# Diagonal matrix in k-space

I'm having some trouble with an integration I hope you guys can help me with. I have that:

${{\mathbf{v}}_{i}}\left( \mathbf{k} \right)=\frac{\hbar {{\mathbf{k}}_{i}}}{m}$ and ${{\mathbf{v}}_{j}}\left( \mathbf{k} \right)=\frac{\hbar {{\mathbf{k}}_{j}}}{m}$

Now according to my teacher when I multiply these two, and integrate the angular part ($\,d\theta \,d\varphi$) over $d{\mathbf{k}}$, with $d\mathbf{k}={{k}^{2}}\,dk\sin \left( \theta \right)\,d\theta \,d\varphi$, I should end up with:

${{\mathbf{v}}_{ij}}{{\left( \mathbf{k} \right)}^{2}}=\frac{{4\pi{\hbar }^{2}}}{3{{m}^{2}}}{{k}^{2}}$

My question is why?I see that when there is no angle dependency I get the $4\pi$, and nothing touches $\hbar$, $m$ and $k$. But where to the $1/3$ come from? That's probably my main concern.

So anyone with a hint or something ?

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It looks like this is really just a math question. I can migrate it to Mathematics, unless there's some reason someone thinks it should stay. –  David Z May 15 '13 at 20:46
I think it's as likely to get answered here as there. Maybe give it one or two days, and see if it does... –  Emilio Pisanty May 15 '13 at 20:47
@Denver, this probably requires a little bit more context than you're giving to have a unique answer. Are $\mathbf{k}_i$ and $\mathbf{k}_j$ different vectors or components of one? What does "multiply" mean? Inner product? dyadic product? –  Emilio Pisanty May 15 '13 at 20:50
I reckon he means the tensor (dyadic) product $k_i k_j$. Then the factor of 3 comes out from taking the trace. Denver, are you at all familiar with tensors? –  Michael Brown May 15 '13 at 20:53
The main equation where this integral appears is: $\mathcal{L}_{\,ij}^{\,\left( \alpha \right)}=\frac{{{e}^{2}}}{4{{\pi }^{3}}}\int{\left( -\frac{\partial f}{\partial \varepsilon } \right)}\,\tau \left( \varepsilon \left( \mathbf{k} \right) \right){{\mathbf{v}}_{i}}\left( \mathbf{k} \right){{\mathbf{v}}_{j}}\left( \mathbf{k} \right){{\left( \varepsilon \left( \mathbf{k} \right)-\mu \right)}^{\alpha }}d\mathbf{k},$ where $\tau$ end up being a constant and $v$ the velocity of the electron with wave vector $\mathbf{k}$. –  Denver Dang May 15 '13 at 20:59

Focus on the integral

$$I_{ij}(k) = \int k_i k_j\ \mathrm{d}\Omega_k.$$

This is a rank 2 symmetric tensor which can only depend on $\vec{k}$ through its magnitude $k^2$, since the direction has been integrated over. So the only possibility is that $I_{ij}$ is proportional to the unit tensor (Kronecker delta):

$$I_{ij}(k) = f(k^2) \delta_{ij},$$

where $f(k^2)$ is some function to be determined. We can find $f(k^2)$ by taking the trace (using the summation convention for repeated indices):

$$\begin{array}{ll} \delta_{ij} I_{ij}(k) &= \delta_{ij} f(k^2) \delta_{ij} \\ &= 3 f(k^2)\\ &= \delta_{ij} \int k_i k_j\ \mathrm{d}\Omega_k \\ &= \int k^2\ \mathrm{d}\Omega_k \\ &= k^2 4\pi. \end{array}$$

Thus

$$I_{ij}(k) = \frac{4\pi}{3} k^2 \delta_{ij}.$$

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I see now :) Thank you very much. –  Denver Dang May 16 '13 at 12:50