Instantaneous doubling of the Infinite Square Well Width

In summary: Then set n=2 and evaluate the integral, which will give you a number. Square that number and that is the probability the particle will be found in the n=2 state when it is measured immediately after the well is expanded.In summary, the conversation discusses a particle of mass m moving in an infinite square well of width a and its energy eigenfunctions. It relates two orthogonal eigenfunctions to each other and shows that the eigenfunctions defined in the problem are orthogonal. It also considers the particle's state when the width of the potential is doubled, and the expansion postulate of quantum mechanics is used to make a prediction about the outcome of a measurement. The terms ψ(x) and n(x) are defined, and a
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Homework Statement


A particle of mass m is moving in an infinite square well of width a. It has the following normalised energy eigenfunctions:

$$u_n (x) = \sqrt{\frac{2}{a}} sin(\frac{n \pi x}{a})$$ (1)

a) Give an expression that relates two orthogonal eigenfunctions to each other and use it to show that the eigenfunctions un(x) defined above are orthogonal.

b) Suppose the particle is in the ground state when the width of the potential is doubled such that the well now extends from x = 0 to x = 2a. Write down the eigenfunctions for the new well.

c) Suppose that we make a measurement of the energy of the particle immediately after the well is extended. The expansion postulate of quantum mechanics can be used to make a prediction regarding the outcome of this measurement. Define the terms ψ(x) and
11px-Greek_phi_Didot.svg.png
n(x) in the equation, giving specific reference to the eigenfunctions defined in your answer to part (b), and also to Eq (1), as appropriate. Write down a general equation for calculating the coefficients an.

d) Calculate the probability that this measurement will determine the particle to be in the first excited state (n = 2)

Homework Equations


[/B]
Expansion postulate:

$$ \psi (x) = \sum_{n=1}^{\infty} a_n \phi _n (x)$$

The Attempt at a Solution



For part a), I know the equation for orthogonality is

$$\int \phi _n*\phi _m d\tau = 0$$

For n≠m.

So I simply put the eigenfunction defined in (1) into this for n and m and showed for all values of n and m that it equals zero.

b) Here, I have just substituted 'a' for '2a' in eq. (1)?

c) I know that ψ is obviously the wave function, and
11px-Greek_phi_Didot.svg.png
the eigenfunction.. where a set of eigenfunctions can form the 'complete set'? As all eigenfunctions of a Hermitian operator form a complete set... The weighting coefficients can be determined by:

$$a_n = \int \phi _n*\ \psi\ d\tau $$

Provided that
11px-Greek_phi_Didot.svg.png
n(x) are orthonormal.

But I'm not sure how to link this to my answer for b) and equation 1.

d) So, I'm led to believe that as the well expands 'instantaneously' the wave function remains the same as before, and 0 between x=0 and x=2a. How do I link this with the 'new' eigenfunctions defined in b)? I know I want to find the weighted coefficient a2 as this can give me the probability for the n=2 state.

I tried to find the coefficient using the equation defined in part c) and the two eigenfunction of the well, before and after, and by splitting the integral up between 0-a and a-2a, you can make the integral between a-2a = 0. But this didn't seem to yield anything and is obviously incorrect. I'm struggling to link some physical interpretation here - am I looking for some overlap of wave functions? Why?
 
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  • #2
You have all the ingredients you need. You need to write the "old" ground state wavefunction ##\psi(x)=\sqrt{2/a)} \sin(\pi x/a)## as a linear combination of the "new" eigenfunctions ##\phi_n(x)## of the expanded well that you find in part (b). In other words find ##a_n## such that $$\psi(x)=\sum_{n=0}^\infty a_n \phi_n(x).$$ You seem to have an expression for the ##a_n##, so crank out an integral to find the general form for arbitrary n.
 

1. What is the concept of instantaneous doubling of the Infinite Square Well Width?

Instantaneous doubling of the Infinite Square Well Width refers to the sudden increase in the size of the potential well in the Infinite Square Well model. This is typically done by doubling the width of the well, resulting in a sudden change in the energy levels and wave function solutions.

2. What is the effect of instantaneous doubling of the Infinite Square Well Width on energy levels?

The effect of instantaneous doubling of the Infinite Square Well Width is a sudden increase in the energy levels of the system. This means that the energy of the particles inside the well will increase, resulting in a larger energy gap between each level.

3. How does instantaneous doubling of the Infinite Square Well Width affect the wave function solutions?

Instantaneous doubling of the Infinite Square Well Width results in a sudden change in the wave function solutions. The wave functions will become more spread out and will have a larger amplitude at the boundaries of the well. This is because the particles have more space to move within the well, resulting in a larger probability of finding them at the edges.

4. What is the significance of instantaneous doubling of the Infinite Square Well Width in physics?

Instantaneous doubling of the Infinite Square Well Width is significant because it allows us to study the behavior of particles in a sudden change in potential energy. This can help us understand how particles react to sudden changes in their environment and can have applications in various fields such as quantum mechanics and solid-state physics.

5. Is instantaneous doubling of the Infinite Square Well Width a realistic scenario?

No, instantaneous doubling of the Infinite Square Well Width is not a realistic scenario. In real systems, changes in potential energy occur gradually and not instantaneously. However, it serves as a useful model for studying the behavior of particles in sudden changes in potential energy and can provide valuable insights into real-world systems.

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