Effective Mass from Band Curvature

Hello! Welcome to the first lesson of our second module, "Semiconductor Physics and Transport."

In our previous module, we established a robust picture of the electronic band structure of crystals. We concluded by exploring how the alignment of band edges in momentum space gives rise to direct and indirect bandgaps, which in turn dictates the optical properties of a semiconductor. You'll recall that transitions on an $E(\mathbf{k})$ diagram are vertical for light absorption/emission.

Now, we shift our focus from these vertical, inter-band transitions to the motion of charge carriers within a single band. How does an electron in the conduction band or a hole in the valence band respond to an external electric or magnetic field? The full quantum mechanical treatment is complex, as the particle is constantly interacting with the periodic potential of the crystal lattice.

Today's lesson introduces a remarkably powerful simplification: the effective mass approximation. We will derive how the intricate effects of the crystal potential can be bundled into a single parameter, $m^{\ast }$, which replaces the free electron mass in our equations of motion. This allows us to treat electrons and holes as quasi-classical particles, paving the way for understanding electrical transport. We will see that this effective mass is determined entirely by the curvature of the energy bands, $E(\mathbf{k})$.

1. The Semiclassical Model of Electron Dynamics

To understand how an electron moves, we consider it not as a simple plane wave, but as a wave packet constructed from Bloch states near a specific crystal momentum $\mathbf{k}$. The velocity of this electron is the group velocity of the wave packet. Given your background in quantum mechanics, you'll know this is given by:

$${\mathbf{v}}_g=\frac{1}{\hslash }{\nabla }_{\mathbf{k}}E(\mathbf{k})$$

This equation immediately tells us something profound: the electron's velocity is determined by the slope of the $E(\mathbf{k})$ dispersion. An electron in a flat band (${\nabla }_{\mathbf{k}}E=0$) has zero velocity and is localized, a concept that aligns with what we saw in the tight-binding model for small hopping parameters.

Now, let's apply an external force $\mathbf{F}$ (e.g., from an electric field, $\mathbf{F}=-e\mathbf{E}$). The work done on the electron over a time $dt$ is $dE=\mathbf{F}\cdot {\mathbf{v}}_{g}dt$. We can also express the change in energy in terms of the change in its wavevector, $d\mathbf{k}$: $dE={\nabla }_{\mathbf{k}}E(\mathbf{k})\cdot d\mathbf{k}$.

Equating these two expressions for $dE$:

$$\mathbf{F}\cdot {\mathbf{v}}_{g}dt={\nabla }_{\mathbf{k}}E(\mathbf{k})\cdot d\mathbf{k}$$

Substituting ${\mathbf{v}}_g=\frac{1}{\hslash }{\nabla }_{\mathbf{k}}E(\mathbf{k})$, we get:

$$\frac{1}{\hslash }\mathbf{F}\cdot ({\nabla }_{\mathbf{k}}E)dt=({\nabla }_{\mathbf{k}}E)\cdot d\mathbf{k}$$

This implies a fundamental relationship for the evolution of the crystal momentum under an external force:

$$\mathbf{F}=\hslash \frac{d\mathbf{k}}{dt}$$

This is a cornerstone of the semiclassical model. It states that the external force changes the crystal momentum $\hslash \mathbf{k}$, not the mechanical momentum $m\mathbf{v}$ as in free space. The internal forces from the crystal lattice are already accounted for in the band structure $E(\mathbf{k})$.

2. Derivation of the Effective Mass

With the semiclassical equation of motion in hand, we can now derive the effective mass by calculating the electron's acceleration $\mathbf{a}=d{\mathbf{v}}_g/dt$ and comparing it to Newton's second law, $\mathbf{F}=m\mathbf{a}$.

