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Fine and hyperfine structure of optical spectra. Theoretical Introduction




, molecules and ions and, accordingly, spectral lines due to the interaction of the magnetic moment of the nucleus with the magnetic field of electrons . The energy of this interaction depends on the possible mutual orientations of the nuclear spin and electron spins.

Respectively, hyperfine splitting- splitting of energy levels (and spectral lines) into several sublevels caused by such an interaction.

According to classical concepts, an electron revolving around the nucleus, like any charged particle moving in a circular orbit, has a magnetic dipole moment. Similarly, in quantum mechanics, the orbital angular momentum of an electron creates a certain magnetic moment. The interaction of this magnetic moment with the magnetic moment of the nucleus (due to the nuclear spin) leads to hyperfine splitting (that is, creates a hyperfine structure). However, the electron also has spin, which contributes to its magnetic moment. Therefore, there is a hyperfine splitting even for terms with zero orbital angular momentum.

The distance between the sublevels of the hyperfine structure is 1000 times smaller in order of magnitude than between the levels of the fine structure (this order of magnitude is essentially due to the ratio of the mass of the electron to the mass of the nucleus).

Anomalous hyperfine structure due to the interaction of electrons with the quadrupole electric moment of the nucleus.

Story

Hyperfine splitting was observed by A. A. Michelson in 1881, but was explained only after V. Pauli in 1924 suggested the presence of a magnetic moment in atomic nuclei.

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Literature

  • Landau L.D., Lifshits E.M. Theoretical physics . Volume 3. Quantum mechanics (non-relativistic theory).
  • Shpolsky E.V. Atomic physics. - M.: Nauka, 1974.

An excerpt characterizing the hyperfine structure

“There is nothing to have fun,” answered Bolkonsky.
While Prince Andrei met with Nesvitsky and Zherkov, on the other side of the corridor Strauch, an Austrian general who was at Kutuzov's headquarters to monitor the food of the Russian army, and a member of the Hofkriegsrat, who had arrived the day before, were walking towards them. There was enough space along the wide corridor for the generals to disperse freely with three officers; but Zherkov, pushing Nesvitsky away with his hand, said in a breathless voice:
- They're coming! ... they're coming! ... step aside, the road! please way!
The generals passed with an air of desire to get rid of troubling honors. On the face of the joker Zherkov suddenly expressed a stupid smile of joy, which he seemed unable to contain.
“Your Excellency,” he said in German, moving forward and addressing the Austrian general. I have the honor to congratulate you.
He bowed his head and awkwardly, like children learning to dance, began to scrape one leg or the other.
The General, a member of the Hofkriegsrath, looked sternly at him; not noticing the seriousness of the stupid smile, he could not refuse a moment's attention. He squinted to show he was listening.
“I have the honor to congratulate you, General Mack has arrived, in perfect health, only a little hurt here,” he added, beaming with a smile and pointing to his head.
The general frowned, turned away, and walked on.
Gott, wie naive! [My God, how simple he is!] – he said angrily, moving away a few steps.
Nesvitsky embraced Prince Andrei with laughter, but Bolkonsky, turning even paler, with an evil expression on his face, pushed him away and turned to Zherkov. That nervous irritation into which the sight of Mack, the news of his defeat, and the thought of what awaited the Russian army had brought him, found its outlet in bitterness at Zherkov's inappropriate joke.
“If you, dear sir,” he spoke piercingly with a slight trembling of his lower jaw, “want to be a jester, then I cannot prevent you from doing so; but I announce to you that if you dare another time to make a fuss in my presence, then I will teach you how to behave.
Nesvitsky and Zherkov were so surprised by this trick that they silently, with their eyes wide open, looked at Bolkonsky.
“Well, I only congratulated you,” said Zherkov.
- I'm not joking with you, if you please be silent! - Bolkonsky shouted and, taking Nesvitsky by the hand, he walked away from Zherkov, who could not find what to answer.
“Well, what are you, brother,” Nesvitsky said reassuringly.

