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Unformatted text preview: The Dependence of Amplitudes on Time 7—1 Atoms at rest; stationary states We want now to talk a little bit about the behavior of probability amplitudes
in time. We say a “little bit,” because the actual behavior in time necessarily
involves the behavior in space as well. Thus, we get immediately into the most
complicated possible situation if we are to do it correctly and in detail. We are
always in the difﬁculty that we can either treat something in a logically rigorous
but quite abstract way, or we can do something which is not at all rigorous but
which gives us some idea of a real situationipostponing until later a more careful
treatment. With regard to energy dependence, we are going to take the second
course. We will make a number of statements. We will not try to be rigorous—but
will just be telling you things that have been found out, to give you some feeling
for the behavior of amplitudes as a function of time. As we go along, the precision
of the description will increase, so don’t get nervous that we seem to be picking
things out of the air. It is, of course, all out of the air—the air of experiment and
of the imagination of people. But it would take us too long to go over the historical
development, so we have to plunge in somewhere. We could plunge into the ab
stract and deduce everything—which you would not understand—or we could
go through a large number of experiments to justify each statement. We choose
to do something in between. An electron alone in empty space can, under certain circumstances, have a
certain deﬁnite energy. For example, if it is standing still (so it has no translational
motion, no momentum, or kinetic energy), it has its rest energy. A more compli
cated object like an atom can also have a deﬁnite energy when standing still, but
it could also be internally excited to another energy level. (We will describe later
the machinery of this.) We can often think of an atom in an excited state as having
a deﬁnite energy, but this is really only approximately true. An atom doesn’t
stay excited forever because it manages to discharge its energy by its interaction
with the electromagnetic ﬁeld. So there is some amplitude that a new state is
generated—with the atom in a lower state, and the electromagnetic ﬁeld in a higher
state, of excitation. The total energy of the system is the same before and after,
but the energy of the atom is reduced. So it is not precise to say an excited atom
has a deﬁnite energy; but it will often be convenient and not too wrong to say that
it does. [Incidentally, why does it go one way instead of the other way? Why does an
atom radiate light? The answer has to do with entropy. When the energy is in the
electromagnetic ﬁeld, there are so many different ways it can be—so many diﬁerent
places where it can wander—that if we look for the equilibrium condition, we
ﬁnd that in the most probable situation the ﬁeld is excited with a photon, and the
atom is deexcited. It takes a very long time for the photon to come back and ﬁnd
that it can knock the atom back up again. It’s quite analogous to the classical
problem: Why does an accelerating charge radiate? It isn’t that it “wants” to lose
energy, because, in fact, when it radiates, the energy of the world is the same as it
was before. Radiation or absorption goes in the direction of increasing entropy.] Nuclei can also exist in different energy levels, and in an approximation which
disregards the electromagnetic effects, we can say that a nucleus in an excited state
stays there. Although we know that it doesn’t stay there forever, it is often useful
to start out with an approximation which is somewhat idealized and easier to
think about. Also it is often a legitimate approximation under certain circum
stances. (When we ﬁrst introduced the classical laws of a falling body, we did not
include friction, but there is almost never a case in which there isn’t some friction.) 7—1 7—1 Atoms at rest; stationary states
7—2 Uniform motion 7—3 Potential energy; energy
conservation 74 Forces; the classical limit 7—5 The “precession” of a spin
onehalf particle Review: Chapter 17, Vol. I, SpaceTime
Chapter 48, Vol. 1, Beats Then there are the subnuclear “strange particles,” which have various masses.
