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The shortest derivation of Landau quantization in the completely free electron gas starts with the minimal-coupling Hamiltonian $$ H = \frac{1}{2m} \left( \vec p + \frac{e}{c}\vec A \right) ^2 \equiv \frac{1}{2m} \vec P^2 $$ and due to $[P_x, P_y]=-i \hbar e B / c$ then proceeds in complete analogy to the derivation of a harmonic oscillator's spectrum: Introduce ladder operators $a \propto P_x + i P_y$ such that $[a^\dagger, a] = 1$, then express the Hamiltonian using these operators to arrive at $$ H = \hbar \omega_c \left( n + \frac{1}{2} \right) + \frac{\hbar^2 k_z^2}{2m} $$ with $n = a^\dagger a$ and $\omega_c = e B /m$.

What this derivation is missing, however, is that in a solid, you get the cyclotron effective mass

$$m_c = \frac{\hbar^2}{2\pi}\frac{\partial S(\epsilon, k_z)}{\partial \epsilon}$$

instead of the plain electron mass $m$, where $S$ is the area of a semiclassical cyclotron orbit.

Is there a way to amend it so the effective mass does show up, while retaining its simplicity?

Also, is there a relation between the cyclotron effective mass and the inertial effective mass known from the semiclassical equations of motion in a solid? I read that they are equal for parabolic bands, yet people seem to conflate them no matter the shape - e.g. when discussing Landau and Pauli susceptibilities $$\chi_{Landau} \propto \left(\frac{m}{m^*}\right)^2 \chi_{Pauli},$$ $m^*$ is usually just taken to be the ordinary, inertial effective mass.

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    $\begingroup$ What is $S(\epsilon, K_z)$? Where does this formula for the cyclotron effective mass $m_c$ come from? $\endgroup$
    – freecharly
    Oct 19 '16 at 17:09
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    $\begingroup$ @freecharly: My bad, I edited the question to make it clearer. In the lecture notes I'm reading, the formula comes from a semiclassical derivation of Landau levels using Bohr-Sommerfeld quantization. I much prefer the "proper" quantum mechanical derivation I outlined above, but I can't find a version of it that results in any kind of effective mass. Yet it clearly has to enter somehow, because the relationship between Landau and Pauli susceptibilities is well-known. $\endgroup$
    – smheidrich
    Oct 19 '16 at 18:12
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    $\begingroup$ Also, here is one example of a book that just asserts it has to be the inertial/band effective mass instead of cyclotron effective mass: "Quantum Theory of the Electron Liquid" by Giuliani and Vignale. They just put $m^*$ instead of $m$ into the Hamiltonian right from the beginning, but that seems a bit hand-wavey to me, because you only arrive at the band effective mass in the first place by making semiclassical approximations. $\endgroup$
    – smheidrich
    Oct 19 '16 at 18:19
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In solids, the electron dynamics is usually treated in the framework of the semiclassical equations of motion using the $E(\vec{k})$ energy-wave vector dispersion relations for Bloch wave solutions of the Schrödinger equation in the periodic potential of the crystal lattice. These equations of motion are $$\vec{F}=\frac{d\vec{p}}{dt}$$ where is the force on an electron and $\vec{p}=\hbar \vec{k}$ is the crystal momentum. The force $\vec{F}$ includes external electric and magnetic fields (Lorentz force). Thus the movement in k-space is fully determined, which also determines the energy of the electron according to the dispersion relation $E(\vec{k})$. If there is only a magnetic field, there is no energy change in the k-space movement and the electron will move in k-space on a closed path (not necessarily a circle) which lies on the intersection of an constant-energy surface with a plane normal to the magnetic field. For the movement of an electron in real space, the group velocity $\vec{v}=\frac{∂E/∂\vec{k}}{\hbar}$ of a superposition of Bloch waves has to be considered giving the expectation value of velocity, and the expectation value of acceleration $\vec{a}=M^{-1}\vec{F}$, where $M^{-1}$ is the tensor of the inverse effective mass. If this tensor is diagonal in the chosen cartesian coordinate system, the diagonal elements are $1/m_i=\frac{∂^2E}{\hbar^2∂k_i^2}$, $i=x,y,z$, where $m_i$ are the effective masses. These effective masses appear in different expressions for the so called conductivity effective mass, the density-of-states effective mass, the optical mass and also the cyclotron effective mass $m_c = \frac{\hbar^2}{2\pi}\frac{\partial S(\epsilon, k_z)}{\partial \epsilon}$, where S is the area enclosed by the path in k-space. This $m_c$ is an effective mass related to the tensor of inverse mass along the closed path in k-space.

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  • $\begingroup$ The last two sentences would have been enough. So, to summarize, you would say that $m_c$ is simply the band effective mass $m^*=m^*(\vec k)$ for movement along a cyclotron orbit? That would answer my second question if it's true, thanks. $\endgroup$
    – smheidrich
    Oct 20 '16 at 16:49
  • $\begingroup$ In the general, the inverse effective mass tensor can cary along the closed path in k-space. Only in the case of rotational symmetry along the B-axis in z-direction (path is a circle), you get a constant effective mass $m^*$ along the path corresponding to the x and y components of the inverse effective mass tensor $1/m_i=\frac{∂^2E}{\hbar^2∂k_i^2}$, i-x,y. $\endgroup$
    – freecharly
    Oct 20 '16 at 18:29
  • $\begingroup$ In the general, the inverse effective mass tensor will vary along the path. The expression for $m_c$ then gives an effective weighted average derived from this tensor along the path. It is instructive to do this e.g. for ellipsoid equal energy surfaces found in silicon and germanium. Then the path is, in general an ellipse. In metals, the k-space path at the Fermi level can be rather complicated. $\endgroup$
    – freecharly
    Oct 20 '16 at 18:29
  • $\begingroup$ That makes a lot of sense, thank you! Do you know any books where Landau quantization in solids is explained to this level of detail? $\endgroup$
    – smheidrich
    Oct 20 '16 at 19:25
  • $\begingroup$ I just saw that there is a short description on page 242 under the title Experimental Methods in Fermi Surface Studies, in Charles Kittel, Introduction to Solid State Physics, 8th edition. I am not sure whether this is sufficient. $\endgroup$
    – freecharly
    Oct 20 '16 at 19:57

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