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Unformatted text preview: 12 The Hyperfine Splitting in Hydrogen 121 Base states for a system with two spin onehalf particles In this chapter we take up the “hyperﬁne splitting” of hydrogen, because
it is a physically interesting example of what we can already do with quantum
mechanics. It’s an example with more than two states, and it will be illustrative of
the methods of quantum mechanics as applied to slightly more complicated prob
lems. It is enough more complicated that once you see how this one is handled
you can get immediately the generalization to all kinds of problems. As you know, the hydrogen atom consists of an electron sitting in the neigh
borhood of the proton, where it can exist in any one of a number of discrete
energy states in each one of which the pattern of motion of the electron is diﬁerent.
The ﬁrst excited state, for example, lies 3/4 of a Rydberg, or about 10 electron
volts, above the ground state. But even the socalled ground state of hydrogen
is not really a single, deﬁniteenergy state, because of the spins of the electron and
the proton. These spins are responsible for the “hyperﬁne structure” in the energy
levels, which splits all the energy levels into several nearly equal levels. The electron can have its spin either “up” or “down” and, the proton can
also have its spin either “up” or “down.” There are, therefore, four possible spin
states for every dynamical condition of the atom. That is, when people say “the
ground state” of hydrogen, they really mean the “four ground states,” and not
just the very lowest state. The four spin states do not all have exactly the same
energy; there are slight shifts from the energies we would expect with no spins.
The shifts are, however, much, much smaller than the 10 volts or so from the
ground state to the next state above. As a consequence, each dynamical state has
its energy split into a set of very close energy levels—the socalled hyperﬁne splitting. The energy differences among the four spin states is what we want to calculate
in this chapter. The hyperﬁne splitting is due to the interaction of the magnetic
moments of the electron and proton, which gives a slightly different magnetic
energy for each spin state. These energy shifts are only about tenmillionths
of an electron volt—really very small compared with 10 volts! It is because of
this large gap that we can think about the ground state of hydrogen as a “four
state” system, without worrying about the fact that there are really many more
states at higher energies. We are going to limit ourselves here to a study of the
hyperﬁne structure of the ground state of the hydrogen atom. For our purposes we are not interested in any of the details about the positions
of the electron and proton because that has all been worked out by the atom so to
speak—it has worked itself out by getting into the ground state. We need know
only that we have an electron and proton in the neighborhood of each other with
some deﬁnite spatial relationship. In addition, they can have various different
relative orientations of their spins. It is only the effect of the spins that we want to
look into. The ﬁrst question We have to answer is: What are the base states for the system?
Now the question has been put incorrectly. There is no such thing as “the” base
states, because, of course, the set of base states you may choose is not unique.
New sets can always be made out of linear combinations of the 01d. There are
always many choices for the base states, and among them, any choice is equally
legitimate. So the question is not what is the base set, but what could a base set
be? We can choose any one we wish for our own convenience. It is usually best
to start with a base set which is physically the clearest. It may not be the solution 121 12—1 12—2 123
12—4
12—5
12—6 Base states for a system with
two spin onehalf particles The Hamiltonian for the ground
state of hydrogen The energy levels
The Zeeman splitting
The states in a magnetic ﬁeld The projection matrix for spin
one ELECTRON
/ Fig. 12—1. A set of base states for
the ground state of the hydrogen atom. to any problem, or may not have any direct importance, but it will generally
make it easier to understand what is going on.
We choose the following four base states: State I : The electron and proton are both spin “up.”
State 2: The electron is “up” and the proton is “down.”
State 3: The electron is “down” and the proton is “up.”
