Rigid Rotator (Quantum Mechanics)

[bon]

Homework Statement

The Hamiltonian for a rigid rotator in the xy plane is H = -hbar^2 / 2I d^2/dphi^2

Find the energy levels and eigenfunctions of H.

The unnormalised wavefn of the rotator at time t=0 is:

psi = 1 + 4sin^2 phi

Find the possible results of a measurement of its energy and their relative probabilities

Homework Equations

The Attempt at a Solution

Ok so I see that the Hamiltonian is basically hbar^2 Lz ^2 / 2I therefore its energy levels are hbar ^2 m^2 / 2I

Also see that its eigenfunctions are psi = A sin mphi + B cos mphi

where normalisation means |A|^2 + |B^2| = 1/pi (is this right..?)

I\'ve decomposed sin^2phi and have written psi (phi) = 3-2cos2phi

I can also normalise this

I see that its a sum of m=0 wavefunction and m=2 wavefunction

but to work out the relative probabilities I need to work out the amplitude <psi m=2|psi>

But my question is: what do I use for |psi m=2>? I don't have the constants A and B..

So how can i work out the amplitude?

bit confused..

thanks!

 

[bon]

Have i got the right normalisation condition etc?

 

[vela]

Ok so I see that the Hamiltonian is basically hbar^2 Lz ^2 / 2I therefore its energy levels are hbar ^2 m^2 / 2I

Also see that its eigenfunctions are psi = A sin mphi + B cos mphi

If you're allowed to use your knowledge of L_z, you can just say what the eigenfunctions and eigenvalues are. If not, you need to derive them, in which case, at this point, you don't know anything about m other than it's some number. You might find it easier by writing the solutions as
\psi = e^{\pm I am \phi}
Now you have to enforce boundary conditions to find the allowed values of m. The condition here is that the wave function must be single valued, i.e. if \phi \to \phi+2\pi, you get the same answer.

where normalisation means |A|^2 + |B^2| = 1/pi (is this right..?)

No. Where did you get 1/\pi from? Also, it looks like you have a typo on the LHS.

 

[bon]

Oh i got the 1/pi from the integral of cos^2 and sin^2 between 0 and 2pi..?

What is the normalisation meant to be?

 

[bon]

If you\'re allowed to use your knowledge of L_z, you can just say what the eigenfunctions and eigenvalues are. If not, you need to derive them, in which case, at this point, you don\'t know anything about m other than it\'s some number. You might find it easier by writing the solutions as
\\psi = e^{\\pm I am \\phi}
Now you have to enforce boundary conditions to find the allowed values of m. The condition here is that the wave function must be single valued, i.e. if \\phi \\to \\phi+2\\pi, you get the same answer.

No. Where did you get 1/\\pi from? Also, it looks like you have a typo on the LHS.

Ignoring the typo, where has my normalisation gome wrong?

I just integrated psi*psi over 2pi and that's what you get,,

 

[vela]

Let me ask you this: what's the difference between a normalized wave function and an unnormalized wave function?

 

[bon]

Let me ask you this: what\'s the difference between a normalized wave function and an unnormalized wave function?

normalised will integrate over all space to give 1, unnormalised wont..

 

[bon]

when i say integrate over all space, i mean integrate psi * psi

 

[vela]

OK, so you had \psi(\phi) = A\sin m\phi+B\cos m\phi, and to normalize it, you want

\int_0^{2\pi} \psi^*(\phi)\psi(\phi)\,d\phi = 1