In this chapter we solved for the first excited state of a two-dimensional harmonic oscillator subject to perturbation W = κ′ x y. How do the three-fold degenerate energy E = 3ћω and the four-fold degenerate energy E = 4ћω separate due to the same perturbation?

Question

In this chapter we solved for the first excited state of a two-dimensional harmonic oscillator subject to perturbation [latex]W=\kappa^{\prime}x y[/latex]. How do the three-fold degenerate energy [latex]E=3\hbar \omega[/latex] and the four-fold degenerate energy [latex]E=4\hbar \omega[/latex] separate due to the same perturbation?

Accepted Answer

A two-dimensional harmonic oscillator with motion in the [latex]x–y[/latex] plane is subject to perturbation [latex]W=\kappa^{\prime}x y[/latex]. We are asked to find howthe three-fold degenerate energy [latex]E=3\hbar \omega[/latex] and the four-fold degenerate energy [latex]E=4\hbar \omega[/latex] separate due to this perturbation.
Separation of variables x and y allows us to write the unperturbed Hamiltonian as

[latex]H^{(0)}={\frac{\hbar\omega}{2m}}\bigl(p_{x}^{2}+p_{y}^{2} \bigr)+{\frac{\kappa}{2}}\bigl(x^{2}+y^{2}\bigr)=\hbar\omega \bigl (\hat{b}_{x}^{\dagger}\hat{b}_{x}+\hat{b}_{y}^{\dagger}\hat{b}_{y}+1\bigr)[/latex]

where [latex]{\hat{x}}={\sqrt{\hbar/2m\omega}} \cdot( {\hat{b} }_{x}+{\hat{b}}_{x}^{\dagger})[/latex] and [latex]\hat{y} =\sqrt {\hbar/ 2m \omega}\cdot(\hat{b}_{y}+\hat{b}_{y}^{\dagger})[/latex]. The eigenstates are of the form [latex]\psi_{n m}^{(0)} =\phi_{n}^{(0)}(x)\phi_{m}^{(0)}(y)=|n m angle,[/latex] and the energy eigenvalues are

[latex]E_{n m}=\hbar\omega\,(n+m+1)[/latex]

where n and m are positive integers. States with eigenenergy [latex] 3\hbar ω \,\,\mathrm{are}\, E_{02},E_{11},\,\mathrm{and}\,\,E_{20},[/latex] and so they are three-fold degenerate. To find the effect of the perturbation [latex]W=\kappa^{\prime}x y[/latex], we start by writing the total Hamiltonian:

[latex]H=H^{(0)}+W=\hbar{}\omega(\hat{b}_{x}^{\dagger}\hat{b}_{x}+\hat{b}_{y}^{\dagger}\hat{b}_{y}+1)+\frac{\hbar{}\kappa^{\prime}}{2m\omega}(\hat{b}_{x}+\hat{b}_{x}^{\dagger})(\hat{b}_{ y} +\hat{b}_{y}^{\dagger})[/latex]

This has eigenfunction solutions that are linear combinations of the unperturbed eigenstates so that

[latex]\Psi_{1}=a_{1}\psi_{1}+a_{2}\psi_{2}+a_{3}\psi_{3}[/latex]

The coefficients [latex]{{a}}_{n}[/latex] may be found by writing the Schrödinger equation in matrix form

[latex]\left[\begin{matrix}\langle02|H|02 angle & \langle02 |H|11 angle & \langle02|H|20 angle \ \langle11|H|02 angle & \langle11 |H|11 angle & \langle11|H|20 angle \ \langle20|H|02 angle & \langle20 |H|11 angle & \langle20|H|20 angle \end{matrix} ight]\left[\begin{matrix} \begin{matrix} a_{1} \ a_{2} \ a_{3} \end{matrix}\end{matrix} ight]=E\left[\begin{matrix} \begin{matrix} a_{1} \ a_{2} \ a_{3} \end{matrix}\end{matrix} ight][/latex]

The diagonal terms have a value that is the unperturbed eigenvalue [latex]3\hbar \omega[/latex] . To show this, consider

