Normalization of Linear Superposition of ψ States

To normalize the wave function, you would need to know the values of A and B, or have additional information about the system.
  • #1
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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
 
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  • #2
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##.
 
  • #3
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.
 
  • #4
The Head said:
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)/[itex]\hbar[/itex].

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.
 
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  • #5
.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
 

1. What is the purpose of normalizing the linear superposition of ψ states?

The purpose of normalization is to ensure that the probability of finding a particle in a certain state is equal to 1. This allows for accurate and consistent calculations of probabilities and measurements in quantum mechanics.

2. How is normalization achieved in the linear superposition of ψ states?

Normalization is achieved by dividing each coefficient in the superposition by the square root of the sum of the squares of all coefficients. This ensures that the total probability of finding the particle in any state is equal to 1.

3. Why is it important to normalize the linear superposition of ψ states?

Normalizing the superposition allows for meaningful comparisons between different quantum states and simplifies calculations involving probabilities and measurements. It also ensures that the wavefunction remains physically meaningful and does not violate the laws of quantum mechanics.

4. Can a superposition of ψ states be normalized if one of the coefficients is negative?

Yes, a superposition can still be normalized if one or more coefficients are negative. The negative sign will be included in the calculation, but it will not affect the overall normalization. The absolute value of the coefficient is what matters in the normalization process.

5. Can the linear superposition of ψ states be normalized if the coefficients are complex numbers?

Yes, the linear superposition can still be normalized if the coefficients are complex numbers. The normalization process remains the same, but the complex conjugate of each coefficient must be used in the calculation of the normalization constant. This ensures that the total probability remains real and positive.

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