okay
so this time we're going to consider the
case E < V_0 send a
plane wave in as before from the left
and by the way when i'm drawing the wave
functions here on this plot
it's really a bit of a mix of of two
different diagrams
in the picture here this is showing the
potential so the y-axis here is the
potential as a function of position
i'm drawing the waves coming in like
this
and of course really i'm here i'm
plotting the amplitude of the
wave function in the region as a
function of position
it's a very mixed notation but
it's something we tend to do
hopefully the idea is clear that this is
a wave coming in and this just happens
to be the potential landscape it's
moving in
now this is phi_in
coming in this way as before we're going
to find
reflected coming back
but this time the because
E < V_0
past the region over here is
what's called classically forbidden
so if we just take a step back for a
second and think about what would happen
classically
the potential here could be something
like a height a height is a
gravitational potential
so we could have a height that's
zero over here and then V_0 over
here
and this would be like some kind of wall
and if we send in a classical particle
as in say throw a tennis ball
in the direction of a wall if the
tennis ball has
this energy a gravitational energy
that's larger than that of the wall as
in it's going over the wall
then sure it can go over here but if
we send in
a tennis ball whose gravitational
potential energy is less than that of
the top of the wall it should bounce
back
of course this is then classically
forbidden tennis ball can't go into the
wall
in fact we'll see that in the
Schroedinger equation it's allowed
the form of the solutions are now our
evanescent type solutions
exponentially growing or decreasing and
in this case
only the decreasing solution can be
relevant
again not worrying too much about the
fact that i'm mixing
my notations with the potential and the
wave function
so we'll have an exponentially
decreasing amplitude over here
rather that's right an exponentially
decreasing
wave function because otherwise
the wave function would have to blow up
at infinity if we allow the positive
solution this barrier goes up to
infinity
okay so let's take a look at solving
this
explicitly and looking at the
classically forbidden region
okay so we'll go a little bit faster
than we did in the last one but the
working is very much the same
so in region one
we'll have phi_1 which is equal to
phi_in + phi_reflected
and this can be written as e to the i k x
plus r e to the minus i k x just as
before nothing's changing in region one
substitute into the time independent
Schroedinger equation
and we'll find that e equals h bar
squared k
squared over 2m just like before in
region 2 however
we'll have phi in region 2 is equal to
phi
transmitted remember t for transmitted
not for
time or anything like that and this is
equal to
t e to the minus
kappa x; kappa used
instead of k just to indicate that it's
really a different quantity to a wave
vector
substitutes into the time independent
Schroedinger equation in region 2
and we find that e equals minus h bar
squared
kappa squared over 2m plus
V_0 so
the minus here coming from the fact
that there was a minus ready in the
time independent Schroedinger equation
and we don't cancel it with a an i
squared as we do
in the case of plane waves okay
so boundary conditions as before in the
first boundary condition is that the
wave function is continuous so phi in
region one
at x=0 is equal to phi in
region two
at x=0 which tells us that
1+r=t just like before
boundary condition 2 the first oh dear
sorry
the first derivative of the wave
function must also
match the first derivative is also
continuous
and in this case what we find this gives
us is that
i k (1 - r) that's what we got before
on
in region 1 is now equal to minus
kappa t in region 2.
so slightly different form in fact
rewriting we have
1-r equals multiply through by minus
i we get i kappa over k
t just as before we have two equations
two unknowns r and t the amplitudes
for reflection and transmission around
the kappa
and k are both already known from
kappa from this equation k from this
equation
and using these two equations
we can derive
the probability of reflection which is k
minus i kappa over k
plus i kappa and the probability of
transmission
is 2 k/(k+i kappa)
in fact if you compare to
the results we got
for the case that energy was greater
than V_0 you'll see that
because all we've really done is we've
changed what was e to the
i k prime x to e to the minus kappa x
we've just changed i k prime to minus
kappa
and if you make those substitutions
you'll find you get the same results
all right so let's flip the page around
and look at the probabilities so those
are the amplitudes
actually let's just comment on this
quickly so we have the amplitude
for the particle to exist in region two
is non-zero
so remember region two is classically
forbidden but there's an
amplitude to find the particle there
what that means is that if we perform a
measurement we could find the particle
in region two
how does this tally
with the fact that there isn't enough
energy for the particle to be there
the answer is that our measurement
device will provide the energy if
it's to observe
the particle and really what we're doing
is changing the boundary conditions on
the problem we're changing the potential
when we make the measurement we'll
see more of that in the quantum
tunnelling
problem in the next video
nevertheless we let's take a look at
the probability
for the probability current densities in
the different regions
so let me just write this down like this
there we go partial subscript
x again means d phi* in this case by
dx
holding time constant
and so what we get is that phi
for phi_in
which equals e to the i k x
substitutes into here and just as before
we find
j_in equals h bar k over
m phi
transmitted equals the
transmission amplitude e to the minus
kappa
x substitute that into here let me just
write it out like this again
so we need this we substitute phi and
phi* into here
evaluating it we find this
so we bring down a minus kappa in both
cases
but now the minus sign between the two
terms remember it's this term
minus the complex conjugate but the two
terms are the same because phi
is now real and so actually we find that
these perfectly cancel out
we find that j_transmitted=0
and so therefore
the probability of transmission
which is equal to |j_t/j_in|
is equal to zero
this is what we call the
probability flux
the probability that the probability
current is going to
a traveling wave in the classically
forbidden region
so even though there's an amplitude to
find the particle over there
the probability current
density is zero so there's no
current propagating through that
region if we look at
the probability for reflection which is
|j_reflected / j_incident|
in fact the expression is the
same as we found
in the last video still
|r|^2
and if we substitute it in we find
for r we had
(k- i kappa)/(k+ i kappa)
and if you do the modulus square of this
you'll find out that it equals one
which must have been the case because
the probability of transmission plus
the probability of reflection is always
equal to one
from conservation of probability okay
thank you for your time