# Constrained Brilloiun zone sum to integral

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.

## 1 Answer

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.

• 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 $V$. – Claudiu Iaru Dec 18 '19 at 14:38
• 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. – Claudiu Iaru Dec 18 '19 at 14:43
• 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. – Ruslan Dec 18 '19 at 16:09
• 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? – Claudiu Iaru Dec 19 '19 at 15:45
• 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. – Ruslan Dec 19 '19 at 17:55