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In Chapter 3 of Kardar's statistical physics of fields, in the context of lower critical dimension, he works out an example about superfluids where starting from the probablity of a particular configuration in the ordered phase, ie $$\mathscr{P}[\theta(\mathbf{x})] \propto \exp\left\lbrace-\frac{K}{2} \int \mathrm{d}^d\mathbf{x} (\nabla\theta)^2\right\rbrace,$$ where $\theta$ is the phase of the wavefunction, he goes on to calculate the correlation functions, obtaining $$ \left\langle \theta(\mathbf{x})\theta(\mathbf{x}')\right\rangle = - \frac{C_d(\mathbf{x}-\mathbf{x}')}{K},$$ where $$ C_d(\mathbf{x}) = - \int \frac{\mathrm{d}^d\mathbf{q}}{(2\pi)^d} \frac{e^{i\mathbf{q}\cdot\mathbf{x}}}{\mathbf{q}^2}$$ is presumebly the Coulomb potential in $d$ dimensions since $$\nabla^2 C_d(\mathbf{x}) = \delta^d(\mathbf{x}).$$ He then uses the Gauss theorem, ie $$ \int \mathrm{d}^d\mathbf{x} \nabla^2 C_d = \oint \mathrm{d}S\cdot\nabla C_d$$ together with $\nabla C_d = (\mathrm{d}C_d/\mathrm{d}r)\hat{r}$, which simply is due to the spherical symmetry, to build the solution \begin{equation} C_d(x) = \frac{x^{2-d}}{(2-d)S_d} + c_0, \end{equation} where $S_d$ is the area of unit sphere in $d$ dimensions and $c_0$ a constant to be determined. Then it is argued that this clearly diverges for $d \leq 2$ dimesnions as $x\to\infty$, while approaches $c_0$ for $d>2$. This is however related to the celebrated Mermin-Wagner theorem.

Now my question is that it seems to me $C_d(\mathbf{x})$ is a divergent integral as the integrand doesn't decay fast enough when $\mathbf{q}\to\infty$, except in $d=1$. So I would naively guess the integral is convergent in $d=1$, while in $d \geq 2$ it diverges, which is exactly opposite what Kardar claims.

I guess there is some type of regularization scheme at work in Kardar's approach (maybe momentum cutoff or dimensional regularization), but I don't understand it exactly and in which of the steps outlined above it's been applied. I tried calculating the integral directly (in the $d$ dimensional sphereical coordinates) hoping that I can apply the momentum cutoff on the final result and compare it with Kardar's solution, but have no clue how to do the calculation since the angle between $\mathbf{q}$ and $\mathbf{x}$ introduced by $e^{i\mathbf{q}\cdot\mathbf{x}}$ makes it complicated.

EDIT

I see the integral needs probably both lower and upper cutoffs (say $\Lambda_l$ and $\Lambda_u$) to converge, depending on the dimensionality of the space so \begin{equation} C_d(\mathbf{x}) = \lim_{\substack{\Lambda_u\to\infty \\ \Lambda_l \to 0}} \left\lbrace\mathcal{F}(|\mathbf{x}|) + \mathcal{G}(\Lambda_{u}, \Lambda_{l}) + c_0\right\rbrace. \end{equation} (The above form probably needs justification)

Then one argues the function $\mathcal{G}$ doesn't affect the correlations meaningfully (it is just a constant that approaches infinity) and we only study the behavior of function $\mathcal{F}$. My question is how to make sure the manipulation in Kardar's book is exactly doing this.