State 4: The electron and proton are both “down.” We need a handy notation for these four states, so we’ll represent them this way: State 1 z I + +); electron up, proton up. State 2: I + —); electron up, proton down. 12.1
State 3: I — +); electron down, proton up. ( ) State 4: I — —); electron dawn, proton down. You will have to remember that the ﬁrst plus or minus sign refers to the electron
and the second, to the proton. For handy reference, we’ve also summarized the
notation in Fig. 12—1. Sometimes it will also be convenient to call these states
I 1), I 2), I 3), and I 4). You may say, “But the particles interact, and maybe these aren’t the right
base states. It sounds as though you are considering the two particles indepen
dently.” Yes, indeed! The interaction raises the problem: what is the Hamiltonian
for the system, but the interaction is not involved in the question of how to describe
the system. What we choose for the base states has nothing to do with what
happens next. It may be that the atom cannot ever stay in one of these base states,
even if it is started that way. That’s another question. That’s the question:
How do the amplitudes change with time in a particular (ﬁxed) base? In choosing
the base states, we are just choosing the “unit vectors” for our description. While we’re on the subject, let’s look at the general problem of ﬁnding a set
of base states when there is more than one particle. You know the base states for
a single particle. An electron, for example, is completely described in real life—not
in our simpliﬁed cases, but in real life—by giving the amplitudes to be in each of
the following states: I electron “up” with momentum p)
or I electron “down” with momentum p). There are really two inﬁnite sets of states, one state for each value of p. That is to say that an electron state I ll!) is completely described if you know all the ampli
tudes <+ip I ‘1’) and (—sp I 1’), where the + and — represent the components of angular momentum along some
axis—usually the zaxis—and p is the vector momentum. There must, therefore,
be two amplitudes for every possible momentum (a multiinﬁnite set of base
states). That is all there is to describing a single particle. When there is more than one particle, the base states can be written in a
similar way. For instance, if there were an electron and a proton in a more com
plicated situation than we are considering, the base states could be of the following
kind: I an electron with spin “up,” moving 'with momentum p1 and a proton with spin “down,” moving with momentum p2). And so on for other spin combinations. If there are more than two particles—
same idea. So you see that to write down the possible base states is really very easy.
The only problem is, what is the Hamiltonian? For our study of the ground state of hydrogen we don’t need to use the full
sets of base states for the various momenta. We are specifying particular m0 1 2—2 mentum states for the proton and electron when we say “the ground state.” The
details of the conﬁguration—the amplitudes for all the momentum base states—
can be calculated, but that is another problem. Now we are concerned only with
the eﬁects of the spin, so we can take only the four base states of (12.1). Our
next problem is: What is the Hamiltonian for this set of states? 12—2 The Hamiltonian for the ground state of hydrogen We’ll tell you in a moment what it is. But first, we should remind you of one
thing: any state can always be written as a linear combination of the base states.
For any state i ¢> we can write l¢>= l++><++l¢>+l+ ~><+ —¢>+— +><— +l¢>
+ l — —><— — lib) (122) Remember that the complete brackets are just complex numbers, so we can also
write them in the usual fashion as C,, where i = 1, 2, 3, or 4, and write Eq. (12.2) as ll): ++>C1++>C2+—+>C3+l— —>C4 (123) By giving the four amplitudes C, we completely describe the spin state I W If
these four amplitudes change with time, as they will, the rate of change in time is
given by the operator 17. The problem is to ﬁnd the H. There is no general rule for writing down the Hamiltonian of an atomic
system, and ﬁnding the right formula is much more of an art than ﬁnding a set of
base states. We were able to tell you a general rule for writing a set of base states
for any problem of a proton and an electron, but to describe the general Hamilton—
ian of such a combination is too hard at this level. Instead, we will lead you to a
Hamiltonian by some heuristic argument—and you will have to accept it as the
correct one because the results will agree with the test of experimental observation. You will remember that in the last chapter we were able to describe the
Hamiltonian of a single, spin onehalf particle by using the sigma matrices—or the
exactly equivalent sigma operators. The properties of the operators are sum
marized in Table 12—1. These operators—which are just a convenient, shorthand
way of keeping track of the matrix elements of the type (+  az  +)——were
useful for describing the behavior of a single particle of spin onehalf. The question
is: Can we ﬁnd an analogous device to describe a system with two spins? The
answer is yes, very simply, as follows. We invent a thing which we will call “sigma
electron,” which we represent by the vector operator a“, and which has the
x, y, and zcomponents, (7,", 0;, (72. We now make the convention that when one
of these things operates on any one of our four base states of the hydrogen atom,
it acts only on the electron spin, and in exactly the same way as if the electron were
all by itself. Example: What is 0':  — +>? Since 0,, on an electron “down”
is —i times the corresponding state with the electron “up”, 0ﬂ~H=—H+H (When 0; acts on the combined state it ﬂips over the electron, but does nothing to
the proton and multiplies the result by —i.) Operating on the other states, a;
would give m+H=H—H m+~bu—x am—a=—n+a.
