13
$\begingroup$

Intro:

To avoid any terminology confusion, this is asked in the context of Solid State Physics and semiconductors.

The canonical definition given for the effective mass is that it is related to the curvature of the conduction and valence bands in the band structure (energy dispersion in terms of k-vector) for electrons and holes respectively. So $m_{eff}$ is given by: $$m_{eff}^{-1} = \hbar^{-2}\frac{\partial^2 E}{\partial k^2} \tag{*}$$

Moreover for the holes, one often has to distinguish between light and heavy ones, as the band structure may contain multiple valence bands with different curvatures. Effective masses are usually tabulated in terms of free electron masses, e.g. for Silicon we know that the heavy holes have $49\%$ the mass of the free electron and light holes $16\%.$

Question:

  • I'm just curious to learn how experimentally, for a given material (GaAs, Ge, ...) one goes about measuring the effective masses? What are the most usually adopted methods? It would be enlightening to learn about the key ideas behind such experimental procedures.
  • Are there both direct and indirect ways of measuring $m_{eff}?$ (indirect in the sense that we measure other observables from which we may be able to infer reliably what the effective masses ought to be, and direct in the sense that an experiment designated solely to measure $m_{eff}.$)
  • If a useful account of ideas involved is difficult to implement in such platform, feel free to refer to relevant literature where a solid exposure is given.
$\endgroup$
10
+50
$\begingroup$

The method I'll describe is called Cyclotron Resonance, and it's a neat way to directly measure $m^*$ by using a fixed magnetic field $\boldsymbol B$.

The equation of motion of the electrons in a certain material, when in presence of a magnetic field $\boldsymbol B$ are

$$ m^*\dot{\boldsymbol v}=-e\boldsymbol v\times \boldsymbol B -\frac{m^*}{\tau}\boldsymbol v\tag{1} $$ where $\tau$ is the relaxation time$^1$ of the electrons (in general, $\tau^{-1}$ is a very small number, so for now we might take $\tau^{-1}=0$; it will be important later). If we take $\tau^{-1}=0$, then the solution of $(1)$ is well-known: the electron moves in a circular orbit, with angular frequency $$ \omega_c=\frac{eB}{m^*} \tag{2} $$

By measuring $\omega_c$ for different values of $B$ we can get a very precise measurement of $m^*$. But, the obvious question, how can we effectively measure $\omega_c$ in a laboratory? The answer is surprisingly easy, as we'll see in a moment.

If, in the situation above, we turn on a monochromatic source of light (say, a laser) with frequency $\omega$, there will be an electric field $\boldsymbol E\;\mathrm e^{-i\omega t}$, and the new equations of motion will be $$ m^*\dot{\boldsymbol v}=-e(\boldsymbol E(t)+\boldsymbol v\times \boldsymbol B) -\frac{m^*}{\tau}\boldsymbol v\tag{3} $$

By using the ansatz $\boldsymbol v(t)=\boldsymbol v_0\;\mathrm e^{-i\omega t}$, and solving for $\boldsymbol v_0$ (left as an exercise), you can easily check that in this case, $\boldsymbol v(t)$ will be proportional to $\boldsymbol E(t)$ (which should be more or less intuitive). For example, if we take $\boldsymbol B$ in the $z$ direction, then $\boldsymbol v$ is given by $$ \boldsymbol v_0=\frac{e}{m^*}\begin{pmatrix} i\omega-1/\tau&\omega_c&0\\-\omega_c&i\omega-1/\tau&0\\0&0&i\omega-1/\tau\end{pmatrix}^{-1}\boldsymbol E \tag{4} $$ where $^{-1}$ means matrix inverse.

This system will absorb energy from the source, so that the transmitted light will be less intense than the incoming light. The absorbed power is just $\text{Re}[\boldsymbol j\cdot\boldsymbol E]$, and as $\boldsymbol j\propto \boldsymbol v$, it's easy to check that $$ P\propto \text{Re}\left[\frac{1-i\omega \tau}{(1-i\omega\tau)^2+\omega_c^2\tau^2}\right]\propto \frac{1}{(1-\omega^2\tau^2+\omega_c^2\tau^2)^2+4\omega^2\tau^2} \tag{5} $$

Now, if we vary $\omega$, the power $P$ changes, and from $(5)$ we can see that there will be resonance when $(\omega^2+\omega_c^2)\tau^2=1$. In practice, $\omega\tau\ll 1$, so the resonant frequency is $\omega\approx\omega_c$:

$\hspace{100pt}$absorbed power

where the lines correspond to $\tau=0.1,\;0.5,\;1,\;3$, from green to blue. As you can see, for $\tau\to 0$, the resonance tends to $\omega_c$, so by measuring the resonant frequency we get the value of $\omega_c$, i.e., the value of $m^*$.


