Re: Is this a common mistake in QM books?..
- From: Roland Franzius <roland.franzius@xxxxxx>
- Date: Thu, 2 Feb 2006 20:27:08 +0000 (UTC)
Cyberkatru wrote:
I have seen this at least a couple of times now (for example on page 210 of
Zettili's book). A 1-d basic QM problem. We have particle of energy E moving
from minus \infinity toward the origin. The potential is zero for negative
x and V_0 for postive x. Here lets say for definiteness that E>V_0. In any
case, the region where the potential is positive is refered to as the
"repulsive region". But why? If the potential is constant then there is no
force! The region on the right is no more "attractive" than that on the
left. The only place that the particle feels anything is at the origin where
there is a step in potential and thus a delta function-like force. Thus we
seem to be modelling a particle penetrating a thin barrier with nothing on
either side (like gold leaf or something??) and not one penetrating deep
into a semiconductor block as is sometimes suggested.
Similarly, if the potential were positive only for x in some interval then
the partical would feel a repulsive kick upon entering and then nothing
while in the interval proper and then a helping push as it exited at the
other end. Am I wrong? Force is minus derivative of potential even in QM
right?
The step potential is indeed a crude approximation leaving only the
energy-momentum relations left and right of the barrier for problems in
the long wavelength limit.
The real problem can be reconstructed using a regularization of the step
potential say a symmetric
V(0)=0 V=+-V0=/2; |x| > a
V = x/a V0/2; |x| < a
The resulting force is stepwise constant
F = 0; |x|>a
F= -V0/(2a);|x|<a
The div of the electric field is the charge density
rho= dE/dx = -d/dx (H(x-a) - H(a-x)) V0/(2a)
= V0/(2a) (-delta(x-a)+delta(a-x)
Now you see, that a classical particle of negative charge crossing this
dipole layer of positive charge on the left plate and negative charge on
the right will be loosing its momentum becuase it is working against the
constant field inside the capacitor. In the regions outside the constant
electric fields of both plates with opposite equal charge densities
anihilate.
The step potential is the idealization: a->0, charge density->oo, with
V0 constant. So in effect you have to fit free waves of different
kinetic energies, momentum, wavelength in the two halfspaces at the
boundary in such a manner that current density is continous.
Probably in quantum mechanics the regions are are called attractive or
repulsive because in a wavepacket concentrated initially in the low
potential region all modes with kinetic energy = 1/2 k^2 difference
below the step energy are reflected. With the effect, that there is as
net current towards infinity in the low potential direction.
But this net current in the one particle state desribes a smaller number
of cases a particle is observed as transmitted to the right and a bigger
number of measurements wirh result, the particle is reflected. This has
no meaning for the unphysical unnormalized wave states.
--
Roland Franzius
.
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