# Time Dependent Wave Function for Particle in Infinite Square Well

1. Apr 9, 2014

### LunaFly

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

A particle is in a bound state of the infinite square well. It is in a state represented by the following wavefunction, written here at t=0:

ψ(x)= -√(2/3)√(2/L) * sin (3πx/L) + i*√(1/3)√(2/L) * sin (2πx/L)

(a)Write the full time-dependent wavefunction for this state, and calculate the time-dependent probability density.
(b)Use a computer program to plot the probability density at a series of time steps to show how the probability density evolves with time.

2. Relevant equations

**Note: I used "h" to represent h-bar below.

Schrodinger equation: -h2/(2m) * ∂ψ(x,t)/∂x + V(x,t) * ψ(x,t) = ih * ∂ψ(x,t)/∂t

From deriving the time-independent wave function of particle in infinite square well: ψ(x,t)= ψ(x)*ø(t),
with ø(t)=e-i*E*t/h

Energy of particle in nth state En = n22*h2/(2*m*L2)

3. The attempt at a solution

I know that the ψ(x) given is a linear combination of energy eigenfunctions, the first being from the n=3 state and the second being from the n=2 state of the particle in an infinite square well.

My first instinct was to multiply each individual eigenfunction by the ø(t) that corresponds to said state. I ended up with a wavefunction of:

ψ(x,t)= -√(2/3)√(2/L) * sin (3πx/L) * e-i*E3*t/h + i*√(1/3)√(2/L) * sin (2πx/L) * e-i*E2*t/h,

with E3= energy of particle in n=3 state, and E2= energy of particle in n=2 state.

This all seemed fine and dandy. It solved the time-dependent Schrodinger equation, so I was pleased.

Where things got weird was when I calculated the probability density, ψ*ψ. I ended up with the equation:

ψ*ψ= (2/L) * (2/3 * sin2(3πx/L) + 1/3 * sin2(2πx/L) + i*√(2)/3 * sin (2πx/L) * sin (3πx/L) * (ei*(E2-E3)*t/h - ei*(E3-E2)*t/h))

This is consistent with the problem statement in that ψ(x, 0) = ψ(x); however I feel like the imaginary term in there is wrong. How can the probablility have an imaginary component? I have tried using Euler's formula to transform the eikt term but still end up with an imaginary term.

This also leads to the problem of plotting this probability density function. Is it possible to make a plot with an imaginary term? I am using Mathematica but haven't found much information on the subject.

Any insight to where I went wrong or how to continue and deal with the imaginary term would be most appreciated.

Last edited: Apr 9, 2014
2. Apr 10, 2014

### electricspit

I think you want this:

$\sin{\theta}=\frac{1}{2i} [e^{i\theta}-e^{-i\theta}]$

See how the energies are mismatched in the second exponential term? If you pull out the negative you will get it in this form. If you use the above formula you will get a real probability (see the $i$ on the bottom?).

Remember also that the modulus squared of the wave function must be real to be physically meaningful.

3. Apr 10, 2014

### LunaFly

Excellent! Thank you for the reply, electricspit. This is exactly what I need to cancel out all of the "i"'s in the expression.

Also thanks for the clarification about the probability density.

Last edited: Apr 10, 2014
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