Question 7.2: Consider a particle of mass m in a two-dimensional oscillato...

In terms of the rotated coordinates, the Hamiltonian is

H=\frac{1}{2m}\left(\frac{\partial^{2}}{\partial \acute{x}^{2} }+\frac{\partial^{2}}{\partial \acute{y}^{2} } \right) +\frac{1}{2}m(1+\epsilon ) \omega ^{2}\acute{x}^{2}+\frac{1}{2}m(1-\epsilon ) \omega ^{2}\acute{y} ^{2}.

This amounts to two independent one-dimensional oscillators. The exact solutions are

\psi _{mn}=\psi ^{+}_{m}(\acute{x} )\psi ^{-}_{n}(\acute{y} ),

where \psi ^{\pm }_{m} are one-dimensional oscillator states with frequencies \omega _{\pm} =\sqrt{1\pm \epsilon}\omega   respectively. The first few exact energies,

E_{mn}=\left(m+\frac{1}{2} \right) \hbar \omega _{+}+\left(n+\frac{1}{2} \right) \hbar \omega _{-}    (7.21)

are shown in Figure 7.5.

The two states which grow out of the degenerate first-excited states as ϵ is increased have m=0 ,n=1 (lower state) and m=1, n=0 (upper state). If we track these states back to ϵ=0 (in that limit \omega_{+}= \omega _{-}=\omega ) we get

\underset{\epsilon \longrightarrow 0}{\lim}\psi _{01}(x)=\underset{\epsilon \longrightarrow 0}{\lim}\psi ^{+}_{0}(\acute{x} ) \psi ^{-}_{1}(\acute{y})=\psi _{0}\left (\frac{x+y}{\sqrt{2} } \right) \psi_{1}\left (\frac{x-y}{\sqrt{2} } \right)=\sqrt{\frac {2}{\pi} } \frac{m \omega }{\hbar } \frac{x-y}{\sqrt{2}}e^{-\frac{m \omega }{2\hbar } (x^{2}+y^{2})}

=\frac{-\psi ^{0}_{a}+ \psi ^{0}_{b}}{\sqrt{2} }

\underset{\epsilon \longrightarrow 0}{\lim}\psi _{10}(x)=\frac{\psi ^{0}_{a}+\psi ^{0}_{b}}{\sqrt{2} }     (7.22)

Therefore the “good” states for this problem are

\psi ^{0}_{\pm }\equiv \frac{1} {\sqrt{2} } (\psi ^{0}_{b}\pm\psi ^{0}_{a})       (7.23)

In this example we were able to find the exact eigenstates of H and then turn off the perturbation to see what states they evolve from. But how do we find the “good” states when we can’t solve the system exactly?