# Quantum harmonic oscillator wave function

Tags:
1. Nov 24, 2016

### Dean Navels

How do you find the wave function Φα when given the Hamiltonian, and the equation:

α(x) = αΦα(x)

Where I know the operator

a = 1/21/2((x/(ħ/mω)1/2) + i(p/(mħω)1/2))

And the Hamiltonian,

(p2/2m) + (mω2x2)/2

And α is a complex parameter.

I obviously don't want someone to do this question for me, just a point in the right direction. Thanks in advance!

2. Nov 24, 2016

### hilbert2

Do a Google search with terms "harmonic oscillator raising and lowering operators" and you will probably find many web sites where this is discussed in detail.

3. Nov 24, 2016

### Dean Navels

I've read a lot of them, what I don't understand is that when you apply a lowering or raising operator Φα becomes Φα+1 or Φα-1 yet here it doesn't change.

4. Nov 24, 2016

### hilbert2

Suppose the function $\Phi_\alpha$ can be expanded in terms of $\psi_n$, the eigenstates of the harmonic oscillator hamiltonian:

$\Phi_\alpha (x) = c_0 \psi_0 (x) + c_1 \psi_1 (x) + c_2 \psi_2 (x) + \dots$.

Now form a recursion equation for the numbers $c_n$ from the condition $a\Phi_\alpha = \alpha\Phi_\alpha$

EDIT: Sorry, I meant "recurrence relation", which is typical English-language term for that.

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

### Dean Navels

Is this required to solve it via ladder operators?

6. Nov 24, 2016

### hilbert2

^ Yes, you have to use the fact that any quantum state can be expanded in the basis of the eigenfunctions of the hamiltonian, and also the knowledge of how the ladder operators act on those functions.

7. Nov 24, 2016

### Dean Navels

After doing extensive reading on and around it, I know how to find the formula for different energies corresponding to different eigenfunctions, but now I need to put Φα(x) in the form of a normalisation constant C multiplied by an exponential.

8. Nov 24, 2016

### hilbert2

The action of a lowering operator $a$ on a QHO eigenstate $\psi_n$ is:

$a\psi_n (x) = \sqrt{n}\psi_{n-1}(x)$ .

Now form a general linear combination from the functions $\psi_n$ and act on it with $a$:

$a\sum_{n=0}^{\infty} c_n \psi_n (x) = \sum_{n=1}^{\infty} \sqrt{n} c_n \psi_{n-1} (x) = \sum_{n=0}^{\infty} \sqrt{n+1} c_{n+1} \psi_{n} (x)$ .

Now you should be able to write an equation for $c_n$ in terms of $c_{n-1}$ and $\alpha$.

9. Nov 25, 2016

### Dean Navels

I don't understand how to involve alpha. Thanks so much for all your help by the way.

10. Nov 25, 2016

### Dean Navels

Would I be correct in saying ∑(n-1)^½ Cn-1 Ψn(x)

11. Nov 25, 2016

### hilbert2

The eigenfunctions $\psi_n$ are linearly independent, therefore $\alpha c_n = \sqrt{n+1}c_{n+1}$ (why?). So what are the first few terms in the expansion of $\Phi_\alpha (x)$ ?

12. Nov 25, 2016

### ehild

See, for example http://quantummechanics.ucsd.edu/ph130a/130_notes/node153.html
The solutions of the Schrödinger equation can be written as a product of a Gaussian with a polynomial. You get recurrence relation among the coeffiients of the polynomials. These polynomials have the name Hermite polynomials.

13. Nov 25, 2016

### hilbert2

^ He's not looking for the eigenfunctions of the hamiltonian, the problem is about the eigenfunctions of the lowering operator.