An electron moves in the finite square well for

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Wave mechanics in one dimension · Wave mechanics in one dimension

Quantum Mechanics: From the Photon to the Hydrogen Atom, Ch Stationary states and bound states in one dimension

Problem

An electron moves in the finite square well

V = -V_0

for

-a < x < a

and

V = 0

for

|x| > a

, with

a = 0.40

nm and

V_0 = 2.0

eV. A bound state has

-V_0 < E < 0

; write

k = \sqrt{2m_e(E + V_0)}/\hbar

for the wavenumber inside and

\kappa = \sqrt{-2m_eE}/\hbar

for the decay constant outside, and put

z = ka

z_0 = (a/\hbar)\sqrt{2m_eV_0}

. Take

\hbar^2/2m_e = 0.0381

eV nm

^2

. ((a)) For an odd bound state take

\psi = A\sin kx

inside and

\psi = Ce^{-\kappa x}

for

x > a

. By matching

\psi

and

\psi'

x = a

, derive the condition

\kappa = -k\cot ka

, and show it becomes

-\cot z = \sqrt{(z_0/z)^2 - 1}

. ((b)) Evaluate

z_0

for this well and state how many odd bound states it has. ((c)) Read the crossing from a sketch of the two curves and refine it by trial: give

z

to two decimal places, and hence the energy of the odd state above the bottom of the well,

E + V_0

, in eV. ((d)) Compare with the corresponding level of an infinite well of the same full width

2a

: give that level in eV (measured from the bottom of the well) and say which lies lower.