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5DUQ3-Y_gX4
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STEVEN G. JOHNSON: Let
me stop the recording.
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5DUQ3-Y_gX4
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ALAN EDELMAN: OK.
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5DUQ3-Y_gX4
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Otherwise, see you on
Friday, same room, 11:00 AM.
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PROFESSOR: All right.
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So let's get now the state n1.
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So I want to make
a general remark.
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You have an equation like this.
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And you want to solve it.
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It's a vector equation.
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Operator and a vector
equal number and a vector--
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a more operator and a vector.
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To make sure you
have solved it, when
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you have a vector equation
you must make sure
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that every component-- you
can write a vector equation
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in the form vector equals zero.
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And then you must make sure that
every component of the vector
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is zero.
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What we did here
is we found what
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happens when I look at
the component along n0.
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And I figure out that,
whoops, this equation,
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when I look at the
component along n0,
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tells me what the energy is.
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So the rest of the
information of this equation
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arises when I look at
it along the components
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on the other states, not n0 but
the k states that we introduced
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from the beginning, the k0's
that run from one to infinity.
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So what we're
going to do is take
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that original--
this second equation
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and form k0 h0 minus Em0 m1 1
is equal to k0 Em1 minus delta H
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n0.
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So I now took the same equation
and I put it in a problem
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with k0.
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And I say, look, k
will be different
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from n, because
when we put k equal
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to n that already we've done.
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And we've learned all about it.
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And, in fact, n1 didn't appear.
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The state that we wanted
didn't appear at all.
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So now we do this
with arbitrary k.
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And we need to figure
out what this gives.
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So you have to look
at these things
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and try to remember a little
of the definitions with both.
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So h0, we know what
it gives from k0.
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It gives you a number,
the energy of that state.
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So this is another number.
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So that's great.
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This simplifies this
to ek0 minus En0
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times the overlap of k0 with n1.
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That's the left hand side.
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How about the right hand side?
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All right.
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Let's see what this is.
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First term.
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The En1 is a number,
so I must ask
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myself is what happens
when k0 meets n0?
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Well, those are our
original orthonormal states.
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And we said that k
is different from n.
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So this term is 0 with an En1.
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This is a number.
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And these two states
are orthogonal.
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So this term gives you a 0,
not because this number is 0,
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but because the overlap is 0.
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And I get here
minus k0 delta H n0.
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And it's good
notation to call this,
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to save writing, delta Hkn
It's a good name for it.
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It's the matrix k--
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the kn-th element of the matrix
delta H. And this is a number
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so I can solve k0 n1
is equal to minus delta
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Hkn divided by Ek0 minus En0.
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And this is true for
every k different for n.
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And here we find, for the first
time, our energy denominators.
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These energy denominators
are the things
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that are going to make life
interesting and difficult.
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And it answers the
question already
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that if you had
degenerate states,
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there would be some
k state that have
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the same energy as this one.
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And this blows up.
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And this is unsolvable
for this component.
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So you start
getting difficulties
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if you have degeneracies.
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As long as every k state--
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all the other states
of the spectrum
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have different energy from En,
nevermind if the other states
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are degenerate.
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They're not degenerate
with the state you care.
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You care just about one
state now, the n-th state.
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And if that's nondegenerate, all
these denominators are non-zero
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and you're OK.
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So here is the solution
for this thing.
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Now I can write the expressions
for the state and the energy.
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So let me do it.
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So I have this n1 like that.
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Now you can say the following.
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Let me do this very
deliberately first.
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n1 is equal to the sum
over all k of k0 k0 n1.
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This is the resolution
of the identity formula.
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That's the unit operator.
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You can always do that.
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And now you know that the
state n1 is orthogonal to n0.
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