9. Compare the obtained value with the theoretical value calculated using the universal constants.

The report must contain:

1. Optical layout of a spectrometer with a prism and a rotating prism;

2. Table of measurements of the angles of deviation of lines - benchmarks of mercury and their average values;

3. Table of measurements of the deviation angles of hydrogen lines and their average values;

4. The values ​​of the found frequencies of the hydrogen lines and the interpolation formulas used for the calculations;

5. Systems of equations used to determine the Rydberg constant by the least squares method;

6. The resulting value of the Rydberg constant and its value calculated from the universal constants.

3.5.2. Spectroscopic determination of nuclear moments

3.5.2.1. Experimental determination of the parameters of hyperfine splitting of spectral lines.

To measure the hyperfine structure of spectral lines, it is necessary to use high-resolution spectral instruments, therefore, in this work, we use a cross-dispersion spectral instrument in which the Fabry-Perot interferometer is placed inside a prism spectrograph (see Fig. 3.5.1 and Section 2.4.3.2,

rice. 2.4.11).

The dispersion of a prism spectrograph is sufficient to separate the spectral emission lines due to transitions of the valence electron in an alkali metal atom, but is completely insufficient to resolve the hyperfine structure of each of these lines. Therefore, using only a prism spectrograph, we would obtain on a photographic plate an ordinary emission spectrum, in which the hyperfine structure components would merge into one line, the spectral width of which is determined only by the resolving power of the ICP51.

The Fabry-Perot interferometer makes it possible to obtain an interference pattern within each spectral line, which is a sequence of interference rings. The angular diameter of these rings θ, as is known from the theory of the FabryPerot interferometer, is determined by the ratio of the thickness of the air layer of the standard t and the wavelength λ:

θk = k

where k is the order of interference for the given ring.

Thus, each spectral line is not just a geometric image of the entrance slit, built by the optical system of the spectrograph in the plane of the photographic plate, each of these images now turns out to be crossed by segments of interference rings. If there is no hyperfine splitting, then one system of rings corresponding to different orders of interference will be observed within a given spectral line.

If, however, within a given spectral line there are two components with different wavelengths (hyperfine splitting), then the interference pattern will be two systems of rings for wavelengths λ and λ ", shown in Fig. 3.5.2 by solid and dotted lines, respectively.

Rice. 3.5.2. Interference structure of a spectral line consisting of two close components.

The linear diameter of the interference rings d in the small angle approximation is related to the angular diameter θ by the relation:

d = θ×F 2 ,

where F 2 is the focal length of the spectrograph camera lens.

Let us obtain expressions relating the angular and linear diameters of the interference rings with the wavelength of the radiation that forms the interference pattern in the Fabry-Perot interferometer.

In the small angle approximation cos θ 2 k ≈ 1− θ 8 k and for two lengths

waves λ and λ ", the conditions for the interference maximum of the kth order are written respectively:

4λ"

θk = 8

−k

θ"k = 8

−k

From here, for the difference in the wavelengths of the two components, we obtain:

dλ = λ" −λ =

(θk 2

− θ" k 2 )

The angular diameter of the (k +1) -th order of the wavelength is determined

ratio:

8 − (k+1)

k+ 1

From (3.5.9) and (3.5.11) we get:

= θ2

− θ2

k+ 1

excluding t

from (3.5.10)-(3.5.12) we get:

d λ =

θk 2 − θ"k 2

k θ2 − θ2

k+ 1

At small angles, the order of interference is given by

k = 2 λ t (see (3.5.8)), so that equality (3.5.13) takes the form:

d λ =

θk 2 − θ"k 2

2 t θ 2

− θ2

k+ 1

Passing to the wave numbers ν =

We get:

1 d k 2 − d "k 2

dv =

− d2

k+ 1

Now, to determine d ~ ν, we need to measure the linear diameters of two systems of interference rings for two components of the hyperfine structure inside the spectral line under study. To improve the accuracy of determining d ~ ν, it makes sense to measure the diameters of the rings, starting from the second and ending with the fifth. Further rings are located closely to each other and the error in determining the difference in the squares of the diameters of the rings grows very quickly. You can average the entire right side (3.5.16), or separately the numerator and denominator.

