# Quantum mechanics, bidimensional harmonic oscillator

1. Nov 1, 2015

### Aler93

1. The problem statement, all variables and given/known data
At t=0 the wave function of a two-dimensional isotropic harmonic oscilator is

ψ(x,y,0)=A(4α^2 x^2+2αy+4α^2 xy-2) e^((-α^2 x^2)/2) e^((-α^2 y^2)/2)

where A its the normalization constant
In wich instant. Wich values of total energy can we find and which probability?
2. Relevant equations
None

3. The attempt at a solution
I dont know how to start it.

2. Nov 1, 2015

### blue_leaf77

Can you set up the Hamiltonian of an isotropic harmonic oscillator?

3. Nov 1, 2015

### Aler93

I find that the hamiltonian of the isotropic harmonic oscillator is

$$H=\frac{\hbar^2}{2m} \nabla^2+\frac{1}{2}m^2w^2p^2$$

that's what you mean ?

Last edited: Nov 1, 2015
4. Nov 1, 2015

### blue_leaf77

That does not seem to be what we are interested in. I just searched in Google and I can find quickly enough in the first few search results. What about 1D harmonic oscillator, do you know how the Hamiltonian looks like?

5. Nov 1, 2015

### Aler93

No, i don´t know, I'm really cinfused about it

6. Nov 1, 2015

### Staff: Mentor

Please use $\#\# \: \#\$# delimiters for inline LaTeX equations and $\\ \: \\$ for displayed equations. I've edited your post accordingly.

7. Nov 1, 2015

### blue_leaf77

The Hamiltonian for 1D oscillator reads
$$H = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac{1}{2}m\omega^2x^2$$
For 2D harmonic oscillator, you clearly need to add two more terms identical to those two above only that the variable is now $y$. If you are still confused, go back to Google.

8. Nov 1, 2015

### Aler93

Developing the Laplacian for a 2D harmonic oscilator, i´ll be

$$H=\frac{\hbar^2}{2m} (\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2)+\frac{1}{2}m^2w^2(x^2+y^2)$$

9. Nov 1, 2015

### blue_leaf77

Correct, except for the absence of a minus sign in front of $\hbar^2/2m$. Now this equation is easy to solve because the variables $x$ and $y$ are separated. Suppose the solutions of the separate Hamiltonians
$$H_1 = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + \frac{1}{2}m\omega^2x^2$$
and
$$H_2 = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial y^2} + \frac{1}{2}m\omega^2y^2$$
are $u_p(x)$ and $u_q(y)$ respectively. What do you think the solution of the total Hamiltonian $H = H_1+H_2$ will look like in terms of $u_p(x)$ and $u_q(y)$?

10. Nov 1, 2015

### Aler93

I´ll be the superposition rigth?
If $U_m(x)$ and $U_n(y)$ are the solutions, the total Hamiltonian is
$$H= U_m(x) + U_n(y)$$

11. Nov 1, 2015

### blue_leaf77

No, its not a superposition. Ok let me rephrase in previous statement, I should have said "solve the Schroedinger equation", not "solve the Hamiltonian". What we want to solve is
$$(H_1+H_2)\psi_{pq}(x,y) = E_{pq}\psi_{pq}(x,y)$$
with the total energy $E_{pq} = E_p+E_q$. Now I will give you a hint. In this problem, we have that
$$H_1\psi_{pq}(x,y) = E_p \psi_{pq}(x,y) \\ H_2\psi_{pq}(x,y) = E_q \psi_{pq}(x,y) \\ H_1 u_{p}(x) = E_p u_{p}(x) \\ H_2 u_{q}(x) = E_q u_{q}(x)$$
Using these, you are ought to find $\psi_{pq}(x,y)$ in terms of $u_{p}(x)$ and $u_{q}(y)$.

12. Nov 1, 2015

### Aler93

A question, ih this correct?

$$H_1 \psi_{pq}(x,y)=H_1u_p(x)$$

As $H_1$ only depends of X, and from your last post you say $u_p(x)$ was the $H_1$ solution, Is it true?

13. Nov 1, 2015

### blue_leaf77

No, it's not like that since it will imply that $\psi_{pq}(x,y)=u_p(x)$.
This problem actually belongs to those which can be solved by separation of variables method. I suggest you read http://tutorial.math.lamar.edu/Classes/DE/SeparationofVariables.aspx, especially Example 1.

14. Nov 1, 2015

### Aler93

15. Nov 2, 2015

### blue_leaf77

Yes, you can find what I asked at the end of page 2.
Yes, that's how we should proceed. The goal at this step is just to make you aware of how the total eigenfunction looks like in terms of the individual oscillator eigenfunctions. In the next step, we are supposed to expand the given wavefunction $\psi(x,y,0)$ in terms of $\psi_{pq}(x,y)$'s.
$$\psi(x,y,0) = \sum_p \sum_q c_{pq} \psi_{pq}(x,y)$$
with $c_{pq}$ a constant. In general, each sum above runs from $0$ to $\infty$. However, if we can relate the appearance of $\psi(x,y,0)$ with the nature of the function $\psi_{pq}(x,y)$ for a given $p$ and $q$, we should find that only a few constants $c_{pq}$ out of these infinite series are non-vanishing. So, your next task is to find out the functional form of $\psi_{pq}(x,y)$. First, you have known how $\psi_{pq}(x,y)$ looks like in terms of $u_p(x)$ and $u_q(y)$, right?