Consider the three-dimensional harmonic oscillator, for which the potential is

$\mathbf{V}\left(r\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m\omega }}^{\mathbf{2}}{\mathbf{r}}^{\mathbf{2}}$

(a) Show that separation of variables in cartesian coordinates turns this into three one-dimensional oscillators, and exploit your knowledge of the latter to determine the allowed energies. Answer:

${\mathbf{E}}_{\mathbf{n}}\mathbf{=}\left(n+3/2\right)\sout{\mathbf{h}}\mathbf{\omega }$

(b) Determine the degeneracyof$\mathbf{d}\left(n\right)\mathbf{}\mathbf{of}\mathbf{}{\mathbf{E}}_{\mathbf{n}}\mathbf{.}$

(1) The quantum harmonic oscillator has an energy eigenvalue of:

$\mathrm{E}=\left(\mathrm{n}+\frac{1}{2}\right)\sout{\mathrm{h}}\mathrm{\omega }\forall \mathrm{n}\in {\mathrm{\mathbb{Z}}}^{+}$

(2) The sum of consecutive number is $\sum _{\mathrm{i}=1}^{\mathrm{n}+1}\mathrm{i}=\frac{\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}{2}=1+2+3+...+\left(\mathrm{n}+1\right)$

(3) The potential is$\mathrm{V}\left(\mathrm{r}\right)=\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{r}}^2$

(a)

Consider the formula for the potential of a three-dimensional harmonic oscillator,

The value of the potential is

$$\begin{gathered} \mathrm{V}\left(\mathrm{r}\right)=\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{r}}^2 \\ =\frac{1}{2}{\mathrm{m\omega }}^2\left({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2\right) \\ =-\frac{\sout{{\mathrm{h}}^2}}{2\mathrm{m}}{\nabla }^2\mathrm{\psi }+\mathrm{V\psi } \\ =\mathrm{E\psi } \\ \Rightarrow -\frac{\sout{{\mathrm{h}}^2}}{2\mathrm{m}}\left(\frac{{\partial }^2\mathrm{\psi }}{\partial {\mathrm{X}}^2}+\frac{{\partial }^2\mathrm{\psi }}{\partial {\mathrm{y}}^2}+\frac{{\partial }^2\mathrm{\psi }}{\partial {\mathrm{Z}}^2}\right)+\mathrm{V\psi }=\mathrm{E\psi } \\ \mathrm{The}\mathrm{variables}\mathrm{can}\mathrm{be}\mathrm{safely}\mathrm{separated}\mathrm{because}\mathrm{x},\mathrm{y},\mathrm{and}\mathrm{z}\mathrm{do}\mathrm{not}\mathrm{depend}\mathrm{on}\mathrm{each} \\ \mathrm{other}.\mathrm{The}\mathrm{value}\mathrm{is}: \\ \mathrm{\psi }\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\mathrm{X}\left(\mathrm{x}\right)\mathrm{Y}\left(\mathrm{y}\right)\mathrm{Z}\left(\mathrm{z}\right)\Rightarrow \\ -\frac{\sout{{\mathrm{h}}^2}}{2\mathrm{m}}\left(\mathrm{YZ}\frac{{\mathrm{d}}^2\mathrm{X}}{{\mathrm{dx}}^2}+\mathrm{XZ}\frac{{\mathrm{d}}^2\mathrm{Y}}{{\mathrm{dy}}^2}+\mathrm{XY}\frac{{\mathrm{d}}^2}{\mathrm{Z}}{\mathrm{dZ}}^2\right)+\mathrm{V\psi }=\mathrm{E\psi } \\ \Rightarrow -\frac{\sout{{\mathrm{h}}^2}}{2\mathrm{m}}\left(\frac{1}{\mathrm{X}}\frac{{\mathrm{d}}^2\mathrm{X}}{{\mathrm{dx}}^2}+\frac{1}{\mathrm{Y}}\frac{{\mathrm{d}}^2\mathrm{Y}}{{\mathrm{dy}}^2}+\frac{1}{\mathrm{Z}}\frac{{\mathrm{d}}^2\mathrm{Z}}{{\mathrm{dZ}}^2}\right)+\frac{1}{2}{\mathrm{m\omega }}^2\left({\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2\right)=\mathrm{E}\Rightarrow \\ \left(-\frac{\sout{{\mathrm{h}}^2}}{2\mathrm{m}}-\frac{1}{\mathrm{X}}\frac{{\mathrm{d}}^2\mathrm{X}}{{\mathrm{dx}}^2}+\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{x}}^2\right) \\ +\left(-\frac{\sout{{\mathrm{h}}^2}}{2\mathrm{m}}-\frac{1}{\mathrm{Y}}\frac{{\mathrm{d}}^2\mathrm{Y}}{{\mathrm{dy}}^2}+\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{y}}^2\right) \\ +\left(-\frac{\sout{{\mathrm{h}}^2}}{2\mathrm{m}}-\frac{1}{\mathrm{Z}}\frac{{\mathrm{d}}^2\mathrm{Z}}{{\mathrm{dZ}}^2}+\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{z}}^2\right)=\mathrm{E}\Rightarrow \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{above}\mathrm{value}, \end{gathered}$$

