Effective Mass and Semiclassical Electron Dynamics

Derivation

Effective Mass and Semiclassical Electron Dynamics

D-252 Home PU-303 Threads energy · waves · fields Depends on Bloch's Theorem from Translational Symmetry, Tight-Binding Band Dispersion

Statement

For a wavepacket built from Bloch states of a single non-degenerate band \(n\), the centre of the packet obeys the semiclassical equations of motion \(\mathbf{v}_n(\mathbf{k}) = \frac{1}{\hbar}\nabla_{\mathbf{k}} E_n(\mathbf{k})\) and \(\hbar\,\dot{\mathbf{k}} = \mathbf{F}_{\text{ext}}\); differentiating the velocity gives an acceleration \(\dot v_i = \sum_j (M^{-1})_{ij} F_j\) whose inverse-mass tensor is the band curvature, \((M^{-1})_{ij} = \frac{1}{\hbar^2}\frac{\partial^2 E_n}{\partial k_i \partial k_j}\).

Why it matters

The effective mass is the single number that lets a crystal electron be treated as a free particle in a modified Newton's law. Sign, magnitude and anisotropy of \(M^{-1}\) fix the sign of Hall coefficients, the shape of cyclotron orbits, carrier mobilities, and the very distinction between electrons and holes — all read directly off the band structure without re-solving the many-body problem.

It is also the bridge between the microscopic Bloch picture and the transport equations of device physics: mobility, effective density of states, and the semiconductor rate equations all inherit \(M\) from \(E_n(\mathbf{k})\).

Assumptions

Single band, no interband transitions.If external fields are strong or the gap is small, Zener tunnelling mixes bands and the one-band \(E_n(\mathbf{k})\) is no longer the relevant energy — \(\mathbf{k}\) still evolves but the velocity formula must sum over bands.

Wavepacket narrow in \(\mathbf{k}\), broad in \(\mathbf{r}\).If the packet is not localised in \(\mathbf{k}\) compared with the Brillouin-zone scale, \(E_n(\mathbf{k})\) and its gradient cannot be replaced by their values at the packet centre and the group-velocity identity fails.

External fields vary slowly on the lattice scale and slowly on the scale of \(\hbar/E_{\text{gap}}\).If the potential varies over a unit cell, the perturbation cannot be pushed outside the periodic problem; the crystal Hamiltonian must be re-diagonalised and no clean force law survives.

Berry curvature neglected.In crystals lacking inversion or time-reversal symmetry the velocity acquires an anomalous term \(-\dot{\mathbf{k}}\times\boldsymbol{\Omega}(\mathbf{k})\); dropping it misses the intrinsic anomalous Hall effect.

Derivation

\[ \psi(\mathbf{r},t) = \int d^3k\; g(\mathbf{k})\, u_{n\mathbf{k}}(\mathbf{r})\, e^{\,i(\mathbf{k}\cdot\mathbf{r} - E_n(\mathbf{k})t/\hbar)} \]

Superpose Bloch states of one band with envelope \(g(\mathbf{k})\) sharply peaked at \(\mathbf{k}_0\); each factor carries phase \(\omega(\mathbf{k})t\) with \(\omega = E_n/\hbar\) (from Bloch's theorem). A

\[ \mathbf{v}_n = \left.\nabla_{\mathbf{k}}\,\omega(\mathbf{k})\right|_{\mathbf{k}_0} = \frac{1}{\hbar}\nabla_{\mathbf{k}} E_n(\mathbf{k}) \]

The centre of a wavepacket moves at the group velocity \(\nabla_{\mathbf{k}}\omega\); substitute \(\omega = E_n/\hbar\). This is the first semiclassical equation. A

\[ \frac{dE_n}{dt} = \mathbf{F}_{\text{ext}}\cdot\mathbf{v}_n \]

Work–energy theorem for the packet: the power delivered by the external force equals the rate of change of the band energy the electron carries. B

\[ \frac{dE_n}{dt} = \nabla_{\mathbf{k}} E_n \cdot \dot{\mathbf{k}} = \hbar\,\mathbf{v}_n\cdot\dot{\mathbf{k}} \]