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  • $\begingroup$ I think the convergence depends non-trivially on the exponential. One way to argue this is to note that the convergent/divergent behavior of $C_d(x)$ flips when you change from taking the limit $x \rightarrow \infty$ to $x \rightarrow 0$. In the latter limit, one can reasonably ignore the exponential (which is also what you do when you argue for the divergent behavior in $d \geq 0$) and then we have no contradiction. Other than Kardar's explanation though, I don't have a good intuition for how exactly the exponential modifies the convergence. $\endgroup$ – Henry Shackleton Feb 23 at 16:34
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    $\begingroup$ Actually, I think you might be able to argue for the divergent behavior for $x\rightarrow \infty$ with the stationary phase approximation. I need to think about it more though. $\endgroup$ – Henry Shackleton Feb 23 at 16:54
  • $\begingroup$ @HenryShackleton For $x \to 0$ the value of the integral only depends on the possible IR and UV cutoffs and hence probably not physically important in this context. $\endgroup$ – dedekindCuttage Feb 23 at 17:11
  • $\begingroup$ I agree that it's not physically important. I was just trying to argue as to why power counting was not sufficient to extract the $x \rightarrow \infty$ limit. $\endgroup$ – Henry Shackleton Feb 23 at 17:15
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When you say the integrals diverge i guess you are refering to the following:

$$ |C_d| \leq \int \frac{d^d\mathbf{q}}{(2\pi)^d} \frac{1}{\mathbf{q}^2} = \text{const.} \times \int_0^\infty q^{d-3} dq $$

which is UV-divergent unless if $d \geq 2$ and IR-divergent if $d \leq 2$.

However, this doesn't mean the integral is ill-defined, just that this way of bounding the integral doesn't produce an useful result.

Let's try to improve this bound:

$$ C_d = -\int \frac{d^d\mathbf{q}}{(2\pi)^d} \frac{e^{i\mathbf{q \cdot x}}}{\mathbf{q}^2} = - \frac{1}{(2\pi)^d} \int_0^\infty q^{d-3} \left[ \int_{S_{d-1}} e^{i q \ \mathbf{\hat{q}\cdot x}} \ d \Omega_{d-1} \right] dq \ ,$$

here $\mathbf{\hat{q}} = \frac{\mathbf{q}}{q}$ is the unit vector in $\mathbf{q}$ direction and $d \Omega_{d-1}$ is the surface element on the $(d-1)$ sphere.

For handling the integral over the $d-1$-sphere, pick spherical coordinates such that the $q_{n}$-axis coincides with the $\mathbf{x}$-axis, so that $\mathbf{q \cdot x} = q x \cos(\theta)$ and

$$ d \Omega_{d-1} = \sin(\theta)^{d-2} d \theta d \Omega_{d-2} \ .$$

For the case $d=2$, one has to look at the integral

$$ \int_{S_{1}} e^{i q \ \mathbf{\hat{q}\cdot x}} \ d \Omega_{1} = \ \int_{-\pi}^\pi e^{i q x \cos(\theta)} d \theta $$

instead. One may look this up to be $2\pi J_0(qx)$, where $J_0$ is the Bessel function of the first kind. It has the asymptotics

$$J_0(qx) \sim \text{const.} \times q^{-1/2} \ , $$

so

$$ C_2 = - \frac{1}{2\pi} \int_{0}^\infty \frac{J_0(qx)}{q} dq $$

does not suffer from UV-divergencies.

Start with the case $d > 2$ even. Then

$$\int_{S_{d-1}} e^{i q \ \mathbf{\hat{q}\cdot x}} \ d \Omega_{d-1} = \text{Vol}(S_{d-2}) \ \int_{0}^\pi e^{i q x \cos(\theta)} \sin(\theta)^{d-2} d \theta = \\ =(2\pi)^{\frac{d}{2}}\frac{J_{\frac{d-2}{2}}(qx)}{(qx)^{\frac{d-2}{2}}} \ . $$

Hence

$$ C_{d} = -\frac{1}{(2\pi)^{\frac{d}{2}}} \frac{1}{x^{d-2}} \int_0^\infty q^{d/2-2} J_{\frac{d}{2}-1}(q) dq \ .$$

for the case of $d=4$ it gives

$$C_4 = -\frac{1}{(2\pi)^2 x^2} \int_0^\infty J_1(q) dq = -\frac{1}{(2\pi)^2 x^2} \ .$$

for all other $d$ the integral is divergent and one has to introduce a regularization.