$$\begin{gathered} -\frac{\sout{{\mathrm{h}}^2}}{2\mathrm{m}}\frac{1}{\mathrm{X}}\frac{{\mathrm{d}}^2\mathrm{X}}{{\mathrm{dx}}^2}+\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{x}}^2={\mathrm{E}}_{\mathrm{x}} \\ -\frac{\sout{{\mathrm{h}}^2}}{2\mathrm{m}}\frac{1}{\mathrm{Y}}\frac{{\mathrm{d}}^2\mathrm{Y}}{{\mathrm{dy}}^2}+\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{y}}^2={\mathrm{E}}_{\mathrm{y}} \\ -\frac{\sout{{\mathrm{h}}^2}}{2\mathrm{m}}\frac{1}{\mathrm{Z}}\frac{{\mathrm{d}}^2\mathrm{Z}}{{\mathrm{dZ}}^2}+\frac{1}{2}{\mathrm{m\omega }}^2{\mathrm{z}}^2={\mathrm{E}}_{\mathrm{z}} \\ \mathrm{so},\mathrm{E}={\mathrm{E}}_{\mathrm{x}}+{\mathrm{E}}_{\mathrm{y}}+{\mathrm{E}}_{\mathrm{z}} \\ {\mathrm{E}}_{\mathrm{x}}=\left({\mathrm{n}}_{\mathrm{x}}+\frac{1}{2}\right)\sout{\mathrm{h}}\mathrm{\omega } \\ {\mathrm{E}}_{\mathrm{y}}=\left({\mathrm{n}}_{\mathrm{y}}+\frac{1}{2}\right)\sout{\mathrm{h}}\mathrm{\omega } \\ {\mathrm{E}}_{\mathrm{z}}=\left({\mathrm{n}}_{\mathrm{z}}+\frac{1}{2}\right)\sout{\mathrm{h}}\mathrm{\omega } \\ \Rightarrow \mathrm{E}=\left({\mathrm{n}}_{\mathrm{x}}+{\mathrm{n}}_{\mathrm{y}}+{\mathrm{n}}_{\mathrm{z}}+\frac{3}{2}\right)\sout{\mathrm{h}}\mathrm{\omega } \\ \Rightarrow \mathrm{E}=\left(\mathrm{n}+\frac{3}{2}\right)\sout{\mathrm{h}}\mathrm{\omega }\forall \mathrm{n}\in \mathrm{{\rm Z}} \\ \mathrm{As},\mathrm{n}={\mathrm{n}}_{\mathrm{x}}+{\mathrm{n}}_{\mathrm{y}}+{\mathrm{n}}_{\mathrm{z}} \\ \mathrm{So},\mathrm{this}\mathrm{is}\mathrm{proved}\mathrm{that}\mathrm{E}=\left(\mathrm{n}+\frac{3}{2}\right)\sout{\mathrm{h}}\mathrm{\omega }. \end{gathered}$$

(b)

Assumtion of that,

$$\begin{gathered} {\mathrm{n}}_{\mathrm{x}}=\mathrm{n}\Rightarrow {\mathrm{n}}_{\mathrm{y}}=0,{\mathrm{n}}_{\mathrm{z}}=0 \\ \Rightarrow \mathrm{d}=1 \\ \mathrm{So},{\mathrm{n}}_{\mathrm{x}}=\mathrm{n}-1\Rightarrow {\mathrm{n}}_{\mathrm{y}}=1,{\mathrm{n}}_{\mathrm{z}}=0 \\ \mathrm{OR} \\ {\mathrm{n}}_{\mathrm{y}}=0,{\mathrm{n}}_{\mathrm{z}}=1\Rightarrow \mathrm{d}=2 \\ \mathrm{Also},{\mathrm{n}}_{\mathrm{x}}=\mathrm{n}-2\to {\mathrm{n}}_{\mathrm{y}}=2{\mathrm{n}}_{\mathrm{z}}=0 \\ \mathrm{OR} \\ {\mathrm{n}}_{\mathrm{y}}=0,{\mathrm{n}}_{\mathrm{z}}=2\mathrm{OR}{\mathrm{n}}_{\mathrm{y}}=1,{\mathrm{n}}_{\mathrm{z}}=1\Rightarrow \mathrm{d}=3 \\ \mathrm{It}\mathrm{is}\mathrm{obtained}\mathrm{that}, \\ \mathrm{d}\left(\mathrm{n}\right)=\underset{\mathrm{i}=1}{\overset{\mathrm{n}+1}{\sum \mathrm{i}}} \\ =\frac{\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}{2} \\ \mathrm{d}\left(\mathrm{n}\right)=\frac{\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}{2} \\ \mathrm{Hence},\mathrm{the}\mathrm{degeneracy}\mathrm{d}\left(\mathrm{n}\right)\mathrm{is}\frac{\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}{2}. \end{gathered}$$