# Hamiltonian and first order perturbation

1. Feb 24, 2016

### gasar8

1. The problem statement, all variables and given/known data

Particle is moving in 2D harmonic potential with Hamiltonian:

$$H_0 = \frac{1}{2m} (p_x^2+p_y^2)+ \frac{1}{2}m \omega^2 (x^2+4y^2)$$

a) Find eigenvalues, eigenfunctions and degeneracy of ground, first and second excited state.
b) How does $$\Delta H = \lambda x^2y$$ split second excited state? (First order of perturbation)

2. Relevant equations
$$x= \frac{x_0}{ \sqrt{2} } (a_x+a_x^\dagger) \\ y= \frac{x_0}{ \sqrt{2} } (a_y+a_y^\dagger)\\ a |n> = \sqrt{n} |n-1>\\ a^\dagger |n> = \sqrt{n+1} |n+1>\\ a a^\dagger - a^\dagger a = 1$$
3. The attempt at a solution

a)
I was trying to solve Schroedinger's equation first with separating x,y, using $$H |\psi> = E |\psi>, |\psi>=X(x)Y(y),$$ but got stuck at:
$$-\frac{\hbar}{2m} \frac{X''}{X}+ \frac{1}{2}m \omega^2 x^2 = E$$,
$$-\frac{\hbar}{2m} \frac{Y''}{Y}+ 2 m \omega^2 y^2 = E$$.
Then, I tried by using anihillation and creation operators (for p and x), but gave up, because all the $$(a_i+a_i^\dagger)^2$$ became too complicated for calculating the ground, first and second state.

b)
I am assuming that the correction is 0, because y in ΔH is even function and if we integrate it across all space it gives us 0, but I can't prove it.
Second excited state has got |11>, |20> and |02>, so I was calculating $$\Delta H |20>, \Delta H |02>,\Delta H |11>,$$:

$$\lambda x^2 y |20> = \frac{\lambda \sqrt{2} x_0^3}{4} (a_x+a_x^\dagger)^2(a_y+a_y^\dagger)|20>$$

using commutator, I got:
$$\lambda x^2 y |20> = \frac{\lambda \sqrt{2} x_0^3}{4} (a_x^2+a_x^{\dagger 2}+1+2 a^\dagger a)(a_y+a_y^\dagger)|20>$$.

Here I have got problems, because if I calculate $$(a+a^\dagger)^2 = a^2+a a^\dagger + a^\dagger a + a^{\dagger 2}= \\ =a^2+a^{\dagger 2}+2 a a^\dagger-1 \\=a^2+a^{\dagger 2}+2 a^\dagger a+1$$
and $$a^\dagger a$$gives different result on |20> as $$a a^\dagger.$$
Thank you very much.

2. Feb 24, 2016

### blue_leaf77

This is by no means an introductory level of harmonic oscillator problem. In other words, this question assumes that you know how to solve 1D harmonic oscillator along with its eigenfunctions and energies. You are ought to use these information to solve part a). So, do you know how the asked quantities look like for 1D harmonic oscillator?
That's a good point.
No, some of them do not correspond to the second excited state. You need to know the expression of the energy eigenvalues to answer this.

3. Feb 24, 2016

### TSny

You meant that y in ΔH is an odd function.

But be careful here. For a degenerate energy level, an odd-function perturbation doesn't necessarily imply that the first-order correction to the degenerate energy level will be zero.

4. Feb 24, 2016

### gasar8

I know that the energies are $$E = \hbar \omega (n + \frac{1}{2}),$$ so I was assuming that E=Ex+Ey, where $$E_x = \hbar \omega_x (n_x + \frac{1}{2})\\ E_y = \hbar \omega_y (n_y + \frac{1}{2}).\\ \omega_x= \omega_y = \omega$$
If this is correct, I would get $$\omega_x= \sqrt{\frac{k_x}{m}}\\ \omega_y=\sqrt{\frac{k_y}{4m}}.$$ But are these two ω really the same?
Now, I get : $$E = \hbar \omega (n_x+n_y+1)$$
Eigenfunctions for harmonic oscillators are some Hermite polynomials? Is this correct?
Wow, I don't understand this. I thought that second excited state is, when (nx+ny+1)=3, so I would get these kets?