[latex]\langle{02}|H|{02} angle=\hbar\omega\langle{02}|{\hat{b}}_{x}^{\dagger}{\hat{b}}_{x}+{\hat{b}}_{y}^{\dagger}{\hat{b}}_{y}+1|{02} angle+{\frac{\hbar\kappa^{\prime}}{2m\omega}}\langle{02}|{\hat{b}}_{x}\hat{b}_{y}+{\hat{b}}_{x}^{\dagger}{\hat{b}}_{y}+{\hat{b}}_{x}{\hat{b}}_{y}^{\dagger}+{\hat{b}}_{x}^{\dagger}{\hat{b}}_{y}^{\dagger}|{02} angle[/latex]

The first term on the right-hand side has value [latex]3\hbar\omega,[/latex] and the second term on the righthand side is zero. Because the perturbation W is linear in x and y, only the off-diagonal terms adjacent to the diagonal are finite. For example,

[latex]\langle11\vert H\vert02 angle=\hbar\omega\langle11 \vert{\hat{b}}_{x}^{\dagger}{\hat{b}}_{x}+{\hat{b}}_{y}^{\dagger}{\hat{b}}_{y}+1\vert02 angle+{\frac{\hbar\kappa^{\prime}}{2m\omega}}\langle11\vert({\hat{b}}_{x}+{\hat{b}}_{x}^{\dagger})({\hat{b}}_{y}+{\hat{b}}_{y}^{\dagger})\vert02 angle[/latex]

The first term on the right-hand side is zero, leaving

[latex]\langle11\vert H\vert02 angle={\frac{\hbar\kappa^{\prime}}{2m\omega}}\langle11\vert\hat{b}_{x}\hat{b}_{y}+\hat{b}_{x}^{\dagger}\hat{b}_{y}+\hat{b}_{x}\hat{b}_{y}^{\dagger}+\hat{b}_{x}^{\dagger}\hat{b}_{y}^{\dagger}\vert02 angle={\frac{\hbar \kappa^{ \prime}}{2m\omega}}\langle11\vert \hat{b}_{x}^{ \dagger}\hat{b}_{ y}\vert02 angle[/latex]

Recalling that [latex]|b^{\dagger}n angle= (n+1)^{1/2}|n+1 angle [/latex] (Eqn (6.73)) and [latex]|\hat{b}n angle=n^{1/2}|n-1 angle [/latex]

[latex]|b^{\dagger}n angle=(n+1)^{1/2}\,|\,n+1\, angle \quad\quad \quad\quad\quad\quad(6.73)[/latex]

(Eqn (6.74)) allows us to conclude that

[latex]|\,\hat{b}n\, angle=n^{1/2}\,|\,n-1\, angle \quad\quad \quad\quad(6.74)[/latex]

[latex]\langle11\vert H\vert02 angle=\frac{\hbar\kappa^{\prime}}{2m\omega}\langle11\vert\hat{b}_{x}^{\dagger}\hat{b}_{y}\vert02 angle=\frac{\sqrt{2}\hbar\kappa^{\prime}}{2m\omega} \langle11 \vert\hat{b}_{x}^{\dagger}\vert01 angle=\frac{\sqrt{2}\hbar\kappa^{\prime}}{2m\omega}\langle11\vert11 angle=\frac{\sqrt{2} \hbar \kappa^{\prime}}{2m\omega}[/latex]

It follows that

[latex]\left[\begin{array}{l l l}{{3\hbar\omega}}&{{\frac{\sqrt {2} \hbar κ^{\prime}{}}{2m\omega}}}&{{0}}\ {{\frac{\sqrt{2 } \hbar κ^{\prime}{}}{2m\omega}}}&{{3\hbar\omega}}&{{\frac{\sqrt{2}\hbar κ^{\prime}}{2m\omega}}}\ \,\,\,\,{{0}}&{{\frac{\sqrt{2}\hbar κ^{\prime}{}}{2m\omega}}}&{{3\hbar\omega}}\end{array} ight] \left[\begin{matrix} \begin{matrix} a_{1} \ a_{2} \ a_{3} \end{matrix}\end{matrix} ight]=E\left[\begin{matrix} \begin{matrix} a_{1} \ a_{2} \ a_{3} \end{matrix}\end{matrix} ight][/latex]

All we need to do now is find the eigenvalues of the matrix. The solutions are

[latex]E_{1}=3\hbar\omega[/latex]

[latex]E_{2}={\frac{3m\hbar\omega^{2}-\hbar\kappa^{\prime}}{m\omega}}[/latex]

[latex]E_{3}={\frac{3m\hbar\omega^{2}+\hbar\kappa^{\prime}}{m\omega}}[/latex]