Chain rule on \(E_n(\mathbf{k}(t))\), then insert \(\nabla_{\mathbf{k}}E_n = \hbar\mathbf{v}_n\) from step 2. A

\[ \mathbf{v}_n\cdot\left(\mathbf{F}_{\text{ext}} - \hbar\,\dot{\mathbf{k}}\right) = 0 \quad\Rightarrow\quad \hbar\,\dot{\mathbf{k}} = \mathbf{F}_{\text{ext}} \]

Equate steps 3 and 4. Because the applied force (e.g. \(-e\mathbf{E}\)) can point in any direction independent of \(\mathbf{v}_n\), the bracket must vanish identically, giving the crystal-momentum equation. C

\[ \dot v_{n,i} = \frac{d}{dt}\!\left(\frac{1}{\hbar}\frac{\partial E_n}{\partial k_i}\right) = \frac{1}{\hbar}\sum_j \frac{\partial^2 E_n}{\partial k_i\,\partial k_j}\,\dot k_j \]

Differentiate the velocity (step 2) in time; \(E_n\) depends on \(t\) only through \(\mathbf{k}(t)\), so apply the chain rule component-wise. A

\[ \dot v_{n,i} = \frac{1}{\hbar^2}\sum_j \frac{\partial^2 E_n}{\partial k_i\,\partial k_j}\,F_{\text{ext},j} \]

Insert \(\hbar\dot k_j = F_{\text{ext},j}\) from step 5. A

\[ \dot v_{n,i} = \sum_j (M^{-1})_{ij}\,F_{\text{ext},j}, \qquad (M^{-1})_{ij} \equiv \frac{1}{\hbar^2}\frac{\partial^2 E_n}{\partial k_i\,\partial k_j} \]

Match to the free-particle form \(\dot v_i = \sum_j (M^{-1})_{ij}F_j\); the coefficient is by definition the inverse effective-mass tensor. B

Result

\[ \mathbf{v}_n=\frac{1}{\hbar}\nabla_{\mathbf{k}}E_n,\qquad \hbar\dot{\mathbf{k}}=\mathbf{F}_{\text{ext}},\qquad (M^{-1})_{ij}=\frac{1}{\hbar^2}\frac{\partial^2 E_n}{\partial k_i\,\partial k_j} \]

Reading. A Bloch electron drifts at the band's group velocity and responds to a force by having its crystal momentum \(\hbar\mathbf{k}\) pushed at exactly the free-particle rate; the entire influence of the lattice is compressed into the curvature of \(E_n(\mathbf{k})\). Where the band curves up (band bottom) the mass is positive; where it curves down (band top) the mass is negative — the origin of holes. Off-diagonal curvature makes the mass a tensor, so acceleration need not be parallel to force.

Units check. \([\partial^2 E/\partial k^2] = \text{J}\,\text{m}^2\) and \([\hbar^2]=\text{J}^2\text{s}^2\), so \([M^{-1}] = \text{J}\,\text{m}^2/(\text{J}^2\text{s}^2)=\text{m}^2\,\text{J}^{-1}\text{s}^{-2}\). With \(\text{J}=\text{kg}\,\text{m}^2\text{s}^{-2}\) this is \(\text{kg}^{-1}\) — an inverse mass, as required. Likewise \([\nabla_k E/\hbar]=\text{J}\,\text{m}/(\text{J}\,\text{s})=\text{m}\,\text{s}^{-1}\), a velocity.

Limiting cases

Free electron: \(E=\hbar^2 k^2/2m\) gives \((M^{-1})_{ij}=\delta_{ij}/m\), recovering ordinary Newtonian mass \(M=m\,\mathbb{1}\).
Parabolic band edge: near a non-degenerate extremum \(E\approx E_0+\tfrac{\hbar^2}{2}\sum_{ij}(M^{-1})_{ij}(k-k_0)_i(k-k_0)_j\), a constant tensor — the standard "effective-mass approximation."
Isotropic minimum: curvature the same in all directions gives a scalar \(m^*\); useful for the \(\Gamma\)-point conduction band of many III–V semiconductors.
Band top: curvature negative, \(m^*<0\); recast as a positive-mass hole with charge \(+e\).
Flat band: \(\partial^2 E/\partial k^2\to 0\) gives \(m^*\to\infty\) — carriers do not accelerate, the localisation limit.

