WEBVTT
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PROFESSOR: So let's write
the Hamiltonian again
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in terms of v and v dagger.
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So for this equation, v
dagger v, from this equation,
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is equal to 2 h bar over
m omega, a dagger a.
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And immediately above
equation v dagger v's, this--
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we substitute into
the Hamiltonian.
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And Hamiltonian becomes the nice
object h bar omega, a dagger
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a plus 1/2 if you want.
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All right.
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We did this hard work
of factorization.
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We have to show what's good for.
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Well, in fact, we're
going to be able to solve
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the harmonic oscillator
without ever talking
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about differential-- almost
ever talking about differential
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equations.
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In fact, we will not talk about
the second order differential
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equation.
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Thanks to our
great work here, we
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will have to talk about a first
order differential equation,
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and a much simpler one.
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And only one, not for
and n 1, 2, 3, infitinty,
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infinity number of polynomials.
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It's a great simplification.
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Other Hamiltonians
admit factorization.
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In fact, there's whole books
of factorizable Hamiltonians,
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because those are the nicest
Hamiltonians to solve.
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Let's see why, though.
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We haven't said why yet.
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Here is the leading
thing that we can do.
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Remember we recalled 5
psi the integral dx phi
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star of x psi of x.
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This is just notation.
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So the expectation value,
calculate the expectation value
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of the Hamiltonian
in some state psi.
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Could be the general state sum.
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So what are you supposed to do?
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You're supposed to do psi h psi.
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This is normalized state,
the expectation value
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is the integral
of psi star h psi.
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That's what this is.
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But now, let's put
in this information.
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And the expectation value
of this would be psi h--
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or let me do it this way-- h
bar ome-- well, let's go slow.
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Psi h omega a dagger a a
psi plus h omega over 2 psi.
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So I just calculated
h on psi, and I
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wrote what it is-- h
omega this, this term.
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So this is two terms--
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h omega psi a dagger a psi
plus h omega over 2 psi psi.
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OK.
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So what did I gain
with the factorization?
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So far, it looks like nothing.
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But here we go--
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this term is equal to 1,
because the wave function
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is normalized.
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And here I can do one thing--
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I can remember my definition
of a Hermitian conjugate.
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I can move an operator and
put its Hermitian conjugate
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on the other side.
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So think of this
operator, a dagger--
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a dagger is acting on
this wave function.
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What is a dagger?
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It's this.
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And p is h bar over idex.
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So this-- you know how to act.
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But if a dagger is here, I
can put it on the first wave
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function, but I must put
the dagger of this operator,
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and the dagger of a dagger is a.
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So this is h omega, a psi
a psi plus h omega over 2.
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Now here comes the next thing.
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If this is an inner
product, any phi phi
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is greater or equal than 0,
because you would have phi star
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phi, and that's positive.
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So any of that [INAUDIBLE]
is greater or equal than 0.
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Note-- here you have
some function, but here
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the same function.
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It is this case.
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That thing is greater
or equal than 0.
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That is the great benefit of
the factorized Hamiltonian--
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if h has a v dagger v, you
can flip the v dagger here
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and it becomes v psi v
psi, and it's positive.
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And you've learned
something very important,
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and you can get
positive energies.
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In fact, from here,
since this is positive,
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this must be greater or
equal than h omega over 2.
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Because this is greater
than equal than 0.
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So the expectation value
of the Hamiltonian--
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if you would be thinking
now of energy eigenstates,
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the energy eigenvalue
is the expectation value
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of the Hamiltonian in
an energy eigenstate
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must be greater than
h bar omega over 2.
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And in some blackboard
that has been erased,
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we remember that the lowest
energy state had energy--
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there it is.
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The lowest energy state
has energy h omega over 2.
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So look at this, and
you say, OK, this
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shows that any eigenstate
must have energy
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greater than h omega over 2.
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But could there be one state
for which the energies exactly
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h omega over 2.
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Yes, if this inner product is 0.
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But for an inner product
of two things to be 0,
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each function must be 0.
