Hamiltonian and first order perturbation

[gasar8]

Homework Statement

[/B]
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)

Homework Equations

x= \frac{x_0}{ \sqrt{2} } (a_x+a_x^\dagger) \\<br /> y= \frac{x_0}{ \sqrt{2} } (a_y+a_y^\dagger)\\<br /> a |n&gt; = \sqrt{n} |n-1&gt;\\<br /> a^\dagger |n&gt; = \sqrt{n+1} |n+1&gt;\\<br /> a a^\dagger - a^\dagger a = 1

The Attempt at a Solution

a)
I was trying to solve Schroedinger's equation first with separating x,y, using H |\psi&gt; = E |\psi&gt;, |\psi&gt;=X(x)Y(y), but got stuck at:
-\frac{\hbar}{2m} \frac{X&#039;&#039;}{X}+ \frac{1}{2}m \omega^2 x^2 = E,
-\frac{\hbar}{2m} \frac{Y&#039;&#039;}{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&gt;, \Delta H |02&gt;,\Delta H |11&gt;,:

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

using commutator, I got:
\lambda x^2 y |20&gt; = \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&gt;.

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 agives different result on |20> as a a^\dagger.
Can anyone please help me?
Thank you very much.

 

[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?

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.

That's a good point.

Second excited state has got |11>, |20> and |02>

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.

 

[TSny]

The Attempt at a Solution

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.

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.

 

[gasar8]

I know that the energies are E = \hbar \omega (n + \frac{1}{2}), so I was assuming that E=E_x+E_y, 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 (n_x+n_y+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>

 

[blue_leaf77]

I was assuming that E=Ex+Ey

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

Eigenfunctions for harmonic oscillators are some Hermite polynomials? Is this correct?

Yes that's correct.

 

[TSny]

Aha, thanks for warning. So, how do I use creation and anihillation operators here in perturbation?

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.

 

[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 (n_x + 4 n_y + 2) = 3 and second excited state is only |10>?

 

[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 $\omega$ .

 

[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})

(n_x + 2 n_y + 3/2) = 3?

 

[blue_leaf77]

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})

(n_x + 2 n_y + 3/2) = 3?

Yes, that's correct.

 

[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:

 

[blue_leaf77]

Sorry but the picture is too small.

antisimetric

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.

 

[gasar8]

Sorry, I thought it can be enlarged. Here's the link: http://shrani.si/f/1s/g6/3Zw6xWnV/4/20160225183031.jpg

Ok, maybe I misunderstood our profesor. :)

 

[TSny]

Overall, looks pretty good. But when writing x and y in terms of creation and destruction operators, is the x_o 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 $a^\dagger_x a^\dagger_x a_y$.

 

[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 ω_x and ω_y again?

Hehe, nice hack, thanks. :)

 

[TSny]

Or do I have to take into account ω_x and ω_y again?

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