# Quantum mechanics problem - Potentials

1. Sep 13, 2011

### RedPsi

1. The problem statement, all variables and given/known data
The problem involves a particle confined to an interval 0 < x < L (in one dimension). It asks to solve for the functional form of the potential V(x) given a wave function (to get to the relevant question, I won't bother providing this). In order to solve for the potential, it was wise to use the fact that the Schrodinger equation is linear - i.e. separate the wave function into two terms and solve for V(x) for each case. In the end, the two potentials should have similar functional form.

My problem begins with the results - how to identify my constants.

2. Relevant equations
First potential found:

V(x) = $\alpha$$\bar{h}$ - ($\bar{h}$2$\pi$2/2mL2)

Second potential found:

V(x) = ($\alpha$ + $\beta$)$\bar{h}$ - (2$\bar{h}$2$\pi$2/mL2)

where alpha and beta are unknowns.

3. The attempt at a solution

From here, it appears that the second term in each expression assumes the form of energy for a typical infinite-square well problem:

$\bar{h}$2$\pi$2n2/mL2

where n corresponds to the quantum number indicating energy state.

My main question is, what is beta? The first term in each represents the energy eigenvalue produced by the Hamiltonian in the SEquation.

My initial thoughts:
First expression -> n = 1 state, Second expression -> n = 2 state.

1.) beta = (n-1)

OR

2.) beta = (n-1)alpha

OR

If it was real nice,
3.) beta = 0
I don't have the background knowledge on how the energy eigenfunction (first term) would look like in this perspective. Any help would be appreciated!

Last edited: Sep 14, 2011
2. Sep 14, 2011

### kuruman

Hi RedPsi and welcome to PF. Note that the two potentials are constant and do not depend on x. This means that the problem is the same as the god old particle in the box problem except that the zero of energy has been shifted in each of the two cases. The parameters α and β determine by how much. Does this help?

3. Sep 14, 2011

### RedPsi

Hi Kuruman, thanks for the reply!

Well, it's definitely a start, but I still don't see how I would go about finding the beta value. The question is clearly indicating that I find some connection between the two potentials in order to find this, but I just don't see it.

hmm.

4. Sep 14, 2011

### kuruman

To help you some more, I will need additional information. How did alpha and beta get into the picture? Can you post the exact statement of the problem? Furthermore, I am not sure what you mean by "separate the wave function into two terms and solve for V(x) for each case." The statement may make sense in the context of the problem, but I don't have that context.

5. Sep 14, 2011

### RedPsi

Sure thing,

A particle of mass m is conﬁned to the interval 0 < x < L in one dimension, and it has
wavefunction

Ψ(x, t) = Aexp(−iαt)sin(πx/L)[1 + exp(−iβt)cos(πx/L)]

(a) What is functional form of the potential energy V (x) in the region 0 < x < L?

(b) Determine the value of A and β. Is it possible to determine the value of α without more
information, and does it matter

This is the exact form of the question. My professor hinted that for part a, we separate psi into two terms (as you can see it is possible), and solve the Schrodinger Eq. for each piece (possible since SE is linear). He stated that the functional form of V(x) for each psi should have similar functional form. From here, he also stated that doing this will help us determine that beta is in the next part of the question.

My problem lies in his last statement.

6. Sep 14, 2011

### kuruman

Now it makes sense to me. Look what happens when you bring exp(−iαt)sin(πx/L) into the square brackets. You get the sum of two terms: sin(πx/L)exp(−iαt) + sin(πx/L)cos(πx/L)exp(−i(α+β)t). The first term is the eigenstate of the Hamiltonian for n=1. Can you recognize the second term also as an eigenstate? Hint: What is sin(2θ) in terms of cosθ and sinθ?

7. Sep 14, 2011

### RedPsi

Yes, it assumes the form 2sin(theta)cos(theta) which is exactly what we have. Thus, the second term is in the second eigenstate (n=2 as mentioned in the starting post). Although this provides me with a new way to look at it, I still don't know what the functional form of the first term is.

Therefore returning to the original problem, how do I find beta :S

8. Sep 14, 2011

### kuruman

You know that the general solution of the time-dependent Schrodinger equation is

$\Psi(x,t)={\sum}_{n} sin\frac{n\pi x}{L}e^{-iE_n t/\hbar}$

Since you know that the given wave function is a linear combination of the first two states, it should follow that α is related to E1 and α+β to E2.

9. Sep 14, 2011

### RedPsi

!! I totally overlooked that! Thank's so much!!

For a generic infinite-square well where we set v(x) = 0 then,

alpha = E1
alpha + beta = E2 => beta = E2 - E1 = 3(hbar)^2(pi)^2/(2mL^2)

hmm, is that so?

10. Sep 14, 2011

### kuruman

It is almost so. Note that there is no hbar in the denominators of the exponentials with alpha and beta. You need to adjust the dimensions a bit.

11. Sep 14, 2011

### RedPsi

beta = 3hbar pi^2 /(2mL^2) in the end.

It's interesting how making both potentials equal to each other yields the same thing.

Anywho, thank you kuruman for your help, it has been invaluable!