What Happens When a Particle Encounters a Step Potential Higher Than Its Energy?

[heardie]

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
$\begin{array}{l}<br /> \psi (x) = \frac{1}{2}\{ (1 + i)e^{ikx} + (1 - i)e^{ - ikx} \} ,x \ge 0 \\ <br /> \psi (x) = e^{ - kx} ,x &lt; 0 \\ <br /> \end{array}$<br />

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=$k = \frac{{\sqrt {2mE} }}{\hbar }$<br />
On the right k= $k = i\kappa = \frac{{i\sqrt {2m(U - E)} }}{\hbar }$<br />
Since the k’s are equal this immediately implies that U=0, does it not? In which case the ratio E/U is infinite…

 

[turin]

Since the k’s are equal this immediately implies that U=0, does it not? In which case the ratio E/U is infinite…

No, but you are thinking along the right lines. You have recognized that the requirement on k puts strong restriction on ... something. Attack the problem firstly as if k were different in the two regions, and then see what conditions have to be met so that the wavefunction can be put in the given form. (Biggest hint: Notice that there is an i in the exponent in one region and not in the other.)

One thing I just noticed: it appears to me that the particle is incident from the right so you may have your suggestions backwards.

 

[heardie]

Ok so if I call the region II solution k'
$k&#039; = \frac{{\sqrt {2m(E - U)} }}{\hbar }$

Since U > E this is complex.
$k&#039; = \frac{{i\sqrt {2m(U - E)} }}{\hbar } = i\kappa $<br />

This leads to the exponential solution
$e^{ - \kappa x} $<br />

If I then impose
$\kappa = k$ i get
$\begin{array}{l}<br /> \frac{{\sqrt {2mE} }}{\hbar } = \frac{{\sqrt {2m(U - E)} }}{\hbar } \\ <br /> \sqrt E = \sqrt {U - E} \\ <br /> E = U - E \\ <br /> 2E = U \\ <br /> E = \frac{U}{2} \\ <br /> \end{array}$<br />

oh hey...i definatly did that yesterday...musta been dropping a negative somewhere!

Ok, seems my latex sucks but how is E=U/2??

Cheers!

 

[turin]

... how is E=U/2??

That was my initial gut feeling about it. Are you asking how it is possible, or are you asking if I think this is correct? I think it is probably correct (without actually combing through your calculations).

 

[heardie]

Turin - I don't think I would ever ask if something is possible in QM, per se, since things that seem impossible always seem to fall out of QM! So much more to learn as well - makes me wonder what other things (classically impossible) will occur