This preview shows pages 1–16. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 17 Symmetry and Conservation Laws 17—1 Symmetry In classical physics there are a number of quantities whlch are conserved—
such a momentum, energy, and angular momentum. Conservation theorems
about corresponding quantities also exist in quantum mechanics. The most beau
tiful thing of quantum mechanics is that the conservation theorems can, in a
sense, be derived from something else, whereas in classical mechanics they are
practically the starting pomts of the laws. (There are ways in clas51cal mechanics
to do an analogous thing to what we will do in quantum mechanics, but it can be
done only at a very advanced level.) In quantum mechanics, however, the conserva
tion laws are very deeply related to the principle of superposition of amplitudes,
and to the symmetry of physical systems under various changes. This 15 the subject
of the present chapter. Although we will apply these ideas mostly to the conserva
tion of angular momentum, the essential point is that the theorems about the
conservation of all kinds of quantities are—in the quantum mechanics——related to
the symmetries of the system. We begin, therefore, by studying the question of symmetries of systems. A
very simple example is the hydrogen molecular ion—we could equally well take the
ammonia molecule—in which there are two states. For the hydrogen molecular
ion we took as our base states one in which the electron was located near proton
number 1, and another In Wthh the electron was located near proton number 2.
The two states—which we called I I) and  2)—are ShOWn again in Fig. 17—1(a).
New, so long as the two nuclei are both exactly the same, then there is a certain
symmetry in this physical system. That is to say, if we were to reﬂect the system
in the plane halfway between the two protons—by which we mean that everything
on one side of the plane gets moved to the symmetric position on the other side—
we would get the situatlons in Fig. 17—l(b). Since the protons are identical, the
operation of reﬂection changes l 1) into I 2) and l 2) into I I). We’ll call this reﬂec
tion operation P and write P11) = 2>, F2>=11>. (17.1) So our P is an operator in the sense that it “does something” to a state to make a
new state. The interesting thing is that P operating on any state produces some
other state of the system. Now 13, like any of the other operators we have described, has matrix elements
which can be deﬁned by the usual obvious notation. Namely, and P12=<IlP2> are the matrix elements we get if we multiply I3  1) and P  2) on the left by (I From Eq. (17.1) they are
[P 1 =P = 1 2 =0
<l>11<>, (In)
(Ilpl2)=P12=<11>=1 In the same way we can get P21 and P22. The matrix of P—with respect to the
base system  I) and  2)——is
0 1
p _ (1 0). We see once again that the words operator and matrix in quantum mechanics are
17—1 (17.3) 17—1 Symmetry 172 Symmetry and conservation
17—3 The conservation laws 17—4 Polarized light 17—5 The disintegration of the A0 17—6 Summary of the rotation
matrices Review: Chapter 52, Vol. I, Symmetry
in Physical Laws Reference: Angular Momentum in
Quantum Mechanics:
A. R. Edmonds, Princeton
University Press, 1957 (b) 5h) on '9 Fig. l7—l. If the states ) I) and I 2)
are reﬂected in the plane PP, they go
into i 2) and l 1), respectively. PROS. é
l'> _ PROB. ? V
a V
l'> l> AFTER TIME AFTER TIME
l2> t l'> l2> l'> l2> t 2
(0) (b)
Fig. 17—2. In a symmetric system, if c: pure I 1) state develops as shown in part (a), a pure [2) state will develop as in part (b). practically interchangeable. There are slight technical differences—like the differ
ence between a “numeral” and a “number”——but the distinction is something
pedantic that we don’t have to worry about. So whether 13 deﬁnes an operation,
or is actually used to deﬁne a matrix of numbers, we will call it interchangeably
an operator or a matrix. Now We would like to point out something. We will suppose that the physics
of the whole hydrogen molecular ion system is symmetrical. It doesn’t have to be
—it depends, for instance, on what else is near it. But if the system is symmetrical,
the following idea should certainly be true. Suppose we start at t = 0 with the
system in the state I I) and ﬁnd after an interval of time t that the system turns
out to be in a more complicated situation—in some linear combination of the two
base states. Remember that in Chapter 8 we used to represent “going for a
period of time” by multiplying by the operator U. That means that the system
would after a while—say 15 seconds to be deﬁnite—be 1n some other state. For
example, it might be x/f/_3 parts of the state  1) and iV/m parts of the state i 2),
and We would write Mat 15 sec) = 005, 0) l 1) = \/2/3]1)+ i\/1/3]2>. Now we ask what happens if we start the system in the symmetric state I 2) and
wait for 15 seconds under the same conditions? It is clear that if the world is
symmetric—as we are supposing—we should get the state symmetric to (17.4): (17.4) [,iatiSsec) = U(15,0)2> = \/2/3\2) + ixﬁﬁlr). (17.5) The same ideas are sketched diagrammatically in Fig. 17—2. So if the physics of a
system is symmetrical with respect to some plane, and we work out the behavior
of a particular state, we also know the behavior of the state we would get by
reﬂecting the original state in the symmetry plane. We would llke to say the same things a litte bit more generallyﬂwhich means
a little more abstractly. Let Q be any one of a number of operations that you
could perform on a system without changing the physics. For instance, for Q we
might be thinking of P, the operation of a reﬂection in the plane between the two
atoms in the hydrogen molecule. Or, in a system with two electrons, we might be
thinking of the operatiOn of interchanging the two electrons. Another possibility
would be, in a spherically symmetric system, the operation of a rotation of the
whole system through a ﬁnite angle around some axis—which wouldn’t change
the physics. Of course, we would normally want to give each special case some
special notation for Speciﬁcally, we will normally deﬁne the R110) to be the
operation “rotate the system about the yaxis by the angle 0”. By Q we mean
just any one of the operators we have described or any other one—which leaves
the basic physical situation unchanged. Let’s think of some more examples. If we have an atom with no external
magnetic ﬁeld or no external electric ﬁeld, and if we were to turn the coordinates
around any axis, it would be the same physical system. Again, the ammonia
molecule is symmetrical with respect to a reﬂection in a plane parallel to that of
the three hydrogens—so long as there is no electric ﬁeld. When there is an electric
ﬁeld, when we make a reﬂection we would have to change the electric ﬁeld also, 1 7—2 and that changes the physical problem. But if we have no external ﬁeld, the
molecule is symmetrical. Now we consider a general situation. Suppose we start with the state I ¢ 1)
and after some time or other under given physical conditions it has become the
state I (02). We can write 1%) = U 1 $1). (17.6) [You can be thinking of Eq. (17.4).] Now imagine we perform the operation Q
on the whole system. The state I ipl) will be transformed to a state I M), which
we can also write as Q I $1). Also the state I $2) is changed into I (05) = Q I 4/2).
