-
Notifications
You must be signed in to change notification settings - Fork 0
/
M3L15p.txt
137 lines (128 loc) · 5.32 KB
/
M3L15p.txt
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
#
# File: content-mit-8-421-3x-subtitles/M3L15p.txt
#
# Captions for 8.421x module
#
# This file has 127 caption lines.
#
# Do not add or delete any lines.
#
#----------------------------------------
Before we talk about some really [? cute ?] and nice aspects
of the fully quantized Hamiltonian,
I want to spend a few minutes talking
about degeneracy factors.
I've already given you my opinion that you should not
think in almost all situations about levels
which have a degeneracy.
Just think about states.
A state is a state, and it counts as one.
And if you have a level, which has triple degeneracy,
well it has three states.
Just count the states and look at the states.
However, there are [? formulae ?] which
involve degeneracy factors.
And just to remind you, when we had the discussion
of Einstein's A and B coefficients,
the Einstein A coefficient was proportion to the B coefficient
responsible for stimulated emission
from the excited to the ground state.
But the Einstein B coefficient for absorption
was related to the Einstein B coefficient
for stimulated emission by involving
these degeneracy factors.
So degeneracies appear and in some formulae
it makes a lot of sense to use them.
Also I've always said for fundamental understanding
you should just assume all degeneracies are one.
This is how you can avoid some baggage in deriving equations.
And I'm still standing to my statement,
I want to show you now a situation
where it becomes useful to consider degeneracy factors.
So let me give you an example.
We can now look at the situation where
we have an excited p-state and the ground state, which is s.
Or I can look at the opposite situation,
where we have an s-state, which can radiate to a p-state.
By symmetry, through the different p-states, n plus 1,
n minus 1, n equals [INAUDIBLE] are just
connected by spatial rotations, so therefore
the lifetime of the p-state, of the three p-states,
and the rate of spontaneous emission, are the same.
But if you now assume that you have absorption,
you go from the s-state to the p-state,
then you find that the Einstein B coefficient for absorption
there are now three possible ways, not just
one polarization, all three polarization.
And you will find that this is proportional to 3 times r.
However, in this situation, it's a reverse,
but let me just finish here.
So here, the natural linewidths, and the rate
of stimulated emission described by the B coefficient
from the excited state to the ground state,
is proportional to r.
Where s in the other situation, if we have absorption now,
each of those levels, there's only one transition one
[? pathway, ?] therefore you will find that the coefficient
for absorption is proportional to r,
wear s gamma and the stimulated emission, which is now Bsp,
is proportional to three r because there are there
[? pathways. ?]
So, depending what the situation is, you have to be careful.
And, you would say, but if it's s to p transition
it may be connected by the same matrix element,
and therefore you would say, shouldn't there
be a line strength which is independent
whether you go from s to p, or p to s, which just describes,
in a natural way, what is really the coupling between the s
and p-state.
And yes, indeed, there is, in the literature,
some definition of line strength where the line strengths, s,
would be proportional to the sum of all [? rates ?]
between an initial and the final state, and do some over all.
So therefore, when you use this formula for the line strengths,
whether you have the situation on the left side,
or on the right side, you would do always
the sum over the three possible transitions,
and so the line strengths are the same for both situations.
It's just generic for an s to p transition.
So if you use this definition, but then you
have the situation that spontaneous emission is always
given by the line strengths, but you
have to multiply now by the multiplicity of the excited
state.
If you have a p-state, the whole line strengths
is distributed over three states.
And each state has only a spontaneous emission
rate which is a third of what the line strengths give you.
I don't want to beat it to death because I
hate degeneracy factors, but I just
thought this example with the p to s
and s to p transition tells you why they necessarily
have to appear in derivations like Einstein's A and B
coefficient.
I hope there are no further questions about degeneracies.
Making these comments also allows me to say, well,
when I derived the Einstein A coefficient, what
we did last class, I did not use any degeneracy factors.
This is correct.
Our derivation assumed that there was only one final state.
We did not include degeneracy factors.
We also assumed that we had a dipole matrix element which
was along the z-axis, and so by those definitions
I have implicitly [? picked ?] geometry which
can be represented by that we have an excited p-state,
in the m equals zero state.
And we have a pi transition with linear polarization
to the s-state.
And by doing that, I did not have
to account for any degeneracies.
But in general if you derive microscopically
an equation for spontaneous emission,
you may have to take into account that your excited
state has different transitions, sigma plus and sigma minus
transitions to different states, and you
have to be careful how you do the sum
over all possible final states.
And this is where degeneracies would eventually matter.
Questions?