2
$\begingroup$

I'm reading some papers on spin-ice models and in a few of them, they state the correlation function of the polarization in momentum space (in 3D) as being

$$\langle P_i(\mathbf{k}) P_j(\mathbf{-k})\rangle = \frac{1}{\kappa}\left(\delta_{ij} - \frac{k_i k_j}{|\mathbf{k}|^2} \right)$$

where $\kappa$ is some constant. It is then stated that Fourier transforming this back to real space gives

$$\langle P_i(0)P_j(\mathbf{r})\rangle = \frac{4\pi}{\kappa}\left(\delta^3(\mathbf{r}) + \frac{1}{r^3}\left(\delta_{ij} - \frac{3 x_i x_j}{|\mathbf{x}|^2} \right) \right).$$

I have two questions -

1) Why is the Fourier transform of $<P_i(k)P_j(-k)> = <P_i(0) P_j(r)>$? I get that the system is translationally invariant and so all correlations will only depend on the difference $x-y$ and that we can set $y = 0$, but I can't explicitly relate these using a Fourier transform.

2) Similarly, where did the $\delta$ function come from after the Fourier transform to real space? I can't reproduce this result.

$\endgroup$
  • 1
    $\begingroup$ Your equation must be wrong since the $\delta^3(\mathbf{r})$ term is missing $i$ and $j$ indices to match the left hand side. $\endgroup$ – JamalS Jul 13 '17 at 19:57
2
$\begingroup$

1) This is a standard QFT theorem, namely "translation invariance = momentum conservation", and it's proven in every QFT textbook. You prove it by computing the Fourier transform of the position space correlator: $$\int d^3x d^3 y \; e^{-ip \cdot x} e^{-i k \cdot y} < P_i(x) P_j(y)>\,.$$ You now use translation invariance and show that the above integral is proportional to $\delta^3(p+k)$. Then you define the momentum space correlator as $$(2\pi)^3 \delta^3(k+p) < P_i(k) P_j(-k)> \;\; = \; \int d^3x d^3y \; e^{-ip \cdot x} e^{-i k \cdot y} < P_i(x) P_j(y)>$$ or $$< P_i(k) P_j(-k)> \;\; =\; \int d^3x \; e^{-i k\cdot x} <P_i(x) P_j(0)>\,.$$ I have probably missed a few factors of $2\pi$ here and there.

2) To see this, take the trace (I'm assuming that there are 3 polarizations, such that $\delta_{ij}^2 = 3$). On top, you get $$ \delta_{ij} < P_i(k) P_j(-k) > = \frac{2}{\kappa} = \text{constant.}$$ On the bottom, this kills the traceless part: $$ \delta_{ij} < P_i(x) P_j(0) > = \frac{4\pi}{\kappa} \delta^3(x)\,.$$ (There is a factor missing in front of the delta function, like $\delta_{ij} \times \text{constant}$, but I've set it to one here.) To show that these expressions are equal, you have to show that the Fourier transform of a constant is a delta function. This is not so difficult. Good luck!

$\endgroup$

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.