Wavepacket Problem: Writing |psi(x,t)|^2

[trash057]

Homework Statement

A wavepacket, psi(x,t), can be expressed as a linear combination of eigenstates. Assuming that only 2 eigenstates, phi0(x) and phi1(x), contribute to the linear combination write down the expression for |psi(x,t)|^2.

Homework Equations

[Boltzmann's constant = 1.38 x10^23 J /K, Planck's constant h = 6.626 x10 ^-34 J s,
Gas constant R = 8.314 J K /mol]

The Attempt at a Solution

I need help with how to start solving this problem.

 

[ppyadof]

This is just an educated guess, but I'd say something like:
\psi = a_0 \psi_0 + a_1 \psi_1
So:
\int \psi \psi^{*} d^3r = \int (a_0 \psi_0 + a_1 \psi_1)(a^{*}_0 \psi^{*}_0 + a^{*}_1 \psi^{*}_1) d^3r
Now all you need to do is expand the RHS and see what you get (the integral is over all space, so what does that tell you about what the integral must equal?). If I were you I would also assume that the orginal wavefunctions (psi0 and psi1) are normalised, and that the constants a0 and a1 have no imaginary component.

That's what I would do for this question anyway, but I'm definitely no expert on the subject. lol.

 

[trash057]

This is just an educated guess, but I'd say something like:
\psi = a_0 \psi_0 + a_1 \psi_1
So:
\int \psi \psi^{*} d^3r = \int (a_0 \psi_0 + a_1 \psi_1)(a^{*}_0 \psi^{*}_0 + a^{*}_1 \psi^{*}_1) d^3r
Now all you need to do is expand the RHS and see what you get (the integral is over all space, so what does that tell you about what the integral must equal?).

The integral must equal 1. But do I need to expand the RHS? It just says to write the expression not to evaluate the expression.

 

[ppyadof]

The integral must equal 1. But do I need to expand the RHS? It just says to write the expression not to evaluate the expression.

oh yeah, my bad.

 

[trash057]

Hey, no problem. Thanks for your help. I'm not sure how to expand this. I can handle numbers but when symbols are used, I start to have trouble. I'm guessing there will be four results, two zeroes and two ones. The ones being psi1 x psi1*, and psi0 x psi0*, but I'm not sure how to expand this and get those results.

 

[ppyadof]

Unless I made a mistake, you should end up with something like:
|\psi|^2 = |a_0 \psi_0|^2 + |a_1 \psi_1|^2 + a_0a^{*}_1\psi_0\psi^{*}_1 + a_1a^{*}_0\psi_1\psi^{*}_0

Obviously the normalisation constants of the first two terms of that expression are going to be real. You can prove that by writing the normalisation constants as a complex exponential a_0 = r_0e^{i\theta_0}. If you do that with a1 aswell, then you could simplify the other terms involving a1*a0 and a0*a1.