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A Computation time for adiabatic evolution

  1. Mar 5, 2016 #1
    Hi,

    my problem: following the adiabatic theorem we get an equation for the coefficients:

    [itex]
    \dot{a}_{m}=-a_{m} \langle{\psi_{m}} | \dot{\psi}_{m}\rangle - \sum_{n \neq m} \frac{\langle \psi_{m} | \dot{H}| \psi_{m}\rangle}{E_{n}-E_{m}} exp(\int_ 0^t E_{n}(t')-E_{m}(t') \, dt')
    [/itex]

    we start in an eigenstate m. When the gap is large and the time derivative of H is small. We can neglect the sum and get for the coefficent m:

    [itex]
    a_m^0= a_m(0) exp(-i \gamma_m (t))
    [/itex]

    so far so good. Now we want to determine the 1st order coefficients and make the ansatz:

    [itex]
    a_m= a_m^0 + a_m^1
    [/itex]

    and we get:

    [itex]
    a_m^1= \sum_{n \neq m} \frac{\langle \psi_{m} | \dot{H}| \psi_{m}\rangle}{(E_{n}-E_{m})^2} a_n^0 exp(\int_ 0^t E_{n}(t')-E_{m}(t') \, dt')
    [/itex]

    the last step is problematic. It's been a while since I used perturbation theory and I don't know how you get this result. The computation time should be:

    [itex]
    t<<\frac{\langle \psi_{m} | \dot{H}| \psi_{m}\rangle}{(E_{n}-E_{m})^2}
    [/itex]

    which is clear once you obtain the result above. Can anyone help?
     
  2. jcsd
  3. Mar 6, 2016 #2
    not sure how to edit posts... a few remarks: it should be t>>..... and gamma is the geometric phase.
     
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