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.

  • 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, 2017 at 19:57

1 Answer 1


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!


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