Breaks when

Strong fields / small gaps: when \(eEa \gtrsim E_{\text{gap}}^2/E_F\) (Zener criterion) interband tunnelling promotes the electron to a neighbouring band and the single-band velocity and mass lose meaning.
At band degeneracies or crossings: \(\partial^2 E/\partial k_i\partial k_j\) diverges or is ill-defined (e.g. the linear Dirac cones of graphene, where \(E\propto|k|\) gives infinite curvature at the node); mass must be replaced by a velocity, and multiband \(\mathbf{k}\cdot\mathbf{p}\) treatment is required.
Rapid time dependence: fields oscillating faster than \(E_{\text{gap}}/\hbar\) drive real interband transitions, invalidating the adiabatic single-band assumption behind step 5.
Broken symmetry with Berry curvature: the velocity gains an anomalous \(-\dot{\mathbf{k}}\times\boldsymbol\Omega\) term, so \(\mathbf{v}=\hbar^{-1}\nabla_k E\) is incomplete even when the mass tensor is finite.

Failure modes

Writing \(m^* = \hbar^2/(\partial^2 E/\partial k^2)\) with a stray sign — dropping the reciprocal relationship and forgetting that \(M^{-1}\), not \(M\), is the curvature, so anisotropic bands get inverted wrongly.
Treating \(\hbar\mathbf{k}\) as true mechanical momentum \(m\mathbf{v}\); crystal momentum is conserved only modulo a reciprocal-lattice vector and is not \(m\mathbf{v}\).
Using \(\mathbf{F}=-e\mathbf{E}\) but forgetting the magnetic part: the full force is \(-e(\mathbf{E}+\mathbf{v}\times\mathbf{B})\) with \(\mathbf{v}=\hbar^{-1}\nabla_k E\), which itself depends on \(\mathbf{k}\).
Evaluating the curvature at the wrong \(\mathbf{k}\): using the zone-centre mass for a carrier sitting near a zone-boundary saddle point gives the wrong sign entirely.
Assuming acceleration is parallel to force in an anisotropic crystal; the off-diagonal tensor elements rotate the acceleration away from \(\mathbf{F}\).
Confusing the density-of-states effective mass, the conductivity effective mass, and the band-curvature mass — they coincide only for an isotropic parabolic band.

Discussion

The deepest content of the derivation is that the lattice does not need to be re-solved every time a field is applied: all of its dynamical influence is packaged into a single function \(E_n(\mathbf{k})\) and its derivatives. The first derivative gives the drift velocity, the second gives the response to force. This is why band structure is the master datum of solid-state physics — once you have \(E_n(\mathbf{k})\), transport follows by differentiation, not by fresh quantum mechanics.

The sign of the curvature carries as much physics as its magnitude. Near a band maximum the mass is negative, meaning the electron accelerates opposite to the applied force. Rather than keep track of negatively-massed electrons in a nearly-full band, one counts the few empty states as positive-mass, positive-charge holes. The effective-mass tensor is thus the formal origin of the electron/hole duality that underpins every semiconductor device.

The clean force law \(\hbar\dot{\mathbf{k}}=\mathbf{F}_{\text{ext}}\) is an adiabatic statement: it holds precisely because the wavepacket stays in band \(n\). Modern treatments derive both equations from a wavepacket Lagrangian and find a companion term the naive argument misses — the Berry curvature \(\boldsymbol{\Omega}_n(\mathbf{k})=i\langle\nabla_{\mathbf{k}}u_{n\mathbf{k}}|\times|\nabla_{\mathbf{k}}u_{n\mathbf{k}}\rangle\), which adds an anomalous velocity \(-\dot{\mathbf{k}}\times\boldsymbol{\Omega}_n\). Our derivation is the \(\boldsymbol{\Omega}=0\) limit, exact when inversion and time-reversal symmetry coexist; the effective-mass tensor is the "diagonal" band-geometric response while the Berry curvature is the "off-diagonal" one.

Common misconceptions. The effective mass is not a property of the electron but of the band at a chosen \(\mathbf{k}\); the same electron has different masses at different points of the same band. It is not bounded by the free-electron mass — narrow bands give masses hundreds of times \(m_e\) (heavy fermions), and steep bands give masses well below \(m_e\) (e.g. \(0.067\,m_e\) in GaAs). And crystal momentum \(\hbar\mathbf{k}\) is not the electron's real momentum: the lattice can absorb momentum in units of \(\hbar\mathbf{G}\).

