About Green's function in time dependent schrodinger equation

Click For Summary
SUMMARY

The discussion focuses on the application of Green's function in the time-dependent Schrödinger equation as presented in Sakurai's textbook. Specifically, the equations (2.5.7) and (2.5.12) are analyzed, revealing that the differential operator applied to equation (2.5.7) does not yield the expected Schrödinger equation. The key insight is that the boundary condition on the Green's function, as stated in equation (2.5.13), necessitates the introduction of the step function θ(t-t0) to ensure validity for both t > t0 and t < t0. This adjustment allows for proper incorporation of the boundary conditions.

PREREQUISITES
  • Understanding of the time-dependent Schrödinger equation
  • Familiarity with Green's functions in quantum mechanics
  • Knowledge of boundary conditions in differential equations
  • Proficiency in mathematical notation and integrals in physics
NEXT STEPS
  • Study the properties of Green's functions in quantum mechanics
  • Learn about the implications of boundary conditions in differential equations
  • Explore the use of the Heaviside step function in physics
  • Investigate the derivation and applications of the time-dependent Schrödinger equation
USEFUL FOR

Students and researchers in quantum mechanics, particularly those studying time-dependent phenomena and the application of Green's functions in solving Schrödinger equations.

wphysics
Messages
27
Reaction score
0
While I was studying Ch 2.5 of Sakurai, I have a question about Green's function in time dependent Schrödinger 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 Schrödinger 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 Schrödinger 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?
 
Physics news on Phys.org
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:

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
29
Views
2K
  • · Replies 8 ·
Replies
8
Views
5K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K