0
$\begingroup$

In order to evaluate the two-phonon Raman scattering cross section I'm trying to compute the following double sum in $ k $-space (See Phys. Rev. B 17, 4951 (1978), or the book Light Scattering in Solids II, by M. Cardona, page 149):

$$ S=\sum_{\vec{k}_1,\vec{k}_2}{|\Delta_{\vec{k}_1}|^2|\Delta_{\vec{k}_2}|^2\delta(\vec{k}_1+\vec{k}_2)} $$

where $|\Delta_\vec{k}|$ is the phonon displacement, and $ \delta(\vec{k}_1+\vec{k}_2)=1 $ when $ \vec{k}_1+\vec{k}_2=0 $, and otherwise $ \delta(\vec{k}_1+\vec{k}_2)=0 $.

This means that only the pairs of vectors $ \vec{k} $ and $ -\vec{k} $ will contribute to the sum.

Note that $ \Delta_\vec{k}=\Delta_{-\vec{k}} $, so when $\delta(\vec{k}_1+\vec{k}_2)=0 $ we have $|\Delta_{\vec{k}_1}|^2|\Delta_{\vec{k}_2}|^2=|\Delta_{\vec{k}_1}|^4$

To evaluate $ S $, I first change the summation over the brillouin zone into an integral:

$$ \sum_\vec{k}{\Delta_\vec{k}}=\frac{V}{(2\pi)^3}\int_\vec{k}{\Delta_\vec{k}d^3\vec{k}} $$

and obtain:

$$ S=\frac{V^2}{(2\pi)^6}\int_{\vec{k}_1}{|\Delta_{\vec{k}_1}|^2d^3\vec{k}_1}\int_{\vec{k}_2}{|\Delta_{\vec{k}_2}|^2d^3\vec{k}_2} $$

At this point one must define the limits of the summation over the spherical coordinates in order to ensure that the two integrals only take the pairs $ \vec{k}_1=\vec{k} $ and $ \vec{k}_2=-\vec{k} $. My only idea on how to do this is to integrate $ |\Delta_{\vec{k}_1}|^2|\Delta_{\vec{k}_2}|^2 $ together over the entire Brillouin zone instead of separately, such that:

$$S\sim\int|\Delta_{\vec{k}}|^2|\Delta_{-\vec{k}}|^2d^3k $$

However this is wrong, as it would change the dimensionality of the result, since the factor $\frac{V}{(2\pi)^3}$ would only appear once in front of the integral. Otherwise I can separate the integrals first so that $\frac{V}{(2\pi)^3}$ does come out squared, but then by using the delta function to reduce the second integral to simply $ |\Delta_{\vec{k}_1}|^2 $ again changes the dimensionality, since I'm not integrating over a volume any more.

I feel like I am missing something trivial. If anyone could explain what that is, or provide any resources (books, papers) which might help, I would appreciate it.

$\endgroup$
0
$\begingroup$

Why do you have to begin with a switch to an integral? You could simply expand your initial sum using the properties of $\delta$ as

$$S=\sum_{\vec{k}_1,\vec{k}_2}{|\Delta_{\vec{k}_1}|^2|\Delta_{\vec{k}_2}|^2\delta(\vec{k}_1+\vec{k}_2)}= \sum_{\vec{k}}{|\Delta_{\vec{k}}|^2|\Delta_{-\vec{k}}|^2}.$$

Then, using your note that $\Delta_\vec{k}=\Delta_{-\vec{k}}$, we can further develop this into

$$S=\sum_{\vec k}|\Delta_{\vec k}|^4.$$

Now you can do the switch to an integral without the need to worry about Dirac delta's dimensionality.

$\endgroup$
7
  • $\begingroup$ Thanks for your reply! The problem is that V is the crystal volume, which should come out of the equation so that the final result doesn't depend on it. In the reference I cite, $ \Delta_{\vec{k}}\sim{}V^{-\frac{1}{2}} $. This means that I must have $V^2$ as a prefactor, which I get by switching both sums to integrals, in order for the result to be independent of $V$. $\endgroup$ – Claudiu Iaru Dec 18 '19 at 14:38
  • $\begingroup$ By the way, the delta function is apparently a Kronecker delta (according to Light Scattering in Solids), which means that its value is 1 instead of $\infty$ when the argument is 0. I believe this changes things a bit with regards to taking $\Delta_{\vec{k}}$ out of the sum, but I'm not really sure how. $\endgroup$ – Claudiu Iaru Dec 18 '19 at 14:43
  • $\begingroup$ Why is it a problem? You still can switch to the integral starting from the expression for $S$ that I gave at the end, and take out your $V$ as you need. And that the delta is a Kronecker delta was actually obvious. It just has to transform to a Dirac delta when you switch from a sum to an integral, because otherwise its contribution to the integral will be zero. But if you do what I suggested in this answer, you don't have to deal with this conversion of deltas at all. $\endgroup$ – Ruslan Dec 18 '19 at 16:09
  • $\begingroup$ I think I'm missing something obvious. As I see it, when I change $\sum_{\vec{k}}|\Delta_{\vec{k}}|^4$ to an integral I will have $\frac{V}{(2\pi)^3}$ as a prefactor, since there is only one summation index. However, $|\Delta_{\vec{k}}|^4\sim{}V^{-2}$, so there remains $V^{-1}$ that doesn't divide out, and my result will depend on the crystal volume. Is this not the case? Does the delta function also produce the volume as a prefactor? $\endgroup$ – Claudiu Iaru Dec 19 '19 at 15:45
  • $\begingroup$ OK, I see your concern. But what makes think that this sum should be independent of crystal volume? As it's formulated in your first equation, the summation itself (ignoring the $\Delta$ terms, but leaving the $\delta$ in) is proportional to $V^{+1}$. If the $\Delta$ terms provide a factor of $V^{-2}$, the full sum can't be independent of volume. What exactly equation from the PhysRev article are you trying to analyze? I've only found $(12)$ to be similar, but it has some extra terms. $\endgroup$ – Ruslan Dec 19 '19 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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