Worked examples

Effective mass at the bottom of a 1D tight-binding band. Take \(E(k)=E_0-2t\cos(ka)\) with hopping \(t=1.0\ \text{eV}\) and lattice constant \(a=3.0\ \text{Å}\). Find \(m^*\) at \(k=0\). B

\[ \frac{\partial^2 E}{\partial k^2} = 2t a^2\cos(ka)\Big|_{k=0}=2t a^2,\qquad m^*=\frac{\hbar^2}{2t a^2} \]

Differentiate the dispersion twice; invert the curvature per the boxed result. Symbols first. A

\[ m^*=\frac{(1.055\times10^{-34})^2}{2(1.602\times10^{-19})(3.0\times10^{-10})^2}=\frac{1.113\times10^{-68}}{2.884\times10^{-38}} \]

Insert \(\hbar=1.055\times10^{-34}\ \text{J s}\), \(t=1.602\times10^{-19}\ \text{J}\), \(a^2=9.0\times10^{-20}\ \text{m}^2\). A

\[ m^*=3.86\times10^{-31}\ \text{kg}\approx 0.42\,m_e \]

Reading. A wider band (larger \(t\)) or a longer lattice constant gives a lighter, more mobile carrier; here the mass is under half the free value. At the band top \(k=\pi/a\) the same formula gives \(-0.42\,m_e\).

Units check. \((\text{J s})^2/(\text{J}\cdot\text{m}^2)=\text{J}\,\text{s}^2\,\text{m}^{-2}=\text{kg}\). Correct.

Semiclassical acceleration and mean free path in a field. An electron with the mass above, \(m^*=0.42\,m_e\), sits at the isotropic \(\Gamma\)-point minimum of a cubic crystal. A uniform field \(E=1.0\times10^{4}\ \text{V m}^{-1}\) is applied. Find its acceleration, and the velocity it reaches in a scattering time \(\tau=0.20\ \text{ps}\). B

\[ \mathbf{F}=-e\mathbf{E},\qquad a=\frac{eE}{m^*},\qquad v=a\tau=\frac{eE\tau}{m^*} \]

Near an isotropic minimum \(M^{-1}=\mathbb{1}/m^*\), so \(\dot{\mathbf v}=\mathbf F/m^*\); integrate over one scattering time for the drift velocity. A

\[ a=\frac{(1.602\times10^{-19})(1.0\times10^{4})}{3.86\times10^{-31}}=4.15\times10^{15}\ \text{m s}^{-2} \]

Insert numbers with \(e=1.602\times10^{-19}\ \text{C}\). A

\[ v=a\tau=(4.15\times10^{15})(0.20\times10^{-12})=8.3\times10^{2}\ \text{m s}^{-1} \]

Multiply by \(\tau=2.0\times10^{-13}\ \text{s}\); this is the drift velocity, cut off by scattering long before Bloch oscillation. B

\[ a=4.2\times10^{15}\ \text{m s}^{-2},\qquad v_{\text{drift}}\approx 8.3\times10^{2}\ \text{m s}^{-1} \]

Reading. The instantaneous acceleration is enormous, but scattering resets \(\mathbf{k}\) after \(\tau\), so the steady drift velocity stays modest. The mobility \(\mu=e\tau/m^*=v/E\approx0.083\ \text{m}^2\text{V}^{-1}\text{s}^{-1}\) follows immediately.

Units check. \(a\): \((\text{C}\cdot\text{V m}^{-1})/\text{kg}=(\text{J m}^{-1})/\text{kg}=\text{m s}^{-2}\). \(v=a\tau\): \(\text{m s}^{-2}\cdot\text{s}=\text{m s}^{-1}\). Correct.

