# Harmonic oscillator's energy levels

1. May 18, 2013

### Rorshach

1. The problem statement, all variables and given/known data
Okay, this one confuses me a bit:

A particle is in a one-dimensional harmonic oscillator. At time t = 0 is given by its wave function

ψ(x)=Nx3exp(-mωx2/2hbarred)

a) At this point you measure the particle's energy. What measurement values ​​are available? Also determine the corresponding probabilities!
b) After the power supply that gave outcome E = 3hbarredω/2 measure the particle's position immediately. What is the probability of finding the particle in the classically forbidden region? (The classically forbidden region is defined by the condition that V (x)> = E_total)

2. Relevant equations

En=(n+1/2)hbarredω

3. The attempt at a solution
I don't really have an idea how to come up with energy levels that are available, or how to calculate the probabilities of those levels.

2. May 18, 2013

### BruceW

"one-dimensional harmonic oscillator" should imply the form of the Hamiltonian. You can almost guess what it should look like. If it is not in your notes, then you could google it if you are having trouble.

3. May 19, 2013

### Rorshach

I got notice on violation of rules of the forum by posting a problem that I did not try to solve on my own- while it is not entirely true, I understand it could be perceived as such. I am sincerely sorry, won't happen again. I tried to come with idea how to find out how many possible energy levels are available for my function, but only thing that comes to my mind is this basic formula for energy E=(n+1/2)hbarredω. Hamiltonian for my function is of course
h=-hbarred2/2m(d/dx)+1/2mω2x2 , but how to connect it to my search for available energy levels?

4. May 19, 2013

### fluidistic

I don't really understand why would the Hamiltonian help to solve for a).
At first glance the psi function they give you is a supperposition of only 2 eigenstates (can you find them?) that have a definite energy. I guess that these energies are the one you could possibly measure.

As for the energy level formula $E_n =\hbar \omega (n+1/2)$, it comes from solving the Schrödinger's equation and assuming that the solution behaves as $f(x)v(x)$ where v(x) is a decaying exponential that solves the Schrödinger's equation when x tends to infinity and the function f(x) are the Hermite polynomials that you can determine when solving the original Schrödinger's equation using a power series "guess function" for f(x). You get that the solution doesn't diverge if and only if the power series is truncated for some n. This condition gives precisely that $E_n =\hbar \omega (n+1/2)$.

5. May 20, 2013

### Rorshach

I don't know how to find those two eigenstates, how can You tell from first glance that there will be two levels?

6. May 20, 2013

### fluidistic

Hint: Look at the first few Hermite polynomials and at the psi(x) they give you.

7. May 20, 2013

### Rorshach

So it is connected to power to which x is taken right? if x was to the power of 10 there would be 6 levels?

8. May 20, 2013

### fluidistic

If you only had x to the power of 10, I think so. But if you had an x^10 as well as say an x^9 in your psi(x), then no, not only necessarily 6 levels.
The idea is that the $\psi _n (x)$ form a basis for any function $\Psi (x,t)$ for a fixed t. In your case t is fixed at $t=0$.
So that you can write your $\Psi (x,0)$ as a linear combination of the $\psi _n(x)$. One writes $\Psi (x,0)=\Psi(x)=\sum _{n=0}^\infty c_n \psi _n (x)$.
We're lucky enough that the from the $\Psi (x)$ they give you, it's not hard to determine the coefficients $c_n$ and the $\psi _n (x)$ involved.
I'll also tell you a "trick". The associated probabilities with each energy levels are simply the values of $|c_n|^2$. But in your exercise, I think you're meant to find them by integration (I've never done it, so I'll have to do it myself too!). Though it's good to know 2 methods so you can compare the results afterwards.

9. May 20, 2013

### Rorshach

okay, but how do I determine the cn and the ψn(x)? It sounds similar to distributing the function into the Fourier series, but I have never had a clue how to perform Fourier series distribution on a function... if I am right and this is the thing to do, could you show me on example how to perform Fourier series distribution?

10. May 20, 2013

### fluidistic

I'm not sure it's a "Fourier expansion" but the idea is pretty similar. We must exploit the orthogonality property of the eigenstates; or the fact that they form a basis for any psi(x).
We have $\Psi(x)=\sum _{n=0}^\infty c_n \psi _n (x)$. Multiply both sides by $\psi ^* _m (x)$ and integrate from - infinity to positive infinity. After a bit of algebra you'll reach that $c_m = \int _{-\infty}^\infty \psi ^* _ m (x) \Psi (x) dx$.

11. May 20, 2013

### Rorshach

I'm sorry, but I still don't know how to get ψn. It seems to me that without it I cannot do anything.