Edit:
Sorry TSny, yes, of course I meant odd. :)
Aha, thanks for warning. So, how do I use creation and anihillation operators here in perturbation? Do I have to use it on a definite state like |20> or do I have to calculate it on a general state |nm>, because as I have mentioned in first post I get different results when I use $$a_x a_x^\dagger \\ a_x^\dagger a_x$$ on |20>

Last edited: Feb 24, 2016
5. Feb 24, 2016

### blue_leaf77

Yes that's true, but $\omega_x \neq \omega_y$. Take a look again at the potential.
Yes that's correct.

6. Feb 24, 2016

### TSny

First, it would be a good idea to get the correct expression for $\omega_y$ in terms of the $\omega$ given in H. Then you can work out the first few energy levels and their degeneracy. After that, you can try working out the effect of the perturbation.

7. Feb 24, 2016

### gasar8

Oh my God, I really don't know what to do with this factor 4 before y2. :D Is this now correct:
$\frac{1}{2} m \omega^2 (x^2+4y^2)= \frac{1}{2} m \omega^2 x^2+ 4 \frac{1}{2} m \omega^2 y^2),$ so $\omega = \omega_x=4 \omega_y$?

And I get $E=\hbar \omega_x (n_x+\frac{1}{2}) + \hbar \omega_y(n_y+\frac{1}{2})=\hbar \omega (n_x+\frac{1}{2}) + 4 \hbar \omega (n_x+\frac{1}{2}) = \hbar \omega (n_x + 4 n_y + 2)$

So (nx + 4 ny + 2) = 3 and second excited state is only |10>?

8. Feb 24, 2016

### TSny

Try to write the y part of the potential energy as $\frac{1}{2}m\omega_y^2 \; y^2$ and identify $\omega_y$ in terms of $\omega$ .

9. Feb 24, 2016

### gasar8

I feel hopeless. :)
$\frac{1}{2} m \omega_y^2 y^2=\frac{1}{2} 4 m \omega^2 y^2$, so $\omega_y^2=4 \omega^2$

$E=\hbar \omega_x (n_x+\frac{1}{2}) + \hbar \omega_y(n_y+\frac{1}{2})=\hbar \omega (n_x+\frac{1}{2}) + 2 \hbar \omega (n_y+\frac{1}{2}) = \hbar \omega (n_x + 2 n_y +\frac{3}{2})$

(nx + 2 ny + 3/2) = 3?

10. Feb 24, 2016

### blue_leaf77

Yes, that's correct.

11. Feb 25, 2016

### gasar8

Ok, thank you so much guys, you helped me a lot up to the present.
I tried to solve the b) part on my own now, the photo is on the bottom of the post. I really hope it is correct, can someone please check it?
I am just wondering about that final matrix down there. We learned that the matrix should be antisimetric, but I don't find any minus anywhere?
Photo:

12. Feb 25, 2016

### blue_leaf77

Sorry but the picture is too small.
Really? Given that the perturbation is Hermitian, this implies that the off-diagonal elements must be purely imaginary. However, the action of the ladder operators onto the kets always results in a real constant prefactor. I don't see why the off-diagonal matrix element should be purely imaginary.

13. Feb 25, 2016

### gasar8

14. Feb 25, 2016

### TSny

Overall, looks pretty good. But when writing x and y in terms of creation and destruction operators, is the xo factor the same for y as it is for x?

To shorten your calculation somewhat, note that when calculating

$\left< 2, 0| \textrm{sum of a bunch of products of creation and destruction operators} | 0, 1 \right>$,

you should be able to see that the only contribution will be from a term that raises 0 to 2 for x and lowers 1 to 0 for y. So, the only term that contributes will be $a^\dagger_x a^\dagger_x a_y$.

Last edited: Feb 25, 2016
15. Feb 25, 2016

### gasar8

Isn't it? We have $x_0=\sqrt{\frac{\hbar}{m \omega}}$, so the ω is the same, and mass too? Or do I have to take into account ωx and ωy again?

Hehe, nice hack, thanks. :)

16. Feb 25, 2016

### TSny

Yes, you do. The y part of the unperturbed Hamiltonian acts like an independent harmonic oscillator with its own ωy.