3.5.2.2. Determination of the nuclear magnetic moment

In this work, we propose to determine the splitting of the ground state 52 S 1 2 of the stable isotope Rb 87 over

Chapter 10

SUPERFINE SPLITTING IN HYDROGEN


§ 1. Basic states for a system of two particles with spin 1/2

§2. Hamiltonian of the ground state of hydrogen

§ 3. Energy levels

§ 6. Projection matrix for spin 1


§ 1. Basic states for a system of two particles with spin 1 / 2

In this chapter, we will deal with the "hyperfine splitting" of hydrogen, an interesting example of what we are already able to do with the help of quantum mechanics. Here we will no longer have two states, but more. This example is instructive in that it will introduce us to the methods of quantum mechanics applied to more complex problems. By itself, this example is quite complicated, and once you understand how to deal with it, it will immediately become clear to you how to generalize it to other possible problems.

As is known, the hydrogen atom consists of an electron and a proton; the electron sits close to the proton and can exist in one of many discrete energy states, in each of which its pattern of motion is different. Thus, the first excited state lies on 3/4 Rydberg, or 10 ev, above the base state. But even the so-called ground state of hydrogen is not really a separate state with a certain energy, because the electron and the proton have spins. These spins are responsible for the "hyperfine structure" in energy levels, which splits all energy levels into several almost identical levels.

The spin of an electron can be either up or down; the proton too his own spin can look up or down. Therefore, for any dynamic state of the atom, there are four possible spin states. In other words, when a physicist talks about the "ground state" of hydrogen, he really means the "four ground states" and not just the lowest of them. The four spin states do not have exactly the same energy; there are slight shifts with respect to what would be observed in the absence of spins. These shifts, however, are many, many times smaller than those 10 ev, which lie between the ground state and the next higher state.

As a result, the energy of each dynamic state is split into a number of very tight levels - this is the so-called superfine splitting.

The energy differences of the four spin states are what we want to calculate in this chapter. Hyperfine splitting is caused by the interaction of the magnetic moments of the electron and proton; it results in slightly different magnetic energies for each spin state. These energy shifts are only about ten millionth of an electron volt, which is indeed much less than 10 ev!

It is precisely because of such a large gap that the ground state of hydrogen can be considered a “four-level system”, without caring that in fact there are much more states at higher energies. We intend to confine ourselves here to the study of the hyperfine structure of only the ground state of the hydrogen atom.

For our purposes, we don't care about the various details. location electron and proton, because all of them, so to speak, have already been worked out by the atom, they all turned out by themselves when the atom got into the ground state. It is enough to know only that the electron and the proton are not far from each other, in a certain spatial relationship. In addition, they can have all sorts of mutual orientations of spins. And we want to consider only spin effects.

The first question to be answered is what basic states for this system? But this question is posed incorrectly. Such a thing as the only one basis does not exist, and any system of basis states you choose will not be unique. It is always possible to compose new systems from linear combinations of the old one. There are always many choices for basic states, and they are all equally valid.

So, we must ask: not “what is the basis?”, but “what kind of can choose?". And you have the right to choose whatever you like, as long as it is convenient for you.

It's usually best to start with a basis that physically the most obvious. It does not have to solve any problem or be directly important in some way, no, it should generally only make it easier to understand what is going on.

We choose the following basic states:

State 1. Both the electron and the proton have their backs facing up.

State 2. The spin of an electron is up, while that of a proton is down.

State 3. For an electron, the spin looks down, and for a proton -

State 4. Both the electron and the proton have spins

To briefly write these four states, we introduce the following notation:

State 1:|+ +>; the electron has a spin up, the proton has a spin up.

State 2:| + ->; the electron has a spin up,

the proton has a spin way down.

State 3:|- + >; the electron has a spin way down, the proton has a spin up.

State 4:|- - >; the electron has a spin way down, the proton has a spin way down. (10.1)

remember, that the first the plus or minus sign refers to an electron, second - to the proton. To keep these notations at hand, they are summarized in Fig. 10.1.


Fig. 10.1. Set of basic states

for the ground state of the hydrogen atom.

We designate these states | + +>, | + ->> |- +>.

At times it will be more convenient to refer to these states as |1>, |2>, |3>, and |4>.

You might say, “But the particles interact, and maybe those states aren't the right basic states at all. It's like you're looking at both particles independently." Yes indeed! Interaction raises the question: what Hamiltonian systems? But the question is how describe system, does not concern interaction. Whatever we choose as a basis has nothing to do with what happens after. It may turn out that the atom is not capable of stay in one of these basic states, even if it all started with it. But that's another matter. This is a question of how the amplitudes change over time in a chosen (fixed) basis. By choosing basis states, we are simply choosing "unit vectors" for our description.

