WEBVTT
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PROFESSOR: OK.
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So let's try to solve this.
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So my classical approximation is
about solving these equations.
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So let's see what we get.
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Well, the first equation
is kind of simple.
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I think everybody has the
temptation there to just take
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the square root, and
that's what we should do,
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s0 prime is equal to
plus minus p of x.
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And therefore s0 of x is equal
to plus/minus the integral up
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to x of p of x prime dx prime.
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You see p of x is
pretty much known.
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If you know the energy
of your particle,
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then it's completely known,
and it depends on e minus v.
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So this is a solution
in terms of p of x.
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We should think of solving
the differential equation
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in terms of p of x.
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Now, as a first order
differential equation,
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there's a constant
of integration.
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And we'll pick it up to
be a number here, x0.
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So we start integrating
from some place.
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If you integrate
from another place,
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you're shifting the
constant of integration.
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The main thing is
that the x derivative
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here acts as the upper
limit and gives you
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the p of x of that equation.
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So this is our solution.
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Even the plus minus
should not disturb us.
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If you have the p
squared, you don't
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know if the particle is moving
to the left or to the right.
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So that ambiguity is
perfectly reasonable.
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Particle can be moving to
the left or to the right.
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Now, we look at the
second equation.
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So s1 prime is
equal to i over 2.
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s0 double prime, if you
have s0 double prime,
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you have here s0 prime, so
you take another derivative.
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So that's a plus minus p prime
of x divided by s0 prime,
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which is plus minus p of x.
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That's kind of nice.
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The sine is going to cancel.
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So we have here i over 2 p
prime of x over p of x or i
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over 2 logarithm
of p of x prime.
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The derivative of the logarithm
is 1 over the function,
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and then by chain rule,
you get the p prime there.
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So if s1, the prime
derivative is the derivative
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of this thing, so
s1 is going to be
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i over 2 log of p of
x plus a constant.
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So let's reconstruct
our solution.
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That's not hard.
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We wrote the answers up there.
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So the wave function is e
to the i over h bar times s,
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and s is what we had there.
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So our wave function is
e to the i over h bar s,
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and s was s0 plus
i plus h bar is 1.
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There's more.
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Is it right?
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But we're going to ignore it.
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We didn't go that far.
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In fact, nobody goes higher
in the WKB approximation.
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So what do we have here?
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I'll write it.
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This term is kind
of interesting.
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We have e to the i h bar
of x times e to the i s1.
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And s1 was i over 2 log
of p of x plus a constant.
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So look at this items
i is minus 1, 2.
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So you have 1/2.
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This becomes e to the minus
1/2 logarithm of b of x.
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And 1/2 the logarithm
of b of x is
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e to the minus log of square
root of b of x, and when you go
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like that e to minus that
is 1 over the function.
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So p of x like that, and then
we have e to the i over h
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bar integral from x0 to
x p of x prime dx prime.
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This is the classic WKB
approximation, classic result.
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So as promised, this is of the
form of a scale factor here,
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a rho, the square root
of rho times a phase.
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So we did begin with what
looked like a pure phase,
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but then we said s of
x is complicit in fact.
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s0 was real, but
s1 was imaginary.
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With s1 imaginary, the rho of
s1 was to provide the magnitude.
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And this is an intuition,
this approximation scheme.
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The first thing you have
to get right is the phase.
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Once you get the phase
right, the next order,
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you get the amplitude
of the wave right.
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That comes to second order.
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That's a next effect.
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So this is our solution.
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When I began today,
I reminded you
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that we have WKB solutions of
the form square root of rho
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e to the is, and we calculated
some things for that.
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So because of the
signs, s0, I dropped
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the sign, this plus or minus.
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Let me write the
general solutions of WKB
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slightly more complete.
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Let's be more complete.
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It's important to see
the whole freedom here.
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So if we have e greater
than v, remember
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when e is greater than v, the
p of x is a real quantity.
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And we wrote, and
we said that p of x
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we would write as h bar k of x.
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You know, I think I should have
probably, for convenience here,
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let's put the constant.
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We're not attempting to
normalize these wave functions.
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We could not attempt
to do it, because we
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don't know what p of x is.
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And this function may
have limited validity
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as we've spoken.
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But I had the constants here.
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This constant could have
a real or imaginary part.
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It would affect this a.
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So let's put it there.
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OK.
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So if p of x is hk of x, we can
have the following solutions,
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psi of x and t equal a, another
constant, square root of p--
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let's go simpler--
square root of k.
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It's a different a.
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And here, e to the i, since
p is h bar, it cancels here.
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So we have a simpler
integral as well,
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x0 to x, k of x prime, dx prime.
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So that's that term.
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I just use the
opportunity to replace p
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for k, which
simplifies your life,
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simplifies all these constants.
