WEBVTT
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PROFESSOR: So, what is the
wave function that we have?
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We must have a wave
function now that
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is symmetric, and built
with e to the k x, kappa x,
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and into the minus kappa x.
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This is the only possibility.
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E to the minus of kappa
absolute value of x.
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This is psi of x for
x different from 0.
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This is-- as you
can quickly see--
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this is e to get minus
kappa x, when x is positive,
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A e to the kappa x,
when x is negative.
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And, both of them decay.
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The first exponential negative
is the standard decaying
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exponential to the right.
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The one with
positive-- well, here x
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is negative as you go
all the way to the left.
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This one decays case as well.
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And, this thing plotted
is a decaying exponential
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with amplitude A, like that.
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And, a decaying exponential with
amplitude A, and a singularity
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there, which is what
you would have expected.
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So, this seems to be
on the right track--
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it's a continuous wave function.
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The wave function cannot
fail to be continuous,
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that's a complete disaster to
show that an equation could not
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be satisfied.
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So, this is our
discontinuous wave function.
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So, at this moment you
really haven't yet used
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the delta function-- the delta
function with intensity alpha
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down.
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I should have made a comment
that it's very nice that alpha
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appeared here in the numerator.
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If it would have appeared
in the denominator,
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I would be telling you that
I think this problem is not
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going to have a solution.
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Why?
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Because if it appears
in the numerator,
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it means that as the delta
function potential is becoming
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stronger and stronger,
the bound state
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is getting deeper and deeper--
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which is what you would expect.
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But, if it would be
in the numerator--
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in the denominator--
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as the potential gets
deeper and deeper,
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the boundary is going up.
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That makes no sense whatsoever.
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So, it's good that
it appeared there,
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it's a sign that things are
in reasonable conditions.
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So, now we really have to
face the delta function.
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And, this is a procedure you
are going to do many times
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in this course.
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So, look at it, and
do it again and again
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until you're very
comfortable with it.
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It's the issue of discovering
what kind of discontinuity
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you can have with
the delta function.
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And, it's a discontinuity
in the derivative,
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so let's quantify it.
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So, here it is-- we begin
with the Schrodinger equation,
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again.
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But, I will write now the
potential term as well.
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The potential is plus v of x
psi of x equals E psi of x.
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And the idea is to
integrate this equation
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from minus epsilon to epsilon.
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And, epsilon is supposed to
be a small positive number.
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So, you integrate
from minus epsilon
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to epsilon the
differential equation,
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and see what it does to you in
the limit as epsilon goes to 0.
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That's what we're
going to try to do.
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So, what do we get?
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If you integrate this, you
get minus h squared over 2 m.
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And now, you have to integrate
the second derivative
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with respect to x, which
is the first derivative,
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and therefore this is
the first derivative
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at x equal epsilon minus the
first derivative at x equals
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minus epsilon.
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This is from the first
term, because you integrate
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d x d second d x squared
psi is the same thing
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as d x d d x of d psi
d x between A and B.
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And, the integral of a
total derivative is d
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psi d x at B A--
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I think people
write it like this--
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A to B. Evaluate it at the
top, minus the evaluation
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at the bottom.
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Now, the next term is the
integral of psi times v of x.
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So, I'll write it
plus the integral
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from minus epsilon to epsilon
d x minus alpha delta of x psi
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of x--
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that's the potential.
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Now, we use the delta function.
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And, on the right hand
side this will be E times
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the integral from minus epsilon
to epsilon of psi of x d x.
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So, that's the differential
equation integrated.
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And now, we're going
to do two things.
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We're going to do some
of these integrals,
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and take the limit
as epsilon goes to 0.
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So, I'll write this minus
h squared over 2 m limit
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as epsilon goes to 0 of d psi
d x at epsilon minus d psi d x
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at minus epsilon plus.
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Let's think of this integral.
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We can do this integral,
it's a delta function.
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So, it picks the value of
the wave function at 0,
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because 0 is inside the
interval of integration.
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That's why we integrate it
from minus epsilon to epsilon,
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to have the delta
function inside.
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So, you get an alpha
out, a psi of 0,
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and that's what
this integral is.
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It's independent of
the value of epsilon
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as long as epsilon
is different from 0.
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So, this gives you
minus alpha psi of 0.
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And now, the last
term is an integral
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from minus epsilon to
epsilon of the wave function.
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Now, the wave function
is continuous--
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it should be continuous--
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that means it's finite.
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And, this integral,
as of any function
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that is not divergent from
minus epsilon to epsilon
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as epsilon goes to 0, is 0.