Problems

For the free-electron band \(E=\hbar^2 k^2/2m\), verify explicitly that \((M^{-1})_{ij}=\delta_{ij}/m\) and hence \(M=m\,\mathbb{1}\).
Solution
\(\partial E/\partial k_i=\hbar^2 k_i/m\); \(\partial^2 E/\partial k_i\partial k_j=\hbar^2\delta_{ij}/m\). Then \((M^{-1})_{ij}=\hbar^{-2}\cdot\hbar^2\delta_{ij}/m=\delta_{ij}/m\). Inverting the tensor (a multiple of the identity) gives \(M_{ij}=m\,\delta_{ij}\). The effective mass reduces to the bare mass, as it must with no lattice.
A 1D band is \(E(k)=E_0-2t\cos(ka)\) with \(t=0.50\ \text{eV}\), \(a=4.0\ \text{Å}\). Find \(m^*\) at the band bottom \(k=0\) and at the top \(k=\pi/a\).
Solution
\(\partial^2 E/\partial k^2=2ta^2\cos(ka)\). At \(k=0\): \(m^*=\hbar^2/(2ta^2)\). Numbers: \(2ta^2=2(0.5\times1.602\times10^{-19})(4.0\times10^{-10})^2=2(8.01\times10^{-20})(1.6\times10^{-19})=2.56\times10^{-38}\). \(m^*=1.113\times10^{-68}/2.56\times10^{-38}=4.35\times10^{-31}\ \text{kg}=0.48\,m_e\). At \(k=\pi/a\), \(\cos=-1\), so \(m^*=-0.48\,m_e\) (a hole of \(+0.48\,m_e\)).
An electron sits at an anisotropic minimum with \(E=E_0+\tfrac{\hbar^2}{2}\big(k_x^2/m_x+k_y^2/m_y\big)\), \(m_x=0.10\,m_e\), \(m_y=0.90\,m_e\). A force \(\mathbf{F}=F\hat{\mathbf{x}}\!\cos45^\circ+F\hat{\mathbf{y}}\!\sin45^\circ\) is applied. Is the acceleration parallel to \(\mathbf{F}\)?
Solution
\(M^{-1}=\text{diag}(1/m_x,1/m_y)\). \(\dot v_x=F\cos45^\circ/m_x\), \(\dot v_y=F\sin45^\circ/m_y\). Ratio \(\dot v_y/\dot v_x=(m_x/m_y)\tan45^\circ=0.10/0.90=0.111\), so the acceleration makes angle \(\arctan(0.111)=6.3^\circ\) with \(\hat{\mathbf{x}}\), not \(45^\circ\). It is pulled toward the light-mass (\(x\)) direction and is not parallel to \(\mathbf{F}\).
Estimate the Zener/Bloch-oscillation period for the band of Problem 2 (\(a=4.0\ \text{Å}\)) in a field \(E=5.0\times10^{5}\ \text{V m}^{-1}\), and comment on whether it is observable given \(\tau=0.1\ \text{ps}\).
Solution
\(\hbar\dot k=-eE\) so \(k(t)=-eEt/\hbar\); one Bloch period sweeps \(\Delta k=2\pi/a\), giving \(T_B=\hbar\cdot2\pi/(a\,eE)=h/(aeE)\). Numbers: \(T_B=6.626\times10^{-34}/[(4.0\times10^{-10})(1.602\times10^{-19})(5.0\times10^{5})]=6.626\times10^{-34}/3.20\times10^{-23}=2.07\times10^{-11}\ \text{s}=21\ \text{ps}\). Since \(T_B\gg\tau=0.1\ \text{ps}\), scattering destroys coherence roughly 200 times per period; Bloch oscillations are not observable in this bulk crystal — they need superlattices (large \(a\)) or ultracold-atom lattices to lengthen \(T_B\) or \(\tau\).
The conduction band of GaAs near \(\Gamma\) is isotropic parabolic with \(m^*=0.067\,m_e\). A magnetic field \(B=1.0\ \text{T}\) is applied. Find the cyclotron frequency \(\omega_c=eB/m^*\) and the cyclotron energy \(\hbar\omega_c\) in meV.
Solution
\(\omega_c=eB/m^*=(1.602\times10^{-19})(1.0)/(0.067\times9.11\times10^{-31})=(1.602\times10^{-19})/(6.10\times10^{-32})=2.63\times10^{12}\ \text{rad s}^{-1}\). \(\hbar\omega_c=(1.055\times10^{-34})(2.63\times10^{12})=2.77\times10^{-22}\ \text{J}=1.73\ \text{meV}\). The small effective mass makes the cyclotron energy about 15 times larger than for a free electron at the same field — a direct, measurable fingerprint of the band curvature via cyclotron resonance.