Now if the phySI'Cs is symmetrical under Q (don’t forget the if; it is not a general
property of systems), then, waiting for the same time under the same conditions,
we should have I‘l’é) = UNI/'1) (177)
[Like Eq. (17.5)] But we can write Q I (01) for I i//{) and Q I 1/12) for I #19 so (17.7)
can also be written A A A
Q I 1P2) = UQI¢1) (178)
If we now replace I .112) by U I ip1)—Eq. (l7.6)—we get
Q‘I71¢1>= 09m. (17.9) It’s not hard to understand what this means. Thinking of the hydrogen ion it
says that: “making a reﬂection and waiting a while”—the expression on the
right of Eq. (17.9)—is the same as “waiting a while and then making a reﬂection”—
the expression on the left of (17.9). These should be the same so long as U doesn’t
change under the reﬂection. Since (17.9) is true for any starting state I ll) 1), it is really an equation about the operators: A .. . A
QU = UQ. (17.10) This is what we wanted to get—it is a mathematical statement of symmetry. When
Eq. (17.10) is true, we say that the operators (7 and Q commute. We can then
deﬁne “symmetry” in the following way: A physical system is symmetric with
respect to the operation Q when Q commutes with [7, the operation of the passage
of time. [In terms of matrices, the product of two operators is equivalent to the
matrix product, so Eq. (17.10) also holds for the matrices Q and U for a system
which is symmetric under the transformation Q.] Incidentally, since for inﬁnitesimal times 6 we have (7 = 1 — iHe/h—where
H is the usual Hamiltonian (see Chapter 8)—you can see that if (17.10) is true, it is also true that    A
QH = HQ, (17.11) So (17.11) is the mathematical statement of the condition for the symmetry of a
physical situation under the operator It deﬁnes a symmetry. 17—2 Symmetry and conservation Before applying the result we have just found, we would like to discuss the
idea of symmetry a little more. Suppose that we have a very speCial situation:
after we operate on a state With Q, we get the same state, This is a very special case,
but let’s suppose it happens to be true for a state Iwo) that Iup’) = Q I $0) is
physically the same state as I 1/20). That means that I 11/) is equal to I (to) except
for some phase factor.‘I' How can that happen? For instance, suppose that we TInCidentally, you can show that Qtis necessarily a unitary operator—which means
that if it operates on I 1,0) to give some number times I t/x), the number must be of the form
e”, where 6 is real. It’s a small pomt, and the proof rests on the followrng observation.
Any operation like a reﬂection or a rotation doesn‘t lose any particles, so the normaliza
tion of I 11/) and l/J) must be the same; they can only differ by a pure imaginary phase
factor. l73 Fig. 17—3. The state I!) and the
state? II) obtained by reﬂecting II) in
the central plane. have an H; ion in the state which we once called I I). For this state there is equal
amplitude to be in the base states I I) and I 2). The probabilities are shown as a
bar graph in Fig. 17—3(a). If we operate on I I) with the reﬂection operator P, it
ﬂips the state over changing I I) to I2) and I2) to I 1)—we get the probabilities
shown in Fig. 17—3(b). But that’s just the state I I) all over again. If we start with
state I II) the probabilities before and after reﬂection look just the same. However,
there is a difference if we look at the amplitudes. For the state I I) the amplitudes
are the same after the reﬂection, but for the state III) the amplitudes have the
opposite sign. In other words, \/§ \/§
(17.12)
Vi V5
lfwewriteplwo) = e’5  tlz()),wehavethate“s = 1forthestateI1)and e15 = —1 for the state I 11). Let’s look at another example. Suppose we have a RHC polarized photon
propagating in the zdirection. If we do the operation of a rotation around the
zaxis, we know that this just multiplies the amplitude by e“ when 4: is the angle
of the rotation. So for the rotation operation in this case, 6 is just equal to the
angle of rotation. Now it is clear that if it happens to be true that an operator Q just changes the
phase of a state at some time, say i = 0, it is true forever. In other words, if the
state I all) goes over into the state I 11/2) after a time t, or 17(t,0)1¢1) = w2> (17.13)
and if the symmetry of the situation makes it so that Q I $1) = e“5 1101), (17.14)
then it is also true that Q I 72> = e“5 l 1&2). (17.15) This is clear, since QI¢2> = QUWO = 0QI'P1),
andifQ I 'Pi) = ewI‘Pi); then
QM) = 1765171) = «2150171) = 65172). [The sequence of equalities follows from (17.13) and (17.10) for a symmetrical
system, from (17.14), and from the fact that a number like em commutes with an
operator.] So with certain symmetries something which is true initially is true for all
times. But isn’t that just a conservation law? Yes‘ It says that if you look at the
original state and by making a little computation on the side discover that an
operation which is a symmetry operation of the system produces only a multiplica
tion by a certain phase, then you know that the same property will be true of the
ﬁnal state—the same operation multiplies the ﬁnal state by the same phase factor.