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So the other solution
is the wave moving
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in a different direction.
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So k of x, e to the
minus i x0 to x,
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k of x prime v of x prime.
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So that is your solution when
you have e greater than v.
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If we have e less than v,
we still have a solution,
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and we said that p
of x, in that case,
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would be equal to
i h bar kappa of x.
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We use that notation.
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If e is less than v, this
is a negative number.
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So p of x is i times
some positive number
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and square root of
a positive number.
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And we called it
kappa last time.
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So this is the
letter we usually use
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for spatial
dependence in regions
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where the wave function
decays exponentially,
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which is what it's
going to happen here.
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So what is the psi of x and
t is equal to a constant c
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over square root of
kappa of x e to the--
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the i will disappear,
and there will
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be two solutions, one
with plus, one with minus.
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That's the reason I don't have
to be very careful in saying
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whether this is i or minus i.
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There's, anyway, two solutions.
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At this stage, we
don't need to worry.
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So this is from x0 to x,
kappa of x prime, dx prime,
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plus d over square root of
kappa of x, e to the minus x0
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to x, kappa of x
prime, dx prime.
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So this is the complete
solution of WKB.
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If you are in the classically
allowed region top
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or in the classically
forbidden region,
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it's important to realize that
this function, the second term
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is the decaying exponential.
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As x increases, the integral
accumulates more and more
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value, and the wave function
gets more and more suppressed.
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This is a growing
kind of exponential.
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In the previous
iteration in your life,
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kappa was a constant, if
you had constant potentials.
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And this would be e to
the kappa x basically.
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But here, you must
think of kappa
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being some positive
number, positive function,
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as you integrate and x grows,
your cube-- the integral
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becomes bigger.
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And this is a
growing exponential.
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So the sign tells you that,
especially because we've
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ordered the limits properly.
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So we have a decaying
and growing exponential.
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At this moment,
we're pretty much
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done with what WKB does for you.
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Although, we have a few
things still to say.
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So this will be in
terms of comments
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about the general validity
of such approximation,
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but first even some comments
about the current and charge
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density.
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So let's consider
this equation I.
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Let's just make the comments,
comments for equation I on one.
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What is the charge density
or the probability density
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in this case, rho
would be psi squared,
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and in case one, is equal
to a squared over k of x?
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You could, if you wish, this
is a perfectly nice formula,
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multiply it by h
bar up and down.
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And that's the momentum.
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So it's h bar a
squared over p of x.
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And you could say
this is h bar over m
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a squared over v of x,
a local velocity, p of x
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is m over a local velocity.
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And this is an intuition
you've had for a long time.
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The probability density or
the amplitude of the wave
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is going to become
bigger in the regions
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where the particle
has smaller velocity.
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That's the regions of
the potential where
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the particle spends
more time, and it's
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an intuition that almost
immediately comes here.
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This k is essentially
the momentum.
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So that's essentially the
square root of the velocity.
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And this coefficient,
therefore, becomes bigger,
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as the velocity is smaller.
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That's part of the intuition
you've had for a long time,
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regarding these quantities.
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The other piece
is the computation
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of the current from this
equation I. Remember,
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the current is h bar over
m times the imaginary part
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of psi star gradient psi.
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So it's a long computation.
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But we did it for the
case we had before.
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We said that the current
is rho times gradient of s
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over m for the case when
the wave function is
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written in rho e to the s form.
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So in here, rho is
already determined
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is a squared over k of x.
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We have the 1 over m,
and the gradient of s--
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s is this quantity, e
to the i h bar times s.
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So the gradient of s is just
p of x, so h bar k of x.
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Now, they cancel.
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And it's very fast.
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They cancel.
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Would have been a major
disaster if they didn't.
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This is a number.
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It's a squared over m.
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The reason it cancels is
that it would have failed
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the conservation law otherwise.
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The rho dt plus the
divergence of j should be 0.
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In our case, rho has
no time dependence.
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The wave function that
we are considering,
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our time independent
Schrodinger equations,
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we're considering
energy eigenstates.
00:18:21.680 --> 00:18:26.030
And the current
must be a constant.
00:18:26.030 --> 00:18:30.260
In an energy eigenstate, the
current cannot be a spatially
00:18:30.260 --> 00:18:33.800
varying constant, because then
the current would accumulate
00:18:33.800 --> 00:18:38.180
in some place, and that's
inconsistent with stationary
00:18:38.180 --> 00:18:38.990
states.
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And in fact, this is 0, and
the versions of j in this case
00:18:45.560 --> 00:18:52.370
would be dj dx, and if it would
have had some x dependence,
00:18:52.370 --> 00:18:54.340
it would have destroyed
this equation.
00:18:54.340 --> 00:19:00.690
So dj dx is also 0, and
that's all consistent.