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Any integral of a function
that doesn't diverge
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as the limits of
integration go to 0,
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the area under
the function is 0.
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So, this is 0--
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the limit.
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And this thing goes to
0, so we put a 0 here.
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So, at this moment we got
really what we wanted.
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I'll write it this way.
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I'll go here, and I'll say
minus h squared over 2 m,
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and what is this?
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This expression says,
calculate the derivative
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of the function a little
bit to the right of 0,
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and subtract the derivative
of the function a little bit
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to the left of 0.
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This is nothing but the
discontinuity in psi prime.
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You're evaluating for any
epsilon greater than 0--
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the psi prime a little to
right, a little too the left,
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and taking the difference.
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So, this is what we should call
the discontinuity delta at 0--
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at x equals 0.
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And, this and this is for
discontinuity of psi prime at x
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equals 0 minus alpha
psi of 0 equals 0.
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And from here, we discover
that delta zero psi prime
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is equal to minus 2 m alpha
over h squared psi of 0.
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This is the
discontinuity condition
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produced by the delta function.
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This whole quantity is what
we call delta 0 of psi prime.
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And, what it says is that
yes, the wave function
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can have a discontinuous
first derivative if the wave
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function doesn't vanish there.
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Once the wave function
doesn't vanish at that point,
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the discontinuity
is in fact even
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proportional to the value of
the wave function at that point.
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And, here are the constants
of proportionality.
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Now, I don't think it's worth
to memorize this equation
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or anything like that,
because it basically
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can be derived in a few lines.
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This may have looked like
an interesting or somewhat
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intricate derivation,
but after you've done
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is a couple of times--
this is something
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you'll do in a minute or so.
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And, you just integrate
and find the discontinuity
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in the derivative--
that's a formula there.
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And, that's a formula
for a potential,
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minus alpha delta of x.
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So, if somebody gives you
a different potential,
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well, you have to change
the alpha accordingly.
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So, let's wrap this up.
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So, we go to our case.
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Here is our situation.
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So, let's apply this.
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So, what is the value?
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Apply this equation
to our wave function.
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So, what is the
derivative at epsilon?
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It's minus kappa A E to
the minus kappa epsilon.
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That's the derivative of
psi on the positive side.
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I differentiated the top
line of this equation
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minus the derivative
on the left side--
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this one, the derivative.
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So, this is kappa A E
to the kappa epsilon--
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no, kappa minus epsilon again.
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So, that's the left hand side.
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The right hand side would be
minus 2 m alpha h squared psi
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at 0.
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Psi at 0 is A, so
that's what it gives us.
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And we should take the
limit as epsilon goes to 0.
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So, this is going
to 1, both of them.
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So, the left hand side
is minus 2 kappa A,
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and the right hand side is
2 m alpha over h squared A.
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So, the 2--
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it's also minus, I'm sorry--
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so the 2s cancel,
the A cancels--
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you never should have expected
to determine A unless you tried
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to normalize the wave function.
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Solving for energy
eigenstates states will never
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determine A. The Schrodinger
equation is linear,
00:13:15.920 --> 00:13:21.150
so A drops out, the
minus 2 drops out,
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and kappa is equal to
m alpha over h squared.
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So, that said that's great
because kappa is just
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another name for the energy.
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So, I have kappa m
alpha over h squared,
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so that's another
name for the energy.
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So, let's go to the energy.
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The energy is h bar
squared kappa minus h bar
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squared kappa squared over 2 m.
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So, it's minus h bar squared.
00:14:02.980 --> 00:14:10.480
Kappa squared would be m squared
alpha squared h to the fourth,
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and there's a two m.
00:14:14.480 --> 00:14:15.920
All these constants.
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So, final answer.
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E, the bound state energy is
minus m alpha squared minus m
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alpha squared.
00:14:38.144 --> 00:14:43.750
The m cancels it over h
squared minus one half.
00:14:46.750 --> 00:14:53.020
So, back here the units worked
out, everything is good,
00:14:53.020 --> 00:14:56.756
and the number was
determined as minus one half.
00:14:56.756 --> 00:15:01.850
That's your bound state
energy for this problem.
00:15:01.850 --> 00:15:05.450
So, this problem is instructive
because you basically
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learn that in delta functions,
with one delta function
00:15:09.170 --> 00:15:10.580
you get a bound state.
00:15:10.580 --> 00:15:14.900
If you have two delta functions,
you may get more bound states--
00:15:14.900 --> 00:15:18.020
three, four-- people
study those problems,
00:15:18.020 --> 00:15:21.880
and you will investigate the
two delta function cases.