But the heavier ones disintegrate into other light particles, so again it is not correct
to say that they have a precisely deﬁnite energy. That would be true only if they
lasted forever. So when we make the approximation that they have a deﬁnite
energy, we are forgetting the fact that they must blow up. For the moment, then,
we will intentionally forget about such processes and learn later how to take them
into account. Suppose we have an atom—~or an electron, or any particle—which at rest
would have a deﬁnite energy E 0. By the energy E 0 we mean the mass of the whole
thing times c2. This mass includes any internal energy; so an excited atom has a
mass which is diﬁerent from the mass of the same atom in the ground state. (The
ground state means the state of lowest energy.) We will call E 0 the “energy at rest.” For an atom at rest, the quantum mechanical amplitude to ﬁnd an atom at a
place is the same everywhere; it does not depend on position. This means, of course,
that the probability of ﬁnding the atom anywhere is the same. But it means even
more. The probability could be independent of position, and still the phase of the
amplitude could vary from point to point. But for a particle at rest, the complete
amplitude is identical everywhere. It does, however, depend on the time. For a
particle in a state of deﬁnite energy E o, the amplitude to ﬁnd the particle at (x, y, 2)
at the time t is ae—1(E0/ﬁ)t , (7.1) where a is some constant. The amplitude to be at any point in space is the same
for all points, but depends on time according to (7.1). We shall simply assume
this rule to be true. Of course, we could also write (7.1) as tie—1"”, (7.2)
with hw = E0 = Mcz, where M is the rest mass of the atomic state, or particle. There are three different
ways of specifying the energy: by the frequency of an amplitude, by the energy in
the classical sense, or by the inertia. They are all equivalent; they are just different
ways of saying the same thing. You may be thinking that it is strange to think of a “particle” which has
equal amplitudes to be found throughout all space. After all, we usually imagine
a “particle” as a small object located “somewhere.” But don’t forget the uncer
tainty principle. If a particle has a deﬁnite energy, it has also a deﬁnite momentum.
If the uncertainty in momentum is zero, the uncertainty relation, Ap Ax = it,
tells us that the uncertainty in the position must be inﬁnite, and that is just what
we are saying when we say that there is the same amplitude to ﬁnd the particle
at all points in space. If the internal parts of an atom are in a different state with a different total
energy, then the variation of the amplitude with time is diﬁerent. If you don’t
know in which state it is, there will be a certain amplitude to be in one state and a
certain amplitude to be in another—and each of these amplitudes will have a dif
ferent frequency. There will be an interference between these diﬁerent components
—like a beatnote—which can show up as a varying probability. Something will
be “going on” inside of the atomeven though it is “at rest” in the sense that its
center of mass is not drifting. However, if the atom has one deﬁnite energy, the
amplitude is given by (7.1), and the absolute square of this amplitude does not
depend on time. You see, then, that if a thing has a deﬁnite energy and if you ask
any probability question about it, the answer is independent of time. Although
the amplitudes vary with time, if the energy is deﬁnite they vary as an imaginary
exponential, and the absolute value doesn’t change. That’s why we often say that an atom in a deﬁnite energy level is in a stationary
state. If you make any measurements of the things inside, you’ll ﬁnd that nothing
(in probability) will change in time. In order to have the probabilities change in 7—2 time, we have to have the interference of two amplitudes at two different frequencies,
and that means that we cannot know what the energy is. The object will have one
amplitude to be in a state of one energy and another amplitude to be in a state of
another energy. That’s the quantum mechanical description of something when
its behavior depends on time. If we have a “condition” which is a mixture of two different states with differ
ent energies, then the amplitude for each of the two states varies with time according
to Eq. (7.2), for instance, as —i(E1/ﬁ)t e—i(E2/if)t e and (7.3) And if we have some combination of the two, we will have an interference. But
notice that if we added a constant to both energies, it wouldn’t make any difference.
If somebody else were to use a different scale of energy in which all the energies
were increased (or decreased) by a constant amount—say, by the amount A—then
the amplitudes in the two states would, from his point of view, be Ai(E1+A)I/ﬁ e—i(E2+A)t/ﬁ e and (74) All of his amplitudes would be multiplied by the same factor e"“’“””, and all
linear combinations, or interferences, would have the same factor. When we take
the absolute squares to ﬁnd the probabilities, all the answers would be the same.
The choice of an origin for our energy scale makes no difference; we can measure
energy from any zero we want. For relativistic purposes it is nice to measure the
energy so that the rest mass is included, but for many purposes that aren’t rela
tivistic it is often nice to subtract some standard amount from all energies that
appear. For instance, in the case of an atom, it is usually convenient to subtract
the energy Mscz, where M, is the mass of all the separate pieces—the nucleus and
the electrons—which is, of course, different from the mass of the atom. For other
problems it may be useful to subtract from all energies the amount Macz, where
M9 is the mass of the whole atom in the ground state; then the energy that appears
is just the excitation energy of the atom. So, sometimes we may shift our zero of
energy by some very large constant, but it doesn’t make any difference, provided
we shift all the energies in a particular calculation by the same constant. So much
for a particle standing still. 72 Uniform motion If we suppose that the relativity theory is right, a particle at rest in one inertial
system can be in uniform motion in another inertial system. In the rest frame of
the particle, the probability amplitude is the same for all x, y, and 2 but varies with
t. The magnitude of the amplitude is the same for all 1, but the phase depends on t.