This is always true even though we may not know anything else about the inner
mechanism of the universe which Changes a system from the initial to the ﬁnal
state. Even if we do not care to look at the details of the machinery by which the
system gets from one state to another, we can still say that if a thing 18 in a state
with a certain symmetry character originally, and if the Hamiltonian for this thing
is symmetrical under that symmetry operation, then the state will have the same
symmetry character for all times. That’s the basis of all the conservation laws of
quantum mechanics. Let’s look at a special example. Let’s go back to the P operator. We would
like ﬁrst to modify a little our deﬁnition of 13. We want to take for P not just a 17—4 mirror reﬂection, because that requires deﬁning the plane in which we put the
mirror. There is a special kind of a reﬂection that doesn’t require the speciﬁcation
of a plane. Suppose we redeﬁne the operation P this way: First you reﬂect in a
mirror in the zplane so that 2 goes to —z, x stays x, and y stays y; then you turn
the system 180° about the zaxis so that x is made to go to —x and y to —— y. The
whole thing is called an inversion. Every point is projected through the origin to the
diametrically opposite position. All the coordinates of everything are reversed.
We Will still use the symbol P for this operation. It is shown in Fig. 17—4. It is a
little more convenient than a simple reﬂection because it doesn't require that you
specify which coordinate plane you used for the reﬂection—you need specify only
the point which is at the center of symmetry.
Now let’s suppose that we have a state l r140) which under the inversion opera
tion goes into e“5 I («w—that is,
Isl/6) = Pi‘ho = emit/o) (1716) Then suppose that we invert again. After two inversions we are right back where
we started from—~nothing is changed at all. We must have that Pit/6) = FISH/0): like)
Pth) = 13min) = ewﬁlm = reams. But It follows that
(e"’)2 = 1. So if the inversion operator is a symmetry operation of a state, there are only two possibilities for 6: e“5 = =kl, which means that Pita) = Wo) or Flt/o) = — l'l’ol (17.17) Classically, if a state is symmetric under an inversion, the operation gives
back the same state. In quantum mechanics, however, there are the two possibilities:
we get the same state or minus the same state. When we get the same state, P I tpo) =
I 4/0), we say that the state I too) has even parity. When the sign is reversed so that
P I 51/0) = — ] t/xo), we say that the state has odd parity. (The inversion operator
P is also known as the parity operator.) The state I I ) of the H; ion has even parity;
and the state {11) has odd parity~see Eq. (17.12). There are, of course, states
which are not symmetric under the operation P; these are states with no deﬁnite
parity. For instance, in the H? system the state I I) has even parity, the state I II)
has odd parity, and the state  I) has no deﬁnite parity. When we speak of an operation like inversion being performed “on a physical
system” we can think about it in two ways. We can think of physically moving
whatever is at r to the inverse point at —r, or we can think of looking at the same
system from a new frame of reference x’, y’, 2’ related to the old by x’ = ——x,
y’ = — y, and z’ = —z. Similarly, when we think of rotations, we can think of
rotating bodily a physical system, or of rotating the coordinate frame with respect
to which we measure the system, keeping the “system” ﬁxed in space. Generally,
the two points of view are essentially equivalent. For rotation they are equivalent
except that rotating a system by the angle 0 is like rotating the reference frame by
the negative of 6. In these lectures we have usually considered what happens when
a projection is made into a new set of axes. What you get that way is the same as
what you get if you leave the axes ﬁxed and rotate the system backwards by the
same amount. When you do that, the signs of the angles are reversedT T In other books you may ﬁnd formulas wrth different signs; they are probably using
a different deﬁnition of the angles. 17—5 Fig. 17—4. The operation of inver
sion, p. Whatever is at the point A at
(x, y, z) is moved to the point A' at
(—x, —y. z). Many of the laws of physics—but not all—are unchanged by a reﬂection or an
inversion of the coordinates. They are symmetric with respect to an inversion.
The laws of electrodynamics, for instance, are unchanged if we change x to —x,
y to — y, and z to —z in all the equations. The same is true for the laws of gravity,
and for the strong interactions of nuclear physics. Only the weak interactions—
responsible for ﬁdecay—do not have this symmetry. (We discussed this in some
detail in Chapter 52, Vol. I.) We will for now leave out any consideration of the
Bdecays. Then in any physical system where Bdecays are not expected to produce
any appreciable effect—an example would be the emission of light by an atom—
the Hamiltonian H and the operator P will commute. Under these circumstances
we have the following proposition. If a state originally has even parity, and if you
look at the physical situation at some later time, it will again have even parity.
For instance, suppose an atom about to emit a photon is in a state known to have
even parity. You look at the whole thing—including the photon—after the emis
sion; it will again have even parity (likewise if you start with odd parity). This
principle is called the conservation of parity. You can see why the words “conserva
tion of parity” and “reﬂection symmetry” are closely intertwined in the quantum
mechanics. Although until a few years ago it was thought that nature always
conserved parity, it is now known that this is not true. It has been discovered to
be false because the ﬁdecay reaction does not have the inversion symmetry which
is found in the other laws of physics. Now we can prove an interesting theorem (which is true so long as we can
disregard weak interactions): Any state of deﬁnite energy which is not degenerate
must have a deﬁnite parity. It must have either even parity or odd parity. (Re
member that we have sometimes seen systems in which several states have the same
energy—we say that such states are degenerate. Our theorem will not apply to
them.) For a state  1/10) of deﬁnite energy, we know that ﬁll 1P0) = E1110), (17.18) where E is just a number—the energy of the state. If we have any operator Q
which is a symmetry operator of the system we can prove that Q1161» = e” i‘I’o) (17.19) so long as  W0) is a unique state of deﬁnite energy. Consider the new state  306)
that you get from operating with If the physics is symmetric, then  1/43) must
have the same energy as I 4’0). But we have taken a Situation in which there is
only one state of that energy, namely ltpo), so lips) must be the same state—it
can only diﬂer by a phase. That’s the phystcal argument. The same thing comes out of our mathematics. Our deﬁnition of symmetry
is Eq'. (17.10) or Eq. (17.11) (good for any state 1p), HQ 1 it) = Q19  in. (17.20) But we are considering only a state [ M) which is a deﬁnite energy state, so that
H  100) = E l 1,00). Since E is just a number that ﬂoats through Q if we want,
we have GEN/0) = 190) = EQ i 1‘0)
H‘iQ i 190)} = E{Q i 100» (1721) So libs) = Q 1 $0) is also a deﬁnite energy state of fit—and with the same E.
