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

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