Introduction to Quantum Mechanics · Problem 5.37

Identical Particles — David J. Griffiths

Physics — Quantum Mechanics · density_of_states

Problem

An important quantity in many calculations is the density of states

G(E)

G(E)\, dE \equiv \text{number of states with energy between } E \text{ and } E + dE.

For a one-dimensional band structure,

G(E)\, dE = 2 \left( \frac{dq}{2\pi/Na} \right),

where

dq/(2\pi/Na)

counts the number of states in the range

dq

(see Equation 5.63), and the factor of 2 accounts for the fact that states with

q

and

-q

have the same energy. Therefore

\frac{1}{Na} G(E) = \frac{1}{\pi} \frac{1}{|dE/dq|}.

(a) Show that for

\alpha = 0

(a free particle) the density of states is given by

\frac{1}{Na} G_{\text{free}}(E) = \frac{1}{\pi \hbar} \sqrt{\frac{m}{2E}}.

(b) Find the density of states for

\alpha \neq 0

by differentiating Equation 5.71 with respect to

q

to determine

dE/dq

. Note: Your answer should be written as a function of

E

only (well, and

\alpha

m

\hbar

a

, and

N

) and must not contain

q

(use

k

as a shorthand for

\sqrt{2mE}/\hbar

, if you like). (c) Make a single plot showing

G(E)/Na

for both

\alpha = 0

and

\alpha = 1

(in units where

m = \hbar = a = 1

). Comment: The divergences at the band edges are examples of van Hove singularities.