Quantum harmonical oscillator with electric field

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SUMMARY

The discussion centers on transforming the Hamiltonian of a quantum harmonic oscillator subjected to an electric field. The Hamiltonian is defined as H = p²/(2m) + (1/2)mω²x² - qE'x. The goal is to find a change of variables from x to u that simplifies the eigenvalue equation Hφ(x) = Eφ(x) into a more manageable form. The suggested approach involves completing the square on the last two terms of the Hamiltonian to clarify the transformation process.

PREREQUISITES
  • Understanding of quantum mechanics, specifically Hamiltonian mechanics.
  • Familiarity with harmonic oscillators and their eigenvalue problems.
  • Knowledge of variable transformations in differential equations.
  • Basic grasp of electric fields and their effects on charged particles.
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  • Study the method of completing the square in Hamiltonians.
  • Learn about eigenvalue problems in quantum mechanics.
  • Research variable transformations in quantum systems.
  • Explore the effects of electric fields on quantum harmonic oscillators.
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Quantum physicists, graduate students in physics, and researchers working on quantum mechanics and harmonic oscillator problems will benefit from this discussion.

Chen
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Hi,

I have a particle of mass m and charge q, which is located in the potential of an harmonic oscillator and also subject to a constant electric field. The Hamiltonian is given as:

[tex]H = \frac{p^2}{2m} + \frac{1}{2}m \omega ^2 x^2 - q E' x[/tex]

And I need to find a change of variables from x to u, so that the eigenvalue equation:

[tex]H \phi (x) = E \phi (x)[/tex]

Becomes:

[tex][-\frac{h^2}{2m}\frac{d^2}{du^2}+\frac{1}{2}m \omega ^2u^2] \phi (u) = (E + \frac{q^2 E'^2}{2m \omega ^2}) \phi (u)[/tex]

(It's an h-bar there, of course.) I don't even know where to start. I tried plugging u(x) into the original eigenvalue equation and find some constraint on u from there, to no avail.

Thanks
 
Last edited:
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Complete the square on the last 2 terms in the Hamiltonian, and the transformation might become a bit more obvious.

Regards,
George
 
Doh... thanks! :smile:
 

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