Question 4.P.17: Consider a particle of mass m and charge q moving under the ...

To find the eigenenergies of the Hamiltonian

\hat{H}=\frac{1}{2m} \hat{P}^{2}+\frac{1}{2} m\omega ^{2}\hat {X}^{2}-q\varepsilon \hat{X},                  (4.334)

it is convenient to use the change of variable y=\hat{X}-q \varepsilon /(m\omega ^{2}). Thus the Hamiltonian becomes

\hat{H}=\frac{1}{2m} \hat{P}^{2}+\frac{1}{2} m\omega ^{2}\hat {y}^{2}-\frac{q^{2}\varepsilon ^{2} }{m\omega ^{2}}.           (4.335)

Since the term q^{2}\varepsilon ^{2}/(2m\omega ^{2}) is a mere constant and \hat{P}^{2} /(2m)+\frac{1}{2} m\omega ^{2}\hat{y}^{2}=\hat{H}_{HO} has the structure of a harmonic oscillator Hamiltonian, we can easily infer the energy levels:

E_{n} =〈n|\hat{H}|n 〉=\hbar \omega \left(n+\frac{1}{2}\right) -\frac{q^{2}\varepsilon ^{2} }{2m\omega ^{2}}.                  (4.336)

The wave function is given by \psi _{n} (y)=\psi _{n}(x-q\varepsilon /(m\omega ^{2})), where \psi _{n} (y) is given in (4.172):

\psi _{n} (y)=\frac{1}{\sqrt{\sqrt{\pi }2^{n}n!x_{0} } } e^{-x^{2}/2x^{2}_{0} } H_{n} \left(\frac{x}{x_{0} } \right).                   (4.172)

\psi _{n} (y)=\frac{1}{\sqrt{\sqrt{\pi }2^{n}n!x_{0} } } e^{-y^{2}/2x^{2}_{0} } H_{n} \left(\frac{y}{x_{0} } \right) .               (4.337)