$^1$ the relaxation time $\tau$ is related to the mean free path: $\ell\sim v\tau$. Taking $\tau^{-1}\approx 0$ means that we assume the electron performs many cyclotron orbits before colliding with anything (ions, impurities,...).

$\endgroup$
  • $\begingroup$ Thanks for this answer, it is indeed of the kind I'm looking for. 1 or 2 questions if I may: i) Why are we allowed to treat the electron classically here? (as Eq. (3) is basically Newton's equation of motion). ii) Does the picture remain the same for holes? (except possibly for a charge sign change) iii) It would be great if you could include some of the physical insights involved, e.g. where does the ansatz of $v(t)$ come from? Thanks a bunch $\endgroup$ – user929304 Mar 3 '16 at 13:47
  • 1
    $\begingroup$ i) this is a very nice question, but I fear it is a question on itself (it deserves a new post): it takes several paragraphs to answer. Long story short: under certain approximations (semiclassical approximation), the mean effect of the lattice is to alter the dynamics of the electrons by changing their mass. This is the meaning of $m^*$: we can describe the motion of the electrons (intrinsically quantum) as if they were classical free particles, and all the quantum effects are reabsorbed into $m^*$. This is not always valid, but in general it's a good approximation, at least for a (1/2) $\endgroup$ – AccidentalFourierTransform Mar 3 '16 at 14:46
  • $\begingroup$ (2/2) qualitative description ii) yes but note that the effective mass of the holes is different from that of electrons iii) I fear there is no physical insight: its just mathematics. This ansatz is ubiquitous in physics: when you have a differential equation where the factors depend on $t$ throught exponentials, we always use the same ansatz. For example, when solving RLC circuits, or a driven harmonic oscillator, the wave equation, etc. It's just a mathematical trick that people found out centuries ago: if you are to solve any ODE, you can always try an exponential solution: it usually works $\endgroup$ – AccidentalFourierTransform Mar 3 '16 at 14:52
  • 1
    $\begingroup$ I'm glad I could help :) actually, I too am waiting for other contributions to the post, because I would like to learn other methods to measure $m^*$. It would be a shame if you don't get more answers, because this is a very interesting question! $\endgroup$ – AccidentalFourierTransform Mar 3 '16 at 15:26
  • 1
    $\begingroup$ Thanks for getting back to me on such short notice. Fair points. I need to ponder more on all this, I ll get in touch with you in case Im still lost in some ways :) $\endgroup$ – user929304 Apr 13 '16 at 15:57
6
$\begingroup$

Another method to measure the effective mass would be to measure the frequency dependent conductivity and the Hall resistance of a sample. Following Drude theory we can get an expression for the longitudinal conductivity $$\sigma(\omega)=\frac{\sigma_0}{1+i\omega\tau}$$ with $$\sigma_0=\frac{nq^2\tau}{m^*}$$ and $n$ is the density of electrons (holes) and $q$ is the charge per electron (hole). The density of electrons can be determined to great accuracy using a Hall resistance measurement $$R_H=\frac{1}{nq}$$ This leaves two unknowns, $\tau$ and $m^*$ which can be obtained by measuring and fitting $\sigma(\omega)$ to the Drude theory expression.

$\endgroup$
  • 2
    $\begingroup$ very interesting! It would be great if you could add a word or two about how the conductivity is measured in such cases (or maybe a relevant reference). $\endgroup$ – user929304 Mar 11 '16 at 15:43
2
$\begingroup$

While not very widely used, but very good for visualization is ARPES technique which directly maps the band structure (Exmpales :)). This might be very useful when one has band renormalization or collective effects, so that it is no longer straightforward how to define effective mass.

$\endgroup$

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.