# [check exercise] Perturbation theory

1. Sep 15, 2015

### 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. The problem statement, all variables and given/known data

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

2. Sep 15, 2015

### blue_leaf77

Check again your value for $\lambda_3$.

3. Sep 15, 2015

### bznm

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

4. Sep 15, 2015

### blue_leaf77

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 $i=1,2,3$.

5. Sep 15, 2015

### bznm

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

6. Sep 15, 2015

### blue_leaf77

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