We can get a kind of a picture of the behavior of the amplitude if we plot lines of
equal phase~say, lines of zero phase—as a function of x and I. For a particle at
rest, these equalphase lines are parallel to the xaxis and are equally spaced in
the t—coordinate, as shown by the dashed lines in Fig. 7—1. In a different frame—x’, y’, z’, t’—that is moving with respect to the particle
in, say, the xdirection, the x’ and t’ coordinates of any particular point in space
are related to ac and t by the Lorentz transformation. This transformation can be
represented graphically by drawing x’ and t’ axes, as is done in Fig. 741. (See
Chapter 17, Vol. I, Fig. 17—2.) You can see that in the x’t’ system, points of equal
phaseT have a different spacing along the t’axis, so the frequency of the time
variation is different. Also there is a variation of the phase with x’, so the prob
ability amplitude must be a function of x’. 1‘ We are assuming that the phase should have the same value at corresponding points
in the two systems. This is a subtle point, however, since the phase of a quantum me
chanical amplitude is, to a large extent, arbitrary. A complete justiﬁcation of this assump
tion requires a more detailed discussion involving interferences of two or more amplitudes. 7—3 Fig. 7].
of the amplitude of a particle at rest in
the xf systems. Relativistic transformation Under a Lorentz transformation for the velocity 1.», say along the negative
xdirection, the time t is related to the time I’ by t = t’ — x’v/c2
V1 —— v2/c2 ’ so our amplitude now varies as e—u/mb'ot _ e—(i/VLMEOt’/\/1—1)2/‘oil—Eo‘uac'tc2 ‘/l—712/(22) In the prime system it varies in space as well as in time. If we write the amplitude as —(i’ﬁ)(E' t’—p'r’)
e ‘ P , we see that E,’, = EU/x/l —— 112/02 is the energy computed classically for a
particle of rest energy E0 travelling at the velocity Z), and p’ = E,§,v/c2 is the
corresponding particle momentum. You know that x, = (t, x, y, z) and 1),, = (E, pr, p1), p,) are fourvectors, and
that pﬂxﬂ = Et — p  x is a scalar invariant. In the rest frame of the particle,
pm, is just El; so if we transform to another frame, E1 will be replaced by Elt! __ pl . x1.
Thus, the probability amplitude of a particle which has the momentum p will be proportional to
e~(i/h)(Ept—px) , (7.5) where E, is the energy of the particle whose momentum is p, that is, E12 2 \/(pc)2 + E5, (7.6)
where E 0 is, as before, the rest energy. For nonrelativistic problems, we can write
E, = M,c'~’ + W,,, (7.7) where W1, is the energy over and above the rest energy M,c2 of the parts of the
atom. In general, W1, would include both the kinetic energy of the atom as well
as its binding or excitation energy, which we can call the “internal” energy. We would write
2 p»;
2M Wp = Wint —l_ and the amplitudes would be [WW—W). (7.9) Because we will generally be doing nonrelativistic calculations, we will use this
form for the probability amplitudes. Note that our relativistic transformation has given us the variation of the
amplitude of an atom which moves in space without any additional assumptions.
The wave number of the space variations is, from (7.9), _ 12.
k _ h, (7.10) so the wavelength is =2I=£ A k p (7.11) This is the same wavelength we have used before for particles with the momentum
p. This formula was ﬁrst arrived at by de Broglie in just this way. For a moving
particle, the frequency of the amplitude variations is still given by hw = W... (7.12)
74 The absolute square of (7.9) is just 1, so for a particle in motion with a
deﬁnite energy, the probability of ﬁnding it is the same everywhere and does not
change with time. (It is important to notice that the amplitude is a complex wave.
If we used a real sine wave, the square would vary from point to point, which
would not be right.) We know, of course, that there are situations in which particles move from
place to place so that the probability depends on position and changes with time.