Since we have already touched on this, let's take a look at the general problem of finding a set of basic states when there is not one particle, but more. You know the basic states for one particle. An electron, for example, is completely described in real life (not in our simplified cases, but in real life) by setting the amplitudes of being in one of the following states:

| Electron spin up with momentum p> or

| Electron spin down with momentum p>.

There are actually two infinite sets of states, one for each value of p. This means that it is possible to say that the electronic state |y> is completely described only when you know all the amplitudes

where + and - represent the angular momentum components along some axis, usually the axis z, a p is the momentum vector. Therefore, for every conceivable impulse there must be two amplitudes (a doubly infinite set of basic states). That's all that is needed to describe a single particle.

In the same way, the basic states can be written when there are more than one particle. For example, if it would be necessary to consider an electron and a proton in a more complex case than ours, then the basic states could be as follows: An electron with a momentum p 1 spins up, and the proton with momentum R 2 moving backwards. And so on for other spin combinations. If there are more than two particles, the idea remains the same. So you see what to paint possible basic states is actually very easy. The only question is what is the Hamiltonian.

To study the ground state of hydrogen, we do not need to apply the complete sets of basis states for different momenta. We stipulate and fix certain momentum states of the proton and electron when we pronounce the words "ground state". The details of the configuration - the amplitudes for all the momentum basis states - can be calculated, but that is another task. And now we are only concerned with the influence of the spin, so we will limit ourselves to only four basic states (10.1). The next question is: what is the Hamiltonian for this set of states?

§ 2. Hamiltonian of the ground state of hydrogen

You'll know in a minute. But first I want to remind you one thing: any a state can always be represented as a linear combination of basic states. For any state |y|> one can write

Recall that full brackets are just complex numbers, so they can be denoted in the usual way by FROM i, where i=l, 2, 3, or 4, and write (10.2) as

Quad amplitude setting FROM i completely describes the spin state |y>. If this quadruple changes in time (as it will in fact), then the rate of change in time is given by the operator H^. The task is to find this operator H^ .

There is no general rule for how to write the Hamiltonian of an atomic system, and finding the correct formula requires more skill than finding a system of basis states. We were able to give you a general rule on how to write down the system of basic states for any problem in which there is a proton and an electron, but it is too difficult to describe the general Hamiltonian of such a combination at this level. Instead, we will bring you to the Hamiltonian by some heuristic reasoning, and you will have to accept it as correct, because the results will agree with experimental observations.

Recall that in the previous chapter we were able to describe the Hamiltonian of a single particle with spin 1/2 using sigma matrices, or exactly equivalent sigma operators. The properties of the operators are summarized in Table. 10.1. These operators, which are just a convenient, shorthand way of remembering the matrix elements of a type, were useful for describing the behavior separate particles with spin 1/2. The question arises whether it is possible to find a similar means for describing a system with two spins. Yes, and it's very simple. Look here. We'll invent a thing we'll call "electron-sigma" and which we'll represent as a vector operator s e with three components s e x , s e y and s e z . Farther let's agree that when one of them works

Table 10.1· PROPERTIES OF SIGMA OPERATORS

on some of our four basic states of the hydrogen atom, then it acts on the spin of the electron alone, moreover, as if the electron were alone, by itself. Example: what is s y e|-+>? Since s y , acting on an electron with a spin down, gives - i multiplied by the state with an electron whose spin is up, then

s e y |-+>=- i|++>.

(When s y e acts on the combined state, it flips the electron without affecting the proton, and multiplies the result by - i.) Acting on other states, s e at will give

Recall again that the operator s e acts only on the first spin symbol, i.e. per spin electron.