Just remember that the operators 0'6 work only on the ﬁrst spin symbol—that is,
on the electron spin. Next we deﬁne the corresponding operator “sigma proton” for the proton
spin. Its three components 05, 0;}, (72 act in the same way as (re, only on the 12—3 Table 12—1 proton spin. For example, if we have a: acting on each of the four base states, we
get—always using Table 12—1— agl++>=l+_>9 As you can see, it’s not very hard. Now in the most general case we could have more complex things. For
instance, we could have products of the two operators like (7317?. When we have
such a product we do ﬁrst what the operator on the right says, and then do what
the other one sayssl' For example, we would have that c§a§l+—)=0§(<7‘Zl+—>)=0§(—l+—))=—0§+—>=> Note that these operators don’t do anything on pure numbers—we have used
this fact when we wrote af,(— 1) = (—1)a§. We say that the operators “commute”
with pure numbers, or that a number “can be moved through” the operator.
You can practice by showing that the product 0202’ gives the following results
for the four states: aiail++>=+l—+>, U§0§+>=l——>,
civil—+>=+I++>,
viaSI——>=—I+—>. If we take all the possible operators, using each kind of operator only once,
there are sixteen possibilities. Yes, sixteen—provided we include also the “unit
operator” T. First, there are the three: a; 0;, «7:. Then the three 02, 6;}, UE—that
makes six. In addition, there are the nine possible products of the form ago;
which makes a total of 15. And there’s the unit operator which just leaves any
state unchanged. Sixteen in all. Now note that for a fourstate system, the Hamiltonian matrix has to be
a fourby—four matrix of coeﬂicients—it will have sixteen entries. It is easily
demonstrated that any fourby—four matrix—and, therefore, the Hamiltonian
matrix in particular—can be written as a linear combination of the sixteen double
spin matrices corresponding to the set of operators we have just made up. There
fore, for the interaction between a proton and an electron that involves only their
spins, we can expect that the Hamiltonian operator can be written as a linear
combination of the same 16 operators. The only question is, how? Well, ﬁrst, we know that the interaction doesn’t depend on our choice of
axes for a coordinate system. If there is no external disturbance—like a magnetic
ﬁeldﬁto determine a unique direction in space, the Hamiltonian can’t depend on
our choice of the direction of the x, y, and zaxes. That means that the
Hamiltonian can’t have a term like of; all by itself. It would be ridiculous, because
then somebody with a ditferent coordinate system would get diﬂ‘erent results. The only possibilities are a term with the unit matrix, say a constant (1 (times
i), and some combination of the sigmas that doesn’t depend on the coordinates—
some “invariant” combination. The only scalar invariant combination of two
vectors is the dot product, which for our a’s is e D a 'a = 020: + 030: + 0201;. (12.4)
This operator is invariant with respect to any rotation of the coordinate system. T For these particular operators, you will notice it turns out that the sequence of the
operators doesn’t matter. 12—4 So the only possibility for a Hamiltonian with the proper symmetry in space is a
constant times the unit matrix plus a constant times this dot product, say, H = E0 + A tre ' up. (12.5) That’s our Hamiltonian. It’s the only thing that it can be, by the symmetry of
space, so long as there is no external ﬁeld. The constant term doesn’t tell us much;
it just depends on the level we choose to measure energies from. We may just
as well take E0 = 0. The second term tells us all we need to know to ﬁnd the
level splitting of the hydrogen. If you want to, you can think of the Hamiltonian in a dilferent way. If there
are two magnets near each other with magnetic moments p,“ and up, the mutual
energy will depend on ya pp—among other things. And, you remember, we
found that the classical thing we call Me appears in quantum mechanics as [1060.
Similarly, what appears classically as up will usually turn out in quantum mechanics
to be ”pap (where up is the magnetic moment of the proton, which is about 1000
times smaller than ye, and has the opposite sign). So Eq. (12.5) says that the
interaction energy is like the interaction between two magnets—only not quite,
because the interaction of the two magnets depends on the radial distance between
them. But Eq. (12.5) could be—and, in fact, is—some kind of an average inter
action. The electron is moving all around inside the atom, and our Hamiltonian
gives only the average interaction energy. All it says is that for a prescribed ar
rangement in space for the electron and proton there is an energy proportional
to the cosine of the angle between the two magnetic moments, speaking classically.
Such a classical qualitative picture may help you to understand where it comes
from, but the important thing is that Eq. (12.5) is the correct quantum mechanical
formula. The order of magnitude of the classical interaction between two magnets
would be the product of the two magnetic moments divided by the cube of the
distance between them. The distance between the electron and the proton in the
hydrogen atom is, speaking roughly, one half an atomic radius, or 0.5 angstrom.
