EFN316G Intro to Quantum Chemistry: Homework Assignment 6, solution

Problem 2:

Consider a particle in a potential field where V (x) becomes infinite at x < 0, is �V0when 0 < x < a and is zero beyond x = a, i.e. V (x) = 0 when x > a.

(a) Sketch the potential energy curve and sketch the wavefunction for the two lowestbound states (assuming there are at least two bound states) and for one of the continuumstates where E > 0.

Solution:

(See lecture notes). The wave functions must go to zero at the hard wall at x = 0.The bound state wave functions look qualitatively similar to the wave functions of theparticle-in-a-box, except that here the wave functions do not go to zero at x = a. Rather,they join smoothly (continuously and with continuous derivative) an exponential tail intothe classically forbidden region. The ground state wave function has no node. The firstexcited state wave function has one node. The wave function for the continuum statewhich corresponds to E > 0 oscillates in the whole region x > 0, but the wavelength islarger when x > a than when 0 < x < a because the kinetic energy is smaller.

(b) Show that the form of the solutions to the time independent Scrodinger equation inthe region 0 < x < a is

(x) = A sin k1x

and give the value of k1 in terms of the energy and the mass of the particle.

Solution:

The general solution to the Schrodinger equation in the region 0 < x < a is

(x) = A sin k1x + B cos k1x

where k1 =p2m(E + V0)/h . The boundary condition (0) = 0, which results from the

fact that the wave function has to vanish at the hard wall, shows that B has to be chosento be B = 0. Therefore,

(x) = A sin k1x .

(c) Show that the form of the general solution in the region a < x can be written as

(x) = Ce

ik2x + De

�ik2x

and give an expression for k2 in terms of the energy and the mass of the particle.

Solution:

Here it is more convenient to write the general solution of the Schrodinger equationin the complex exponential form rather than as sines and cosines

(x) = Ce

ik2x + De

�ik2x

where k2 =p2mE/h. When E > 0, the wave function oscillates and both terms need in

general to be retained.

(d) The form of the wave function given in part (c) is most convenient for the region a < x

when E > 0 because the particle is then not bound but rather scatters from the wall andwell represented by the and the potential function V (x). One of the terms corresponds toan incoming wave traveling in the negative x�direction and the other term correspondsan outgoing wave traveling in the positive x�direction. Which term corresponds to theincoming wave and which term corresponds to the outgoing wave?

Solution:

Assuming k > 0, the first term corresponds to a wave traveling in the positive x-direction, while the second term corresponds to a wave traveling in the negative x-direction.

(e) The form of the wave function given in part (c) can also be used for the region a < x

when E < 0, but then k2 is an imaginary number and one of the two coe�cients, C or D,must be zero. Which one? Explain!

Solution:

When E < 0, the wave number becomes purely imaginary k2 = i

p2m|E|/h = i↵

where ↵ is a real number. The form of the wave function then becomes

(x) = Ce

�↵x + De

↵x

.

The second term goes to 1 as x ! 1. Since the wave function can never go to 1, onemuch choose D = 0. Then, (x) ! 0 as x ! 1 as is appropriate when going deep into aclassically forbidden region.

(f) Now apply the condition that the wavefunction needs to be continuous and that thederivative of the wavefunction needs to be continuous at x = a and explain why anypositive value of the energy is allowed, but only some negative values (if any). Sketcha wave function corresponding to a positive value of the energy and a wave functioncorresponding to a negative value of the energy (assuming one such solution exists).

Solution (a more detailed solution than what is asked for ...):

The general solution in the region 0 < x < a needs to be matched with the generalsolution in the region a < x. There are two matching conditions, (1) that the wave functionis continuous, and (2) that the derivative of the wave function is continuous. These twoconditions are applied at the joining point of the two regions, x = a.

First consider the case when E > 0. Condition (1) becomes

A sin k1a = Ce

ik2a + De

�ik2a

and condition (2) becomes

k1A cos k1a = ik2Ce

ik2a � ik2De

�ik2a.

There are two equations with three unknowns. There exists a solution for all values ofE > 0. This represents a scattering problem where a particle is coming in from the rightand reflects from a hard wall after dropping into a well at x = a. The coe�cient D givesthe incident flux. The reflected flux is given by C and since the two must be equal (nothing

goes through the hard wall), |C| = |D|. This magnitude is arbitrary, it depends on theexperimental conditions. The two conditions can be used to find the remaining coe�cients,the phase of C/D and the coe�cient A.

Secondlly, consider the case when E < 0. Then, condition (1) becomes

A sin k1a = Ce

�↵a

and condition (2) becomesA sin k1a = Ce

�↵a

.

In addition to these two, one can also invoke here a normalization condition since (x) ! 0as x ! 1, namely Z 1

�1| (x)|2dx = 1 .

Now we have three equations and only two unknowns. There can, therefore, only be asolution under special circumstances, that is only for certain values of E. Those are thebound state energy levels. The energy is quantized when E < 0 since the particle isconfined in that case ( (x) ! 0 as x ! 1 and (x) = 0 for x < 0). An equation thatgives, implicitly, the allowed values of the energy can be obtained by dividing (2) into (1)

A sin k1a

k1A cos k1a=

Ce

�↵a

(�↵)Ce

�↵a

which can be simplified as1

k1tan k1a = � 1

↵

.

After rewriting k1 and ↵ in terms of E, this becomes

hp2m(V0 + E)

tan

p2m(V0 + E)

h

a

!= � hp

2m|E|

or, with further simplification

tan

p2m(V0 + E)

h

a

!= �

sV0 + E

|E| .

When E < 0, the particle is largely confined to the potential well and this leads to quan-tization of the energy.