12. May 20, 2013

### fluidistic

You get them when you solve the -normalized- Schrödinger's equation ($\psi '' + (2\varepsilon - y^2)\psi =0$ where $y=x(m \omega / \hbar )^{1/2}$ and $\varepsilon = E/(\hbar \omega )$) and assume that the solutions you're looking for are of the form $\psi (y)=v(y)\exp (-y^2/2)$. And in order to calculate v(y) you propose a power series of the form $v(y)=\sum _{k=0}^\infty C_k y^k$.
In order for $\psi (y)$ to converge you'll get that the power series for v(y) must be truncated at some value for k=n.
You'll also get the recurrence relation $C_{k+2}=C_k \frac{2k+1-2\varepsilon }{(k+2)(k+1)}$. So for any value of n, you get that v(y) is a Hermite polynomial.
And thus that the $\psi (y)$ turns out to be of the form $\psi _n (y)= H_ n (y)e^{-y^2/2}$.
After replacing the variables back and normalizing, one reaches that $\psi _n (x)=\left ( \frac{m\omega }{\pi \hbar 2^{2n}n!^2} \right ) ^{1/4} \exp \left ( - \frac{m\omega x^2}{2 \hbar} \right ) H_n \left [ \left ( \frac{m\omega }{\hbar} \right ) ^{1/2 }x \right ]$.

So yes, in your exercise I don't know how you could solve it without knowing the eigenfunctions $\psi _n (x)$. Either you look at them in a book/Internet, either you find them yourself (which is a pain but maybe necessary).

13. May 22, 2013

### Rorshach

Ok, but how do I know into which polynomials I am supposed to distribute my wave function? I only know that the hermite polynomial of the third power of x is equal to 8x3-12x. Am I supposed to look for some coefficients that will convert that polynomial into my wave function?

14. May 23, 2013

### vela

Staff Emeritus
What are the first few eigenfunctions for the one-dimensional harmonic oscillator?

15. May 23, 2013

### Rorshach

Those are the functions that follow the formula $$ψ_n=C_nexp(\frac{-\xi^2}{2})H_n(\xi)$$, right?

16. May 23, 2013

### vela

Staff Emeritus
Right, assuming you have the correct expressions for the normalization constant and $\xi$. I'm going to use $\phi_n$ to denote the eigenfunctions, so you don't confuse it with the given wavefunction $\psi$.

You want to find the appropriate coefficients $a_0, a_1, a_2, \dots$ such that
$$\psi(x) = a_0 \phi_0(x) + a_1 \phi_1(x) + a_2 \phi_2(x) + a_3\phi_3(x) + \cdots$$ Substitute in for $\psi$, $\phi_0$, etc., and see if you can figure out how to solve for the coefficients. You should easily be able to see why $a_n=0$ for $n\ge 4$.

17. May 24, 2013

### Rorshach

$a_n=0$, because the highest power in the formula is 3, right? I got to the point where I am supposed to match the coefficients, but I am somewhat confused by hermitian $H_n(\xi)=\sum_{n=0}^\infty c_n\xi^n$, namely: I know that hermitian with the power of $\xi$ being equal to 3 is $H_3(\xi)=8\xi^3-12\xi$ am I supposed to fix the constants for the elements of the hermitian sum so it would match the formula for the hermitian of third power $exp(-\frac{1}{2}\xi^2)C_3(8\xi^3-12\xi)$, or I am supposed to fix the constants so it would only leave $\xi^3$?

Last edited: May 24, 2013
18. May 24, 2013

### vela

Staff Emeritus
I don't understand what you're asking. Write it out explicitly in math.

19. May 24, 2013

### Rorshach

ok, so my wavefunction is $\psi(x)=Nx^3exp(-\frac{m\omega x^2}{2\hbar})$. Now we substitute certain values: $\omega=\sqrt{\frac{k}{m}}, \alpha=\sqrt{\frac{m\omega}{\hbar}}, \lambda=\frac{2E}{\hbar\omega}, \xi=x\alpha$. Then we can write the formula for the eigenfunctions: $\psi(\xi)=C_nH(\xi)exp(-\frac{\xi^2}{2})$. We can write that $H_n(\xi)\sum_{n=0}^\infty\ c_n\xi^n$, right? Now I know that hermit polynomial $H(\xi)$for $\xi^3$ is equal to $8\xi^3-12\xi$. And now I don't know what coefficients I am supposed to be looking for: the ones that will give me just $\xi^3$, or the ones that will give me $8\xi^3-12\xi$.

Last edited: May 24, 2013
20. May 24, 2013

### vela

Staff Emeritus