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PROFESSOR: All right.

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We have this, so let's complete
the Einstein discussion.

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And that's practical
formulas that

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are relevant to some
of the exercises

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you will find in homework.

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So when we set up the rates
in Einstein's argument,

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we said that for
spontaneous emission,

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Einstein's spontaneous emission,
he put a rate that was--

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he said the number
of transitions

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that occur due to
spontaneous emission

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is proportional to
the number of atoms,

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Nb, the number of
photons present,

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times some coefficient
that I don't know, Bab.

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That's how the rate of
atoms changing and going

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from the top level
to the low level was.

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But this is the one per atom.

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This is the transition
rate per atom, multiplied

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by the number of atoms present
in the top level gave you

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the rate for
spontaneous emission.

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This quantity is the one that
is equal to that omega ab.

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So this is equal to
omega ba, like this.

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Therefore, if you compare
the Bab of Einstein,

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it's all except the
U. So Bab is 4 pi

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squared over 3h squared
dab vector squared.

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So Einstein could not
do that because he

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didn't have quantum theory.

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But we do, and we now can
calculate from first principles

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this quantity in an
atomic transition.

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And the other good thing
was that even though there

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was a rate for
spontaneous emission,

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that rate was fixed
by Einstein's argument

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to be h bar omega ba
cubed over pi c cubed Bab.

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So if you knew
the b coefficient,

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you knew the a coefficient for
spontaneous emission-- many

35
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times were interested
in spontaneous emission.

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So we calculate the stimulated
emission coefficient, here,

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and that's given in terms
of this matrix element.

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We might as well give
the general formula.

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So you just substitute
the Bab, and now we

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have that A is equal to 4/3
omega ba cubed over h bar

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c cubed Bab squared
4/3 omega ba cubed.

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There's one h bar in the bottom.

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There's the c cubed.

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Yep.

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So that completes that argument.

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We now have combined
our development

47
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in time dependent
perturbation theory

48
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with a statistical
physics argument.

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[INAUDIBLE] to
determine those rates.

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A little more is
needed if you have

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to compute a practical
rate, as you will

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have to do for the homework.

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It's just the concept
of a lifetime.

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And a couple of selection rules.

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So let me make a few
comments about it.

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We get, for example,
a decay rate,

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A. It's a rate for
spontaneous emission.

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You have an atom.

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There's a decay rate, A. So if
you have a population of atoms,

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N atoms, at t equals 0, and A
being the decay rate per atom,

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it means that the
number of states

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that are changing in a little
time, dt, is minus A times dt--

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the rate multiplied
by the time per atom--

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times the number of atoms.

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So dN dt is minus AN.

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And the number of atoms as
a function of t go N at t

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equals 0 times e
to the minus At.

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So since we always
call processes

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that go e to the minus t over
tau-- tau is the lifetime.

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That is nothing
else than 1 over A.

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So you get the decay rate.

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It's synonymous to a lifetime.

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You just invert it.

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No big problem.

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Nevertheless, if
you have an object,

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sometimes can decay
in different ways.

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It can decay maybe
to one type of state,

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another type of state,
a third type of state.

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So the total A might
be equal to A1 plus A2

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when there are various
decay channels.

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You can decay-- a system can
decay from an excited state.

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It can go to one state to
another state to a third state.

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Each one contributes
a total decay rate

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because a total
disappearance of those atoms

85
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is because they either go into
one or the other or the third.

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So this is true that the
total decay rate is that.

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So this gives you a relation.

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The total lifetime,
which is 1 over this,

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is the sum of the inverses
of the partial lifetimes.

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So if you have a process
that can go in various ways,

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you may want to combine them
to get a total lifetime.

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So the last thing we
should say about this--

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the last thing
we're going to do is

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mention what happens
when you really

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try to calculate these
things for hydrogen levels.

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You're going to have
the ability to calculate

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the lifetime of an excited
state of hydrogen now.

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That's quite the thing,
to be able to get

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a number that is correct up to
factors of 2 and is measurable.

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So what do you have to consider?

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Well, the only thing
we haven't quite done

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is computed this matrix
element of the dipole operator.

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You see, everything else
that could be done was done.

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We integrated over the
incoherent radiation,

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we found the Fermi golden rule,
we have everything simplified.

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But still, the matrix
element is something

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that you have to calculate.

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So there's some selection rules.

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Rules.

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And this will be
discussed in recitation.

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It's pretty important stuff.

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I'll just mention the results.

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It's very readable in Griffiths.

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And it has to do with
the matrix elements.

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So d is the dipole operator,
so it boils down to r.

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So if you really knew n l
prime m prime, the matrix

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elements of nlm, if you knew
all of this, you would be done.

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You could calculate anything.

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So the question is, when
do these elements exist,

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and when are they nonzero?

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You need them to be nonzero to
get a nonzero transition rate.

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So when do they not vanish?

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So the only cases these--

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I'll summarize the answer--

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are nonzero, these
can be nonzero--

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there can still be accidents
and they could cancel--

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only if delta m, the change
in m from one side to another,

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which is m prime minus m,
is either equal to 0 or plus

129
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or minus 1.

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You cannot change that quantum
state of a system with a dipole

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interaction unless m
changes at most by 1.

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And moreover, the one
that is quite interesting

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as well is that delta l,
which is l prime minus l,

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must only be plus or minus 1.

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So those changes have to happen.

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So now you can at least get
a qualitative understanding

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of the hydrogen atom stability.

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You have the ground
state here, 1S.

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Here is 2S, 2P, 3S, 3P, and 3D.

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Let's see who can go to who.

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Well, the ground state
cannot go anywhere.

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That's at the bottom.

143
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So OK.

144
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2S going to 1S.

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Is it possible?

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AUDIENCE: No.

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PROFESSOR: No.

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Very good.

149
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It violates delta l.

150
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So this doesn't happen.

151
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So if this doesn't
happen, this 2S

152
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can almost go nowhere
because it doesn't have--

153
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it has to go by
spontaneous emission.

154
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Something has to go
to something lower,

155
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so it almost
doesn't go anywhere.

156
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So actually, it has
a long lifetime,

157
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and it can decay via two photons
or more complicated decays.

158
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And we will not consider it.

159
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But this is fairly stable.

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Now, this one, 2P,
can go here because it

161
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can go from l equal
1 to l equal 0,

162
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and it can change
m by some value.

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The 3S you would say, oh, it
cannot go here, cannot go here,

164
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but it can go laterally in that
direction because it can go

165
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change l from 1 to 0.

166
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So it's a change of l,
and that is possible.

167
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The 3P can go here.

168
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And my diagram starting
to get cluttered.

169
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And it could go here as
well, all the way down.

170
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And the 3D can only go here.

171
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Cannot go up to here because
that would be l equal 2.

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00:13:43,460 --> 00:13:51,520
So that is your diagram of
possible decays of hydrogen.

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And you're going to be
computing a couple of those.