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Infinite effective mass and its E-k explanation

Short answer

Infinite effective mass occurs where d²E/dk²=0; external force gives zero acceleration. Effective mass m* = ħ²/(d²E/dk²). In an E-k diagram, lower band has m*>0, inflection point m*=∞, upper band m*<0.

What do you understand by infinite effective mass? Derive the expression for effective mass of electron and show that it can be positive as well as negative, explain with the help of E-k diagram.

Effective mass m∗=ℏ2/(d2E/dk2)m^* = \hbar^2/(d^2E/dk^2) is the mass an electron in a crystal appears to have under an external force. Infinite effective mass means d2E/dk2=0d^2E/dk^2 = 0 (the inflection point of the E–k curve): there the external force produces no acceleration at all, because the force from the lattice exactly balances it.

Derivation of m∗m^*

The electron is a wave packet moving with group velocity (E=ℏωE = \hbar\omega):

vg=dωdk=1ℏdEdkv_g = \frac{d\omega}{dk} = \frac{1}{\hbar}\frac{dE}{dk}

An external force FF acting for time dtdt does work dE=Fvg dtdE = F v_g\,dt:

dEdkdk=F⋅1ℏdEdkdt⇒F=ℏdkdt\frac{dE}{dk}dk = F\cdot\frac{1}{\hbar}\frac{dE}{dk}dt \quad\Rightarrow\quad F = \hbar\frac{dk}{dt}

Acceleration:

a=dvgdt=1ℏd2Edk2dkdt=1ℏ2d2Edk2F\begin{aligned} a &= \frac{dv_g}{dt} = \frac{1}{\hbar}\frac{d^2E}{dk^2}\frac{dk}{dt} = \frac{1}{\hbar^2}\frac{d^2E}{dk^2}F \end{aligned}

Comparing with F=m∗aF = m^*a:

m∗=ℏ2d2E/dk2m^* = \frac{\hbar^2}{d^2E/dk^2}

Positive, negative and infinite m∗m^* from the E–k diagram

In a crystal the band has a cosine-like shape. Taking E=E0−2γcos⁡kaE = E_0 - 2\gamma\cos ka,

dEdk=2γasin⁡ka,d2Edk2=2γa2cos⁡ka\frac{dE}{dk} = 2\gamma a\sin ka, \qquad \frac{d^2E}{dk^2} = 2\gamma a^2\cos ka
   E                  (a) E vs k
   ^  __                              __
   |    \  m*<0                m*<0  /
   |     \                          /
   |      * <- inflection,  ->     *
   |       \   m* = infinite      /
   |        \                    /
   |         `--.__ m*>0 __.--'
   +----|------------|------------|----> k
      -pi/a          0          +pi/a

   v_g ~ dE/dk        (b) velocity
   ^           max at +pi/2a
   |             .--.
   |           /      \
   +----------*--------*---------> k
   |  \      /  0      +pi/a
   |    '--'
  • Lower half of band (∣k∣<π/2a|k| < \pi/2a): curve concave up, d2E/dk2>0d^2E/dk^2 > 0, m∗>0m^* > 0. Velocity rises with kk; the electron accelerates in the direction of the force, like a normal particle.
  • Inflection point (∣k∣=π/2a|k| = \pi/2a): d2E/dk2=0d^2E/dk^2 = 0, m∗=∞m^* = \infty. Velocity is maximum; increasing kk does not increase vgv_g. The applied force gives zero acceleration.
  • Upper half of band (π/2a<∣k∣<π/a\pi/2a < |k| < \pi/a): curve concave down, d2E/dk2<0d^2E/dk^2 < 0, m∗<0m^* < 0. Velocity decreases as kk increases, so the electron accelerates opposite to the applied force. Physically, the electron is being Bragg-reflected by the lattice and the lattice pushes back harder than the external force.
  • At k=±π/ak = \pm\pi/a (band edge), vg=0v_g = 0: a standing wave.

Significance

  • Near the bottom of the conduction band electrons have positive m∗m^* and behave like free electrons.
  • Near the top of the valence band electrons have negative m∗m^*. An electron missing from there behaves as a hole with positive charge and positive mass, which is the basis of hole conduction in semiconductors.

✓ Answer checked by Claude Opus 5.5 (high effort)

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