How do we describe such situations? We can do that by considering amplitudes
which are a superposition of two or more amplitudes for states of deﬁnite energy.
We have already discussed this situation in Chapter 48 of Vol. I——even for prob
ability amplitudes! We found that the sum of two amplitudes with different wave
numbers k (that is, momenta) and frequencies w (that is, energies) gives inter
ference humps, or beats, so that the square of the amplitude varies with space
and time. We also found that these beats move with the socalled “group velocity”
given by where Ak and Aw are the differences between the wave numbers and frequencies
for the two waves. For more complicated waves—made up of the sum of many
amplitudes all near the same frequency—the group velocity is dw
v” = a}. (7.13)
Taking w = Ep/h and k = p/f’i, we see that
dB
1),, = (7.14)
Using Eq. (7.6), we have
dEp _ 2 i.
71; — c Ep (7.15)
But Ep : Mc2, so
dEp P
Up— A—l (7.16) which is just the classical velocity of the particle. Alternatively, if we use the non
relativistic expressions, we have W
w=7p and k=%,
and
dw_dW_ d p2 _ p
at — :1;— a; (in) — 17! (7'17) which is again the classical velocity. Our result, then, is that if we have several amplitudes for pure energy states
of nearly the same energy, their interference gives “lumps” in the probability that
move through space with a velocity equal to the velocity of a classical particle
of that energy. We should remark, however, that when we say we can add two
amplitudes of ditTerent wave number together to get a beatnote that will corre
spond to a moving particle, we have introduced something new—something that
we cannot deduce from the theory of relativity. We said what the amplitude did
for a particle standing still and then deduced what it would do if the particle were
moving. But we cannot deduce from these arguments what would happen when
there are two waves moving with different speeds. If we stop one, we cannot stop
the other. So we have added tacitly the extra hypothesis that not only is (7.9) a
possible solution, but that there can also be solutions with all kinds of p’s for the
same system, and that the different terms will interfere. 7—5 ; +
Fig. 7—2. A particle of mass M and momentum p in a region of constant
potential. Re (Am p) >
DlST a x. e were [FOR ¢2‘ 4") Fig. 7—3. The amplitude for a par
ticle in transit from one potential to
another. 7—3 Potential energy; energy conservation Now we would like to discuss what happens when the energy of a particle
can change. We begin by thinking of a particle which moves in a force ﬁeld de
scribed by a potential. We discuss ﬁrst the efl‘ect of a constant potential. Suppose
that we have a large metal can which we have raised to some electrostatic potential
(1). as in Fig. 7—2. If there are charged objects inside the can, their potential energy
will be qu, which we will call V, and will be absolutely independent of position.
Then there can be no change in the physics inside, because the constant potential
doesn’t make any difference so far as anything going on inside the can is concerned.
Now there is no way we can deduce/what the answer should be, so we must make
a guess. The guess which works is more or less what you might expect: For the
energy, we must use the sum of the potential energy V and the energy Ep—which
is itself the sum of the internal and kinetic energies. The amplitude is proportional to
e»ri/ﬁ>t<Ep+V) t—Pxr (7.18) The general principle is that the coefﬁcient of t, which we may call co, is always given by the total energy of the system: internal (or “mass”) energy, plus kinetic
energy, plus potential energy: hw=Ep+ V. Or, for nonrelativistic situations, (7.19) 0 m = Wint + + V. (7.20) Now what about physical phenomena inside the box? If there are several diﬁercnt energy states, what will we get? The amplitude for each state has the
same additional factor
ear/mm over what it would have with V = 0. That is just like a change in the zero of our
energy scale. It produces an equal phase change in all amplitudes, but as we have
seen before, this doesn’t change any of the probabilities. All the physical phenomena
are the same. (We have assumed that we are talking about different states of the
same charged object, so that qd; is the same for all. If an object could change its
charge in going from one state to another, we would have quite another result,
but conservation of charge prevents this.) So far, our assumption agrees with what we would expect for a change of
energy reference level. But if it is really right, it should hold for a potential energy
that is not just a constant. In general, V could vary in any arbitrary way with