But by our hypothesis, there is only one such state; it must be that I 1116) = e125 1 1110). What we have just proved is true for any operator Q that is a symmetry opera
tor of the physrcal system. Therefore, in a situation in which we cons1der only
electrical forces and strong interactions—and no ﬂdecay—so that inversion sym
metry is an allowed approx1mation, we have that P [ 1/1) = 615 I it). But we have
also seen that e“5 must be either +1 or — 1. So any state of a deﬁnite energy (which
is not degenerate) has got either an even parity or an odd parity. 17—6 So 17—3 The conservation laws We turn now to another interesting example of an operation: a rotation.
We conSider the special case of an operator that rotates an atomic system by angle
:1: around the zaxis. We Wlll call this operator‘l' R2012). We are going to suppose
that we have a physical situation where we have no inﬂuences lined up along the
x and yaxes. Any electric ﬁeld or magnetic ﬁeld is taken to be parallel to the
zaxisft so that there will be no change in the external conditions if we rotate the
whole physical system about the zaxis. For example, if we have an atom in empty
space and we turn the atom around the zaxis by an angle 4;, we have the same
physical system. Now then, there are special states which have the property that such an opera
tion produces a new state which is the original state multiplied by some phase
factor. Let us make a quick Side remark to show you that when this is true the
phase change must always be proportional to the angle 4). Suppose that you would
rotate twice by the angle 4). That’s the same thing as rotating by the angle 2¢>. If a
rotation by ¢> has the eﬂect of multiplying the state No) by a phase e“s so that R440 l ¢0> = (‘15 l $0), two such rotations in succession would multiply the state by the factor (e‘5)2 =
em, since R.(¢)Rz(¢) I wo> = R.(¢)e”l¢o> = e'5Rz(¢) I to) = ewe” Ito). The phase change 6 must be proportional to gill We are considering then those
special states I ‘00) for which [ ‘p0) = elmd) i ¢0>: where m is some real number. We also know the remarkable fact that if the system is symmetrical for a rota
tion around 2 and if the original state happens to have the property that (17.22)
is true, then it will also have the same property later on. So this number m is a
very important one. If we know its value initially, we know its value at the end of
the game. It is a number which is conserved—m is a constant of the motzon. The
reason that we pull out m is because it hasn’t anything to do with any special angle
¢, and also because it corresponds to something in classical mechanics. In quantum
mechanics we choose to call mh—for such states as  l//0>—‘the angular momentum
about the zaxis. If we do that we ﬁnd that in the limit of large systems the same
quantity is equal to the zcomponent of the angular momentum of claSSical me
chanics. So if we have a state for which a rotation about the ZaXlS Just produces
a phase factor em", then we have a state of deﬁnite angular momentum about that
axis—and the angular momentum is conserved. It is mft now and forever. Of
course, you can rotate about any axis, and you get the conservation of angular
momentum for the various axes. You see that the conservation of angular
momentum is related to the fact that when you turn a sysrem you get the same
state with only a new phase factor. We would like to show you how general this idea is. We will apply it to two
other conservation laws which have exact correspondence in the physical ideas
to the conservation of angular momentum. In class1cal physics we also have
conservation of momentum and conservation of energy, and it is interesting to
see that both of these are related in the same way to some phys1cal symmetry. Jr Very preCisely, we Will deﬁne R441) as a rotation of the phySical system by —¢ about
the ZaXlS, which is the same as rotating the coordinate frame by +¢. i We can always choose 2 along the direction of the ﬁeld provided there is only one
ﬁeld at a time, and its direction doesn’t change. ii For a fancier proof we should make this argument for small rotations 5 Since any
angle :1: is the sum of a suitable n number of these,¢ = Me, R2615) = [Rz(e)]" and the total
phase change is n times that for the small angle 6, and is, therefore, proportional to ¢ 1777 Suppose that we have a physical system—an atom, some complicated nucleus,
or a molecule, or something—and it doesn’t make any difference if we take the
whole system and move it over to a different place. So we have a Hamiltonian
which has the property that it depends only on the internal coordinates in some
sense, and does not depend on the absolute position in space. Under those cir
cumstances there is a special symmetry operation we can perform which is a
translation in space. Let’s deﬁne Dx(a) as the operation of a displacement by the
distance (1 along the xaxis. Then for any state we can make this Operation and
get a new state. But again there can be very special states which have the property
that when you displace them by it along the xaxis you get the same state except
for a phase factor. It’s also possible to prove,just as we did above, that when this
happens, the phase must be proportional to a. So we can write for these special
states  1/20) 1595(0) l 1.00) = 9”“ l ¢o> (1723) The coefﬁcient k, when multiplied by h, is called the xcomponent of the momentum.
And the reason it is called that is that this number is numerically equal to the
classical momentum p; when we have a large system. The general statement is
this: If the Hamiltonian is unchanged when the system is displaced, and if the
state starts with a deﬁnite momentum in the xdirection, then the momentum in
the xdirection will remain the same as time goes on. The total momentum of a
system before and after collisions—or after explosions or what not—will be the
same. There is another operation that is quite analogous to the displacement in
space: a delay in time. Suppose that we have a physical situation where there is
nothing external that depends on time, and we start something oil” at a certain
moment in a given state and let it roll. Now if we were to start the same thing
off again (in another experiment) two seconds later—or/say, delayed by a time
T—and if nothing in the external conditions depends on the absolute time, the
development would be the same and the ﬁnal state would be the same as the
other ﬁnal state, except that it will get there later by the time 7. Under those
circumstances we can also ﬁnd special states which have the property that the
development in time has the special characteristic that the delayed state is just
the old, multiplied by a phase factor. Once more it is clear that for these special
states the phase change must be proportional to 7'. We can write 5&7) l lilo) = 9—1.“ l 11’0> (1724) It is conventional to use the negative sign in deﬁning co; with this convention
wh is the energy of the system, and it is conserved. So a system of deﬁnite energy is
one which when displaced 7' in time reproduces itself multiplied by e’“‘”. (That’s
what we have said before when we deﬁned a quantum state of deﬁnite energy, so
we’re consistent with ourselves.) It means that if a system is in a state of deﬁnite
energy, and if the Hamiltonian doesn’t depend on t, then no matter what goes on,
the system will have the same energy at all later times. You see, therefore, the relation between the conservation laws and the sym
metry of the world. Symmetry With respect to displacements in time implies the
conservation of energy; symmetry with respect to position in x, y, or 2 implies
the conservation of that component of momentum. Symmetry with respect to
rotations around the x, y, and zaxes implies the conservation of the x, y—, and
zcomponents of angular momentum. Symmetry With respect to reﬂection implies
the conservation of parity. Symmetry with respect to the interchange of two elec
trons implies the conservation of something we don‘t have a name for, and so on.