The corresponding eigenfunctions are given by the coefficients

[latex]\left[\begin{matrix} a_{1} \ a_{2} \ a_{3} \end{matrix} ight] =\frac{1}{\sqrt{2} } \left[\begin{matrix} -1 \ 0 \ 1 \end{matrix} ight] [/latex]

[latex]\left[\begin{matrix} a_{1} \ a_{2} \ a_{3} \end{matrix} ight] =\frac{1}{2 } \left[\begin{matrix} 1\ -\sqrt{2} \ 1 \end{matrix} ight] [/latex]

[latex]\left[\begin{matrix} a_{1} \ a_{2} \ a_{3} \end{matrix} ight] =\frac{1}{2 } \left[\begin{matrix} 1\ \sqrt{2}\ 1 \end{matrix} ight] [/latex]

We may follow a similar procedure to find how the four-fold degenerate levels of a two-dimensional harmonic oscillator with motion in the x–y plane subject to perturbation [latex]W=\kappa^{\prime}x y [/latex] change. The perturbed Schrödinger equation matrix is

[latex]\left[\begin{array}{l l l l l}{{4\hbar\omega}}&{{\frac{\sqrt {3}\hbar\kappa^{\prime}}{2m\omega}}}&{{0}}&{{0}}\ {{\frac{\sqrt {3}\hbar\kappa^{\prime}}{2m\omega}}}&{{4\hbar\omega}}&{{\frac{ \hbar\kappa^{\prime}}{m\omega}}}&{{0}}\ {{0}}&{{\frac{\hbar \kappa^{\prime}}{m\omega}}}&{{4\hbar\omega}}&{{\frac{\sqrt{3} \hbar\kappa^{\prime}}{2m\omega}}}\ {{0}}& {{0}}&{{\frac{ \sqrt {3}\hbar\kappa^{\prime}}{2m\omega}}}&{{4\hbar\omega}} \end {array} ight] \left[\begin{matrix} a_{1} \ a_{2} \ a_{3} \ a_{4} \end{matrix} ight] = E \left[\begin{matrix} a_{1} \ a_{2} \ a_{3} \ a_{4} \end{matrix} ight] [/latex]

which has eigenvalues

[latex]E_{1}={\frac{8m\hbar\omega^{2}-3\hbar\kappa^{\prime}}{2m\omega}}[/latex]

[latex]E_{2}={\frac{8m\hbar\omega^{2}-\hbar\kappa^{\prime}}{2m\omega}}[/latex]

[latex]E_{3}={\frac{8m\hbar\omega^{2}+\hbar\kappa^{\prime}}{2m\omega}}[/latex]

[latex]E_{4}=\frac{8m\hbar\omega^{2}+3\hbar\kappa^{\prime}}{2m\omega}[/latex]

The corresponding eigenfunctions are given by the coefficients

[latex]\left[{\begin{array}{c}{a_{1}}\ {a_{2}}\ {a_{3}}\ {a_{4}}\ \end{array}} ight]={\frac{1}{2{\sqrt{2}}}}\left[{\begin{array}{c}{-1}\ {{\sqrt{3}}}\{-{\sqrt{3}}}\ {1}\end{array}} ight][/latex]

[latex]\left[{\begin{array}{c}{a_{1}}\ {a_{2}}\ {a_{3}}\ {a_{4}}\ \end{array}} ight]={\frac{\sqrt{3}}{2{\sqrt{2}}}}\left[{\begin{array}{c}{1}\-1/{\sqrt{3}}\ -1/{\sqrt{3}}\ {1}\end{array}} ight][/latex]

[latex]\left[{\begin{array}{c}{a_{1}}\ {a_{2}}\ {a_{3}}\ {a_{4}}\ \end{array}} ight]={\frac{\sqrt{3}}{2{\sqrt{2}}}}\left[{\begin{array}{c}{-1}\-1/{\sqrt{3}}\ 1/{\sqrt{3}}\ {1}\end{array}} ight][/latex]

[latex]\left[{\begin{array}{c}{a_{1}}\ {a_{2}}\ {a_{3}}\ {a_{4}}\ \end{array}} ight]={\frac{1}{2{\sqrt{2}}}}\left[{\begin{array}{c}{1}\ {{\sqrt{3}}}\{{\sqrt{3}}}\ {1}\end{array}} ight][/latex]

Question 10.6: In this chapter we solved for the first excited state of a t......

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