[check exercise] Perturbation theory

[bznm]

I have solved this exercise, but I'm not sure that it is good. Please, can you check it? A lot of thanks!

1. Homework Statement
The hamiltonian is ##H_0=\epsilon |1><1|+5/2 \epsilon (|2><2|+|3><3|)##
The perturbation is given by ##\Delta(|2><3|+|3><2|)##

Discuss the degeneration of H0.
Using perturbation theory, find energy-level shift and their eigenvectors.
Solve *exaclty* the problem with ##H=H_0+V##
Consider ##\Delta=\epsilon/2## . In t=0, the system is described by ##|s(0)>=|2>##,. Find ##|s(t)>##2. The attempt at a solution
H is degenerative: the eigenvalue relative to the vector |1> is epsilon. But vectors |2> and |3> have the same eigenvalue.

The energy level related to |1> hasn't energy shift produced by V.
Energy shifts for |2> and |3> are ##\pm \Delta##

For ##\lambda=\Delta## the eigenvector is ##|v_1>=1/ \sqrt 2 (|2>+|3>)##
For ##\lambda=- \Delta## the eigenvector is ##|v_2>=1/ \sqrt 2 (|2>-|3>)##

The exactly resolution of the problem, give me the eigenvalues: ##\lambda_1= \epsilon##, ##\lambda_2= 5/2 \epsilon + \Delta##, ##\lambda_3= \epsilon##. The corresponding eigenvectors are ##v_0=|1>##, ##|v_1>=1/ \sqrt 2 (|2>+|3>)## ,##|v_2>=1/ \sqrt 2 (|2>-|3>)##

The last point is the one that I'm not so sure that could be good:

I have written |2> as linear combination of |v_1> and v_2>:
##|2>= 1/ \sqrt 2 (|V_1>+|v_2>##, so
##|s(t)>=e^{-iHt/ \hbar} 1/ \sqrt 2 (|V_1>+|v_2>)= 1/ \sqrt 2 e^{-i2 \epsilon t / \hbar} (|v_2>+e^{-i \epsilon t / \hbar} |v_1>)##

 

[blue_leaf77]

Check again your value for ##\lambda_3##.

 

[bznm]

Oh, I'm sorry! I have done a typo.. :(
## \lambda_3 =5/2 \epsilon - \Delta ##

 

[blue_leaf77]

##|s(t)>=e^{-iHt/ \hbar} 1/ \sqrt 2 (|V_1>+|v_2>)= 1/ \sqrt 2 e^{-i2 \epsilon t / \hbar} (|v_2>+e^{-i \epsilon t / \hbar} |v_1>)##

The eigenvalues of the new Hamiltonian give you the new energy level corresponding to ##|v_1\rangle##, ##|v_2\rangle##, and ##|v_3\rangle##. Therefore ##e^{-iHt/\hbar}|v_i \rangle = e^{-iH\lambda_i t/\hbar}|v_i \rangle ## for ##i=1,2,3##.

 

[bznm]

mmh.. I haven't understood.. where was I wrong?

 

[blue_leaf77]

Ah sorry I just realized we are using different indexing for the new states. Yes your answer is correct.