both time and space, and the complete result for the amplitude must be given in
terms of a differential equation. We don’t want to get concerned with the general
case right now, but only want to get some idea about how some things happen,
so we will think only of a potential that is constant in time and varies very slowly
in space. Then we can make a comparison between the classical and quantum ideas. Suppose we think of the situation in Fig. 7—3, which has two boxes held at
the constant potentials ¢1 and ¢2 and a region in between where we will assume
that the potential varies smoothly from one to the other. We imagine that some
particle has an amplitude to be found in any one of the regions. We also assume
that the momentum is large enough so that in any small region in which there are
many wavelengths, the potential is nearly constant. We Would then think that in
any part of the space the amplitude ought to look like (7.18) with the appropriate
V for that part of the space. Let’s think of a special case in which ¢1 = 0, so that the potential energy
there is zero, but in which q¢2 is negative, so that classically the particle would
have more energy in the second box. Classically, it would be going faster in the
second box—it would have more energy and, therefore, more momentum. Let’s/
see how that might come out of quantum mechanics. 7—6 With our assumption, the amplitude in the ﬁrst box would be proportional to e—(i/hmWm+pt/2M+V1>t—m~1 , (7.21) and the amplitude in the second box would be proportional to
e—(i/ﬁ)l(Wint+P§/2M+V2)1—P2"] I (722) (Let’s say that the internal energy is not being changed, but remains the same in
both regions.) The question is: How do these two amplitudes match together
through the region between the boxes? We are going to suppose that the potentials are all constant in time~so that
nothing in the conditions varies. We will then suppose that the variations of the
amplitude (that is, its phase) have the same frequency everywhere—because, so
to speak, there is nothing in the “medium” that depends on time. If nothing in
the space is changing, we can consider that the wave in one region “generates”
subsidiary waves all over space which will all oscillate at the same frequency——
just as light waves going through materials at rest do not change their frequency.
If the frequencies in (7.21) and (7.22) are the same, we must have that p2 p2
Wine + 2—1;! + V1 = Wint + 7A2? + V2 (723) Both sides are just the classical total energies, so Eq. (7.23) is a statement of the
conservation of energy. In other words, the classical statement of the conservation
of energy is equivalent to the quantum mechanical statement that the frequencies
for a particle are everywhere the same if the conditions are not changing with time.
It all ﬁts with the idea that ho.) = E. In the special example that V1 = 0 and V2 is negative, Eq. (7.23) gives that
p2 is greater than 121, so the wavelength of the waves is shorter in region 2. The
surfaces of equal phase are shown by the dashed lines in Fig. 7—3. We have also
drawn a graph of the real part of the amplitude, which shows again how the
wavelength decreases in going from region 1 to region 2. The group velocity of
the waves, which is p/M, also increases in the way one would expect from the
classical energy conservation, since it is just the same as Eq. (7.23). There is an interesting special case where V2 gets so large that V2 — V1 is
greater than pf/ZM. Then 17%, which is given by 2 pa = 2M[2% — V2 + V1]: (7.24) is negative. That means that p2 is an imaginary number, say, ip’. Classically, we
would say that the particle never gets into region Ziit doesn‘t have enough energy
to climb the potential hill. Quantum mechanically, however, the amplitude is still
given by Eq. (7.22); its space variation still goes as e(i/ﬂ)P2'”. But if p2 is imaginary, the space dependence becomes a real exponential. Say that
the particle was initially going in the +xdirection; then the amplitude would
vary as eW’”. (7.25) The amplitude decreases rapidly with increasing x. Imagine that the two regions at diﬁerent potentials were very close together,
so that the potential energy changed suddenly from V1 to V2, as shown in Fig.
7—4(a). If we plot the real part of the probability amplitude, we get the dependence
shown in part (b) of the ﬁgure. The wave in the ﬁrst region corresponds to a
particle trying to get into the second region, but the amplitude there falls off 7—7 (ol//<I/// \\\>\ \ (pi/2m<0 / (pa/2m>0l ' I
' l
eta %se Fig. 7—4. a strongly repulsive potential. l (bl rRe(Amp) Fig. 7—6. (a) The potential function
for an aparticle in a uranium nucleus.