Some of these princ1ples have classical analogs and others do not. There are more
conservation laws in quantum mechanics than are useful in claSSical mechanics——
or, at least, than are usually made use of. In order that you will be able to read other books on quantum mechanics,
we must make a small technical aside—to describe the notation that people use.
The operation of a displacement with respect to time is, of course, Just the opera 17—8 tion (7 that we talked about before:
15.0) = 0(t + r, I). (17.25) Most people like to discuss everything in terms of inﬁnitesimal displacements in
time, or in terms of inﬁnitesimal displacements in space, or in terms of rotations
through inﬁnitesimal angles. Since any ﬁnite displacement or angle can be ac
cumulated by a succession of inﬁnitesimal displacements or angles, it is often easier
to analyze ﬁrst the inﬁnitesimal case. The operator of an inﬁnitesimal displacement
At in t1me is—as we have deﬁned it in Chapter 8—
mm) = 1 — I—flmﬁ. (17.26) Then H is analogous to the classical quantity we call energy, because if 1? I III)
happens to be a constant times I up) namely, 17 I up) = EI 11/), then that constant
is the energy of the system. The same thing is done for the other operations. If we make a small displace
ment in x, say by the amount Ax, a state I up) will, in general, go over into some other
state I 4/). We can write IW) = fame) I 10> = (I + Ax) It), (1727) smce as Ax goes to zero, the I W) should become Just I w) or 151(0) = l, and for
small Ax the change of Dzmx) from 1 should be proportional to Ax. Deﬁned this
way, the operator [3,. is called the momentum operator—for the xcomponent, of
course. For identical reasons, people usually write for small rotations RAM) I 10> = (I + M) I1») (1728) and call L the operator of the zcomponent of angular momentum. For those
special states for which Rz(¢) I 1,00) = e”"" I 11/0), we can for any small angle—say
A¢~expand the righthand side to ﬁrst order in A4: and get RAM) = em“ I 'Po) = (1 + imA¢) I 1’0)
Comparing this with the deﬁnition of L in Eq. (17.28), we get that L I 11/0) = mh I 1/10). (17.29) In other words, if you operate with .72 on a state with a deﬁnite angular momentum
about the z—axis, you get mh times the same state, where mh is the amount of
zcomponent of angular momentum. It is quite analogous to operating on a
deﬁnite energy state with H to get E I it). We would now like to make some applications of the ideas of the conservation
of angular momentumﬁto show you how they work. The point is that they are
really very simple. You knew before that angular momentum is conserved. The
only thing you really have to remember from this chapter is that if a state I wo)
has the property that upon a rotation through an angle ¢ about the zaxis, it be
comes em‘l’ I 11/0); it has a zcomponent of angular momentum equal to mh. That’s
all we will need to do a number of interesting things. 17—4 Polarized light First of all we would like to check on one idea. In Section 11—4 we showed
that when RHC polarized light is viewed in a frame rotated by the angle 4: about
the zaxis’r it gets multiplied by e‘d’. Does that mean then that the photons of light T Sorry! This angle is the negative of the one we used in Section 11—4.
17—9 ly xi (0) lb) Fig. l7—5. (o) The electric ﬁeld 8
in a circularly polarized light wove. lb)
The motion of an electron being driven
by the circularly polarized light. that are right circularly polarized carry an angular momentum of one unit’r along
the zaxis? Indeed it does. It also means that if we have a beam of light containing
a large number of photons all circularly polarized the same way—as we would
have in a classical beam—it will carry angular momentum. If the total energy
carried by the beam in a certain time is W, then there are N = W/hw photons. Each
one carries the angular momentum it, so there is a total angular momentum of Jz=Nh=LV.
on (17.30) Can we prove classically that light which is right circularly polarized carries
an energy and angular momentum in proportion to W/w? That should be a classical
proposition if everything is right. Here we have a case where we can go from the
quantum thing to the classical thing. We should see if the classical physics checks.
It will give us an idea whether we have a right to call m the angular momentum.
Remember what right circularly polarized light is, classically. It’s described by
an electric ﬁeld with an oscillating xcomponent and an osc111ating ycomponent
90° out of phase so that the resultant electric vector 8 goes in a circle—as drawn in
Fig. 17—5(a). Now suppose that such light shines on a wall which is going to
absorb it—or at least some of it—and consider an atom in the wall according to
the classical physics. We have often described the motion of the electron in the
atom as a harmonic oscillator which can be driven into oscillation by an external
electric ﬁeld. We’ll suppose that the atom is isotropic, so that it can oscillate
equally well in the x or ydirections. Then in the circularly polarized light, the
xdisplacement and the ydisplacement are the same, but one 18 90° behind the
other. The net result is that the electron moves in a circle, as shown in Fig. l7—5(b).
