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## 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 E

_{0}=p

^{2}

_{0}/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?