(b) The qualitative form of the probability
amplitude. The amplitude for a particle approaching Re(Amp.) Fig. 775.
a potential barrier. The penetration of the amplitude through rapidly. There is some chance that it will be observed in the second region—where
it could never get classically——but the amplitude is very small except right near
the boundary. The situation is very much like what we found for the total internal
reﬂection of light. The light doesn’t normally get out, but we can observe it if we
put something within a wavelength or two of the surface. You will remember that if we put a second surface close to the boundary where
light was totally reﬂected, we could get some light transmitted into the second piece
of material. The corresponding thing happens to particles in quantum mechanics.
If there is a narrow region with a potential V, so great that the classical kinetic
energy would be negative, the particle would classically never get past. But quan
tum mechanically, the exponentially decaying amplitude can reach across the
region and give a small probability that the particle will be found on the other side
where the kinetic energy is again positive. The situation is illustrated in Fig. 7—5.
This effect is called the quantum mechanical “penetration of a barrier.” The barrier penetration by a quantum mechanical amplitude gives the ex
planation—or description—of the aparticle decay of a uranium nucleus. The
potential energy of an aparticle, as a function of the distance from the center, is
shown in Fig. 7—6(a). If one tried to shoot an aparticle with the energy E into
the nucleus, it would feel an electrostatic repulsion from the nuclear charge 2 and
would, classically, get no closer than the distance rl where its total energy is equal
to the potential energy V. Closer in, however, the potential energy is much lower
because of the strong attraction of the shortrange nuclear forces. How is it then
that in radioactive decay we ﬁnd aparticles which started out inside the nucleus
coming out with the energy E? Because they start out with the energy E inside
the nucleus and “leak” through the potential barrier. The probability amplitude
is roughly as sketched in part (b) of Fig. 7—6, although actually the exponential
decay is much larger than shown. It is, in fact, quite remarkable that the mean
life of an aparticle in the uranium nucleus is as long as 4% billion years, when the
natural oscillations inside the nucleus are so extremely rapid—about 1022 per see!
How can one get a number like 109 years from 10‘22 sec? The answer is that the
exponential gives the tremendously small factor of about e‘45—which gives the
very small. though deﬁnite, probability of leakage. Once the aparticle is in the
nucleus. there is almost no amplitude at all for ﬁnding it outside; however, if you
take many nuclei and wait long enough, you may be lucky and ﬁnd one that has
come out. 7—8 Fig. 78.
with a transverse potential gradient. Fig. 7—7. The deﬂection of a particle by a
transverse potential gradient. 7—4 Forces; the classical limit Suppose that we have a particle moving along and passing through a region
where there is a potential that varies at right angles to the motion. Classically, we
would describe the situation as sketched in Fig. 7—7. If the particle is moving
along the x—direction and enters a region where there is a potential that varies
with y, the particle will get a transverse acceleration from the force F = —0V/6y.
If the force is present only in a limited region of width w, the force will act only for
the time w/v. The particle will be given the transverse momentum w
Pu=F'v" The angle of deﬂection 60 is then Fw ,
pv 59=£i1~ P wherep is the initial momentum. Using —6V/6y for F, we get (7.26) It is now up to us to see if our idea that the waves go as (7.20) will explain
the same result. We look at the same thing quantum mechanically, assuming that
everything is on a very large scale compared with a wavelength of our probability
amplitudes. In any small region we can say that the amplitude varies as —<i/m1<W+p2/2M+V)c—wi (7.27) e Can we see that this will also give rise to a deﬂection of the particle when V has
a transverse gradient? We have sketched in Fig. 7—8 what the waves of prob
ability amplitude will look like. We have drawn a set of “wave nodes” which you
can think of as surfaces where the phase of the amplitude is zero. in every small
