# Instantaneous doubling of the Infinite Square Well Width

1. May 1, 2017

### carrlr

1. The problem statement, all variables and given/known data
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 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)

2. Relevant equations

Expansion postulate:

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

3. 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 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 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?

2. May 1, 2017

### kuruman

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.