The integral was convergent for $d=4$ because the oscillations of the Bessel function are leading to destructive interference. It is very similar to how $\sum_{n>0}n^{-1}$ is diverging while $\sum_{n>0} (-1)^n n^{-1}$ is not.

Now for $d = 3$ odd:

$$\int_{S_{2}} e^{i q \ \mathbf{\hat{q}\cdot x}} \ d \Omega_{2} = 2\pi \ \int_{0}^\pi e^{i q x \cos(\theta)} \sin(\theta) d \theta = 4\pi \frac{\sin(qx)}{qx} \ . $$

Thus

$$ C_3 = - \frac{1}{2\pi^2 x} \int_0^\infty \frac{\sin(q)}{q} dq = - \frac{1}{4\pi x} \ . $$

$d = 1$:

$$C_1 = \frac{1}{\pi} \int_0^\infty \frac{e^{iqx}}{q^2} d q $$

This has no UV divergences, but clearly suffers from an IR divergence which will be discussed below.

The general $d$ case is a bit awkward, so i am not discussing it here.

So we see that there are no UV-divergencies really for $d \leq 4$. One should remark that in $d=2$ there is however an IR divergence. Naively, one could perform a change of variables $qx =p$, which would lead to $C_2$ being constant. However, the integral as written diverges since $J_0(0) \neq 0$ hence this manipulations are not allowed.

Now turn to the question of the IR-divergence in two dimension. This is real, so one needs to regularize the integral, for example by introducing a cutoff in the following way:

$$C_{2,\mu} := - \frac{1}{2\pi} \int_{0}^\infty \frac{J_0(qx)-f(qx)}{q} dq - \frac{1}{2\pi} \int_{\mu}^\infty \frac{f(qx)}{q} dq \ ,$$

where $f$ is some function such that $f(q) \sim 1 $ as $q\rightarrow 0$. Then the first integral is well-defined since $J_0(qx)-f(qx) \in \mathcal{O}(q^2)$. Furthermore, it does not depend on $x$ by a simple change of variables. Hence

$$C_{2,\mu} = - \frac{1}{2\pi} \int_{\mu}^\infty \frac{f(qx)}{q} dq + C(f) \ , \quad C(f) := - \frac{1}{2\pi} \int_{0}^\infty \frac{J_0(q)-f(q)}{q} dq \ ,$$

We may now pick any function. A convenient choice is $f(x) = 1$ for $x < \Lambda$ and zero otherwise. Then:

$$C_{2,\mu} = -\frac{1}{2\pi} \int_\mu^{\Lambda/x} \frac{dq}{q} + C(f) = \frac{1}{2\pi} \log(x) + \frac{1}{2\pi} \log(\mu/\Lambda) + C(f) \ .$$

One can now check that the $x$-dependence of $C_2(x)$ is in fact universal in the limit $\mu \rightarrow 0$: it does not depend on the choice of $f$. Indeed, consider two different such functions $f_1,f_2$. Then their difference goes to zero at $q=0$. Hence in the integral

$$C_{2,f_1} - C_{2,f_2} = \frac{1}{2\pi} \int_0^\infty \frac{f_1(qx)-f_2(qx)}{q} dq + \text{const.} $$

the lower bound of integration may be extended to $\mu = 0$ and the resulting integral does not depend on position $x$.

Now in one dimensions: let f(q) be a function with $f(0) = 1$.

Then the following is well-defined:

$$ C_{1,f} = \frac{1}{\pi} \int_\mu^\infty \frac{f(qx)}{q^2} dq + \frac{1}{\pi} \int_0^\infty \frac{\cos(qx) - f(qx)}{q^2}dq \ . $$

The first term is in fact independent of $x$ while in the second you introduce a factor sign$(x)$ when changing variables $qx\rightarrow q$. Hence

$$ C_{1,f} = A_f + \text{sign}(x) B_f \ ,$$

with $A_f,B_f$ some constants depending on $f$.

