- #1
AxiomOfChoice
- 533
- 1
Suppose I know
[tex]
H \psi(x) = \left( -\frac{1}{2m} \Delta_x + V(x) \right) \psi(x) = E\psi(x).
[/tex]
Then
[tex]
\psi(x,t) = e^{-iEt}\psi(x)
[/tex]
solves the time-dependent Schrodinger equation
[tex]
\left( i \frac{\partial}{\partial t} + \frac{1}{2m} \Delta_x - V(x) \right)\psi(x,t) = 0.
[/tex]
I've done some computations, and it looks like
[tex]
\Psi(x,t) = e^{-imvx}e^{-imv^2t/2}\psi(x+vt)
[/tex]
is a solution to the time-dependent Schrodinger equation
[tex]
\left( i \frac{\partial}{\partial t} + \frac{1}{2m} \Delta_x - V(x+vt) \right)\Psi = 0.
[/tex]
I have a couple of questions about this:
[tex]
H \psi(x) = \left( -\frac{1}{2m} \Delta_x + V(x) \right) \psi(x) = E\psi(x).
[/tex]
Then
[tex]
\psi(x,t) = e^{-iEt}\psi(x)
[/tex]
solves the time-dependent Schrodinger equation
[tex]
\left( i \frac{\partial}{\partial t} + \frac{1}{2m} \Delta_x - V(x) \right)\psi(x,t) = 0.
[/tex]
I've done some computations, and it looks like
[tex]
\Psi(x,t) = e^{-imvx}e^{-imv^2t/2}\psi(x+vt)
[/tex]
is a solution to the time-dependent Schrodinger equation
[tex]
\left( i \frac{\partial}{\partial t} + \frac{1}{2m} \Delta_x - V(x+vt) \right)\Psi = 0.
[/tex]
I have a couple of questions about this:
- What is going on here physically? That is, what are those two phase factors telling me?
- What does this mean the propagator is? When [itex]H[/itex] is time-independent, [itex]U(t) = e^{-iEt}[/itex]...but what is it in the time-dependent case? Is there a neat little formulation of it?
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