# How to measure transverse current-current correlation at arbitrary wave vector and frequency

Let $\hat J^{\mu}(t,\boldsymbol r) \equiv (c \hat \rho(t,\boldsymbol r), \hat{\boldsymbol J}(t,\boldsymbol r))$ be the density-current operator at spacetime coordinate $(t,\boldsymbol r)$, in the interaction picture. Define the retarded correlation function \begin{eqnarray} D^{\mu\nu}(t-t',\boldsymbol r - \boldsymbol r') = \Theta(t-t') \left\langle \left[ \hat J^{\mu}(t,\boldsymbol r), \hat J^{\nu}(t',\boldsymbol r') \right] \right\rangle \end{eqnarray} where we assume translational symmetry, and the $\langle\cdots\rangle$ could be either thermal or ground-state average. Fourier-transforming we get \begin{eqnarray} D^{\mu\nu}(\omega,\boldsymbol q) = \int dt d\boldsymbol r e^{-i(\boldsymbol q \cdot \boldsymbol r - \omega t)} \Theta(t) \left\langle \left[ \hat J^{\mu}(t,\boldsymbol r), \hat J^{\nu}(0,\boldsymbol 0) \right] \right\rangle \end{eqnarray} In particular, the spatial components are \begin{eqnarray} D^{ij}(\omega,\boldsymbol q) = \int dt d\boldsymbol r e^{-i(\boldsymbol q \cdot \boldsymbol r - \omega t)} \Theta(t) \left\langle \left[ \hat J^{i}(t,\boldsymbol r), \hat J^{j}(0,\boldsymbol 0) \right] \right\rangle \end{eqnarray} where $i,j=x,y,z$. Assuming isotropy (more precisely, $O(3)$ symmetry), we can decompose this into a longitudinal part and a transverse part: \begin{eqnarray} D^{ij}(\omega, \boldsymbol q) = D_L(\omega,\boldsymbol q) \frac{q^i q^j}{q^2} + D_T(\omega,\boldsymbol q) \left( \delta^{ij} - \frac{q^i q^j}{q^2} \right) \end{eqnarray}

Question: Is there any experiment that can in principle access $D_T(\omega,\boldsymbol q)$ (in other words, the transverse current-current correlation) at arbitrary $\omega$ and $\boldsymbol q$, with $\omega$ and $\boldsymbol q$ independent?

Why I think this is non-trivial:

1. One convenient probe is optical conductivity. Unforunately it is determined by $\underset{q\rightarrow 0}{\lim} D^{ij}(\omega, \boldsymbol q)$ and thus does not access finite-$q$ information.

2. Actually, I believe the optical conductivity is technically determined by $D^{ij}(\omega, \boldsymbol q)$ at some nonzero $\boldsymbol q$, and $q\rightarrow 0$ just happens to be a good approximation. Thus optical conductivity measurement does access finite $q$. However, in experiments we still cannot vary $\omega$ and $\boldsymbol q$ independently due to the existence of dispersion relations for transverse modes.

3. It is possible to create an electric field $\boldsymbol E(t,\boldsymbol r) \sim e^{i (\boldsymbol q \cdot \boldsymbol r - \omega t)}$ at arbitrary independent $\omega$ and $\boldsymbol q$ provided $\boldsymbol E$ is parallel to $\boldsymbol q$, so that $\boldsymbol q \times \boldsymbol E = \boldsymbol 0$. But this only measures $D_L(\omega,\boldsymbol q)$, not $D_T(\omega, \boldsymbol q)$.

• There are electric fields satisfying $\mathbf{q} \times \mathbf{E} = 0$. They are called linear electrostatic waves. They are just a linearly polarized, longitudinal mode, so nothing fancy. The associated current (i.e., displacement current) from such modes is very small, however. – honeste_vivere Nov 15 '15 at 16:25

Not an expert in this but I can try to answer some parts of your question. As you pointed out, one can find the optical conductivity to measure the current-current correlation. For that if you use linear response then you can conclude the following (for an isotropic material):

$$\sigma^{ij} = \sigma_L \frac{q^i q^j}{q^2} + \sigma_T \left( \delta^{ij} - \frac{q^i q^j}{q^2} \right)$$ After some manipulations (for derivation see sec. 3.7.2 here), $$\sigma_L = \frac{i}{\omega} \left[ \frac{D_L - e^2 \rho/mc^2}{D_L - e^2 \rho/mc^2 - \omega^2} \right]$$ and $$\sigma_T = - \frac{i}{\omega} \left[ D_T - \frac{e^2 \rho}{mc^2} \right] \left[ 1 - \frac{D_T - e^2 \rho/mc^2}{D_T - e^2 \rho/mc^2 - \omega^2 + q^2} \right]$$

First of all there is no reason why $\omega$ and $q$ should depend on each other. So naively speaking one can measure the full $\sigma(\omega)$ for $q \rightarrow 0$. The DC conductivity can be obtained under static limit, $q \rightarrow 0, \omega \rightarrow 0$. You can see under this limit $\sigma_L$ gives rise to the so-called "Drude peak" which is a limiting measure of $D_L$. The transverse conductivity also give rise to a "fake" Drude peak, a measurement of $D_T$ (fake because the width of this peak scales with temperature depending on the dimensionality).

On some other contexts, superfluid density can also be related to $D_T$, in a superconductor the london penetration depth $\lambda_L$ depends on the (transverse) magnetic response (for derivation see sec. 3.5.5 here): $$\frac{1}{\lambda^2_L} = \lim_{q \rightarrow 0} \, D_T(q, 0) \, .$$

I'm hoping for some better answers.

• Thanks for your answer! I think I can agree that nothing fundamentally prohibits a measurement of the current-current correlation at all $(\omega,\boldsymbol q)$. It just seems to me that you can't do that by measuring optical conductivity, since you cannot set up an EM wave at arbitrary $(\omega,\boldsymbol q)$ with the two chosen independently. I was hoping there be some other experiments for which they can be chosen independently. Do you know of any? – user46652 Nov 11 '15 at 17:54
• Sorry @user46652 I don't really know any experiment on that. – Jon Snow Nov 12 '15 at 0:46
• Thank you still! It was helpful. I'll wait to see what others say. – user46652 Nov 12 '15 at 7:03

This is a really good question. I have also thought about this quite a bit. At present, it does not seem like it is possible with a standard experimental probe. Inelastic x-ray scattering and electron energy loss spectroscopy only measure the longitudinal response function at finite frequency and momentum. As you said though, if you are only seeking small momentum measurements, then in principle optical conductivity measurements will suffice. One of the possible solution to this problem is to pattern a sample, effectively turning it into a diffraction grating and measure the dispersion of transverse modes that way, but even this wouldn't get you everything you want.

Sorry for this sort of non-answer, but such a measurement doesn't seem possible at present.