Energy Eigenstates of a Perturbed Quantum Harmonic Oscillator

  • Thread starter PChar
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  • #1
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Homework Statement



(See attachment)

Homework Equations



[itex] x = \sqrt{\frac{\hbar}{2m \omega}} ( a + a^{\dagger} )[/itex]

[itex] x = i \sqrt{\frac{\hbar m \omega}{2}} ( a^{\dagger} - a )[/itex]


The Attempt at a Solution



In part a) I was able to construct a separable Hamiltonian for the harmonic oscillators in the x and y direction.

The x Hamiltonian includes the term [itex] - \frac{λ x'^{2}}{2}[/itex]

and the y Hamiltonian includes the term [itex] \frac{λ y'^{2}}{2}[/itex]

Before moving on to part b), at my professor's advice, I collected the squared terms like so:

(for x): [itex]\frac{m}{2} ( \omega^{2} - \frac{λ}{m} ) x'^{2}[/itex]

Calling the term within the brackets α (for y I called it β since there is a + instead of a -)

moving on to part b) I attempted to solve for the energy states by expressing all of the position and momentum operators in terms of the raising and lowering operators.

(for x) after expanding:

[itex]H_{x'} = -\frac{\hbar \omega}{4}(a_{x'}^{\dagger 2} - a_{x'}a_{x'}^{\dagger}- a_{x'}^{\dagger}a_{x'} + a_{x'}^{2}) + \frac{\hbar \omega}{4α}(a_{x'}^{2} + a_{x'}a_{x'}^{\dagger} + a_{x'}^{\dagger}a_{x'} + a_{x'}^{\dagger 2})[/itex]

After simplifying using the commutator between a and a dagger and a few steps of algebra:

[itex]H_{x'} = \frac{\hbar \omega}{4α} [ (2n_{x} + 1)(α + 1) + (1 - α)(a_{x'}^{\dagger 2} + a_{x'}^{2})][/itex]

I'm pretty sure I can't have those raising and lowering operators in my energy eigenvalues but I can't see any way to eliminate them, I know that in the unperturbed oscillator, the squared terms from the position and momentum operators will cancel, but the alpha and beta are causing problems.

Thanks in advance.
 

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Answers and Replies

  • #2
TSny
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Hi, PChar.

In terms of the primed quantities, doesn't the Hamiltonian have the form of two independent harmonic oscillators? The x'-oscillator will have ω' = √(ω2-λ/m) and the y-oscillator will have ω'' = √(ω2+λ/m). If so, then you should be able to write out the total energy of the system without really doing any more work.

If you do want to introduce raising and lowering operators, I think they would have the form that you gave below for the unprimed variables except ω would be replaced by ω' for the x’-oscillator and ω'' for the y’-oscillator.

[itex] x = \sqrt{\frac{\hbar}{2m \omega}} ( a + a^{\dagger} )[/itex]

[itex] p = i \sqrt{\frac{\hbar m \omega}{2}} ( a^{\dagger} - a )[/itex]
 
  • #3
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Ah, figures I would try do it the hard way.

So it seems like my energy eigenvalues will be:

[itex]E_{n} = \hbar \omega ' (n_{x} + \frac{1}{2}) + \hbar \omega '' (n_{y} + \frac{1}{2} )[/itex]

Which should also answer the question about degeneracy, there is none due to the ω' and ω" indicating which component (x and y) of the oscillator is at which energy level.
 
  • #4
TSny
Homework Helper
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Yes, that looks right. Good.
 

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