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About Green's function in time dependent schrodinger equation

  • Thread starter wphysics
  • Start date
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?
 

TSny

Homework Helper
Gold Member
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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:

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