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So from this, we conclude that
if there is a ground state,
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it's a state for which a phi--
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or a psi is equal to 0.
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So this is a very
nice conclusion.
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So if the lower
bound is realized,
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so that you get a state with
energy equal h bar over 2,
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then it must be true
that a psi is equal to 0.
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And a psi equal to 0 means x
plus ip over m omega on psi
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is equal to 0, or x plus
p is h bar over i ddx,
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so this is h bar over m omega
d/dx on psi of x is equal to 0.
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And that was the promised fact.
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We have turned the second
order differential equation
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into a first order
differential equation.
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Think of that magic that
has happened to do that.
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You had a second order
differential equation
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because the Hamiltonian
has x squared b squared.
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By factorizing, you go two first
order differential operators.
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And by Hermeticity, you
were led to the condition
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that the lowest energy
state had to be killed by a.
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That's why a is called
the annihilation operator.
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It should be killed.
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And now you have to solve
a first order differential
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equation, which is a game.
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An easy game compared with
a second order differential
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equation.
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So let's, of course, solve it.
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It doesn't take any time.
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Let's call this
the ground state.
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If it exists.
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And this gives you d psi 0 v x
is equal to minus m omega over
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h bar x psi .
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0.
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This can be degraded easily
or you can guess the answer.
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It's an exponential.
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Anything that differentiates
that you should extend
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the same function
as an exponential--
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e to the minus m omega
2 h squared x squared
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is the solution.
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Psi 0 of x is equal to
some number times that.
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This was-- the number is
the Hermit polynomials
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sub 0, and that exponential,
this exponential,
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we wrote a few blackboards ago.
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It's a good exponential.
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It's a perfect Gaussian.
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It's our ground state.
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And 0, if you want
to normalize it,
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m 0 is equal to m omega
over phi h bar to the 1/4.
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And that is the ground state.
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And it has energy,
h omega over 2.
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You could see what
the energy is by doing
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this very simple calculation.
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Look, get accustomed
to these things.
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H hat psi 0.
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What is h?
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Is h omega a dagger a
plus 1/2 acting on psi 0.
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The a acting on psi
0 already kills it.
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Because that's the
defining equation.
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Well that's 0.
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And you get 1/2 h bar omega,
confirming that you did
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get this thing to be correct.
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So this is only the
beginning of the story.
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We found the ground
state, and now we
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have to find the excited states.
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Let me say a couple
of words to set up
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this discussion for next time.
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The excited states appear
in a very nice way as well.
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So first a tiny bit
of language, of h bar.
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This equal h omega,
a dagger a is usually
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called the number operator.
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We'll explain more
on that next time.
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So n number operator
is a dagger a.
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It's a permission operator,
and it's pretty much
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the Hamiltonian.
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It's the number, it's called.
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Why is it called the number
is what we have to figure out.
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It is a counting operator--
it just looks at the state
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and counts things.
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So what does this give us?
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Well, we also know that
the number operator kills
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phi 0, because a kills psi 0.
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A kills it.
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So that's what we have.
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So we did say that
a was a destruction
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operator, annihilation
operator, because it annihilates
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the ground state.
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So if a annihilates
the ground state,
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a dagger cannot annihilate
the ground state.
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Why?
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Because a dagger with a
computator is equal to 1.
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Look at this.
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This is a a dagger
minus a dagger a.
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Act on the ground state.
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That's it.
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Now this term kills it.
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But this term
better not kill it,
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because it has to give
you back the ground
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state if this is true.
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And this is true.
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So a dagger doesn't
kill the ground state.
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Since it doesn't kill it, it's
called a creation operator.
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So you have this
state, but now there's
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also this state a dagger
acting on the vacuum.
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And there's a state a dagger
a dagger acting on the vacuum.
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And all those.
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And what we will
figure out next time
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is that, yes, this
is the ground state.
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And this is the
first excited state.
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And this is the
second excited state.
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And goes on forever.
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So we'll have a very compact
formula for the excited states
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of the harmonic oscillator.
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They're just creation
operators acting on the ground
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state or the [INAUDIBLE].