Let's start with the 1D case for simplicity:

$$a=\frac{d{v}_g}{dt}=\frac{d}{dt}\left(\frac{1}{\hslash }\frac{dE}{dk}\right)$$

Using the chain rule, since $E$ is a function of $k$, and $k$ is a function of $t$:

$$a=\frac{1}{\hslash }\frac{d^{2}E}{d{k}^2}\frac{dk}{dt}$$

Now, we substitute our result from the previous section, $dk/dt=F/\hslash$:

$$a=\left(\frac{1}{{\hslash }^2}\frac{d^{2}E}{d{k}^2}\right)F$$

Comparing this with Newton's law, $a=\frac{1}{m}F$, we can immediately identify the term in the parenthesis as the inverse of a mass. This is the inverse effective mass, $1/m^{\ast }$:

$$\frac{1}{m^{\ast }}=\frac{1}{{\hslash }^2}\frac{d^{2}E}{d{k}^2}⟹m^{\ast }={\hslash }^2{\left(\frac{d^{2}E}{d{k}^2}\right)}^{-1}$$

This is the central result of our lesson. The effective mass is inversely proportional to the curvature of the $E(k)$ band.

To see this derivation in action and build some intuition, the following video provides a clear explanation.

3. Interpreting the Effective Mass

The formula $m^{\ast }={\hslash }^2/(d^{2}E/d{k}^2)$ has profound physical implications.

Curvature and Inertia

• High Curvature (Sharp Band): A large value of $d^{2}E/d{k}^2$ implies a small, positive $m^{\ast }$. The electron has low inertia and is easily accelerated by an external field. This corresponds to mobile electrons in wide bands.
• Low Curvature (Flat Band): A small value of $d^{2}E/d{k}^2$ implies a very large $m^{\ast }$. The electron is "heavy" and difficult to accelerate. This corresponds to nearly localized electrons in narrow bands, as we saw in the tight-binding model.

Sign, Electrons, and Holes

The sign of the curvature is also critically important.

• Near the Conduction Band Minimum (CBM): The band is parabolic and concave up, so $d^{2}E/d{k}^2>0$. This means $m^{\ast }>0$. An electron here behaves as expected, accelerating in the direction opposite to an electric field.

• Near the Valence Band Maximum (VBM): The band is concave down, so $d^{2}E/d{k}^2<0$. This yields a negative effective mass, $m^{\ast }<0$. An electron at the top of the valence band responds to a force as if its mass were negative—it accelerates in the wrong direction!

This strange behavior is the origin of holes. The collective motion of all electrons in a nearly full valence band with one missing electron (at state ${\mathbf{k}}_e$) is equivalent to the motion of a single particle with:

• Charge: $+e$ (opposite to the missing electron)
• Crystal momentum: ${\mathbf{k}}_h=-{\mathbf{k}}_e$
• Effective mass: $m_h^{\ast }=-m_e^{\ast }$. Since $m_e^{\ast }$ was negative near the VBM, the hole's effective mass $m_h^{\ast }$ is positive.

Thus, the seemingly paradoxical negative mass of an electron is elegantly replaced by the much more intuitive concept of a positively charged hole with a positive mass.

4. Generalization to 3D: The Effective Mass Tensor

In a real 3D crystal, the band structure $E(\mathbf{k})$ is not necessarily isotropic. The curvature can be different in different crystallographic directions. This means an electron's inertia can depend on the direction it is accelerated.

To capture this, we generalize the scalar effective mass to a tensor. The acceleration $\mathbf{a}$ is related to the force $\mathbf{F}$ by the inverse effective mass tensor:

$$a_i=\sum _{j=x,y,z}{\left(\frac{1}{m^{\ast }}\right)}_{ij}{F}_j$$

where the components of the tensor are defined by the second partial derivatives of the energy dispersion:

$${\left(\frac{1}{m^{\ast }}\right)}_{ij}\equiv \frac{1}{{\hslash }^2}\frac{{\partial }^{2}E(\mathbf{k})}{\partial k_i\partial k_j}$$

This tensor is symmetric. By diagonalizing it, we can find the principal axes and corresponding principal effective masses. For example, in silicon, the conduction band minima are ellipsoidal "valleys," characterized by a longitudinal mass ($m_l^{\ast }$) and a transverse mass ($m_t^{\ast }$).

Test your understanding!

The figure below shows the conduction band minima for three different semiconductors (A, B, C) along a certain direction in k-space.

1. Rank the effective masses of the electrons ($m_A^{\ast },m_B^{\ast },m_C^{\ast }$) in these three bands from smallest to largest.
2. Which of these materials would you expect to have the highest electron mobility, assuming other scattering factors are equal? Why?

