-
Notifications
You must be signed in to change notification settings - Fork 2
/
M2L9b.txt
167 lines (161 loc) · 6.32 KB
/
M2L9b.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
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
#
# File: content-mit-8-421-2x-subtitles/M2L9b.txt
#
# Captions for 8.421x module
#
# This file has 158 caption lines.
#
# Do not add or delete any lines.
#
#----------------------------------------
If you have a deformation of a nucleus
or if you have any kind of anisotropic shape of an object,
what is the minimum angular momentum in order
to observe it?
And I know a lot of you got confused about it.
So I want to discuss the same thing again but by now focusing
on two different frames, the lab frame and the body-fixed frame.
And I hope you find this discussion insightful.
So let's assume we have an object.
It can be a molecule.
Actually, you have a homework assignment
where you can observe the permanent dipole
moment of a molecule.
And this will lead you into a discussion of lab
frame versus body-fixed frame.
So but let's assume we have an object-- assume
it's a nucleus-- which has a really odd shape.
However, if it has angular momentum of 0,
all you have is one level.
If you have an angular momentum of one half,
you have two levels, and you can now define in the laboratory
that the energy difference is due to the magnetic dipole
moment, for instance, or in the electric dipole moment
if you put in electric field.
If you have I equals 1, you can have three levels--
E1, E0, E minus 1.
And let's say you have put the atoms into an electric field
gradient, you can then ask if E0 is
in the middle between up and down, plus 1 or minus 1,
or whether it's displaced or not.
And depending if this is larger or smaller than 0,
you would say there is a quadrupole moment, which
is larger or smaller than 0.
In other words, what I'm telling you
is, if you have only one level, I equals 0,
you can't say anything about the shape of the object.
If you have two levels, you can determine the dipole moment.
If you have three levels, by the deviation
from the equidistance, you can find a higher order moment.
OK.
But now comes a point.
If you assume you can now take two positions,
you can say that for low I, the deformation.
of the magnetic moment for I equals 0 cannot be measured.
Therefore, it is 0.
Or, you can say deformation exists
but only not in the lab frame but in the body-fixed frame.
So in other words, you would say that the deformation exists
always.
It's just a measurement problem.
At low angular momentum, I cannot measure it.
Or, the other statement would be, well,
it you can't measure it in quantum mechanics,
it means it doesn't exist.
So which statement or which conclusion is correct?
So this is the body-fixed frame argument and this is more
based on the lab frame.
I would say the lab frame argument is always correct
because you can't measure it.
You can't determine it in the lab.
But let me go from there.
Let's assume we have an object.
And we think it has a deformation,
but it has a low angular momentum state.
Can we still say that in its body-fixed frame
we have an object which has a deformation or not?
Now, my personal opinion is the following.
It really depends.
If you have a system where you can add angular momentum
without changing the internal structure-- for example
if I have this stick and I can just spin it up.
If I can spin it up, I can get an angular momentum
wave packet.
I can orient it and measure its deformation.
Then I think I would say even if this state has 0 angular
momentum, it has a deformation.
And the way I know it because if I add angular momentum,
I can measure the deformation.
I cannot measure it at low angular momentum.
So then you would say in the body-fixed system
there is a deformation, but it only
manifests itself in the lab frame
if I add angular momentum.
However, you may have an object, let's say a molecule,
which is so weakly bound that one
quantum of angular momentum due to centrifugal forces
rips it apart.
That exists, an extremely weakly bound state,
which cannot provide enough binding force to withstand even
one unit of angular momentum.
So that's an object which you cannot rotate--
you cannot transfer any angular momentum without destroying it,
without ripping it apart.
And now to say that this molecule has a dipole moment
or has some anisotropy it's like if you are making a statement
which cannot be tested at all.
so at that point, you should rather say,
what matters is what I can measure in the lab,
but I will never be able to measure
any deformation in the lab.
So the first approach with the stick-- the first example
with the stick will apply to a very stable molecule,
which may have a dipole moment.
And you assume it has the same dipole moment
whether it's rotating or not.
And then you would say at 0 angular momentum,
I cannot measure the dipole moment,
but I know it exists in the body-fixed frame.
If you have nuclei-- if you have a nucleus
and you have sort of a wave function of the protons
and neutrons, and it's I equals 0,
you will not find any moment.
And at least for the ground state of nulcei,
when you add angular momentum, you really
change the internal structure.
You have to promote nucleons to higher orbits.
So you cannot add angular momentum and still have
the same object.
So therefore you have to say the ground state with I
equals 0 has no deformation because there is no way
to ever find if there's a deformation,
so it doesn't exist.
But there are excited state of nuclei,
which have a deformation.
And these deformed nuclei can be put into a multiplet
of angular momentum states.
So then you would see I have the same kind of nucleus
but at different angular momentum states.
And at higher angular momentum states
you can, with high accuracy determine quadrupole moments
and things like this.
And then you may say the same internal state has now
a non-rotating state, and you would still be tempted--
and you are correct with that, to associate a deformation even
at the non-rotating state.
I hope those remarks help you to sort of reconcile the two
aspects, whether you have an object which
is stable enough to be spun up.
And then I think you can always talk
about the body-fixed frame.
But if you have an object where you change the internal
structure when you add angular momentum,
I think for fundamental reasons, you cannot associate any
deformation with it.
Any questions about that?
So let me just write down one sentence of a summary
so that the definition of a deformation
in the body-fixed frame makes sense
only if you can add or change angular
momentum without significantly changing
the internal structure.