Time evolution of spherical harmonics

  • Thread starter misterme09
  • Start date
  • #1
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Homework Statement


At t=0, a given wavefunction is:

[tex]\left\langle\theta,\phi|\psi(0)\right\rangle = \frac{\imath}{\sqrt{2}}(Y_{1,1}+Y_{1,-1})[/tex]

Find [tex]\left\langle\theta,\phi|\psi(t)\right\rangle[/tex].

Homework Equations


[tex]\hat{U}(t)\left|\psi(0)\right\rangle = e^{-\imath\hat{H}t/\hbar}\left|\psi(t)\right\rangle[/tex]

[tex]
\hat{H}\left|\ E,l,m\right\rangle = E\left|\ E,l,m\right\rangle
[/tex]
[tex]
\hat{L^{2}}\left|\ E,l,m\right\rangle = l(l+1)\hbar^{2}\left|\ E,l,m\right\rangle
[/tex]
[tex]
\hat{L_{z}}\left|\ E,l,m\right\rangle = m\hbar\left|\ E,l,m\right\rangle
[/tex]

The Attempt at a Solution


I know that you can use the above operator to make time evolution of an energy eigenstate, but I can't figure out what energy to use for the two spherical harmonics in the given state at t=0.
 
Last edited:

Answers and Replies

  • #2
Avodyne
Science Advisor
1,396
90
It depends on what the hamiltonian is.
 
  • #3
18
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Oh, okay, I should have realized this.

This is for a rigid rotator with Hamiltonian

[tex]\hat{H}=\hat{L^{2}}/2I[/tex]

So this means that the energies are both:

[tex]E =1(1+1)\hbar^{2}/2I = \hbar^{2}/I[/tex]

And

[tex]\left\langle\theta,\phi|\psi(t)\right\rangle=\frac{i}{\sqrt{2}}e^{-\imath\hbar t/I}(Y_{1,1}+Y_{1,-1})[/tex]
 
Last edited:
  • #4
Avodyne
Science Advisor
1,396
90
Yep.
 

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