# Finding eigenstates and eigenvalues of hamiltonian

1. Mar 12, 2013

### beans73

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$_{a'}$> and |E$_{a''}$>. In order to find the eigenvalues of the energy, I constructed the matrix:

H = $$\begin{pmatrix} 0 & ∂\\ ∂ & 0 \end{pmatrix}$$

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'|)

2. Mar 12, 2013

### CFede

Yes to both, you are headed in the right direction

3. Mar 14, 2013

### beans73

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$_{0}$=0> = |a'>

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

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

am i allowed to do this??? i'm probably ignoring something important....

4. Mar 15, 2013

### vela

Staff Emeritus
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. Mar 15, 2013

### beans73

oh ok then. would this be the right plan of action then?

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

||E$_{1}$> = |a'> + |a''> and
|E$_{2}$> = |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$_{1}$>+|E$_{2}$>, and then use E$_{1}$ & E$_{1}$ in the exponential? ==>

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

then sub |a'> and |a''> back into this equation?

6. Mar 15, 2013

### vela

Staff Emeritus
Yes, that's exactly what you want to do.

7. Mar 15, 2013

### beans73

yay! thanks for your help :)