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I'm reading this paper and want to prove eq (8):

The field $\psi(\mathbf{x}) \in \mathbb{C}$ exists in a finite periodic 2D square box (of side length $L$), and has a Fourier series expansion, and corresponding inversion formula for the Fourier coefficients $\hat{\psi}_\mathbf{k}$: $$ \psi(\mathbf{x}) = \sum_\mathbf{k} \hat{\psi}_\mathbf{k} e^{i\mathbf{k}\cdot\mathbf{x}} \qquad \iff \qquad \hat{\psi}_\mathbf{k}=\frac{1}{L^2}\int_{Box}\psi(\mathbf{x})e^{-i\mathbf{k}\cdot\mathbf{x}}d\mathbf{x}. \tag{1a,b} $$ The equation I want to prove is an expression for the 2-point correlation function in Fourier space $$ \langle \psi(\mathbf{0})\psi^*(\mathbf{r})\rangle = \sum_\mathbf{k}\langle \hat{\psi}_\mathbf{k}\hat{\psi}^*_\mathbf{k} \rangle e^{i\mathbf{k} \cdot \mathbf{r}} \tag{2}\label{2} $$ where angle brackets mean ensemble average and star means complex conjugate. This has the nice interpretation that the Fourier transform of the 2-point correlation function is just a Fourier series with the power in the Fourier modes as coefficients.

I can't quite do it.

Inspired by the answer to this question I can try using translational invariance to write $\langle \psi(\mathbf{0})\psi^*(\mathbf{r})\rangle = \langle \psi(\mathbf{x})\psi^*(\mathbf{x}+\mathbf{r})\rangle$ and plug in the Fourier series (1a): $$ \sum_{\mathbf{k},\mathbf{k'}}\langle \hat{\psi}_\mathbf{k}\hat{\psi}^*_\mathbf{k'} \rangle \left( e^{-i\mathbf{k'} \cdot \mathbf{r}} - e^{i\mathbf{k}\cdot\mathbf{x}} e^{-i\mathbf{k'} \cdot(\mathbf{x}+\mathbf{r})} \right) =0 $$ so if $\mathbf{k}=\mathbf{k'}$ the term in brackets vanishes and if $\mathbf{k}\neq\mathbf{k'}$ the term in brackets can't vanish for general $\mathbf{x},\mathbf{r},\mathbf{k},\mathbf{k'}$, which says that the two-Fourier-mode correlation must vanish. So only the diagonal terms survive.

Fine, but then this gives me $\langle \psi(\mathbf{0})\psi^*(\mathbf{r})\rangle = \sum_\mathbf{k}\langle \hat{\psi}_\mathbf{k}\hat{\psi}^*_\mathbf{k} \rangle e^{-i\mathbf{k} \cdot \mathbf{r}} $ with the minus sign in the exponent c.f. plus sign in equation \eqref{2}. If it's a minus sign then it probably loses the nice interpretation of just being the obvious Fourier series.

Can anyone please help me with this non-trivial sign error!? N.B. it isn't a matter of just changing $\mathbf{k}\to\mathbf{-k}$ in the sum as the field is complex.

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It appears that the ensemble average gives a system that is both translational and reflection invariant. Using translational invariance you can integrate your right-hand side expression of $\langle \psi(\mathbf{0})\psi^*(\mathbf{r})\rangle = \langle \psi(\mathbf{x})\psi^*(\mathbf{x}+\mathbf{r})\rangle$ over $\mathbf{x}$ and divide by $L^2$ to get a kronecker delta that makes $\mathbf{k}=\mathbf{k'}$. Reflection invariance means you can set $\mathbf{r} \rightarrow -\mathbf{r}$.

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  • $\begingroup$ Thanks, @user200143. Just to make it crystal clear (for myself and future readers!), what you're saying is reflection invariance gives $$\langle \psi(\mathbf{0}) \psi^*(\mathbf{r}) \rangle = \langle \psi(\mathbf{0}) \psi^*(\mathbf{-r}) \rangle $$ but the RHS of this, writing the fields out in their Fourier expansions, is $\sum_\mathbf{k} \langle \hat{\psi}_\mathbf{k}\hat{\psi}^*_\mathbf{k} \rangle e^{i\mathbf{k}\cdot\mathbf{r}}$ which is what I want. Cheers! $\endgroup$ – jms547 Jul 9 '18 at 16:19

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