Grad 1D scattering from step function

[frogjg2003]

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?

 

[mark726]

Thank you for your question. The time difference that is being referred to in this problem is the difference between the arrival time of the incident wave packet at the step and the departure time of the reflected wave packet from the step. This can also be thought of as the time it takes for the wave packet to travel through the step region.

To calculate this time difference, you can use the time evolution operator as you mentioned. The time evolution operator acts on the wavefunction and gives the time-dependent wavefunction, as you have correctly written in your post. To calculate the time difference, you can take the difference in the phase of the wavefunction at two different points in space, say x=0 and x=L, where L is the length of the step region.

The phase of the wavefunction at x=0 is given by k_0x-E_0t/\hbar and the phase at x=L is given by kx+i(\theta-E_0t/\hbar). Taking the difference between these two phases, we get kL+i(\theta-\phi). This can be related to the distance traveled by the wave packet within the step region by using the relation k=\sqrt{2m(V-E_0)/\hbar^2}. This gives us the final expression for the time difference as \frac{\sqrt{2m(V-E_0)}}{\hbar}(L-\tan^{-1}(\frac{2kk_0}{k^2-k_0^2})).

I hope this helps you in your calculations. Please let me know if you have any further questions or if you need any clarifications.