- #1
heardie
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Hi...I've come across this question on a past exam, and I can't seem to resolve it!
A particle is incident on a step potential at x=0. The total energy of the particle E, is less then the height of the step, U. The particle has wavefunctions
[tex]$\begin{array}{l}
\psi (x) = \frac{1}{2}\{ (1 + i)e^{ikx} + (1 - i)e^{ - ikx} \} ,x \ge 0 \\
\psi (x) = e^{ - kx} ,x < 0 \\
\end{array}$
[/tex]
Note that k is the same on both sides of the step
a) ….
b.) How must k be related to E and U on both sides of the step and determine the ration E/U
Well on the left, it is just a free wave-function where k=[tex]$k = \frac{{\sqrt {2mE} }}{\hbar }$
[/tex]
On the right k= [tex]$k = i\kappa = \frac{{i\sqrt {2m(U - E)} }}{\hbar }$
[/tex]
Since the k’s are equal this immediately implies that U=0, does it not? In which case the ratio E/U is infinite…
A particle is incident on a step potential at x=0. The total energy of the particle E, is less then the height of the step, U. The particle has wavefunctions
[tex]$\begin{array}{l}
\psi (x) = \frac{1}{2}\{ (1 + i)e^{ikx} + (1 - i)e^{ - ikx} \} ,x \ge 0 \\
\psi (x) = e^{ - kx} ,x < 0 \\
\end{array}$
[/tex]
Note that k is the same on both sides of the step
a) ….
b.) How must k be related to E and U on both sides of the step and determine the ration E/U
Well on the left, it is just a free wave-function where k=[tex]$k = \frac{{\sqrt {2mE} }}{\hbar }$
[/tex]
On the right k= [tex]$k = i\kappa = \frac{{i\sqrt {2m(U - E)} }}{\hbar }$
[/tex]
Since the k’s are equal this immediately implies that U=0, does it not? In which case the ratio E/U is infinite…
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