3
$\begingroup$

Context

Section $16.2$ of Kerson Huang's Statistical Mechanics ($2$nd edition) deals with a derivation of two-point correlation function $\Gamma({\bf r})$, defined in terms of an order parameter density $m({\bf r})$ as $$\Gamma({\bf r})\equiv \big\langle m({\bf r})m(0)\big\rangle-\big\langle m({\bf r})\big\rangle\big\langle m(0)\big\rangle\tag{1}$$ where $\langle..\rangle$ denotes the ensemble average. To be explicit, for example, $$\langle m({\bf r})\rangle\equiv \sum\limits_{m({\bf r})}m({\bf r})e^{-\beta\mathcal{H}}$$ where $\mathcal{H}$ is the Hamiltonian of the system. All this is fine but I am stuck with something that is usually a pretty trivial step!


He uses the Fourier transform and inverse transform convention $$m({\bf r})=\int \frac{d^3k}{(2\pi)^3} e^{+i{\bf k}\cdot{\bf r}}\tilde{m}({\bf k}),~~ \tilde{m}({\bf k})=\int d^3x e^{-i{\bf k}\cdot{\bf r}}m({\bf r}).\tag{2}$$ With this, he makes the problematic claim that (see just above Eq. $16.11$), $$\boxed{\big\langle\tilde{m}({\bf k})\tilde{m}({\bf p})\big\rangle=(2\pi)^3\delta^{(3)}({\bf k}+{\bf p})|\tilde{m}({\bf k})|^2.}\tag{3}$$ To derive Eq.(3), one would usually proceed $$\big\langle \tilde{m}({\bf k})\tilde{m}({\bf p})\big\rangle=\big\langle\int d^3x \int d^3x^\prime e^{-i{\bf k}\cdot{\bf r}}e^{-i{\bf p}\cdot{\bf r}^\prime} m({\bf r})m({\bf r}^\prime)\big\rangle\tag{4}$$ if the two momenta were equal. But here, I don't see any way standard way of reducing it to the expression $(3)$. So the question is, how does he get Eq.$(3)$?

$\endgroup$
  • $\begingroup$ It seems Huang assumes, system is homogeneous, i.e., $\langle m(r) m(r^\prime) \rangle = \langle m(r-r^\prime) m(0) \rangle$. $\endgroup$ – Sunyam May 8 at 5:56
  • $\begingroup$ That is fine. Yes, he does. But how do you get Eq.$(3)$? @Sunyam $\endgroup$ – SRS May 8 at 6:00
  • $\begingroup$ If one fourier transforms with respect to $r$ and followed by using fourier shift theorem leads to eq. (3). $\endgroup$ – Sunyam May 8 at 6:07
  • $\begingroup$ I could not follow. Can you supply some steps? $\endgroup$ – SRS May 8 at 6:15
  • $\begingroup$ $\int d t_{1} \int d t_{2} e_{}^{-i (\omega_{1}^{}t_{1}^{} + \omega_{2}^{} t_{2}^{})} G(t_1,t_2) = \int dT \int d\tau e_{}^{-i\left((\omega_{1}^{}+\omega_{2}^{})T + (\omega_{1}^{}-\omega_{2}^{})\frac{\tau}{2}\right)} G(T+\frac{\tau}{2},T-\frac{\tau}{2})=\int dT \int d\tau e_{}^{-i(\omega_{1}^{}+\omega_{2}^{})T}e_{}^{-i (\frac{\omega_{1}^{}-\omega_{2}^{}}{2})\tau} G(\frac{\tau}{2},-\frac{\tau}{2})$. $\endgroup$ – Sunyam May 8 at 6:33
5
+50
$\begingroup$

We have \begin{align} \langle \tilde{m}({\bf k})\tilde{m}({\bf p})\rangle &= \bigg\langle\int d\mathbf{r} \int d\mathbf{r}' e^{-i{\bf k}\cdot{\bf r}}e^{-i{\bf p}\cdot{\bf r}^\prime} m({\bf r})m(\mathbf{r}')\bigg\rangle \\ &= \int d\mathbf{r} \int d\mathbf{r}'' e^{-i{\bf k}\cdot{\bf r}}e^{-i{\bf p}\cdot (\mathbf{r} + \mathbf{r}'')} \langle m({\bf r})m(\mathbf{r} + \mathbf{r}'') \rangle\\ &= \int d\mathbf{r} \int d\mathbf{r}'' e^{-i (\mathbf{k} + \mathbf{p}) \cdot{\bf r}}e^{-i{\bf p}\cdot \mathbf{r}''} \langle m({\bf 0})m(\mathbf{r}'')\rangle \\ &= (2 \pi)^3 \delta(\mathbf{k} + \mathbf{p}) \int d\mathbf{r}'' e^{-i{\bf p}\cdot \mathbf{r}''} \langle m({\bf 0})m(\mathbf{r}'')\rangle\\ \end{align} where I shifted the integration variable to $\mathbf{r}'' = \mathbf{r}' - \mathbf{r}$ and used linearity of expectation, then used translational symmetry, then did the $\mathbf{r}$ integral.

Presumably, this means that Huang defines $$|\tilde{m}(\mathbf{k})|^2 = \int d\mathbf{r} \, e^{-i{\bf k}\cdot \mathbf{r}} \langle m({\bf 0})m(\mathbf{r})\rangle.$$ This is very misleading notation, because the right-hand side is not actually equal to the norm squared of $\tilde{m}(\mathbf{k})$, as one can see from dimensional analysis. (It is closely related, as one can see by adapting the argument above, but differs by a factor of $(2\pi)^3 \delta(\mathbf{0})$.) If Huang were being careful, he would have used a different notation, such as $$S(\mathbf{k}) = \int d\mathbf{r} \, e^{-i{\bf k}\cdot \mathbf{r}} \langle m({\bf 0})m(\mathbf{r})\rangle$$ which is the standard notation for a power spectral density, in which case the final result would be $$\langle \tilde{m}({\bf k})\tilde{m}({\bf p})\rangle = (2 \pi)^3 \delta(\mathbf{k} + \mathbf{p}) S(\mathbf{k}).$$ But why would Huang use such bad notation? My impression is that it's just sloppy in general, which explains the exceedingly poor reviews of the book all throughout the internet. Careful books, such as Kardar, would never make such mistakes.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.