I got a question concerning the scattering of phonons and electrons. I read an introductory explanation to this process that is somehow not very satisfactory. It goes like this:

Let $\psi_{k}$ and $\psi_{k'}$ be Bloch-waves within a solid. We denote the probability of transmission between these two states by $P_{k,k'}$. This probability is according to quantum mechanical perturbation theory proportional to $|<k'|H'|k>|^{2}$ where H' denotes the perturbed Hamiltonian or the perturbed potential (caused by either phonons or impurities). Now we assume that our two wave function are of the standard Bloch-form. Hence we obtain: $|<k'|H'|k>| = \int{ d^{3}r u_{k'}(r)^{*} H'(r,t) u_{k}(r) e^{i(k-k')r}}$ (Eq. 1) where $u_{k}$ and $u_{k'}$ have lattice-periodicity.

Now in an inelastic collision with of an electron and a phonon we have by energy conservation: $E(k') - E(k) = \pm \hbar \omega(q)$ where $E(k)$ and $E(k')$ denote the electron energies and on the rights hand side there is the energy of a phonon with a wavevector $q$.

Now comes the tricky part of the analysis:

Now they say that the disturbed potential must include a dependency on $e^{iqr}$ Hence (Eq. 1) (which is the scattering probability) must include a matrix element of the form $<k'|e^{iqr} |k>=\int{d^{3}r u_{k'}(r)^{*} u_{k}(r) e^{i(k-k'+q)r}}$ . (Eq. 2)

Why is that? I don't see their point here. I want to remark here that I do have heard an introductory quantum mechanics class and class in linear algebra. However I neither understand this from a mathematical point of view nor from a physical point of view. Can anyone give me a better explanation of this? Please note that the course I'm taking is not a course in theoretical solid state physics.

Now the derivation goes on:

The say that since $u_{k}u_{k'}$ can be expanded into a Fourier series of reciprocal lattice vectors. Fine - I agree that's legitimate. So assume: $(u_{k}u_{k'})(r) = \sum_{G}f_{g} e^{-iGr}$. Plugging this into this equation they claim the matrix element above (Eq. 2) only does NOT vanish is $k'-k+q=G$ Well this doesn't seem plausible. I mean then $e^{i(k-k'+q-G)r}=1$. Integrating over all space leads to infinity. Well I may agree that the integral over this term vanishes if the equality $k'-k+q=G$ does not hold because of periodicity/symmetry. Maybe one has to restrict the integral to the solid itself to keep it finite.

Can anyone give me a more detailed explanation??

I'm looking forward to your responses. Thanks in advance.

  • 1
    $\begingroup$ Your second question is really one about the Dirac delta function and Fourier transforms on a lattice. You may want to look in the appendix of whatever book you're looking at (or if you're not looking at a book yet, you may want to pick up Ashcroft and Mermin's text and start reading). $\endgroup$ – j.c. Jan 15 '11 at 20:04

Sorry I can't be more specific, but you will probably find the answer to the question in Fundamentals of Carrier Transport By Mark Lundstrom. I also recommend that you listen to his lecture series which includes phonon scattering on iTunesU or from the NanoHub.org ECE 656 Lecture 23: Phonon Scattering 1. Lundstrom makes logical arguments about assuming forms of solutions to integrals; when you hear him explain it seems obvious! Good luck.

** Sorry I couldn't embed that link. The self importance of an administrator is currently limiting me to 1 link per answer! Furthermore, I actually intended this to be a comment to your question, but it seems they have crippled that too.


everything you write is perfectly OK, except for the statement that "it doesn't seem plausible". Yes, it is plausible and yes, the exponential is equal to one. That's the whole reason why the contribution for $G=k'-k+q$ to the interaction amplitude is nonzero. All the terms with different values of $G$ lead to oscillating functions in the complex plane that sum up to zero.

Yes, the integral of 1 over the whole space is infinite. And yes, you also know the way to make it finite: restrict the interval to the solid itself. It is not surprising - and it is physically correct - that the integral is infinite for an infinite solid. It's because the interaction Hamiltonian between the two plane wave states is proportional to the volume of the solid as well - the greater volume, the greater chance that the interaction will occur somewhere.

If you were calculating the things for a finite volume of the solid - e.g. a box - you would get manifestly finite numbers everywhere. However, some formulae would be unnecessary awkward and they would depend on the size and shape of the solid. That's why adult physicists learned from Paul Dirac how to use the distributions such as $\delta(G-G_0)$. For your problem, $G_0=k'-k+q$ is treated as a constant. The defining property of the $\delta$-function is that $$\int_{-\infty}^\infty dG\,f(G) \delta(G-G_0) = f(G_0)$$ for any function $f(G)$. So the integral of $f(G)$ over $G$, weighted by the delta-function, only picks the value at $G=G_0$. That's because $\delta(G-G_0)$ vanishes for all values for which $G\neq G_0$. But for $G=G_0$, it is infinite and so large that the integral of $\delta(G)$ over $G$ is equal to one, and if you insert $f(G)$, it simply picks $f(G_0)$.

The object $\delta(G-G_0)$ is not a function in the usual sense but it is extremely helpful and consistent in dealing with the integrals that appear in the Fourier transformations. The delta-function may be approximated by a function that is equal to $1/\epsilon$ for the argument being between $-\epsilon/2$ and $\epsilon/2$, and otherwise is equal to zero, in the limit where $\epsilon$ goes to zero. The delta-function can also be written as the Fourier transform of the function $1$ divided by $2\pi$: $$\delta(G-G_0) = \frac{1}{2\pi}\int_{-\infty}^\infty dR\,\exp[i(G-G_0)R] $$ You may always imagine that the delta-function exercises are translated to a calculation at a finite volume of the space or solid; then the momenta such as $G$ become discrete - number of zeros of a standing wave, for example, times $1/R_{solid}$ - and the integral over $G$ is replaced by a summation. The function $\delta(G-G_0)$ is then replaced by a simple Kronecker delta symbol $\delta_{G,G_0}$ which is only nonzero - equal to one - if $G=G_0$. But you will pay the price that most formulae will contain powers of the volume and other things, and you need to remember the spacing of the momenta etc.

All these extra things in the formulae will depend on the size and shape of the solid - or space(time) - that you picked. But at the end, you know very well that there exists a sensible physics in the infinite-volume limit that should be independent of the volume (because it was sent to infinity). For you, to learn how to deal with the Dirac distributions is important to do all these calculations effectively because the volume $V$ and all the problematic features about the spacing of the momentum evaporate from the formulae, and the fact that the formulae hold for any large piece of the solid (or space) becomes self-evident.

Best wishes Lubos

  • $\begingroup$ Thanks for your response. Let's use your notation. Fix $G_{0}=k-k'+q$. Then we obtain $<k'|e^{iqr}|k>=\sum_{G} f_{G} \int{d^{3}r e^{i(G_{0}-G)r}}=\sum_{G} f_{G} \delta(G-G_{0})$ so $<k'|e^{iqr}|k>$ is still infinite for G=$G_{0}$ as long as you don't integrate over it (which you obviously don't do). In addition to that I couldn't find a reason why a matrix element of the specific form <k'|e^{iqr}|k> even exists? (That was my first question) This would mean that the disturbed hamiltonian H' contains a (linear) term e^{iqr}. But I don't see why that is the case. $\endgroup$ – Solidz Jan 15 '11 at 23:13

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