region, the wavelength—the distance between successive nodes—is where p is related to V through 2
W + 2%! + V = const. (7.28)
in the region where V is larger, p is smaller, and the wavelength is longer. SO the
angle of the wave nodes gets changed as shown in the ﬁgure. To ﬁnd the change in angle of the wave nodes we notice that for the two
paths (1 and b in Fig. 7—8 there is a difference of potential AV 2 (6V/6y)D, so
there is a difference Ap in the momentum along the two tracks which can be 7—9 The probability amplitude in a region obtained from (7.28): P2 *1 __
A(2 >—MAp— AV. (7.29) The wave number p/h is, therefore, different along the two paths, which means
that the phase is advancing at a dilferent rate. The difference in the rate of increase
of phase is Ak : Ap/h, so the accumulated phase difference in the total distance
w is _ AP. M w : — — AV w. (7.30) A(phase) = Ak w h ph This is the amount by which the phase on path I) is “ahead” of the phase on path
a as the wave leaves the strip. But outside the strip, a phase advance of this amount
corresponds to the wave node being ahead by the amount )\ h
Ax — 2—7; A(phase) — ‘3 A(phase)
or
M
Ax = — F AV w. (7.31) Referring to Fig. 7—8, we see that the new wavefronts will be at the angle 66
given by Ax = D 66; (7.32)
so we have
M ,
D60: ~ﬁAVw, (7.53) This is identical to Eq. (7.26) if we replace p/m by v and AV/D by aV/By. The result we have just got is correct only if the potential variations are slow
and smoothiin what we call the classical limit. We have shown that under these
conditions we will get the same particle motions we get from F = ma, provided
we assume that a potential contributes a phase to the probability amplitude equal
to Vt/fi. In the classical limit, the quantum mechanics will agree with Newtonian
mechanics. 7—5 The “precession” of a spin onehalf particle Notice that we have not assumed anything special about the potential energy——
it is just that energy whose derivative gives a force. For instance, in the Stern
Gerlach experiment we had the energy U = ~pt  B, which gives a force if B has a
spatial variation. If we wanted to give a quantum mechanical description, we
would have said that the particles in one beam had an energy that varied one way
and that those in the other beam had an opposite energy variation. (We could
put the magnetic energy U into the potential energy V or into the “internal”
energy W; it doesn’t matter.) Because of the energy variation, the waves are
refracted, and the beams are bent up or down. (We see now that quantum me
chanics would give us the same bending as we would compute from the classical
mechanics.) From the dependence of the amplitude on potential energy we would also
expect that if a particle sits in a uniform magnetic ﬁeld along the zdirection, its
probability amplitude must be changing with time according to e—(i/ﬁX—ﬂzB) t. (We can consider that this is, in eﬁect, a deﬁnition of M.) In other words, if we
place a particle in a uniform ﬁeld B for a time T, its probability amplitude will be
multiplied by e—KﬂﬁK—szﬁ 7—10 over what it would be in no ﬁeld. Since for a spin onehalf particle, M can be
either plus or minus some number, say ,u, the two possible states in a uniform
ﬁeld would have their phases changing at the same rate but in opposite direc
tions. The two amplitudes get multiplied by d:(1'/ﬂ)uB‘r e (7.34) This result has some interesting consequences. Suppose we have a spin one
half particle in some state that is not purely spin up or spin down. We can describe
its condition in terms of the amplitudes to be in the pure up and pure down states.
But in a magnetic ﬁeld, these two states will have phases changing at a different
rate. So if we ask some question about the amplitudes, the answer will depend
on how long it has been in the ﬁeld. As an example, we consider the disintegration of the muon in a magnetic
ﬁeld. When muons are produced as disintegration products of 7rmesons, they are
polarized (in other words, they have a preferred spin direction). The muons, in
turn, disintegrate—in about 2.2 microseconds on the average—emitting an electron
and two neutrinos: ,u,—)c+VIl7. In this disintegration it turns out that (for at least the highest energies) the electrons
are emitted preferentially in the direction opposite to the spin direction of the muon. Suppose then that we consider the experimental arrangement shown in Fig.
