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Feynman Lectures on Physics Volume 3 Chapter 12

Feynman Lectures on Physics Volume 3 Chapter 12 - 12 The...

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Unformatted text preview: 12 The Hyperfine Splitting in Hydrogen 12-1 Base states for a system with two spin one-half 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 so-called ground state of hydrogen is not really a single, definite-energy 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 so-called 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 ten-millionths 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 12-1 12—1 12—2 12-3 12—4 12—5 12—6 Base states for a system with two spin one-half 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 z-axis—and p is the vector momentum. There must, therefore, be two amplitudes for every possible momentum (a multi-infinite 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- (12-3) 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 1-7. 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 one-half 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 one-half. 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 z-components, (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 four-state system, the Hamiltonian matrix has to be a four-by—four matrix of coeflicients—it will have sixteen entries. It is easily demonstrated that any four-by—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 z-axes. 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 flip-flop 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+">: (12-15) «”-«"|— +>=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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