• Support PF! Buy your school textbooks, materials and every day products Here!

Grad 1D scattering from step function

  • Thread starter frogjg2003
  • Start date
  • #1
265
0

Homework Statement


Consider the 1D potential V(x) such that V(x)=0 for x<0 and V(x) = V for x>0 and assume that a wave packet with energy E0=p20/2m<V is incident on the barrier from the left. Calculate in terms of E0 and V the difference in time between the arrival of the incident packet at the step and the departure of the reflected packet from the step. Relate this time delay to a “distance of travel” within the step.

Homework Equations



time independent Schrödinger equation

The Attempt at a Solution



I've calculated the time independent wavefunction to be
[tex]
\Psi(x) = \begin{cases}
A_Ie^{ik_0x} + B_Ie^{-ik_0x} & x<0\\
A_{II}e^{kx} & x>0 \end{cases}
\\
k = \sqrt{\frac{2mE_0}{\hbar^2}}
\\
k = \sqrt{\frac{2m(V-E_0)}{\hbar^2}}
[/tex]
with the relations
[tex]
\frac{B_I}{A_I} = \frac{ik_0+k}{ik0-k}
\\
\frac{A_{II}}{A_I} = \frac{2ik_0}{ik_0-k}.
[/tex]
From that, I was thinking about writing the wavefunction as
[tex]
\Psi(x) = \begin{cases}
e^{ik_0x} + e^{-i(k_0x-\phi)} & x<0\\
ce^{kx+i\theta} & x>0 \end{cases}
[/tex]
where I've removed the normalization constant and use the two phases [itex]\phi,\theta[/itex] given by
[tex]
\tan(\phi) = \frac{2kk_0}{k^2-k_0^2}
\\
c = \frac{2k_0}{k'}
\\
\tan(\theta) = \frac{k}{k_0}
\\
k'^2=k_0^2+k^2=\frac{2mV}{\hbar^2}
[/tex]

I'm stuck with what the question means by calculate the time difference.
One of my classmates said to apply the time evolution operator [itex]e^{-i\hat{H}t/\hbar}[/itex], which because we're acting on one eigenfunction, becomes [itex]e^{-iE_0t/\hbar}[/itex] and the wavefunction is
[tex]
\Psi(x,t) = \begin{cases}
e^{i(k_0x-E_0t/\hbar)} + e^{-i(k_0x-\phi+E_0t/\hbar)} & x<0\\
ce^{kx+i(\theta-E_0t/\hbar)} & x>0 \end{cases}
[/tex]

Where would I go from here?
 

Answers and Replies

Related Threads on Grad 1D scattering from step function

Replies
1
Views
379
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
3
Views
5K
Replies
4
Views
3K
  • Last Post
Replies
4
Views
775
  • Last Post
Replies
4
Views
1K
Replies
2
Views
20K
Replies
1
Views
2K
Top