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

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?

• Direct absorption in an indirect semiconductor is 'vertical' (modulo the photon momentum). You can have phonon-mediated absorption as well. For emission, you need the phonon to provide the momentum change to get from the conduction band minimum to the valence band maximum, since there are not empty states to transition to deep in the valence band. Commented Apr 1, 2021 at 14:31
• @JonCuster I stand corrected. Thanks for the clarification. Commented Apr 1, 2021 at 14:40
• "since photons are massless, we have that the momentum of the interacting photon is zero" No, the momentum of a photon is not "mv." The photon has some momentum, it is just assumed small compared to the electron's momentum.
– hft
Commented Oct 20, 2021 at 7:51
• @hft You're right; again, I stand corrected. Commented Nov 4, 2021 at 16:57

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

energy conservation $$E_\omega = h \nu = \hbar \omega = E_c - E_v,$$

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

The dispersion of the photon is $$\omega = c k_\omega$$, where $$c$$ is light speed and $$k$$ wave number. The photon energy $$E_\omega = p_{\omega}c = \hbar k_\omega c = \hbar \omega$$. If the photon's wavelength $$\lambda = 5000 \ \dot{A}$$: $$k_\omega = \frac{\omega}{c} = \frac{E_\omega}{\hbar c} = 2 \pi \frac{h \nu}{h c} = \frac{2\pi}{\lambda} \approx \frac{2\pi}{5000 \ \dot{A}} \approx 10^{-3} \ \dot{A}^{-1}$$

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

\begin{align} k_{c,v} & \propto \frac{\pi}{a} \\ & \propto \frac{\pi}{4 \ \dot{A}} \\ k_{c,v} & \propto 1 \ \dot{A}^{-1} \end{align}

Thus, the momentum of the photon is about $$\frac{1}{1000}$$ of the transitioned electron's momentum. Therefore, we assume a direct energy transition for the case of absorption.

• Thanks for the answer. What's $p$ and $\nu$? And how did you get that $$\frac{E_\omega}{\hbar c} = 2 \pi \frac{h \nu}{h c} = \frac{2\pi}{\lambda} \approx \frac{2\pi}{5000 \dot A} \approx 10^{-3} \frac{1}{\dot{A}}$$? Commented Apr 1, 2021 at 12:40
• p is the momentum of photon, $E = pc$ for photon. $\nu$ is the frequency, $E= h\nu$ is also the energy of photon. $\hbar = h/2\pi$. What else need to elucidate? You may use $k = 2\pi/\lambda$, which can also be treated as the definition of wave number (wave vector).
– ytlu
Commented Apr 2, 2021 at 0:44
• @ThePointer you should up vote, at least, and accept if it answers your question. Seems like a great answer to me Commented Apr 2, 2021 at 7:55
• $\lambda$ is the wavelength of a typical light, 4000 angstrom to 7000 angstrom, and for IR the wavelength is longer, makes the photon momentum even smaller. The lattice constant $a$ is the side length of a unit cell, typical 3-5 angstroms. The wave vector of an electron is ranging from $-\frac{\pi}{a} \to \frac{\pi}{a}$. These are some basic knowledges for undertood band structure in a solid.
– ytlu
Commented Apr 2, 2021 at 8:14
• The typical momentum of a photon involed in the absorption process is about $10^{-3}$ of the momentum of the excited electron. Therefore, the momentum of photon is usually neglected, and the absorption is treated $\vec{k}_c = \vec{k}_v$, a direct transition.
– ytlu
Commented Apr 2, 2021 at 10:18