Charged harmonic oscillator in an electric field

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SUMMARY

The discussion focuses on finding the eigenvalues and eigenstates of a charged harmonic oscillator in an external electric field, represented by the Hamiltonian H = p²/(2m) + (1/2)mω²x² - qεx. The transformation u = x - (q²ε²)/(2mω²) is utilized to simplify the Hamiltonian. The energy eigenvalues are derived as E' = E + (q²ε²)/(2mω²), leading to E_n = ħω(n + 1/2). The discussion emphasizes the importance of correctly completing the square to transition back to the variable x.

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bjogae
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Homework Statement



A charged harmonic oscillator is placed in an external electric field [tex]\epsilon[/tex] i.e. its hamiltonian is [tex]H = \frac{p^2}{2m} + \frac{1}{2}m \omega ^2 x^2 - q \epsilon x[/tex] Find the eigenvalues and eigenstates of energy

Homework Equations





The Attempt at a Solution



By completing the square i get
[tex][-\frac{\hbar^2}{2m}\frac{d^2}{du^2}+\frac{1}{2}m \omega ^2u^2] \phi (u) = (E + \frac{q^2 \epsilon^2}{2m \omega ^2}) \phi (u)[/tex]
where
[tex]u=x-\frac{q^2\epsilon^2}{2m\omega^2}[/tex].

Then usually for Hamiltonians of this kind the energy eigenvalues are
[tex]E_n=\hbar\omega(n+\frac{1}{2})[/tex]
but how do I obtain them in this case? Or is this the right way to go?
Do i call
[tex]E + \frac{q^2 \epsilon^2}{2m \omega ^2}=E'[/tex]
which would give me
[tex]E'_n=\hbar\omega(n+\frac{1}{2})[/tex]
And how do I swich back to x?
 
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bjogae said:
By completing the square i get
[tex][-\frac{\hbar^2}{2m}\frac{d^2}{du^2}+\frac{1}{2}m \omega ^2u^2] \phi (u) = (E + \frac{q^2 \epsilon^2}{2m \omega ^2}) \phi (u)[/tex]
where
[tex]u=x-\frac{q^2\epsilon^2}{2m\omega^2}[/tex].

That doesn't look quite right...
 

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