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About Green's function in time dependent schrodinger equation

  1. Jul 18, 2012 #1
    While I was studying Ch 2.5 of Sakurai, I have a question about Green's function in time dependent schrodinger equation. (Specifically, page 110~111 are relevant to my question)

    Eq (2.5.7) and Eq (2.5.12) of Sakurai say
    [itex]\psi(x'',t) = \int d^3x' K(x'',t;x',t_0)\psi(x',t_0)[/itex]
    and
    [itex]\left(H-i\hbar\frac{\partial}{\partial t}\right)K(x'',t,x',t_0) = -i\hbar\delta^3(x''-x')\delta(t-t_0)[/itex]

    We know from the basic Schrodinger equation
    [itex]\left(H-i\hbar\frac{\partial}{\partial t}\right)\psi(x,t) = 0 [/itex]

    So, I applied the differential operator to Eq (2.5.7) and use Eq(2.5.12). But, I couldn't get the right Schrodinger equation like this.
    [itex]\left ( H - i \hbar \frac{\partial}{\partial t}\right ) \psi (x'',t) = \left ( H - i \hbar \frac{\partial}{\partial t}\right ) \int dx' K(x'',t;x',t_0) \psi(x',t_0) = \int \left [ \left ( H - i \hbar \frac{\partial}{\partial t}\right ) K(x'',t;x',t_0) \right ] \psi(x',t_0) dx' = -i \hbar \int \psi(x', t_0) \delta(x''-x') \delta(t-t_0)dx'[/itex]
    [itex]=-i \hbar \psi(x'',t_0) \delta(t-t_0)[/itex]
    which is non zero at t=t_0

    What is the point that I am missing?
     
  2. jcsd
  3. Jul 18, 2012 #2

    TSny

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    Homework Helper
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    Note the boundary condition on K as given in equation (2.5.13). Thus the equation (2.5.7) is only valid for t > to. For t < to, the right hand side of equation (2.5.7) will yield 0 because of (2.5.13). To make (2.5.7) valid for both t > to and t < to, you can introduce the step function θ(t-t0) and write (2.5.7) as

    [itex]\theta[/itex](t-to) [itex]\psi(x'', t)[/itex]= [same right hand side as before]

    This equation now incorporates the boundary condition on K.

    See if everything works out if you apply (H-i[itex]\hbar[/itex][itex]\frac{\partial}{\partial t}[/itex]) to both sides of this equation.
     
    Last edited: Jul 18, 2012
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