Perturbation of a degenerate isotropic 2D harmonic oscillator

In summary, the conversation revolves around finding the energies and eigenkets of a two-dimensional isotropic harmonic oscillator with a perturbation. The conversation includes a discussion of different methods for solving the problem and calculations for the matrix elements of the perturbation operator. The value of the matrix element W_{aa} is determined to be zero.
  • #1
carllacan
274
3

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.
 
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  • #2
carllacan said:
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
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
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
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
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
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.
 
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  • #8
What is the value of <10|01>?... can anyone explain please
 
Last edited:
  • #9
carllacan said:
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
THEYOPHY KYO said:
What is the value of <10|01>?... can anyone explain please
Zero.
THEYOPHY KYO said:
Finally what is the value of #w_{aa}#?
Please make a separate thread if you have an elaborate question.
 

Related to Perturbation of a degenerate isotropic 2D harmonic oscillator

1. What is a degenerate isotropic 2D harmonic oscillator?

A degenerate isotropic 2D harmonic oscillator is a physical system that exhibits periodic motion in two dimensions, with a restoring force that is directly proportional to the displacement from equilibrium. It is referred to as "degenerate" because it has multiple energy states with the same energy level, and "isotropic" because it has the same characteristics in all directions.

2. How does perturbation affect a degenerate isotropic 2D harmonic oscillator?

Perturbation refers to a small disturbance or change in a system. In the case of a degenerate isotropic 2D harmonic oscillator, perturbation can affect the system by altering the restoring force or introducing external forces. This can result in changes in the period and amplitude of the oscillations, as well as the energy levels of the system.

3. What factors can cause perturbation in a degenerate isotropic 2D harmonic oscillator?

Perturbation in a degenerate isotropic 2D harmonic oscillator can be caused by various factors, including external forces such as friction, changes in the mass or stiffness of the system, or changes in the initial conditions (e.g. initial displacement or velocity). Additionally, perturbation can also be caused by interactions with other systems or external influences such as magnetic fields.

4. How is perturbation of a degenerate isotropic 2D harmonic oscillator mathematically modeled?

Perturbation of a degenerate isotropic 2D harmonic oscillator can be mathematically modeled using the perturbation theory, which involves expanding the equations of motion into a series of terms and solving for the effects of each term on the system. This allows for the prediction of changes in the oscillation period, amplitude, and energy levels of the system due to perturbation.

5. What are the practical applications of studying perturbation in a degenerate isotropic 2D harmonic oscillator?

The study of perturbation in a degenerate isotropic 2D harmonic oscillator has practical applications in various fields such as physics, engineering, and chemistry. It can help in understanding the behavior of systems under external influences and in predicting the effects of perturbations on the system. This knowledge can be applied in the design and optimization of various devices and processes, such as precision instruments, electronic circuits, and chemical reactions.

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