more out of him than this. $\endgroup$
    – mike stone
    Jan 13 at 0:23
  • $\begingroup$ I don't see how softening the shell condition can change the $\delta s^3$ scaling. Do you see that? $\endgroup$ Jan 13 at 0:25
  • $\begingroup$ @mikestone I also don't see how the $p^2 + t\Lambda^2 \approx \Lambda^2(1+t)$ approximation becomes invalid for a gaussian shell. That only assumes that the shell is really thin, regardless of whether its a hard or a soft cutoff. (Fradkin's book clearly mentions the thin shell limit.) The exact nature of the cutoff then can only change the higher order corrections above this zeroth order approximation, which I feel like we can neglect in the thin shell limit. Do you agree? $\endgroup$ Jan 13 at 1:54

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