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.


1 Answer 1


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 $\mathbf{r} \rightarrow -\mathbf{r}$.

  • $\begingroup$ 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, 2018 at 16:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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