Question 10.6: In this chapter we solved for the first excited state of a t......
In this chapter we solved for the first excited state of a two-dimensional harmonic oscillator subject to perturbation \hat{W} =\kappa ^{\prime } \hat{x} \hat{y}. How do the three-fold degenerate energy E=3\hbar \omega the four-fold degenerate energy E=4\hbar \omega separate due to the same perturbation?
A two-dimensional harmonic oscillator with motion in the x–y plane is subject to perturbation \hat{W} =\kappa ^{\prime } \hat{x} \hat{y}. We are asked to find how the three-fold degenerate energy E=3\hbar \omega and the four-fold degenerate energy E=4\hbar \omega separate due to this perturbation.
Separation of variables x and y allows us to write the unperturbed Hamiltonian as
\hat{H}^{(0)} =\frac{(\hat{p}^{2}_{x}+\hat{p}^{2}_{y})}{2m} +\frac{\kappa }{2} (\hat{x}^{2}+\hat{y}^{2} )=\hbar \omega (\hat{b}^{\dagger }_{x} \hat{b}_{x}+\hat{b}^{\dagger }_{y} \hat{b}_{y}+1 )where \hat{x}=\sqrt{\hbar /2m\omega }(\hat{b}_{x}+\hat{b}^{\dagger }_{x} ) and \hat{y}=\sqrt{\hbar /2m\omega }(\hat{b}_{y}+\hat{b}^{\dagger }_{y} ). The eigenstates are of the form \psi ^{(0)}_{nm} =\phi ^{(0)}_{n} (x)\phi ^{(0)}_{m} (y)=\mid nm〉, and the energy eigenvalues are
E_{nm} =\hbar \omega (n+m+1)where n and m are positive integers. States with eigenenergy 3\hbar \omega are E_{02},E_{11}, and E_{20}, and so they are three-fold degenerate. To find the effect of the perturbation \hat{W} =\kappa ^{\prime } \hat{x} \hat{y}, we start by writing the total Hamiltonian:
\hat{H} =\hat{H}^{(0)}+ \hat{W}=\hbar \omega (\hat{b}^{\dagger }_{x}\hat{b}_{x}+ \hat{b}^{\dagger }_{y}\hat{b}_{y}+1)+\frac{\hbar \kappa ^{\prime } }{2m\omega } (\hat{b}_{x}+\hat{b}^{\dagger }_{x})(\hat{b}_{y}+\hat{b}^{\dagger }_{y})This has eigenfunction solutions that are linear combinations of the unperturbed eigenstates so that
\psi _{j} =a_{1}(j)\psi ^{(0)}_{11} +a_{2} (j)\psi ^{(0)}_{20} +a_{3}(j)\psi ^{(0)}_{02}The coefficients a_{n} may be found by writing the Schrödinger equation in matrix form
The diagonal terms have a value that is the unperturbed eigenvalue 3\hbar \omega. To show this, consider
\left\langle02\left|\hat{H} \right|02 \right\rangle=\hbar \omega \left\langle02\left|\hat{b}^{\dagger }_{x}\hat{b}_{x} + \hat{b}^{\dagger }_{y}\hat{b}_{y}+1\right|02 \right\rangle +\frac{\hbar \kappa ^{\prime } }{2m\omega } \left\langle02\left|\hat{b}_{x} \hat{b}_{y}+\hat{b}^{\dagger }_{x} \hat{b}_{y}+\hat{b}^{\dagger }_{y} \hat{b}_{x}+\hat{b}^{\dagger }_{x}\hat{b}^{\dagger }_{y} \right|02 \right\rangle
The first term on the right-hand side has value 3\hbar \omega, and the second term of 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,
\left\langle11\left|\hat{H} \right| 02\right\rangle =\hbar \omega \left\langle11\left|\hat{b} ^{\dagger }_{x} \hat{b} _{x} +\hat{b} ^{\dagger }_{y} \hat{b} _{y} +1\right| 02\right\rangle +\frac{\hbar \kappa ^{\prime } }{2m\omega } \left\langle11\left|(\hat{b}_{x} + \hat{b} ^{\dagger }_{x})(\hat{b}_{y} + \hat{b} ^{\dagger }_{y})\right| 02\right\rangleThe first term on the right-hand side is zero, leaving
\left\langle11\left|\hat{H} \right| 02\right\rangle =\frac{\hbar \kappa ^{\prime } }{2m\omega } \left\langle11\left|\hat{b} _{x}\hat{b}_{y}+\hat{b} ^{\dagger }_{x}\hat{b}_{y}+\hat{b}_{x}\hat{b} ^{\dagger }_{y}+\hat{b} ^{\dagger }_{x}\hat{b} ^{\dagger }_{y}\right|02 \right\rangle =\frac{\hbar \kappa ^{\prime } }{2m\omega }\left\langle11\left|\hat{b} ^{\dagger }_{x}\hat{b}_{y}\right|02 \right\rangleRecalling that \mid \hat{b}^{\dagger } n〉 = (n+1)^{1/2} \mid n+1〉 (Eq. (6.64)) and \mid \hat{b}n〉=n^{1/2} \mid n-1〉 (Eq. (6.65)) allows us to conclude that
