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Time dependant perturbation theory

  1. Dec 29, 2009 #1
    Hi, im basically trying to put a wavefunction into the Time Dependant Schrodinger Eqn, as shown in my lecture notes, but i dont understand one of the steps taken...

    [tex]|\right \Psi (t)\rangle=\sum c_n (t) |\right u_n\rangle e^-(\frac{E_n t}{\hbar})[/tex]
    into
    [tex]i\hbar \frac{\delta}{\delta t}|\right \Psi\rangle = H |\right \Psi \rangle[/tex]

    gives the LHS of the TDSE as:

    [tex]i \hbar \sum \left[ c^. _n (t) - \frac{i E_n}{\hbar}c_n \right] |\right \psi \rangle = ... [/tex]
    however, I dont understand the steps taken to get:

    [tex]\left[ c^. _n (t) - \frac{i E_n}{\hbar}c_n \right][/tex]

    i.e. how the time dependant/independant coefficients are separated.

    if anyone could help it would be greatly appreciated, thanks!
     
  2. jcsd
  3. Dec 29, 2009 #2

    cepheid

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    Gold Member

    Hi DickThrust, welcome to PF,

    This is just the product rule for differentiation. The two functions of time being multiplied together (under the summation sign) are cn(t) and |un〉exp(-(iEnt)/ℏ). Thus, applying the product rule:

    [tex] i\hbar \frac{\partial \Psi}{\partial t} = i\hbar \frac{\partial}{\partial t} \sum_n c_n(t) u_n e^{-i\frac{E_n t}{\hbar}} = i\hbar \sum_n \frac{\partial}{\partial t} (c_n(t) u_n e^{-i\frac{E_n t}{\hbar}}) [/tex]

    [tex] = i\hbar \sum_n \left[c_n(t) \frac{\partial}{\partial t}(u_n e^{-i\frac{E_n t}{\hbar}}) + u_n e^{-i\frac{E_n t}{\hbar}} \frac{\partial}{\partial t}(c_n(t)) \right] [/tex]

    [tex] = i\hbar \sum_n \left[c_n(t) \left(\frac{-iE_n}{\hbar}u_n e^{-i\frac{E_n t}{\hbar}}\right) + u_n e^{-i\frac{E_n t}{\hbar}} \frac{\partial}{\partial t}(c_n(t)) \right] [/tex]

    [tex] = i\hbar \sum_n \left[-c_n \left(\frac{iE_n}{\hbar}\right) + \frac{\partial c_n}{\partial t} \right]u_n e^{-i\frac{E_n t}{\hbar}} [/tex]​


    If you add back in the bra-ket notation, (| 〉 which I excluded to remove clutter), and use the the "dot" notation for the time derivative of c, and my expression becomes exactly like yours.
     
  4. Jan 1, 2010 #3
    typical of me, always forgetting the product rule...

    thanks for your help!
     
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