Show answer

1. The effective mass is inversely proportional to the curvature. Band B has the highest curvature (sharpest parabola), while Band A has the lowest curvature (flattest parabola). Therefore, the ranking of effective masses is: $$m_B^{\ast }<m_C^{\ast }<m_A^{\ast }$$
2. Material B would be expected to have the highest electron mobility. Mobility is inversely proportional to effective mass. A smaller effective mass means the electron has less inertia and can be accelerated more easily by an electric field, leading to a higher drift velocity for a given field strength, and thus higher mobility.

5. Theoretical Origin: k·p Perturbation Theory

We've established that effective mass comes from the parabolic approximation of bands near an extremum. But why are the bands parabolic in the first place? A powerful method called k·p perturbation theory provides the answer.

Given your familiarity with quantum mechanical perturbation theory from courses like QFT, you can appreciate the approach. The method treats the $\mathbf{k}$-dependent part of the Hamiltonian as a perturbation around a high-symmetry point (like $\mathbf{k}=0$, the $\Gamma$ point).

The Hamiltonian for a Bloch function $u_{n\mathbf{k}}$ is:

$$\left[H_0+\frac{\hslash }{m_0}\mathbf{k}\cdot \mathbf{p}+\frac{{\hslash }^2{k}^2}{2m_0}\right]u_{n\mathbf{k}}=E_n^{\prime }(\mathbf{k})u_{n\mathbf{k}}$$

Here, $H_0$ is the Hamiltonian at $\mathbf{k}=0$, and the $\mathbf{k}\cdot \mathbf{p}$ term is the perturbation. Applying second-order non-degenerate perturbation theory, the energy $E_n(\mathbf{k})$ for band $n$ near $\mathbf{k}=0$ is found to be:

$$E_n(\mathbf{k})\approx E_n(0)+\frac{{\hslash }^2{k}^2}{2m_0}+\frac{{\hslash }^2}{m_0^2}\sum _{n^{\prime }\ne n}\frac{|⟨u_{n0}|\mathbf{k}\cdot \mathbf{p}|u_{n^{\prime }0}⟩{|}^2}{E_n(0)-E_{n^{\prime }}(0)}$$

This expression is quadratic in $k$, which mathematically justifies the parabolic band approximation. By rearranging this into the form $E_n(\mathbf{k})\approx E_n(0)+\frac{{\hslash }^2{k}^2}{2m^{\ast }}$, we can derive an expression for the effective mass. The key insight is that the effective mass $m^{\ast }$ is not equal to the free electron mass $m_0$ because of the summation term, which represents the "pushing" interaction between band $n$ and all other bands $n^{\prime }$. Bands that are close in energy (small denominator $E_n-E_{n^{\prime }}$) and have a strong momentum matrix element coupling them contribute the most to modifying the effective mass.

Conclusion

In this lesson, we have developed one of the most important concepts in semiconductor physics. By analyzing the motion of an electron wave packet under an external force, we have shown that the complex interaction with the periodic lattice can be effectively encapsulated in a single quantity: the effective mass, $m^{\ast }$.

Key Takeaways:

• An electron's velocity in a crystal is the group velocity, ${\mathbf{v}}_g=\frac{1}{\hslash }{\nabla }_{\mathbf{k}}E$, determined by the slope of the band.
• An external force $\mathbf{F}$ changes the crystal momentum according to the semiclassical equation $\mathbf{F}=\hslash d\mathbf{k}/dt$.
• The effective mass is derived by comparing the electron's acceleration to Newton's second law, yielding $m^{\ast }={\hslash }^2/(d^{2}E/d{k}^2)$. It is inversely proportional to the band's curvature.
• The sign of the curvature determines the sign of $m^{\ast }$. A negative $m^{\ast }$ for electrons at the top of the valence band motivates the concept of holes as positively charged particles with positive effective mass.
• In 3D, this generalizes to the effective mass tensor, which accounts for anisotropy in the band structure.

Preview of Next Lesson:
The effective mass is not just a theoretical curiosity; it is a critical input parameter for calculating nearly all transport properties. In our next lesson, we will immediately put it to use. We will see how $m^{\ast }$ enters into the expression for the density of states, which we will then use to calculate the intrinsic and extrinsic carrier concentrations in a semiconductor as a function of temperature and doping.