It is, therefore, possible to make a crude estimate that the constant A should be
about equal to the product of the two magnetic moments ,u.e and up divided by
the cube of l / 2 angstrom. Such an estimate gives a number in the right ball park.
It turns out that A can be calculated accurately once you understand the complete
quantum theory of the hydrogen atom—which we so far do not. It has, in fact,
been calculated to an accuracy of about 30 parts in one million. So, unlike the
ﬂipﬂop constant A of the ammonia molecule, which couldn’t be calculated at
all well by a theory, our constant A for the hydrogen can be calculated from a more
detailed theory. But never mind, we will for our present purposes think of the A
as a number which could be determined by experiment, and analyze the physics
of the situation. Taking the Hamiltonian of Eq. (12.5), we can use it with the equation ihC‘, = Z H,,C, (12.6)
i to find out what the spin interactions do to the energy levels. To do that, we need
to work out the sixteen matrix elements H“ = (i l H  j) corresponding to each
pair of the four base states in (12.1). We begin by working out what H  j) is for each of the four base states.
For example, 191+ +> = Aw» £1 + +) = Maia: + Use; + 17:02}  + +). (12.7) Using the method we described a little earlier—it’s easy if you have memorized
Table 12—l—we ﬁnd what each pair of a"s does on I —— +). The answer is ¢§0§l++>=+——),
173051 + +> = — l — —>, (12.8) a2a§++>=+1++>
12—5 Table 12—2 Spin operators for the hydrogen atom
[______________ aiail++>= +I— —>
«2031+ —> = +I— +>
Uiail— +)= ++—)
oiail— —>= ++ +> v30§++>=—I——>
aZaZl+—>=+1—+>
aiaZI—+>=+I+—> 0:03>=—++>
aia2++>=+1++>
0:05+)= —+>
aiaEI—+>=**+)
0202' —)=+) So (12.7) becomes ﬁ++>=A{I——>—l——>+I++>}=AI++>. (129) Since our four base states are all orthogonal, that gives us immediately that <++IH++>=A<++1++>=A,
<+—H++>=A<+I++>=o,
<—+IHI++>=A<—+I++>=0,
<——H++>=A<——l++>=0. (12.10) Remembering that (j l H  i) = (1‘  H I j)*, we can already write down the differ
ential equation for the amplitudes C1: ihC1= H11C1+ H12C2 + H13C3 + H14C4
or
1720'; = AC1. (12.11) That’s all! We get only the one term. Now to get the rest of the Hamiltonian equations we have to crank through
the same procedure for H operating on the other states. First, we will let you
practice by checking out all of the sigma products we have written down in Table
12—2. Then we can use them to get: Hl+ —>=A{2I—+>—I+ —>1,
HI— +>= A{2+ —>—I— +>}, (12.12)
HI——>=AI——>. Then, multiplying each one in turn on the left by all the other state vectors, we
get the following Hamiltonian matrix, H”: J' "l A o o o
H”: 0 ‘A 2A 0 (12.13) 0 2A A o o 0 o A It means, of course, nothing more than that our diﬂerential equations for the four
amplitudes C, are ihCl = AC1, ihc'2 = —AC2 + 2AC3, (12.14)
ihC'3 = 2AC2 — AC3, ma, = AC4. Before solving these equations we can’t resist telling you about a clever
rule due to Dirac—it will make you feel that you are really advanced—although
we don’t need it for our work. We have—from the equations (12.9) and (12.12)—
that ae'a‘pl++)=++>, ae'apI+—)=2—+)"i+">: (1215)
«”«"— +>=2+ —>—I~+>,
a°a‘°l——>=——). 12—6 Look, said Dirac, I can also write the ﬁrst and last equations as ce'ap++)=2l++>—I++>. «°'a”——>=2l——>———): then they are all quite similar. Now I invent a new operator, which I will call
Pm" em, and which I deﬁne to have the following propertieszi‘ Pepin exch I + +) = I + “H,
Pepin cxch I + *> = I — +>,
Pspinexch I — +) = I + —).
P spin excl] I _ —) = I '_ —)' All the operator does is interchange the spin directions of the two particles. Then
I can write the whole set of equations in (12.15) as a simple operator equation: Ge ' 0,1) = 2Papin exch — 1 (12'16) That’s the formula of Dirac. His “spin exchange operator” gives a handy
rule for ﬁguring out 0""  a”. (You see, you can do everything now. The gates
are open.) 12—3 The ene...
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