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BARTON ZWIEBACH: After
this long detour,

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you must think that
one is just trying

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to avoid doing the real
computation, so here comes,

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the real computation.

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The real computation is
taking that right hand side

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on the top of the blackboard
and trying to just calculate

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this right hand side.

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So back to the calculation.

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The calculation dN/dt
is equal to this thing

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over there, integral
dx i over h-bar.

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I'll still copy it here-- h
psi-star psi minus psi-star h

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psi.

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OK.

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Well, let's do this.

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This whole quantity is d-rho/dt,
and let's see how much it is.

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Well, you would have of the
following-- i over h-bar h

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psi-star.

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Well, h in detail is over
there, so I'll put it here.

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Minus h squared over 2m d
second dx squared of psi-star.

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So I'm beginning h psi star--

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that's from the first term in
the Hamiltonian-- times psi.

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And then from the other
term in the Hamiltonian

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is the potential, so it would be
plus V of x and t psi-star psi.

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V of x and t times
psi-star times psi.

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This other term would
be minus psi-star h psi,

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so it's going to be opposite
sign to here, so plus h squared

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over 2m psi-star d second
dx squared psi, and then

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minus psi-star V of x and t psi.

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Here, there was a little
thing that I probably

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should have said before
is that the potential is

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real, that's why it didn't
get complex conjugated here.

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H psi would have a term V psi
and we just conjugate the psi.

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OK, this is not so bad.

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In particular, you see that
these two terms cancel.

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So that's neat.

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And now, this becomes
the following--

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this d-rho/dt has become minus
ih over 2m d second psi-star dx

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squared times psi minus psi-star
d second psi dx squared.

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OK.

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That's what d-rho/dt is and
that's the thing that should be

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0 when you integrate--

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it doesn't look like
anything equal to 0,

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and that was pretty
much to be expected.

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So what do we have
to do with this?

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Well, we have to
simplify it more,

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and what could save us
is, and it's usually

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the same thing that saves you
all the time when you want

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to show an integral vanishes,
many times, what you show

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is that it is a
total derivative.

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So remember, we're
computing here d-rho/dt,

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which is all this
thing circled here,

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and it's to be
integrated over x.

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So if I could show this is
a derivative with respect

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to x, the total x
derivative, then the integral

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would go to the
boundaries and I would

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have a chance to make it 0.

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So what do we have?

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That derivative is
indeed at boundaries,

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so d-rho/dt is equal to
minus i h-bar bar over 2m.

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And look, this can be
written as d/dx of something

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and what is that something?

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It's d psi-star dx times psi
minus psi-star d second psi--

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no not d second--
d first psi dx.

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The nice thing that
happens here is

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that if you act
with this d/dx, you

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get the second derivative
terms that you had in there.

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But you also get derivatives
acting here on d psi

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and here on d psi-star,
but those will cancel.

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So it's a very
lucky circumstance,

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it had better happen, but
this is a total derivative

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with respect to x.

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And that's just
very a good deal.

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So we're going to
rewrite it a little more.

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I'll write it as
the following way--

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this whole factor is h over 2im,
that's with its sign, output

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the d/dx outside--

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I'll put an extra minus
sign, so I will flip

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the order of these two terms--

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psi-star d psi dx minus
psi d psi-star dx.

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OK.

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Well, in many ways,
the most difficult part

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of the calculation
is over and it's now

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a matter of giving
proper names to things.

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Why do I say that?

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Because look, want to
see the finish line?

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It's here.

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We've shown this whole integrand
is d/dx of that right hand

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side.

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Therefore, when you
do the integral,

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you will have to go to the
boundary with that thing,

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so you just need to see what
happens to these quantities

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as x goes to infinity.

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And as x goes to
infinity, we said

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that psi must go to
0 from the beginning.

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And d psi dx must not blow
up, so if psi goes to 0

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and d psi dx doesn't blow up,
this whole thing goes to 0

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and dN/dt is equal
to 0 and you're done.

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So you're done
with the conditions

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that we mention that
the wave function must

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satisfy these conditions.

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But let's clean up this,
because we've actually

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discovered an important
quantity over there that

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is going to play a role.

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So here you see that you
have a complex number

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minus its complex conjugate.

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So this is like z
minus z-star, which

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is equal to 2i I times
the imaginary part of z.

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If you subtract from a complex
number its complex conjugate,

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you get the imaginary part only
survives, but it's twice of it.

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So from here, this whole thing
is 2i times the imaginary part

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of psi-star d psi dx.

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So d-rho/dt is equal
to minus d/dx of what?

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Of 2i times the imaginary
part of that, cancels the 2i,

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you get h-bar over m imaginary
part of psi-star d psi dx.

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And this quantity is going to
be called the current density.

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So the current density, you
say, why the current density?

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We'll see in a minute.

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But let's write it here because
it'll be very important.

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J of x and t is h-bar
over m imaginary part

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of psi-star d psi dx.

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So if this is called
the current density,

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you would have an equation
here d-rho/dt is equal to minus

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dJ/dx d/dx of J dx, or d-rho/dt
plus dJ/dx is equal to 0.

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Now this is called
current conservation.

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You've seen it before
in electromagnetism

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and we'll review it here
in a second as well.

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So look what has happened.

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You began with the introduction
of a charged density, which

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was a probability density,
but you were led now

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to the existence of a current.

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And you've seen that
in three dimensions,

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more than in one dimension--

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I think probably
in one dimension

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it doesn't look that
familiar to you,

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but let me make sure you will
recognize it in a few seconds.

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So think units here first.

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Units.

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What are the units
of the wave function?

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Well, the wave function,
you integrate over x squared

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and it gives you 1.

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So the integral of psi
squared dx is equal to 1,

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so this has units of length,
this must have units of 1

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over square root of length.

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And what are therefore the
units of psi-star d/dx psi,

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which is part of the
current formula ?

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Well, 1 of the square
root of length--

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1 over square root of length is
one over length and another 1

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over length is 1
over length squared.

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OK.

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And then you have
h-bar, which has

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units of mL squared over T.
Probably done that before

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already.

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And therefore, h over m has
units of L squared over T.

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So the current has
units of h over m--

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the units of current
has units of h over m,

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which is L squared over T--

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times units of this whole
thing, which is 1 over L

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squared, so at
the end, 1 over T.

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And this means just
probability per unit time.

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That's the units of current.

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Probability has no units, so
we're dealing probability,

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those are pure numbers, but this
is probability per unit time.

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So probability per unit time.