The electron is displaced at some displacement r from its equilibrium position at the
origin and goes around with some phase lag with respect to the vector 8. The
relation between 8 and r might be as shown in Fig. 17—5(b). As time goes on, the
electric ﬁeld rotates and the displacement rotates with the same frequency, so
their relative orientation stays the same. Now let’s look at the work being done
on this electron. The rate that energy is being put into this electron is 0, its velocity,
times the component of 48 parallel to the velocity: dW 27 = qStv. (17.31)
But look, there is angular momentum being poured into this electron, because
there is always a torque about the origin. The torque is qatr, which must be
equal to the rate of change of angular momentum sz/dt: dig E— = qszr. (1732)
Remembering that v = cor, we have that all: _ 1. dW _ (.0 Therefore, if we integrate the total angilar momentum which is absorbed, it is
proportional to the total energy—the constant of proportionality being l/w,
which agrees with Eq. (17.30). Light does carry angular momentum—l unit
(times it) if it is right circularly polarized along the zaxis, and —1 unit along the
zaxis if it is left circularly polarized. Now let’s ask the following question: If light is linearly polarized in the
xdirection, what is its angular momentum? Light polarized in the xdirection
can be represented as the superposition of RHC and LHC polarized light. There
fore, there is a certain amplitude that the angular momentum is +h and another T It is usually very convenient to measure angular momentum of atomic systems in
units of h. Then you can say that a spin one—half particle has angular momentum =+=1/2
with respect to any axis. Or. in general. that the zcomponent of angular momentum
is m. You don’t need to repeat the I; all the time. 17—10 amplitude that the angular momentum is wit, so it doesn’t have a deﬁnite angular
momentum. It has an amplitude to appear with +ft and an equal amplitude to
appear with —h. The interference of these two amplitudes produces the linear
polarization, but it has equal probabilities to appear with plus or minus one unit
of angular momentum. Macroscopic measurements made on a beam of linearly
polarized light will show that it carries zero angular momentum, because in a large
number of photons there are nearly equal numbers of RHC and LHC photons
contributing opposite amounts of angular momentum—the average angular
momentum is zero. And in the classical theory you don’t ﬁnd the angular mo
mentum unless there is some circular polarization. We have said that any spinone particle can have three values of J,, namely
+1, 0, —1 (the three states we saw in the Stern—Gerlach experiment). But light is
screwy; it has only two states. It does not have the zero case. This strange lack
is related to the fact that light cannot stand still. For a particle of spin j which is
standing still, there must be the 2j + 1 possible states with values of jZ going in
steps of 1 from —j to +j. But it turns out that for something of spin j with zero
mass only the states with the components +j and — j along the direction of motion
exist. For example, light does not have three states, but only two—although a
photon is still an object of spin one. How is this consistent with our earlier proofs—
based on what happens under rotations in space—that for spinone particles three
states are necessary” For a particle at rest, rotations can be made about any
axis without changing the momentum state. Particles with zero rest mass (like
photons and neutrinos) cannot be at rest; only rotations about the axis along the
direction of motion do not change the momentum state. Arguments about rota
tions around one axis only are insufﬁcient to prove that three states are required,
given that one of them varies as 6” under rotations by the angle ¢.'l' One further side remark. For a zero rest mass particle, in general, only one
of the two spin states with respect to the line of motion (+ j, — j) is really necessary.
For neutrinoswhich are spin onehalf particles—only the states with the com—
ponent of angular momentum opposite to the direction of motion (—h/2) exist
in nature [and only along the motion (+ h/ 2) for antineutrinos]. When a system has
inversion symmetry (so that parity is conserved, as it is for light) both components
(+ j, and —j) are required. 17—5 The disintegration of the A0 Now we want to give an example of how we use the theorem of conservation
of angular momentum in a specifically quantum physical problem. We look at
breakup of the lambda particle (A0), which disintegrates into a proton and a 7r—
meson by a “weak” interaction: A°—>p+7r”. Assume we know that the pion has spin zero, that the proton has spin onehalf,
and that the A0 has spin onehalf. We would like to solve the following problem:
Suppose that a A0 were to be produced in a way that caused it to be completely
polarized—by which we mean that its spin is, say “up,” with respect to some suit
ably chosen zaxis—see Fig. 17—6(a). The question is, with what probability will it
disintegrate so that the proton goes off at an ang 0 with respect to the zaxis—as
in Fig. l7~6(b)? In other words, what is the ang lar distribution of the disintegra
tions? We will look at the disintegration in the coordinate system in which the
A0 is at rest—we will measure the angles in this rest frame; then they can always
be transformed to another frame if we want. 1' We have tried to ﬁnd at least a proof that the component of angular momentum
along the direction of motion must for a zero mass particle be an integral multiple of
fi/Z—and not something like It/ 3. Even using all sorts of properties of the Lorentz
transformation and what not, we failed. Maybe it’s not true. We’ll have to talk about
it with Prof. Wigner, who knows all about such things. 17—11 BEFORE AFTER
f2 [2
l e
l P
 , /‘
  /
J / 71..
A0 / (
l / i
/ l
I 1r_/ /’41 I
I / O ,
l I
(a) (b) Fig. 17—6. A A0 with spin "up"
decays into a proton and a pion (in the
CM system). What is the probability that
the proton will go off at the angle 6? BEFORE lz lz AFTER I
I
I
I
I

I
A‘
I
I

I
l
I
I C) or I
' I
l l
lvrr 'Ithr l "'O 7‘0
I
l YES N0 (0) (b) (C) Fig. 17—7. Two possibilities for the
decay of a spin “up” A0 with the proton
going along the +zoxis. Only (b)
conserves angular momentum. BEFORE AFTER \ Iv, I
l
l
I I
A f or Q
I I
I 
I
I I
l I
l l l

I

I . r5,
‘ I
' I
NO YES
(0) (b) (C) Fig. 17—8. The decay along the
zoxis for CI A" with spin "down." We begin by looking at the special circumstance in which the proton is emitted
into a small solid angle A9 along the zaxis (Fig. 17—7). Before the disintegration
we have a A0 with its spin “up,” as in part (a) of the ﬁgure. After a short time—for
reasons unknown to this day, except that they are connected With the weak decays—
the A0 explodes Into a proton and a pion. Suppose the proton goes up along the
+zaxis. Then, from the conservation of momentum, the pion must go down.
