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5DUQ3-Y_gX4 | STEVEN G. JOHNSON: Let
me stop the recording. |
5DUQ3-Y_gX4 | ALAN EDELMAN: OK. |
5DUQ3-Y_gX4 | Otherwise, see you on
Friday, same room, 11:00 AM. |
_bTZbn7M2Hc | PROFESSOR: All right. |
_bTZbn7M2Hc | So let's get now the state n1. |
_bTZbn7M2Hc | So I want to make
a general remark. |
_bTZbn7M2Hc | You have an equation like this. |
_bTZbn7M2Hc | And you want to solve it. |
_bTZbn7M2Hc | It's a vector equation. |
_bTZbn7M2Hc | Operator and a vector
equal number and a vector-- |
_bTZbn7M2Hc | a more operator and a vector. |
_bTZbn7M2Hc | To make sure you
have solved it, when |
_bTZbn7M2Hc | you have a vector equation
you must make sure |
_bTZbn7M2Hc | that every component-- you
can write a vector equation |
_bTZbn7M2Hc | in the form vector equals zero. |
_bTZbn7M2Hc | And then you must make sure that
every component of the vector |
_bTZbn7M2Hc | is zero. |
_bTZbn7M2Hc | What we did here
is we found what |
_bTZbn7M2Hc | happens when I look at
the component along n0. |
_bTZbn7M2Hc | And I figure out that,
whoops, this equation, |
_bTZbn7M2Hc | when I look at the
component along n0, |
_bTZbn7M2Hc | tells me what the energy is. |
_bTZbn7M2Hc | So the rest of the
information of this equation |
_bTZbn7M2Hc | arises when I look at
it along the components |
_bTZbn7M2Hc | on the other states, not n0 but
the k states that we introduced |
_bTZbn7M2Hc | from the beginning, the k0's
that run from one to infinity. |
_bTZbn7M2Hc | So what we're
going to do is take |
_bTZbn7M2Hc | that original--
this second equation |
_bTZbn7M2Hc | and form k0 h0 minus Em0 m1 1
is equal to k0 Em1 minus delta H |
_bTZbn7M2Hc | n0. |
_bTZbn7M2Hc | So I now took the same equation
and I put it in a problem |
_bTZbn7M2Hc | with k0. |
_bTZbn7M2Hc | And I say, look, k
will be different |
_bTZbn7M2Hc | from n, because
when we put k equal |
_bTZbn7M2Hc | to n that already we've done. |
_bTZbn7M2Hc | And we've learned all about it. |
_bTZbn7M2Hc | And, in fact, n1 didn't appear. |
_bTZbn7M2Hc | The state that we wanted
didn't appear at all. |
_bTZbn7M2Hc | So now we do this
with arbitrary k. |
_bTZbn7M2Hc | And we need to figure
out what this gives. |
_bTZbn7M2Hc | So you have to look
at these things |
_bTZbn7M2Hc | and try to remember a little
of the definitions with both. |
_bTZbn7M2Hc | So h0, we know what
it gives from k0. |
_bTZbn7M2Hc | It gives you a number,
the energy of that state. |
_bTZbn7M2Hc | So this is another number. |
_bTZbn7M2Hc | So that's great. |
_bTZbn7M2Hc | This simplifies this
to ek0 minus En0 |
_bTZbn7M2Hc | times the overlap of k0 with n1. |
_bTZbn7M2Hc | That's the left hand side. |
_bTZbn7M2Hc | How about the right hand side? |
_bTZbn7M2Hc | All right. |
_bTZbn7M2Hc | Let's see what this is. |
_bTZbn7M2Hc | First term. |
_bTZbn7M2Hc | The En1 is a number,
so I must ask |
_bTZbn7M2Hc | myself is what happens
when k0 meets n0? |
_bTZbn7M2Hc | Well, those are our
original orthonormal states. |
_bTZbn7M2Hc | And we said that k
is different from n. |
_bTZbn7M2Hc | So this term is 0 with an En1. |
_bTZbn7M2Hc | This is a number. |
_bTZbn7M2Hc | And these two states
are orthogonal. |
_bTZbn7M2Hc | So this term gives you a 0,
not because this number is 0, |
_bTZbn7M2Hc | but because the overlap is 0. |
_bTZbn7M2Hc | And I get here
minus k0 delta H n0. |
_bTZbn7M2Hc | And it's good
notation to call this, |
_bTZbn7M2Hc | to save writing, delta Hkn
It's a good name for it. |
_bTZbn7M2Hc | It's the matrix k-- |
_bTZbn7M2Hc | the kn-th element of the matrix
delta H. And this is a number |
_bTZbn7M2Hc | so I can solve k0 n1
is equal to minus delta |
_bTZbn7M2Hc | Hkn divided by Ek0 minus En0. |
_bTZbn7M2Hc | And this is true for
every k different for n. |
_bTZbn7M2Hc | And here we find, for the first
time, our energy denominators. |
_bTZbn7M2Hc | These energy denominators
are the things |
_bTZbn7M2Hc | that are going to make life
interesting and difficult. |
_bTZbn7M2Hc | And it answers the
question already |
_bTZbn7M2Hc | that if you had
degenerate states, |
_bTZbn7M2Hc | there would be some
k state that have |
_bTZbn7M2Hc | the same energy as this one. |
_bTZbn7M2Hc | And this blows up. |
_bTZbn7M2Hc | And this is unsolvable
for this component. |
_bTZbn7M2Hc | So you start
getting difficulties |
_bTZbn7M2Hc | if you have degeneracies. |
_bTZbn7M2Hc | As long as every k state-- |
_bTZbn7M2Hc | all the other states
of the spectrum |
_bTZbn7M2Hc | have different energy from En,
nevermind if the other states |
_bTZbn7M2Hc | are degenerate. |
_bTZbn7M2Hc | They're not degenerate
with the state you care. |
_bTZbn7M2Hc | You care just about one
state now, the n-th state. |
_bTZbn7M2Hc | And if that's nondegenerate, all
these denominators are non-zero |
_bTZbn7M2Hc | and you're OK. |
_bTZbn7M2Hc | So here is the solution
for this thing. |
_bTZbn7M2Hc | Now I can write the expressions
for the state and the energy. |
_bTZbn7M2Hc | So let me do it. |
_bTZbn7M2Hc | So I have this n1 like that. |
_bTZbn7M2Hc | Now you can say the following. |
_bTZbn7M2Hc | Let me do this very
deliberately first. |
_bTZbn7M2Hc | n1 is equal to the sum
over all k of k0 k0 n1. |
_bTZbn7M2Hc | This is the resolution
of the identity formula. |
_bTZbn7M2Hc | That's the unit operator. |
_bTZbn7M2Hc | You can always do that. |
_bTZbn7M2Hc | And now you know that the
state n1 is orthogonal to n0. |
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