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  • $\begingroup$ Thank you for your answer. I think I'm not following when you introduce variable $u$. Should the integrand not be $(1-u^2)^{(d-3)/2} \, du$ instead? $\endgroup$ – dedekindCuttage Feb 26 at 17:33
  • $\begingroup$ Thanks for the edits. Nice note on alternating series. For the $d=2$ case, should we not recover a logarithmic potential? Also, $d=4$ case is interesting. I have to brush up my Bessel functions education:) Btw, do you have any ideas as for how to develop this calculation for higher dimensions? $\endgroup$ – dedekindCuttage Feb 27 at 16:15
  • $\begingroup$ This is getting perfect. So when you say "for all other d the integral is divergent and one has to introduce a regularization" I guess you mean UV divergence right? Then "So we see that there are no UV-divergencies really" is not accurate anymore. $\endgroup$ – dedekindCuttage Feb 28 at 11:03
  • $\begingroup$ There are no UV divergences in the physically interesting cases of d=1,2,3,4. $\endgroup$ – Lorenz Mayer Feb 28 at 11:04
  • $\begingroup$ I see. So the initial bound that you were trying to improve wasn't a meaningful one for d=2,3,4. For higher dimensions, however, the integral is indeed UV divergent and needs regularization, and the intuition that the oscillations cancel each other doesn't really work. Is that correct? $\endgroup$ – dedekindCuttage Feb 28 at 11:10
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Indeed, if one uses textbook measure theory, this is nonsense since the function is not integrable. The proper mathematical setting for understanding the correctness of this kind of computations with full rigor is the theory of Schwartz distributions.

See for example https://math.stackexchange.com/questions/3120284/how-to-calculate-c-a-where-leftf-mapsto-int-mathbbr-fracft-f0t where a rigorous computation of the Fourier transform of the OP's question is done.

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  • $\begingroup$ I looked at your answer there but couldn't really follow the steps. My question here is not exactly about the mathematical rigor (that would also help if I can grasp it) but rather about understanding the steps in Kardar's approach (which I assume is quite common in physics community) and its relation to a more physically understandable regularisation like momentum cutoff. $\endgroup$ – dedekindCuttage Feb 23 at 21:52
  • $\begingroup$ Also, I think we should have some sort of justification as to why 'regularising' the original divergent integrals is allowed, while the convergence / divergence of the final result is reliable to draw conclusions about the correlation functions. $\endgroup$ – dedekindCuttage Feb 23 at 21:55
  • $\begingroup$ Behind the rigor are in fact some physical ideas. There is no such things as functions $T(x)$ which you can evaluate at a point: think temperature field. No matter how thin your thermometer is you will not measure the temperature exactly at $x$ but some weighted average $\langle T,f\rangle=\int T(y)f(y)dy$ for some function that is peaked around $x$. There is no harm in allowing weights $f(y)$ of any sign, as well as $\int f\neq 1$ for more generality. This change of point of view means that you should think of $T$ in terms of its local averages $\langle T,f\rangle$. Then your issues are gone. $\endgroup$ – Abdelmalek Abdesselam Feb 23 at 22:10
  • $\begingroup$ I see. I am not familiar with the distributions except maybe in the context of delta functions. I guess the reason is they are not often used/mentioned in the mainstream literature/textbooks in condensed matter physics. But what do you think of the validity of manipulations like the one I mentioned in the question? Is it only giving the right result by chance or is there an underlying reason why it works? $\endgroup$ – dedekindCuttage Feb 24 at 1:39
  • $\begingroup$ All Kardar does is compute the Fourier transform of the function (or rather distribution) $1/p^2$. This is also what my answer does, but with a slightly different method. It is possible to render Kardar's argument completely rigorous, but only in the setting of distributions (I might do it if I have time). He writes $\Delta C=\delta$. I think we agree that the RHS is not a function but a distribution. One never writes apple$=$orange, so the LHS also has to be undertood as a distribution. Also the integral you mentioned always diverges in Lebesgue sense. In $d=1$, it diverges at the origin. $\endgroup$ – Abdelmalek Abdesselam Feb 24 at 15:04

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