Transitions between the conduction and valence band must have the same $k$-vector?

Question

I am currently studying Diode Lasers and Photonic Integrated Circuits, second edition, by Coldren, Corzine, and Mashanovitch. In chapter 1.2 ENERGY LEVELS AND BANDS IN SOLIDS, the authors say the following:

Although Fig. 1.4 suggests that many conduction—valence band state pairs may interact with photons of energy $E_{21}$, Appendix 1 shows that the imposition of momentum conservation in addition to energy conservation limits the interaction to a fairly limited set of state pairs for a given transition energy. This situation is illustrated on the electron energy versus $k$-vector ($E$ − $k$) plot shown schematically in Fig. 1.5. (Note that $\text{momentum} \equiv \hbar \mathbf{k}$.) Because the momentum of the interacting photon is negligibly small, transitions between the conduction and valence band must have the same $k$-vector, and only vertical transitions are allowed on this diagram. This fact will be very important in the calculation of gain.

My question relates to this part:

Because the momentum of the interacting photon is negligibly small, transitions between the conduction and valence band must have the same $k$-vector, and only vertical transitions are allowed on this diagram. This fact will be very important in the calculation of gain.

With regards to the momentum of the interacting photon, my understanding is that, since momentum is mass times velocity, and since photons are massless, we have that the momentum of the interacting photon is zero (or, as the authors say, negligible). Furthermore, I have done some study of semiconductor physics outside of this textbook, so I understand that the vertical transitions shown in figure 1.5 are for a direct band gap semiconductor (my understanding is that indirect band gap transitions are not vertical). What I don't understand is why the momentum of the interacting photon being negligible implies that transitions between the conduction and valence band must have the same $k$-vector. So why does the momentum of the interacting photon being negligible imply that transitions between the conduction and valence band must have the same $k$-vector?

Answer 1

For an absorption process, the electron absorbed a photon of energy $E_\omega$ and transition to conduction band $E_c(k_c)$ from valance band $E_v(k_v)$. This process satisfies both energy and momentum conservation:

energy conservation $$ E_\omega = E_c - E_v, $$

and momentue conservation $$ \hbar k_\omega = \hbar k_c - \hbar k_v. $$

The dispersion of a photon is $\omega = c k$, where $c$ is light speed and $k$ wave number. The energy $E_\omega = pc = \hbar \omega$. $$ k = \frac{\omega}{c} = \frac{E_\omega}{\hbar c} = 2 \pi \frac{h \nu}{h c} = \frac{2\pi}{\lambda} \approx \frac{2\pi}{5000 \dot A} \approx 10^{-3} \frac{1}{\dot{A}} $$

The wave vector of a electron in solid is inversely propotional to the lattice constant $a\approx 4 \dot A$

$$ k_{c,v} \approx \frac{\pi}{a} \approx \frac{\pi}{4 \dot A} \approx 1 \frac{1}{\dot A} $$

The momentum of photon is about 1/1000 of the electron's momentum. Therefore, we assume a direct transition for absorption.