Introduction To Quantum Mechanics
Introduction To Quantum Mechanics
3rd Edition
ISBN: 9781107189638
Author: Griffiths, David J., Schroeter, Darrell F.
Publisher: Cambridge University Press
Question
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Chapter 4, Problem 4.46P

(a)

To determine

The energy of three-dimensional harmonic oscillator using separation of variable.

(a)

Expert Solution
Check Mark

Answer to Problem 4.46P

Three-dimensional harmonic oscillator can be rewritten as three one-dimensional oscillators and its energy is (n+32)ω.

Explanation of Solution

Write the expression for Schrodinger equation for harmonic oscillator in terms of polar co-ordinates

2m(2ψx2+2ψy2+2ψz2)2+12mω2(x2+y2+z2)=Eψ        (I)

Here, x,y,z is the Cartesian co-ordinates, m is the mass of the particle, ω is the angular frequency, E is the total energy and ψ is wave function.

Write the expression for the change of variable

ψ(x,y,z)=X(x)Y(y)Z(z)        (II)

Rewrite expression (II) using (I)

(2m1Xd2Xdx2+12mω2x2)+(2m1Yd2Ydy2+12mω2y2)+(2m1Zd2Zdz2+12mω2z2)=E        (III)

The first term is a function of x only, the second of y only and the third of z only.

Rewrite expression (III) as three separate differential equations

2m1Xd2Xdx2+12mω2x2=ExX        (IV)

2m1Yd2Ydy2+12mω2y2=EyY        (V)

2m1Zd2Zdz2+12mω2Z2=EzZ        (VI)

Rewrite the expression (III)

    Ex+Ey+Ez=E        (VII)

Conclusion:

Expressions (IV), (V) and (VI) are of the form of one-dimensional oscillator

Write the expression for energy of one-dimensional oscillators

    Ex=(nx+12)ωEy=(ny+12)ωEz=(nz+12)ω        (VIII)

Here, nx,ny,nz are quantum numbers.

Rewrite expression (VII)

    E=(nx+12)ω+(ny+12)ω+(nz+12)ω

Let nnx+ny+nz. Here

    E=(n+32)ω   

Therefore, the energy of the three-dimensional harmonic oscillator is (n+32)ω.

(b)

To determine

The degeneracy of the energy.

(b)

Expert Solution
Check Mark

Answer to Problem 4.46P

The degeneracy of energy is (n+1)(n+2)2.

Explanation of Solution

Consider the case when nx=n, the possible values of ny and nz are 0 and zero respectively.

Consider the case when nx=n1, the possible combination of ny and nz are 0,1 and 1,0. This means there are two possible ways.

Consider the case when nx=n2, the possible combination of ny and nz are 0,2, 2,0 and 1,1. This means there are three possible ways.

Conclusion:

Generalise the number of ways in which the energy can be distributed among the three levels

    d(n)=1+2+3+....+(n+1)=(n+1)(n+2)2

Here, d(n) is the degeneracy of energy.

Therefore, the degeneracy of energy is (n+1)(n+2)2.

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