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Time-dependent Perturbation theory -maths problems

  1. Oct 11, 2015 #1

    Hey guys,

    I signed up here because I needed some information on some quantum physics problems.

    My question is related to quantum physics, and more precisely the derivation of time dependent perturbation theory. First of all, I am not able to understand all the maths structures and formulas, and it is in my opinion the biggest difficult I need to overcome to fully understand it. I checked some books and scripts just to find some more information about it, but unfortunately my doubts weren't stirred up. I hope that here I could find some more help and I really appreciate it.

    So to begin with:

    We get a hamiltonian, that can be written as
    [tex]H(t)=H _{0}+V(t) [/tex], where [tex]V(t)[/tex] is a small time-dependent perturbation.

    Let [tex]\psi ^{(0)} _{k} [/tex] be the wave functions of stationary states of unperturbed system with time multiplier. Any solution to the unperturbed wave equation can be expressed as the sum :
    [tex]\psi= \sum_{k}a _{k} \psi ^{(0)} _{k}[/tex]

    The solutions to perturbed equation:
    [tex]ih \frac{\delta}{\delta t}\psi=(H _{0}+V)\psi[/tex] (1) (where h is Dirac constant, Planck constant divided by 2 Pi)
    are in the form of:

    [tex]\psi= \sum_{k}a _{k}(t)\psi ^{(0)} _{k}[/tex] (2)

    Substituting (2) to the (1) and knowing that functins [tex] \psi^{(0)}_{k}[/tex] satisfy and equation :

    [tex]ih \frac{\delta}{\delta t }\psi _{k} ^{(0)}=H_{0}\psi _{k} ^{(0)}[/tex]

    somehow we obtain: (here is the problem for me. I don't understand how they solve below mentioned equations step by step)

    [tex]ih \sum_{k}\psi _{k} ^{(0)} \frac{d}{dt}a _{k}(t)= \sum_{k} a_{k}V\psi _{k} ^{(0)}[/tex]

    and then multiplying this equation from left side by [tex]\psi _{m} ^{(0)}* [/tex] and integrating we get :

    [tex]ih \frac{d}{dt}a _{m}(t) = \sum_{k}V _{mk}(t)a _{k}(t) [/tex]

    where [tex]V _{mk}(t)=\int \psi _{m} ^{(0)*} V\psi _{k} ^{(0)}dq=V _{mk}e ^{i\omega _{mk}t}[/tex]

    and [tex]\omega _{mk}= \frac{E _{m} ^{(0)}-E _{k} ^{(0)} }{h}[/tex]

    I will be very thankful if someone could explain it to me. I want to mention, that the majority of this text come from Landau book "Quantum mechanics. Non relativistic theory". At the end I want to add that I am Polish, so my English may be not perfect, but I still hope it is enough communicative ;)

     
    Last edited: Oct 11, 2015
  2. jcsd
  3. Oct 11, 2015 #2

    blue_leaf77

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    It's just product rule of differentiation
    $$\frac{\partial\psi}{\partial t} = \sum_{k} \frac{\partial }{\partial t} \left(a _{k}(t)\psi ^{(0)} _{k}\right) = \sum_{k} a _{k}(t) \frac{\partial \psi^{(0)} _{k}}{\partial t} + \psi ^{(0)} _{k} \frac{\partial a _{k}(t)}{\partial t}$$
    Is that not what you can't understand?
     
  4. Oct 11, 2015 #3
    Ye,it is obvoius relation from mathematial analysis ;)
    the derivative of a sum is the sum of the derivatives.
    So it looks like that (step by step):

    [tex]ih\frac{\delta}{\delta t}\sum_{k}a _{k}(t)\psi _{k}^{(0)}=(H _{0}+V)\sum_{k}{a _{k}(t)\psi _{k}^{(0)}[/tex]


    [tex]\Leftrightarrow ih\left( \sum_{k}a_{k}(t) \frac{\delta}{\delta t} \psi_{k}^{(0)}+\sum_{k} \frac{d}{dt} a_{k}(t) \psi_{k}^{(0)} \right)=H_{0} \sum_{k}a_{k}(t)\psi_{k}^{(0)}+V\sum_{k}a_{k}(t)\psi_{k}^{(0)}[/tex]

    From [tex]ih\frac{\delta}\delta t\psi_{k}^{(0)}=H_{0}\psi_{k}^{(0)}[/tex] we obtain by multiplying boths sides by[tex]\left(\frac{-i}{h}\right) \Rightarrow \frac{\delta}{\delta t}\psi_{k}^{(0)}=-\frac{i}{h}H_{0}\psi_{k}^{(0)}[/tex]. Substituting this, gives us the final result
    [tex]ih \sum_{k}\psi_{k}^{(0)}\frac{d}{d t}a_{k}(t)=\sum_{k}a_{k}V\psi_{k}^{0}[/tex]

    But the problem is to multiply this equation by [tex]\psi_{m}^(0*)[/tex] and then taking integral:

    [tex]ih\int\sum_{k}\psi_{m}^{(0*)}\psi_{k}^{(0)}\frac{d}{dt}a_{k}(t)=\int\sum_{k}\psi_{m}^{(0)*}V\psi_{k}^{(0)}[/tex]
     
  5. Oct 11, 2015 #4
    Ok, so I would retype this onto form [tex]ih\sum_{k}\int\psi_{m}^{(0)*}\psi_{k}^{(0)}\frac{d}{dt}a_{k}(t)=\sum_{k}\int\psi_{m}^{(0)*}V\psi_{k}^{(0)}[/tex]
    and then knowing that the integral [tex] \int \psi_{m}^{(0)*}\psi_{k}^{(0)} [/tex] can be expressed as the inner product [tex](\psi_{m}^{(0)*},\psi_{k}^{(0)})=\delta_{mn}[/tex]
    So for all [tex]k\neq m[/tex] components of the sum are zero, and finally we obtain: [tex] ih\frac{d}{dt}a_{m}(t)=\sum_{k}\int\psi_{m}^{(0)*}V\psi_{k}^{(0)}[/tex]

    What do you think about it ?
     
  6. Oct 11, 2015 #5

    blue_leaf77

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    There should be ##a_k(t)## in the right hand side, other than that I don't see anything wrong with your derivation.
     
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