(a) Verify that the matrices representing the operators \hat{X} and \hat{P} in the N-space for a harmonic oscillator obey the correct commutation relation [\hat{X},\hat{P}]=i\hbar.
(b) Show that the energy levels of the harmonic oscillator can be obtained by inserting the matrices of \hat{X} and \hat{P} into the Hamiltonian \hat{H}=\hat{P}^{2}/(2m)+\frac{1}{2}m\omega ^{2}\hat{X}^{2}.
(a) Using the matrices of \hat{X} and \hat{P} in (4.181)
\tan \left(k\frac{a}{2} \right) =\frac{\hbar ^{2}k}{mV_{0} } \Longrightarrow \tan \left(\sqrt{\frac{ma^{2}|E| }{2\hbar ^{2}} } \right) =\sqrt{\frac{2\hbar ^{2}|E|}{mV^{2}_{0} } }and (4.182),
E(a)\simeq \frac{1}{2m}\left(\frac{\hbar }{2a} \right) ^{2} +\frac{1}{2}m\omega ^{2}a^{2}we obtain
\hat{X}\hat{P}=i\frac{\hbar }{2}\left(\begin{matrix} 1 & 0 & -\sqrt{2} & … \\ 0 & 1 & 0 & … \\ \sqrt{2} & 0 & 1 & … \\ \vdots & \vdots & \vdots & \ddots \end{matrix} \right), \hat{X}\hat{P}=i\frac{\hbar }{2}\left(\begin{matrix} -1 & 0 & -\sqrt{2} & … \\ 0 & -1 & 0 & … \\ \sqrt{2} & 0 & -1 & … \\ \vdots & \vdots & \vdots & \ddots \end{matrix} \right); (4.292)
hence
\hat{X}\hat{P}-\hat{P}\hat{X}=i\hbar \left(\begin{matrix} 1 & 0 & 0 & … \\ 0 & 1 & 0 & … \\ 0 & 0 & 1 & … \\ \vdots & \vdots & \vdots & \ddots \end{matrix} \right) (4.293)
or [\hat{X},\hat{P}]=i\hbar I, where I is the unit matrix.
(b) Again, using the matrices of \hat{X} and \hat {P} in (4.181) and (4.182), we can verify that
\hat{X}^{2}=\frac{\hbar }{2m\omega } \left(\begin{matrix} 1 & 0 & \sqrt{2} & … \\ 0 & 3 & 0 & … \\ \sqrt{2} & 0 & 5 & … \\ \vdots & \vdots & \vdots & \ddots \end{matrix} \right), \hat{P}^{2}=-\frac{m\hbar \omega }{2} \left(\begin{matrix} -1 & 0 & \sqrt{2} & … \\ 0 & -3 & 0 & … \\ \sqrt{2} & 0 & -5 & … \\ \vdots & \vdots & \vdots & \ddots \end{matrix} \right); (4.294)
hence
\frac{\hat{P}^{2}}{2m} +\frac{1}{2}m\omega ^{2}\hat{X}^{2} =\frac{\hbar \omega }{2} \left(\begin{matrix} 1 & 0 & 0 & … \\ 0 & 3 & 0 & … \\ 0 & 0 & 5 & … \\ \vdots & \vdots & \vdots & \ddots \end {matrix} \right). (4.295)
The form of this matrix is similar to the result we obtain from an analytical treatment, E_{n}=\hbar \omega (2n+1)/2, since
H_{n\prime n} =〈n^{\prime }|\hat{H}|n 〉=\frac{\hbar \omega }{2}(2n+1) \delta _{n\prime n}. (4.296)