Perturbation of a degenerate isotropic 2D harmonic oscillator

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Homework Statement



A two-dimensional isotropic harmonic oscillator of mass μ has an energy of 2hω. It experiments a perturbation V = xy. What are its energies and eigenkets to first order?

Homework Equations



The energy operator / Hamiltonian: H = -h²/2μ(Px² + Py²) + μω(x² + y²)

The Attempt at a Solution



The only eigenstates with such an energy are |1 0> and |0 1>, so now I have to find an operator that
a) conmutes with H and V and
b) has non-degenerate eigenvalues whose eigenkets are linear combination of these two states.

I've been trying some operators, but none seems to fulfill all the conditions. Which one would you use?

Thank you for your time.
 

Answers and Replies

  • #2
Orodruin
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The only eigenstates with such an energy are |1 0> and |0 1>, so now I have to find an operator that
a) conmutes with H and V and
b) has non-degenerate eigenvalues whose eigenkets are linear combination of these two states.

I've been trying some operators, but none seems to fulfill all the conditions. Which one would you use?

Treat the problem in degenerate perturbation theory. Your state space is built out of |0 1> and |1 0> as you say. The problem is equivalent to diagonalizing V in the constrained state space constructed from these two states.
 
  • #3
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Yep, that's what I'm doing, but Griffiths mentions another method which consists on finding an operator with the mentioned conditions and using its eigenkets in non-degenerate perturbation theory. According to him ihis approach can be used very often and it is faster and easier, so I'm doing the problem both ways.

Since you've mentioned it can you check if I've made this calculations for the matrix elemnts of V correctly?
We have
[itex] a_x = Ax+Bp_x[/itex]
(for some constants A and B; I wrote them like this so simplify manipulation)
[itex] x = \frac{a_x + a_x^{\dagger}}{2A}[/itex]

and also
[itex] [a_x, x] = i\hbar B[/itex],
[itex] [a_x, a_x^{\dagger}] = 1[/itex]
So what I've done is:

[itex]W_{aa} = \langle 10 \vert xy \vert 10\rangle = \langle 0 \vert a_x xy a_x^{\dagger}\vert 0\rangle =[/itex]
[itex]\langle 0 \vert x a_x y a_x^{\dagger}\vert 0\rangle+i\hbar B\langle 0 \vert y a_x^{\dagger}\vert 0\rangle =[/itex]
[itex]\langle 0 \vert x y a_x a_x^{\dagger}\vert 0\rangle+i\hbar B\langle 0 \vert a_x^{\dagger}y\vert 0\rangle =[/itex]
[itex]\langle 0 \vert x y a_x^{\dagger} a_x\vert 0\rangle+\langle 0 \vert x y \vert 0\rangle+0 = 0 + \langle 0 \vert x y \vert 0\rangle[/itex]

(And similar for ##W_{bb}##)

I've used the fact that [itex]a_x\vert 0\rangle = 0[/itex] and [itex]\langle 0\vert a_x^{\dagger} = 0[/itex]. Is it right?
 
  • #4
Orodruin
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You would also need the off-diagonal terms. The diagonal terms should be zero by virtue of ##x## and ##y## commuting. Thus, the interesting matrix element is
$$
W_{ab} = \langle 1 0 | xy | 0 1\rangle = \langle 1_x |x| 0_x\rangle \langle 0_y |y| 1_y\rangle
$$
as well as its hermitian conjugate.
 
  • #5
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Yep, I get zero on diagonal terms. I get non-zero on the off-diagonals, but I didn't do that. How did you separate |01> and xy like that?

the method I used consisted on writing x and y in terms of the ##a## and ##a^{\dagger}## and then conmute until I got ##a## at the right-most end of product of operators or ##a^{\dagger}## at the left-most end. Then I used [itex]a_x\vert 0\rangle = a_x\vert 0 1\rangle = 0[/itex] and [itex]\langle 0\vert a_x^{\dagger} =\langle 0 1\vert a_x^{\dagger} = 0[/itex] (and viceversa for y). Is this right?
 
  • #6
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Also, if I get null diagonal terms then the energies turn out to be [itex]E^1 = \pm W_{ab}[/itex]. This means the perturbed energies are equal in magnitude and of oposite sign... It feels weird....
 
  • #7
Orodruin
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Well, it is really not so strange. The energies split from the unperturbed level by the same amount, but in different directions. This is related to the perturbation being traceless.

The separation of x and y comes from knowledge about what is really meant by ##|1 0\rangle## etc., namely
$$
|n_x n_y\rangle = |n_x\rangle \otimes |n_y\rangle, \quad
\hat x = \hat x \otimes 1
$$
and so on. The x and y operators commute with each other (and so do their corresponding creation/annihilation operators).

I would also express x and y in terms of the creation and annihilation operators. This is how I immedeately see that the diagonal terms disappear.
 
  • #8
What is the value of <10|01>?... can any one explain please
 
Last edited:
  • #9
Yep, that's what I'm doing, but Griffiths mentions another method which consists on finding an operator with the mentioned conditions and using its eigenkets in non-degenerate perturbation theory. According to him ihis approach can be used very often and it is faster and easier, so I'm doing the problem both ways.

Since you've mentioned it can you check if I've made this calculations for the matrix elemnts of V correctly?
We have
[itex] a_x = Ax+Bp_x[/itex]
(for some constants A and B; I wrote them like this so simplify manipulation)
[itex] x = \frac{a_x + a_x^{\dagger}}{2A}[/itex]

and also
[itex] [a_x, x] = i\hbar B[/itex],
[itex] [a_x, a_x^{\dagger}] = 1[/itex]
So what I've done is:

[itex]W_{aa} = \langle 10 \vert xy \vert 10\rangle = \langle 0 \vert a_x xy a_x^{\dagger}\vert 0\rangle =[/itex]
[itex]\langle 0 \vert x a_x y a_x^{\dagger}\vert 0\rangle+i\hbar B\langle 0 \vert y a_x^{\dagger}\vert 0\rangle =[/itex]
[itex]\langle 0 \vert x y a_x a_x^{\dagger}\vert 0\rangle+i\hbar B\langle 0 \vert a_x^{\dagger}y\vert 0\rangle =[/itex]
[itex]\langle 0 \vert x y a_x^{\dagger} a_x\vert 0\rangle+\langle 0 \vert x y \vert 0\rangle+0 = 0 + \langle 0 \vert x y \vert 0\rangle[/itex]

(And similar for ##W_{bb}##)

I've used the fact that [itex]a_x\vert 0\rangle = 0[/itex] and [itex]\langle 0\vert a_x^{\dagger} = 0[/itex]. Is it right?
Finally what is the value of #w_{aa}#?
 
  • #10
blue_leaf77
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What is the value of <10|01>?... can any one explain please
Zero.
Finally what is the value of #w_{aa}#?
Please make a separate thread if you have an elaborate question.
 

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