# Time Dependence of the Infinite Square Well

1. Dec 27, 2009

### Odysseus

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

The question comes straight from Intro to QM by Griffiths (pg 29, Q 2.6).

A wave equation is given representing an even mixture of the first two energy levels of the infinite square well. The task is to normalize the wave function, state it explicitly and then derive some expectation values.

2. Relevant equations

The wave equation is
Ψ(x,o)=A[ψ_1+ψ_2]

where ψ_1 and ψ_2 are the first and second stationary states in the one dimensional infinite square well. ψ_n=sqrt(2/a)*sin((n*pi*x)/a), where a is the length of the interval.

3. The attempt at a solution

For convenience I set β=pi/a

(1) the normalization:
(A^2) ∫|Ψ(x,o)|^2 dx (x=0 to a) = 1

Which works out to

(A^2)[∫(sin(βx)^2)dx + ∫(sin(βx)*sin(2βx))dx + ∫(sin(2βx)^2)dx] = 1

The different wave functions are orthonormal which cancels the middle integral and the other two are just the normalization integrals for the individual energy levels, thus.

(A^2)[(a/2)+(a/2)] = 1

A=sqrt(1/a)

(2) Explicit statement of wave function and |Ψ(x,t)|^2

Ψ(x,t)= sqrt(1/a)[ψ_1+ψ_2]e^(-iωt)

Now here is my main question, the problem calls for |Ψ(x,t)|^2 to be written in terms of sinusoidal functions of time, but

|Ψ(x,t)|^2 = ΨΨ*

so

ΨΨ*= (1/a)[(ψ_1+ψ_2)^2](e^(-iωt))(e^(iωt))

ΨΨ*= (1/a)[(ψ_1+ψ_2)^2]

because the exponentials cancel. There is no time dependence. There must be though because in the question immediately following, which asks for <x>, says that the value oscillates in time.

The math in the book is a little out of my comfort zone so it is completely possible that I've missed something trivial.

Thanks for any help.

2. Dec 28, 2009

### EricAngle

Your expression for Ψ(x,t) is incorrect. The wavefunction Ψ(x,t) is a sum over the stationary states ψ_1(x) and ψ_2(x), with both states multiplied by their respective exponential time factors, exp(- i E_n t / hbar), where E_n is the energy of the n-th stationary state.

Last edited: Dec 29, 2009
3. Dec 29, 2009

### Odysseus

Thanks a lot, Eric. It makes much more sense now.