Normalization of Linear Superposition of ψ States

[The Head]

Homework Statement

An electron in an infinitely deep potential well of thickness 4 angstroms is placed in a linear superposition of the first and third states. What is the frequency of oscillation of the electron probability density?

Homework Equations

E=hω

The Attempt at a Solution

My main problem right now is with the normalization of the wave function:

I have ψTotal=Aψ1+ Bψ3
∫|ψT|^2=1=∫(|A|^2*sin^2(pi*z/Lz) +|B|^2*sin^2(3pi*z/Lz))dz (the other terms with A*B and B*A become zero after integration)
=(Lz(|A|^2+|B|^2))/2=1

So
|A|^2+|B|^2=2/Lz

What I'd like to do next is calculate the probability density of the electrons, but I end up with |A|^2, |B|^2, A*B, & B*A terms and don't know how to get actual numbers for all these constants, or otherwise to properly normalize this. I know what to do if the superposition of states was equally split, but it does not say that here.

Any thoughts on what I'm doing wrong? Thanks

 

[TSny]

You don't need to worry about the values of A and B to answer the question. The frequency of oscillation of the probability distribution does not depend on A or B. So, you can leave A and B unspecified. Since you are interested in the time dependence of the probability distribution, you will need to include the time dependence of each of the two stationary states when you calculate $|\Psi_{total}(t)|^2$.

 

[The Head]

So would the only thing that really matters be the exp(+/-(E3-E1)it/h), where ω=(E3-E1)/h? It just seems strange to me that whether one eigenstate dominates or not does not affect the frequency of electron probability density.

Also, I have difficulty in general with normalization. If I needed to normalize this sort of function, would that be possible with the information given, or would I need something else?

Thank you & I appreciate your help.

 

[TSny]

So would the only thing that really matters be the exp(+/-(E3-E1)it/h), where ω=(E3-E1)/h? It just seems strange to me that whether one eigenstate dominates or not does not affect the frequency of electron probability density.

Yes, ω=(E3-E1)/$\hbar$.

When you write out $|\Psi_{total}(t)|^2$, you should be able to see the effect of changing the relative sizes of A and B. In particular, you should think about the case where A = B and the case where A >> B or A << B.

Also, I have difficulty in general with normalization. If I needed to normalize this sort of function, would that be possible with the information given, or would I need something else?

From the information given in the problem, you cannot determine the magnitudes of A and B. As you showed, all you can determine is the value of |A|² + |B|².

 

[paranormal]

.I would first like to clarify that the concept of normalization is a crucial aspect of quantum mechanics. It ensures that the total probability of finding a particle in any location is equal to 1. In other words, a normalized wave function represents a physically meaningful state.

In the given scenario, the normalization condition is given by:

∫|ψT|^2dz = ∫(|A|^2*sin^2(pi*z/Lz) +|B|^2*sin^2(3pi*z/Lz))dz = 1

This can be rewritten as:

(Lz/2)*(|A|^2+|B|^2) = 1

Since the thickness of the potential well is given as 4 angstroms, we can substitute Lz = 4 angstroms in the above equation to get:

2*(|A|^2+|B|^2) = 1

Now, we know that the linear superposition of states ψ1 and ψ3 is given as:

ψTotal = Aψ1+ Bψ3

To find the values of A and B, we can use the normalization condition along with the fact that the states are orthogonal to each other. This means that:

∫ψ1*ψ3dz = 0

Using this condition, we can write:

∫(Aψ1+ Bψ3)*ψ3dz = 0

Simplifying this, we get:

|B|^2 = - (A*B)/3

Substituting this value in the normalization condition, we get:

2*(|A|^2+|B|^2) = 1

2*(|A|^2 - (A*B)/3) = 1

|A|^2 - (A*B)/3 = 1/2

Substituting the value of B from the previous equation, we get:

|A|^2 + (A^2)/3 = 1/2

Solving this quadratic equation, we get two solutions:

A = ±sqrt(3/8) and B = ∓sqrt(1/8)

Since the values of A and B can be either positive or negative, we can choose either of the solutions. For simplicity, let's choose A = sqrt(3/8) and B = -sqrt(1/8).

Now, to find the frequency of oscillation