Finding eigenstates and eigenvalues of hamiltonian

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  • #1
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Hey there, the question i'm working on is written below:-

Let |a'> and |a''> be eigenstates of a Hermitian operator A with eigenvalues a' and a'' respectively. (a'≠a'') The Hamiltonian operator is given by:

H = |a'>∂<a''| + |a''>∂<a'|

where ∂ is just a real number.

Write down the eigenstates of the hamiltonian. what are their energy eigenvalues?




Was feeling a bit confused by the question at first, and was just wondering if someone could let me know if my thoughts so far are on the right track?

Essentially, I just wrote the eigenstates as the kets: |E[itex]_{a'}[/itex]> and |E[itex]_{a''}[/itex]>. In order to find the eigenvalues of the energy, I constructed the matrix:

H = [tex]
\begin{pmatrix}
0 & ∂\\
∂ & 0
\end{pmatrix}
[/tex]


so that I could use the general det(H - λI) = 0 to find the eigenvalues. is this reasoning vaguely in the correct direction?

Also, was just wondering, if in the original hamiltonian equation, i'm allowed to take the ∂ symbol out (written below) because it is just a real number?

H = ∂(|a'><a''| + |a''><a'|)
 

Answers and Replies

  • #2
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Yes to both, you are headed in the right direction
 
  • #3
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thanks for that. i have actually continued on with this problem, and i've come across another question.

i found the eigenvalues to be ±∂. the problem then asks

b) suppose the system is known to be in state |a'> at t=0. write down the schrodinger picture for t>0.

Which my working out is:|α, t[itex]_{0}[/itex]=0> = |a'>

|α, t[itex]_{0}[/itex]=0; t> = U(t,t[itex]_{0}[/itex])|a'>
=U(t,0)|a'>
=exp( (-iE[itex]_{a'}[/itex]t) / h-bar )|a'>
=exp( -i∂t / h-bar )|a'>

question c)
what is the probability for finding the system in state |a''> for t>0, if the system is known to be in state |a'> at t=0?

for this, i was thinking that for |a''>= U(t,t[itex]_{0}[/itex])|a'> = =exp( -i∂t / h-bar )|a'>

and for the probability, i need to calculate |<a''|a'>|[itex]^{2}[/itex] = |(exp( -i∂t / h-bar ) <a'| )(|a'>)|[itex]^{2}[/itex], where <a'|a'> = 1

am i allowed to do this??? i'm probably ignoring something important....
 
  • #4
vela
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thanks for that. i have actually continued on with this problem, and i've come across another question.

i found the eigenvalues to be ±∂. the problem then asks

b) suppose the system is known to be in state |a'> at t=0. write down the schrodinger picture for t>0.

Which my working out is:|α, t[itex]_{0}[/itex]=0> = |a'>

|α, t[itex]_{0}[/itex]=0; t> = U(t,t[itex]_{0}[/itex])|a'>
=U(t,0)|a'>
=exp( (-iE[itex]_{a'}[/itex]t) / h-bar )|a'>
=exp( -i∂t / h-bar )|a'>
The state ##\lvert a'\rangle## isn't an eigenstate of the Hamiltonian, so it's not correct to say it has an energy ##E_{a'}##. You need to express ##\lvert a'\rangle## in terms of the eigenstates of H.
 
  • #5
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oh ok then. would this be the right plan of action then?

using the eigenvectors i found for the hamiltonian|E[itex]_{1}[/itex]> = (1,1) and |E[itex]_{2}[/itex]> = (1,-1). i then constructed:

||E[itex]_{1}[/itex]> = |a'> + |a''> and
|E[itex]_{2}[/itex]> = |a'> - |a''> ( i have left out the normalization constant here)

then i can have |a'> and |a''> written in terms of the eigenstates of H. back in the schrodinger picture, can i then substitute |a'> for |E[itex]_{1}[/itex]>+|E[itex]_{2}[/itex]>, and then use E[itex]_{1}[/itex] & E[itex]_{1}[/itex] in the exponential? ==>

|α, t[itex]_{0}[/itex]=0; t> = U(t,t[itex]_{0}[/itex])|a'>
=U(t,0)|a'>
=exp((-iE[itex]_{a}[/itex]′t) / h-bar)|a'>
=(exp(-i∂t/h-bar)||E[itex]_{1}[/itex]> + exp(i∂t/h-bar)||E[itex]_{2}[/itex]>

then sub |a'> and |a''> back into this equation?
 
  • #6
vela
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Yes, that's exactly what you want to do.
 
  • #7
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yay! thanks for your help :)
 

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