Introduction to Quantum Mechanics · Problem 6.6

Symmetries and Conservation Laws — David J. Griffiths

Physics — Quantum Mechanics · periodic_potential

Problem

Consider a particle of mass

m

moving in a potential

V(x)

with period

a

. We know from Bloch's theorem that the wave function can be written in the form of Equation 6.12. Note: It is conventional to label the states with quantum numbers

n

and

q

\psi_{nq}(x) = e^{-iqx} u_{nq}(x)

where

E_{nq}

is the

n

th energy for a given value of

q

. (a) Show that

u

satisfies the equation

-\frac{\hbar^2}{2m} \frac{d^2 u_{nq}}{dx^2} - \frac{i \hbar^2 q}{m} \frac{d u_{nq}}{dx} + V(x) u_{nq} = \left(E_{nq} - \frac{\hbar^2 q^2}{2m}\right) u_{nq}. \quad (6.14)

(b) Use the technique from Problem 2.61 to solve the differential equation for

u_{nq}

. You need to use a two-sided difference for the first derivative so that you have a Hermitian matrix to diagonalize:

\frac{d\psi}{dx} \approx \frac{\psi_{j+1} - \psi_{j-1}}{2\Delta x}

. For the potential in the interval 0 to

a

let

V(x) = \begin{cases} -V_0 & a/4 < x < 3a/4 \\ 0 & \text{else} \end{cases}

with

V_0 = 20\,\hbar^2/2ma^2

. (You will need to modify the technique slightly to account for the fact that the function

u_{nq}

is periodic.) Find the lowest two energies for the following values of the crystal momentum:

qa = -\pi, -\pi/2, 0, \pi/2, \pi

. Note that

q

and

q + 2\pi/a

describe the same wave function (Equation 6.12), so there is no reason to consider values of

qa

outside of the interval from

-\pi

\pi

. In solid state physics, the values of

q

inside this range constitute the first Brillouin zone. (c) Make a plot of the energies

E_{1q}

and

E_{2q}

for values of

q

between

-\pi/a

and

\pi/a

. If you've automated the code that you used in part (b), you should be able to show a large number of

q

values in this range. If not, simply plot the values that you computed in (b).

Related theory

Electric Potential And Potential Difference

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