Effective Mass

For derivations, response-specific masses, experimental extraction, and failure tests, see the canonical Effective Mass article.

Formula

For a band $E_n(\mathbf k)$ expanded near $\mathbf k_0$,

$$E_n(\mathbf k) \approx E_n(\mathbf k_0) + \frac{\hbar^2}{2} \sum_{ij} (m^{-1})_{ij} \Delta k_i\Delta k_j,$$

where $\Delta\mathbf k=\mathbf k-\mathbf k_0$ and

$$(m^{-1})_{ij} = \frac{1}{\hbar^2} \frac{\partial^2 E_n} {\partial k_i\,\partial k_j} \biggr\rvert_{\mathbf k_0}.$$

In one dimension,

$$m^\ast = \frac{\hbar^2} {d^2E/dk^2}.$$

The band velocity used in semiclassical wave-packet dynamics is

$$\mathbf v_n(\mathbf k) = \frac{1}{\hbar} \nabla_{\mathbf k}E_n(\mathbf k).$$

Assumptions

• A smooth isolated band is being approximated locally in $\mathbf k$.
• The expansion point and coordinate axes are specified.
• The effective-mass tensor describes curvature of a band dispersion, not the bare particle mass.
• The simple parabolic approximation is local and may fail far from the expansion point.

Validity

Effective mass is most useful near band extrema, where the linear term in $E_n(\mathbf k)$ vanishes. At a band minimum the effective mass tensor is positive in stable directions. At a band maximum, the curvature can be negative; this is one reason hole language is useful in semiconductors.

In anisotropic bands, the tensor form matters. Reducing it to one scalar is an additional approximation tied to a direction, symmetry, or averaged response.

Common Mistakes

• Treating $m^\ast$ as a universal material constant independent of band, direction, density, and energy.
• Forgetting that negative curvature gives negative electron effective mass in the electron-band description.
• Using the parabolic approximation through a van Hove singularity or band crossing.
• Confusing transport, cyclotron, optical, and density-of-states effective masses.

Quick Check

For a one-dimensional parabolic band $E(k)=E_0+\hbar^2k^2/(2m_0)$, what is $m^\ast$?

Solution

The second derivative is $d^2E/dk^2=\hbar^2/m_0$. Therefore

$$m^\ast = \frac{\hbar^2}{\hbar^2/m_0} = m_0.$$

Canonical Links

• Effective Mass derives the curvature tensor and distinguishes density-of-states, conductivity, cyclotron, optical, and quasiparticle masses.
• Tight-Binding Model
• Fermi Liquid Theory Preview for the quasiparticle mass, self-energy renormalization, backflow, and Galilean $F_1^s$ relation.
• Polarons Preview for the curvature mass of a mobile particle whose environmental cloud recoils and rearranges.
• Bloch Theorem
• Density of States
• Math Needed for Quantum Matter
• Condensed Matter Roadmap
• Effective Hamiltonians in Quantum Matter

References

• N. W. Ashcroft and N. D. Mermin, Solid State Physics, Holt, Rinehart and Winston, 1976.
• C. Kittel, Introduction to Solid State Physics, 8th ed., Wiley, 2004.
• M. P. Marder, Condensed Matter Physics, 2nd ed., Wiley, 2010.