Chapter 4, Problem 4.46P

(a)

To determine

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

Answer to Problem 4.46P

Three-dimensional harmonic oscillator can be rewritten as three one-dimensional oscillators and its energy is $\left(n+\frac{3}{2}\right)\hslash \omega$.

Explanation of Solution

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

$\frac{-\hslash }{2m}{\left(\frac{{\partial }^2\psi }{\partial x^2}+\frac{{\partial }^2\psi }{\partial y^2}+\frac{{\partial }^2\psi }{\partial z^2}\right)}^2+\frac{1}{2}m{\omega }^2\left(x^2+y^2+z^2\right)=E\psi$        (I)

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

Write the expression for the change of variable

$\psi \left(x,y,z\right)=X\left(x\right)Y\left(y\right)Z\left(z\right)$        (II)

Rewrite expression (II) using (I)

$\left(\frac{-\hslash }{2m}\frac{1}{X}\frac{d^{2}X}{d{x}^2}+\frac{1}{2}m{\omega }^2{x}^2\right)+\left(\frac{-\hslash }{2m}\frac{1}{Y}\frac{d^{2}Y}{d{y}^2}+\frac{1}{2}m{\omega }^2{y}^2\right)+\left(\frac{-\hslash }{2m}\frac{1}{Z}\frac{d^{2}Z}{d{z}^2}+\frac{1}{2}m{\omega }^2{z}^2\right)=E$        (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

$\frac{-\hslash }{2m}\frac{1}{X}\frac{d^{2}X}{d{x}^2}+\frac{1}{2}m{\omega }^2{x}^2=E_{\text{x}}X$        (IV)

$\frac{-\hslash }{2m}\frac{1}{Y}\frac{d^{2}Y}{d{y}^2}+\frac{1}{2}m{\omega }^2{y}^2=E_{\text{y}}Y$        (V)

$\frac{-\hslash }{2m}\frac{1}{Z}\frac{d^{2}Z}{d{z}^2}+\frac{1}{2}m{\omega }^2{Z}^2=E_{\text{z}}Z$        (VI)

Rewrite the expression (III)

    $E_{\text{x}}+E_{\text{y}}+E_{\text{z}}=E$        (VII)

Conclusion:

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

Write the expression for energy of one-dimensional oscillators

    $\begin{array}{l}{E}_{\text{x}}=\left(n_{\text{x}}+\frac{1}{2}\right)\hslash \omega \\ E_{\text{y}}=\left(n_{\text{y}}+\frac{1}{2}\right)\hslash \omega \\ E_{\text{z}}=\left(n_{\text{z}}+\frac{1}{2}\right)\hslash \omega \end{array}$        (VIII)

Here, $n_{\text{x}},n_{\text{y}},n_{\text{z}}$ are quantum numbers.

Rewrite expression (VII)

    $E=\left(n_{\text{x}}+\frac{1}{2}\right)\hslash \omega +\left(n_{\text{y}}+\frac{1}{2}\right)\hslash \omega +\left(n_{\text{z}}+\frac{1}{2}\right)\hslash \omega$

Let $n\equiv n_{\text{x}}+n_{\text{y}}+n_{\text{z}}$. Here

    $E=\left(n+\frac{3}{2}\right)\hslash \omega$   

Therefore, the energy of the three-dimensional harmonic oscillator is $\left(n+\frac{3}{2}\right)\hslash \omega$.

(b)

To determine

The degeneracy of the energy.

Answer to Problem 4.46P

The degeneracy of energy is $\frac{\left(n+1\right)\left(n+2\right)}{2}$.

Explanation of Solution

Consider the case when $n_{\text{x}}=n$, the possible values of $n_{\text{y}}$ and $n_{\text{z}}$ are 0 and zero respectively.

Consider the case when $n_{\text{x}}=n-1$, the possible combination of $n_{\text{y}}$ and $n_{\text{z}}$ are $0,1$ and $1,0$. This means there are two possible ways.

Consider the case when $n_{\text{x}}=n-2$, the possible combination of $n_{\text{y}}$ and $n_{\text{z}}$ 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

    $\begin{array}{c}d\left(n\right)=1+2+3+\mathrm{....}+\left(n+1\right)\\ =\frac{\left(n+1\right)\left(n+2\right)}{2}\end{array}$

Here, $d\left(n\right)$ is the degeneracy of energy.

Therefore, the degeneracy of energy is $\frac{\left(n+1\right)\left(n+2\right)}{2}$.