Since the proton is a spin onehalf particle, its spin must be either “up” or “down”—
there are, in principle, the two possibilities shown in parts (b) and (c) of the ﬁgure.
The conservation of angular momentum, however, requires that the proton have
spin “up.” This is most easily seen from the followmg argument. A particle moving
along the zaxis cannot contribute any angular momentum about this axis by virtue
of its motion; therefore, only the spins can contribute to J2. The spin angular
momentum about the zaxis is +fI/2 before the disintegration, so it must also be
+h/2 afterward. We can say that since the pion has no spin, the proton spin
must be “up.” If you are worried that arguments of this kind may not be valid in quantum
mechanics, we can take a moment to show you that they are. The initial state
(before the disintegration), which we can call ] A0, spin +2) has the property that
if it IS rotated about the zaxis by the angle 4:, the state vector gets multiplied by
the phase factor eW 2. (In the rotated system the state vector is 3““ 2  A0, spin +2).)
That’s what we mean by spin “up” for a spin onehalf particle. Since nature’s
behavior doesn’t depend on our choice of axes, the ﬁnal state (the proton plus
pion) must have the same property. We could write the ﬁnal state as, say, I proton going +2, spin +2; pion going ——z). But we really do not need to specify the pion motion, since in the frame we have
chosen the pion always moves opposite the proton; we can simplify our description
of the ﬁnal state to  proton going +2, spin +2). Now what happens to this state vector If we rotate the coordinates about the
zaxis by the angle 4;? Since the proton and pion are moving along the zaxis, their motion isn’t
changed by the rotation. (That’s why we picked this special case; we couldn’t
make the argument otherwise.) Also, nothing happens to the pion, because it is
spin zero. The proton, however, has spin onehalf. If its spin is “up” it will con
tribute a phase change of 6”” in response to the rotation. (If its spin were
“down” the phase change due to the proton would be e_”’/ 2.) But the phase change
with rotation before and after the excitement must be the same if angular mo
mentum is to be conserved. (And It will be, Since there are no outside inﬂuences in
the Hamiltonian.) So the only possibility is that the proton spin Will be “up.”
If the proton goes up, its spin must also be “up.” We conclude, then, that the conservation of angular momentum permits the
process shown in part (b) of Fig. 17—7, but does not permit the process shown In
part (c). Since we know that the disintegration occurs, there is some amplitude
for process (b)—proton gomg up with spin “up.” We’ll let a stand for the amplitude
that the disintegration occurs In this way in any 1nﬁnitesrmal interval of timet‘ Now let’s see what would happen If the A0 spin were initially “down.” Again
we ask about the decays In which the proton goes up along the zaxis, as shown in
Fig. 17—8. You Will apprecrate that in this case the proton must have spin “down”
If angular momentum is conserved. Let’s say that the amplitude for such a dis
integration is b. We can’t say anything more about the two amplitudes a and I). They depend
on the inner machinery of A0, and the weak decays, and nobody yet knows how to TWe are now assuming that the machinery of the quantum mechanics is suﬁiCIently
familiar to you that we can speak about things In a physwal way Without taking the time
to write down all the mathematical details. In case what we are saying here is not clear
to you, we have put some of the missmg details in a note at the end of the section. l7~12 calculate them. We’ll have to get them from experiment. But with just these
two amplitudes we can ﬁnd out all we want to know about the angular distribution
of the disintegration. We only have to be careful always to deﬁne completely the
states we are talking about. We want to know the probability that the proton will go off at the angle 0
with respect to the zaxis (into a small solid angle A9) as drawn in Fig. 17—6.
Let’s put a new zaxis in this direction and call it the z’axis. We know how to
analyze what happens along this axis. With respect to this new axis, the A° no
longer has its spin “up,” but has a certain amplitude to have its spin “up” and
another amplitude to have its spin “down.” We have already worked these out
in Chapter 6, and again in Chapter 10, Eq. (10.30). The amplitude to be spin
“up” is cos 0/2, and the amplitude to be spin “down” is‘l‘ —sin 9/2. When the
A° spin is “up” along the z’axis it will emit a proton in the +z’direction with the
amplitude a. So the amplitude to ﬁnd an “up”spinning proton coming out along
the z’direction is 9
acos— 2 (17.33) Similarly, the amplitude to ﬁnd a “down”spinning proton coming along the posi
tive z’axis is —bsin2 2 (17.34) The two processes that these amplitudes refer to are shown in Fig. 17—9. 2 91 l 1 I; 1’ 1  p
(a) i I ,t.‘ l / sl"! 1)
l 9 / I
\’ vp
A+ _' 2’4" A+ —>
/ I I g? i // : b” V” 1 5/4  1 I l

I /

 Amplitude 0 cos 6/2 Fig. 17—9. Two possible decay states for the A°. Let’s now ask the following easy question. If the A° has spin up along the
zaxis, what is the probability that the decay proton will go off at the angle 0?
The two spin states (“up” or “down” along 2') are distinguishable even though
we are not going to look at them. Soto get the probability we square the amplitudes
and add. The probability f (0) of ﬁnding a proton in a small solid angle A9 at 9 is m) = [alzcosz g + "2125.11123 2 (17.35) Remembering that sin2 6/2 = §(l — cos 0) and that cos2 0/2 = 45(1 + cos 0),
we can write f(9) as f(0) = + cos a. (17.36) 1' We have chosen to let 2’ be in the xzplane and use the matrix elements for Ry(0).
You would get the same answer for any other choice. 17—13 Amplitude b cos 9/2 The angular distribution has the form f(0) = 6(1 + 0: cos 0). (17.37) The probability has one part that is independent of 0 and one part that varies
linearly with cos 0. From measuring the angular distribution we can get a and 3,
and therefore, IaI and IbI. Now there are many other questions we can answer. Are we interested only
in protons with spin “up” along the old zaxis? Each of the terms in (17—33) and
(17—34) will give an amplitude to ﬁnd a proton with spin “up” and with spin
“down” with respect to the z’axis (+z’ and ——z’). Spin “up” with respect to the
old axis I +2) can be expressed in terms of the base states I +z’) and I —z’).
