I am trying to understand the Kubo formula, written, for example as $$ \sigma_H \sim \sum_{E_\alpha<E_F<E_\beta} \frac{\epsilon_{ab}}{(E_\alpha-E_\beta)^2} \langle\alpha|[\hat H,\hat x^a]|\beta\rangle \langle\beta|[\hat H,\hat x^b]|\alpha\rangle $$ where $$ \hat H|\alpha\rangle=E_\alpha |\alpha\rangle. $$ Now suppose (or explcitly construct a Hamiltonian $H$) that $$ |\alpha\rangle = \phi_\alpha (k_\alpha) |k_\alpha\rangle $$ where $|k\rangle$ is ordniary momentum eigen-ket, and $\phi_\alpha$ some (spinor) functions. Now we have a problem, $$ \langle\beta|[\hat H,\hat x^b]|\alpha\rangle = (E_\beta-E_\alpha) \phi_\beta^+ \phi_\alpha \langle k_\beta|\hat x^b|k_\alpha\rangle = (E_\beta-E_\alpha) \phi_\beta^+ \phi_\alpha i \partial_{k_b} \delta(k_\beta-k_\alpha) $$ plugging it into the Kubo formula we have a product of $$ \partial_{k_b} \delta(k_\beta-k_\alpha) \partial_{k_a} \delta(k_\beta-k_\alpha) $$ how to get rid of this product of derivatives (!) of two delta-functions? As appropriately mentioned in the comments, a product of two simple deltas would give a $\delta(0)$ (i.e. volume of the system) upon integration in $k$, but derivatives spoil this =(

PS I personally believe that I am simply making some error, and there should not appear such constructions.

  • $\begingroup$ What is $\phi_\alpha$ ? $\endgroup$ – lcv Jan 7 at 12:13
  • $\begingroup$ @lcv some spinor functions in the aprticular case of my interest, but actualy can be whatever you want. $\endgroup$ – Ignat Fialkovskiy Jan 7 at 12:15
  • $\begingroup$ Where did the subscripts of a and b come from in your derivatives? Are we to assume they are actually $\alpha$ and $\beta$ respectively? $\endgroup$ – Triatticus Jan 7 at 17:35
  • $\begingroup$ @Triatticus $a,b$ are the space indices, equal to $1,2$, summed with epsilon symbol. $\endgroup$ – Ignat Fialkovskiy Jan 7 at 17:39
  • $\begingroup$ Ah ok I wasn't looking at the top formula, you could use that the derivative, so to speak, of the delta function is related to the heavyside step function probably. $\endgroup$ – Triatticus Jan 7 at 19:28

The fact that you have a delta function as the matrix element between the states implies that you have a continuum of states. Then there is not a sum but a double integral over energies ($E_\alpha$ and $E_\beta$) which using the dispersion relations can be translated into a double integral over momenta $k_\alpha$ and $k_\beta$. So one of the delta functions (ignoring the derivatives for now - this will come later) will be eliminated by the integral, and you will remain with $\delta(0)$. This has a clear meaning in the context of this integral - it is proportional to the volume of the system, or the number of states - anyway a macroscopic extrinsic quantity. This should not necessarily surprise you, as the conductance (which is what the Kubo formula gives) could and sometimes should very well scale with the number of charge carriers, and be an extrinsic property. A way to regularize it is to confine the system to a box of size $L$, and then everything are sums, and take $L\to\infty$, (e.g. $2\pi/L \to dk$) and then see exactly what grows with $L$ and what stays fixed, and also get the correct number of $2\pi$ etc.

A separate problem is the derivative of the delta functions. You should probably carefully eliminate them using integration by parts, within each integral. I'm not sure that will be easy but I think this is a crucial step in evaluating the integrals.

  • $\begingroup$ You are absilutely right, the problem is not in delta-functions themselves, but in their derivatives. And the problem is that integrating by parts you pass a derivative from one delta function to the other. $\endgroup$ – Ignat Fialkovskiy Jan 7 at 12:30
  • $\begingroup$ Sometimes it helps writing the delta function using some representation, at the cost of introducing another integral $\delta(x) = \int dz e^{ixz}/(2\pi)$ and then you can take the derivative explicitly, hoping that the integrals will make sense at the end. It doesn't always work, of course $\endgroup$ – yu-v Jan 7 at 12:48

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