7—9. If polarized muons enter from the left and are brought to rest in a block of
material at A, they will, a little while later, disintegrate. The electrons emitted
will, in general, go off in all possible directions. Suppose, however, that the muons
all enter the stopping block at A with their spins in the x—direction. Without a
magnetic ﬁeld there would be some angular distribution of decay directions; we
would like to know how this distribution is changed by the presence of the mag
netic ﬁeld. We expect that it may vary in some way with time. We can ﬁnd out
what happens by asking, for any moment, what the amplitude is that the muon
will be found in the (+x) state. We can state the problem in the following way: A muon is known to have
its spin in the +xdirection at t = 0; what is the amplitude that it will be in the
same state at the time 7'? Now we do not have any rule for the behavior of a spin
onehalf particle in a magnetic ﬁeld at right angles to the spin, but we do know what
happens to the spin up and spin down states with respect to the ﬁeld—their ampli
tudes get multiplied by the factor (7.34). Our procedure then is to choose the
representation in which the base states are spin up and spin down with respect
to the zdirection (the ﬁeld direction). Any question can then be expressed with
reference to the amplitudes for these states. Let’s say that Mt) represents the muon state. When it enters the block A, its
state is (1/(0), and we want to know (12(7) at the later time 7. If we represent the two
base states by (+2) and (—2) we know the two amplitudes (+2 I 1(0)) and
(—z  ¢(0))——we know these amplitudes because we know that MO) represents a state with the spin in the (+x) state. From the results of the last chapter, these
amplitudes areT <+Z +X> = C+ = SI and (7.35) <~Z+x)= C_=l— x/i They happen to be equal. Since these amplitudes refer to the condition at t = 0,
let’s call them C+(0) and C_(0). T If you skipped Chapter 6, you can just take (7.35) as an underived rule for now. We will give later (in Chapter 10) a more complete discussion of spin precession, including
a derivation of these amplitudes. 7—11 Fig. 7—9. A
ment. muondecoy experi Fig. 7—10. Time dependence of the
probability that a spin onehalt particle
will be in a (—H state with respect to the o 1r 2.” xaxis. Now we know what happens to these two amplitudes with time. Using
(7.34), we have : C+(O)e—(1T/ﬁ)IABI
and (7.36) C_(t) = C_(0)e+(z‘/1‘L)u1n But if we know C+(t) and C#(t), we have all there is to know about the condition
at t. The only trouble is that what we want to know is the probability that at t
the spin will be in the +xdirecti0n. Our general rules can, however, take care of
this problem. We write that the amplitude to be in the (+x) state at time t. which
we may call A+(z), is A+(l) = <+XIP(1)> = <+x l +Z><+Z I \W» + <+xi —Z><ZH/(t)> 01‘ AM = <+x1+z>C+o> + <+xl —z>c_(z). (7.37) Again using the results of the last chapter—or better the equality (qb I )0 =
(X  ¢)* from Chapter 5§we know that <+xl+z>=vi§ <+xI—z>=\/i§ So we know all the quantities in Eq. (7.37). We get A+(t) = %e(i/ﬁ)ulit + Jzeﬂi/MM‘ or
A+(t) = cos 9—1—3 t.
h A particularly simple result! Notice that the answer agrees with what we expect
for t = 0. We get A+(O) = 1. which is right, because we assumed that the muon
was in the (+x) state at t r 0. The probability P+ that the muon will be found in the (+x) state at t is
(14+)2 or _ 2&5.
P+—COS h The probability oscillates between zero and one. as shown in Fig. 740. Note
that the probability returns to one for uBt/h : 7r (not 27r). Because we have
squared the cosine function, the probability repeats itself with the frequency
2pB/h. PROS, TO HAVE
SPIN IN +1 DIR. Thus, we ﬁnd that the chance of catching the decay electron in the electron
counter of Fig. 7—9 varies periodically with the length of time the muon has been
sitting in the magnetic ﬁeld. The frequency depends on the magnetic moment a.
The magnetic moment of the muon has, in fact, been measured in just this way. We can, of course, use the same method to ansWer any other questions about
the muon decay. For example, how does the Chance of detecting a decay electron 7—12 in the ydirection at 90° to the xdireetion but still at right angles to the ﬁeld depend
on I? If you work it out, the amplitude to be in the (+y) state varies as
0082 {(MBr/h) — 7r/4} , which oscillates with the same period but reaches its max
imum onequarter cycle later, when ,uBt/h = 7r/4. In fact, what is happening is
that as time goes on, the muon goes through a succession of states which correspond
to complete polarization in a direction that is continually rotating about the zaxis.
We can describe this by saying that the spin is precessing at the frequency _2LB h (7.38) “p You can begin to see the form that our quantum mechanical description
will take when we are describing how things behave in time. 7—13 ...
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This note was uploaded on 06/21/2009 for the course SSD 34 taught by Professor Dfdedf during the Spring '09 term at Abraham Baldwin Agricultural College.
 Spring '09
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