Rigid Rotator (Quantum Mechanics)

In summary: Right?But you messed up the left-hand side. When you multiply \psi by its complex conjugate, you get something that depends on both \sin and \cos. You should get a coefficient in front that depends on m, and you may also get some terms that depend on m and \phi. That's fine. You should end up with something of the formA^2 \int_0^{2\pi} \sin^2 m\phi + B^2 \cos^2 m\phi + AB \sin m\phi \cos m\phi\,d\phiwhich you should be able to find an antiderivative for using trig identities, and then evaluate at 2\pi
  • #1
bon
559
0

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!
 
Physics news on Phys.org
  • #2
Have i got the right normalisation condition etc?
 
  • #3
bon said:
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 Lz, 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
[tex]\psi = e^{\pm I am \phi}[/tex]
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 [itex]\phi \to \phi+2\pi[/itex], you get the same answer.
where normalisation means |A|^2 + |B^2| = 1/pi (is this right..?)
No. Where did you get [itex]1/\pi[/itex] from? Also, it looks like you have a typo on the LHS.
 
  • #4
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?
 
  • #5
vela said:
If you\'re allowed to use your knowledge of Lz, 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
[tex]\\psi = e^{\\pm I am \\phi}[/tex]
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 [itex]\\phi \\to \\phi+2\\pi[/itex], you get the same answer.

No. Where did you get [itex]1/\\pi[/itex] 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,,
 
  • #6
Let me ask you this: what's the difference between a normalized wave function and an unnormalized wave function?
 
  • #7
vela said:
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..
 
  • #8
when i say integrate over all space, i mean integrate psi * psi
 
  • #9
OK, so you had [itex]\psi(\phi) = A\sin m\phi+B\cos m\phi[/itex], and to normalize it, you want

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

1. What is a rigid rotor in quantum mechanics?

A rigid rotor is a model used in quantum mechanics to describe the rotational motion of a molecule or atom. It assumes that the molecule or atom is a rigid body with no internal vibrations or deformations, and only the overall rotational motion is considered.

2. How is the rigid rotor model used in quantum mechanics?

The rigid rotor model is used to calculate the energy levels and wavefunctions of a molecule or atom in rotational motion. This can help in understanding the rotational spectra of molecules and can provide insights into their structure and properties.

3. What is the difference between a classical and quantum rigid rotor?

In classical mechanics, rotational motion is described by Newton's laws of motion and the conservation of angular momentum. In quantum mechanics, the rigid rotor model takes into account the quantization of angular momentum, where the angular momentum can only take certain discrete values.

4. How do you solve the Schrödinger equation for a rigid rotor?

The Schrödinger equation for a rigid rotor can be solved by using the separation of variables technique, where the wavefunction is expressed as a product of radial and angular components. The angular component can then be further solved using spherical harmonics and the resulting equations can be solved using appropriate boundary conditions.

5. What are the applications of the rigid rotor model in quantum mechanics?

The rigid rotor model has various applications in quantum mechanics, including the study of rotational spectra of molecules, understanding the rotational motion of atoms and molecules in different environments, and in the development of quantum technologies such as quantum computing and quantum sensing.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
230
  • Introductory Physics Homework Help
Replies
1
Views
659
Replies
8
Views
1K
Replies
3
Views
282
  • Quantum Physics
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Quantum Physics
Replies
6
Views
1K
  • Advanced Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
958
Back
Top