Time-dependent wavefunction

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Homework Statement



If given, for instance, psi(phi, 0)=[1/sqrt(2pi)](cos^2(phi/2) + isin(phi)), which is the wavefunction at t=0, how do you go about finding the wavefunction at time t, psi(phi,t)??

Homework Equations





The Attempt at a Solution



Would it simply be psi(phi, t)=psi(phi,0)e^(-iEt/hbar)??

The given wavefunction is for a particle on a ring.... and for that, E_n=(hbar^2)(m^2)/(2I) where I=mr^2. Is that the value of "E" I plug into the above equation??

Is there any other constant that I have to multiply the psi(phi,t) equation by as a result of adding on the e^(-iEt/hbar) term??

Any help would be great, I'm not entirely sure if I get this :-/
 

Answers and Replies

  • #2
Avodyne
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In general, you have to write the initial wave function as a sum of energy eigenfunctions with suitable coefficients. Then, as time evolves, each term gets multiplied by e^(-iE_n t/hbar), where E_n is the eigenenergy of that eigenfunction.
 
  • #3
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I'm not sure how to express psi (phi, 0) as a sum of energy eigenfunctions in this case...
 
  • #4
Avodyne
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What are the energy eigenfunctions?

If it's a free particle on a ring, they should be e^(i n phi), n=...,-1,0,+1,..., up to normalization. (The m^2 in your E_n should be n^2, no?)

Your psi(phi,0) is some trig functions; can you express these in terms of complex exponentials?
 
  • #5
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yes and yes. (1/sqrt(2pi))e^(i n phi) is the eigenfunction. and I really meant n^2 when I said m^2 previously...

I just expressed my psi(phi, 0) in exponential terms. The psi I gave in the original post left out some constants, but when I do take into account those:

psi(phi,0)=sqrt(1/4pi) (e^(i phi) + 1)

but how does this get into a summation form..? :-/
 
  • #6
Avodyne
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It already is in a summation form. You have

psi(phi,0) = (1/sqrt(2))( psi_1(phi) + psi_0(phi) ),

where psi_n(phi) is the eigenfunction.
 

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