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[check exercise] Perturbation theory

  1. Sep 15, 2015 #1
    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>)##
     
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  3. Sep 15, 2015 #2

    blue_leaf77

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    Check again your value for ##\lambda_3##.
     
  4. Sep 15, 2015 #3
    Oh, I'm sorry! I have done a typo.. :(
    ## \lambda_3 =5/2 \epsilon - \Delta ##
     
  5. Sep 15, 2015 #4

    blue_leaf77

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    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##.
     
  6. Sep 15, 2015 #5
    mmh.. I haven't understood.. where was I wrong?
     
  7. Sep 15, 2015 #6

    blue_leaf77

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