We can then combine the two amplitudes (17.33) and (17.34) with the proper
coefﬁcients (cos 0/2 and —sin 9/2) to get the total amplitude 29 .20
(acos §+bs1n 5) Its square is the probability that the proton comes out at the angle 6 with its spin
the same as the A0 (“up” along the zaxis). If parity were conserved, we could say one more thing. The disintegration
of Fig. 17—8 is just the reﬂection—in say, the y zplane of the dismtegration of
Fig. 17—7.1' If parity were conserved. b would have to be equal to a or to —a.
Then the coefﬁcient a of (17.37) would be zero, and the disintegration would be
equally likely to occur in all directions. The experimental results show, however, that there is an asymmetry in the
disintegration. The measured angular distribution does go as cos 6 as we predict—
and not as cos2 0 or any other power. In fact, since the angular distribution has
this form, we can deduce from these measurements that the spin of the A0 is 1/2.
Also, we see that parity is not conserved. In fact, the coefﬁcient a is found experi
mentally to be —0.62 =t= 0.05, so b is about twice as large as a. The lack of sym
metry under a reﬂection is quite clear. You see how much we can get from the conservation of angular momentum.
We will give some more examples in the next chapter. Parenthetical note. By the amplitude a in this section we mean the amplitude that the
state I proton going +z, spin +2) is generated in an inﬁnitesrmal time dt from the state
I A, spin +2), or, in other words, that (proton going +2, spin +z I H I A, spin +z) = zha, (17.38) where H is the Hamiltonian of the world—or, at least, of whatever is responsrble for the
Adecay. The conservation of angular momentum means that the Hamiltonian must
have the property that (proton going +z, spin —z I H I A, spin +2) = 0. (17.39)
By the amplitude b we mean that (proton going +2, spin —zI H I A, spin —z) = ihb. (17.40)
Conservation of angular momentum implies that (proton going +2, spin +zI H I A, spin —z) = 0. (17.41) If the amplitudes written in (17.33) and (17.34) are not clear, we can express them
more mathematically as follows. By (17.33) we intend the amplitude that the A With
spm along +z will disintegrate into a proton moving along the +z’direction with its
spin also in the +z’direction, namely the amplitude (proton going +z’, spin +z’ I H I A, spin +2). (17.42)
By the general theorems of quantum mechanics, this amplitude can be written as Z (proton going +z’, spin +z’ I H l A, z‘)(A, iIA, spin +2), (17.43) I Remembering that the spin is an axial vector and ﬂips over in the reflection. 17—14 where the sum is to be taken over the base states  A, i) of the Aparticle at rest. Since the
Aparticle is spin one—half, there are two such base states which can be in any reference
base we wish. If we use for base states spin “up” and spin “down” with respect to z'
(+z’, —z’), the amplitude of (17.43) is equal to the sum (proton going +z’, spin +z’ ] H! A, +z’)(A, +z’ I A, +2)
+(proton going +z’, spin +z’ I HIA, —z’)(A, ——z’ 1 A, +z). (17.44) The ﬁrst factor of the ﬁrst term is a, and the ﬁrst factor of the second term is zero—from
the deﬁnition of (17.38), and from (17.41), which in turn follows from angular momentum
conservation. The remaining factor (A, +z’ IA, +2) of the ﬁrst term is just the amplitude
that a spin onehalf particle which has spin “up” along one axis will also have spin “up”
along an axis tilted at the angle 6, which is cos 6/2ﬁsee Table 62. So (17.44) is Just
a cos 0/2, as we wrote in (17.33). The amplitude of (17.34) follows from the same kind
of arguments for a spin “down” Aparticle. 17—6 Summary of the rotation matrices We would like now to bring together in one place the various things we have
learned about the rotations for particles of spin onehalf and spin one—so they will
be convenient for future reference. On the next page you Will ﬁnd tables of the two
rotation matrices Rz(¢) and Ry(6) for spin onehalf particles, for spinone particles,
and for photons (spinone particles with zero rest mass). For each spin we will
give the terms of the matrix < j  R l i) for rotations about the zaxis or the yaxis.
They are, of course, exactly equivalent to the amplitudes like (+T  0 S) we have
used in earlier chapters. We mean by R,(¢) that the state is projected into a new
coordinate system which is rotated through the angle 4) about the zaXIS—using
always the righthand rule to deﬁne the positive sense of the rotation. By Ry(9)
we mean that the reference axes are rotated by the angle 0 about the y—axis. Know
ing these two rotations, you can, of course, work out any arbitrary rotation. As
usual, we write the matrix elements so that the state on the left is a base state of
the new (rotated) frame and the state on the right is a base state of the old (un
rotated) frame. You can interpret the entries in the tables in many ways. For
instance, the entry e_”” 2 in Table 17—1 means that the matrix element (— R —) =
e‘w’z. It also means that R} —) = 8—1‘1’”  —), or that (— [R = (—  e—“l’”.
It’s all the same thing. 17—15 Table 17—1 Rotation matrices for spin onehalf Two states: I +), “up” along the zaxis, m =
I —), “down” along the zaxis, m 1—) cos 0/2
—sin 0/2 sin 0/2
cos 0/2 Table 172 Rotation matrices for spin one
Three states: I +), W = +1 I0),m=0
l—>:m 1 +1/2
—1/2 £0 + cos a) — isin0 x/E
i0 — cos 0) 5(1 — cos a) + $sin0 §(l + cos 0) Table 17—3
Photons Two states: IR) = ~13 (lx) + ilY», m — \/— IL)=\/— 17—16 +1 (RHC polarized) ‘15 (l x) — fly», m = —1 (LHC polarized) ...
View Full
Document
 Spring '09
 dfdedf

Click to edit the document details