\left\langle11\left|\hat{H} \right| 02\right\rangle =\frac{\hbar \kappa ^{\prime } }{2m\omega }\left\langle11\left|\hat{b}^{\dagger }_{x} \hat{b}_{y} \right|02 \right\rangle =\frac{\sqrt{2}\hbar \kappa ^{\prime } }{2m\omega } \left\langle11\left|\hat{b}^{\dagger }_{x} \right|01 \right\rangle=\frac{\sqrt{2}\hbar \kappa ^{\prime } }{2m\omega }\left\langle11\mid 11\right\rangle =\frac{\sqrt{2}\hbar \kappa ^{\prime } }{2m\omega }It follows that
\left [ \begin{matrix} 3\hbar \omega & \frac{\sqrt{2}\hbar \kappa ^{\prime } }{2m\omega } &0 \\ \frac{\sqrt{2}\hbar \kappa ^{\prime } }{2m\omega } & 3\hbar \omega & \frac{\sqrt{2}\hbar \kappa ^{\prime } }{2m\omega } \\ 0 & \frac{\sqrt{2}\hbar \kappa ^{\prime } }{2m\omega } & 3\hbar \omega \end{matrix} \right ] \left [ \begin{matrix} a_{1} \\ a_{2} \\ a_{3} \end{matrix} \right ] =E\left [ \begin{matrix} a_{1} \\ a_{2} \\ a_{3} \end{matrix} \right ]All we need to do now is find the eigenvalues of the matrix. The solutions are
E_{1} =3\hbar \omegaE_{2} =\frac{3m\hbar \omega ^{2}-\hbar \kappa ^{\prime } }{m\omega }
E_{3} =\frac{3m\hbar \omega ^{2}+\hbar \kappa ^{\prime } }{m\omega }
The corresponding eigenfunctions are given by the coefficients
\left [ \begin{matrix} a_{1} \\ a_{2} \\ a_{3} \end{matrix} \right ] =\frac{1}{\sqrt{2} } \left [ \begin{matrix} -1 \\ 0 \\ 1 \end{matrix} \right ]\left [ \begin{matrix} a_{1} \\ a_{2} \\ a_{3} \end{matrix} \right ] =\frac{1}{2} \left [ \begin{matrix} 1 \\ -\sqrt{2} \\ 1 \end{matrix} \right ]
\left [ \begin{matrix} a_{1} \\ a_{2} \\ a_{3} \end{matrix} \right ] =\frac{1}{2} \left [ \begin{matrix} 1 \\ \sqrt{2} \\ 1 \end{matrix} \right ]
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 \hat{W} =\kappa ^{\prime } \hat{x} \hat{y}. The perturbed Schrödinger equation matrix is
\left [ \begin{matrix} 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{matrix} \right ] \left [ \begin{matrix} a_{1} \\ a_{2} \\ a_{3} \\ a_{4} \end{matrix} \right ] =E\left [ \begin{matrix} a_{1} \\ a_{2} \\ a_{3} \\ a_{4} \end{matrix} \right ]which has eigenvalues
E_{1} =\frac{8m\hbar \omega ^{2}-3\hbar \kappa ^{\prime } }{2m\omega }E_{2} =\frac{8m\hbar \omega ^{2}-\hbar \kappa ^{\prime } }{2m\omega }
E_{3} =\frac{8m\hbar \omega ^{2}+\hbar \kappa ^{\prime } }{2m\omega }
E_{4} =\frac{8m\hbar \omega ^{2}+3\hbar \kappa ^{\prime } }{2m\omega }
The corresponding eigenfunctions are given by the coefficients
\left [ \begin{matrix} a_{1} \\ a_{2} \\ a_{3} \\ a_{4} \end{matrix} \right ] =\frac{1}{2\sqrt{2} }\left [ \begin{matrix} -1 \\ \sqrt{3} \\ -\sqrt{3} \\ 1 \end{matrix} \right ]\left [ \begin{matrix} a_{1} \\ a_{2} \\ a_{3} \\ a_{4} \end{matrix} \right ] =\frac{\sqrt{3} }{2\sqrt{2} }\left [ \begin{matrix} 1 \\ -1/\sqrt{3} \\ -1/\sqrt{3} \\ 1 \end{matrix} \right ]
\left [ \begin{matrix} a_{1} \\ a_{2} \\ a_{3} \\ a_{4} \end{matrix} \right ] =\frac{\sqrt{3} }{2\sqrt{2} }\left [ \begin{matrix} -1 \\ -1/\sqrt{3} \\ 1/\sqrt{3} \\ 1 \end{matrix} \right ]
\left [ \begin{matrix} a_{1} \\ a_{2} \\ a_{3} \\ a_{4} \end{matrix} \right ] =\frac{1 }{2\sqrt{2} }\left [ \begin{matrix} 1 \\ \sqrt{3} \\ \sqrt{3} \\ 1 \end{matrix} \right ]