Let us now define the corresponding proton-sigma operator for the proton spin. Its three components s p x , s p y, s p z , act in the same way as s e, but only on proton spin. For example, if s p x acts on each of the four basic states, then it will turn out (again using Table 10.1)

As you can see, nothing difficult. In the general case, things can be more complicated. For example, the product of operators s e y s p z . When there is such a product, then what the right operator wants is done first, and then what the left one requires. For example,

Note that these number operators do nothing; we used this when we wrote s e x (-1)=(-1) s e x . We say that operators "commutate" with numbers, or that numbers "can be dragged" through an operator. Practice and show that the product s e X s p z gives the following result for four states:

If you go through all the valid operators, each one at a time, then there can be 16 possibilities in total. Yes, sixteen, if we also include the "single operator" 1. First, there is a triple s e X, s e y, s e z, then the triple s p x , s p y , s p z , for a total of six. In addition, there are nine products of the form s e X s p y , for a total of 15. And a single operator that leaves all states intact. That's all sixteen!

Notice now that for a four-state system, the Hamilton matrix would be a 4x4 coefficient matrix, with 16 numbers. It is easy to show that any 4X4 matrix, and in particular the Hamilton matrix, can be written as a linear combination of sixteen double spin matrices, corresponding to the system of operators we have just compiled. Therefore, for an 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.

But first, we know that the interaction does not depend on our choice of axes for the coordinate system. If there is no external perturbation - something like a magnetic field that selects some direction in space - then the Hamiltonian cannot depend on our choice of axis directions x, y and z. This means that the Hamiltonian cannot have terms like s e x by itself. It would look ridiculous, because someone in a different coordinate system would have come up with different results.

Only a term with an identity matrix is ​​possible, say a constant a(multiplied by 1^), and some combination of sigmas that does not depend on coordinates, some "invariant" combination. the only scalar the invariant combination of two vectors is their scalar product, which for our sigmas has the form

This operator is invariant with respect to any rotation of the coordinate system. So the only possibility for a Hamiltonian with a suitable symmetry in space is a constant multiplied by the identity matrix plus a constant multiplied by this dot product, i.e.

This is our Hamiltonian. This is the only thing that, based on symmetry in space, it can equal, as long as there is no external field. A permanent member won't tell us much; it simply depends on the level that we have chosen to read the energies. One could equally well take E 0 =0. And the second term will tell us everything we need to find the splitting of levels in hydrogen.

If you like, you can think about the Hamiltonian differently. If two magnets with magnetic moments m e and m p are close to each other, then their mutual energy depends, among other things, on m e · m R. And we, as you remember, found out that the thing that we called in classical physics m e, appears in quantum mechanics under the name m e s e . Similarly, what looks like m p in classical physics usually turns out to be equal to m p s p in quantum mechanics (where m p is the magnetic moment of the proton, which is almost 1000 times smaller than m e and has the opposite sign). Hence, (10.5) states that the interaction energy is similar to the interaction of two magnets, but not completely, because the interaction of two magnets depends on the distance between them. But (10.5) can be considered (and in fact is) kind of middle ground interaction. The electron somehow moves inside the atom, and our Hamiltonian gives only the average interaction energy. In general, all this suggests that for the prescribed arrangement of the electron and proton in space, there is an energy proportional to the cosine of the angle between the two magnetic moments (classically speaking). Such a classical qualitative picture can help you understand where everything comes from, but the only thing that matters is that (10.5) is the correct quantum mechanical formula.

The order of magnitude of the classical interaction between two magnets would have to be given by the product of the two magnetic moments divided by the cube of the distance between them. The distance between an electron and a proton in a hydrogen atom, roughly speaking, is equal to half the atomic radius, i.e. 0.5 A. Therefore, we can roughly estimate that the constant BUT must be equal to the product of the magnetic moments m e and m p divided by the cube of half an angstrom. Such zeroing leads to numbers that fall just in the right area. But it turns out that BUT can be calculated more accurately, once we understand the full theory of the hydrogen atom, which we are not yet able to do. In fact BUT was calculated to the nearest 30 millionths. As you can see, in contrast to the constant flip BUT ammonia molecule, which theoretically cannot be calculated well, our constant BUT for hydrogen may be calculated from a more detailed theory. But nothing can be done, for our present purposes we will have to count BUT number that can be determined from experience, and analyze the physics of the matter.

Taking the Hamiltonian (10.5), we can substitute it into the equation

and see what the spin interaction does to the energy levels. To do this, you need to count sixteen matrix elements H ij = i| H|j> corresponding to any two of the four basic states (10.1).

Let's start by calculating what is equal to H^ |j> for each of the four basic states. For example,

Using the method described a little earlier (remember Table 10.1, it makes things very easy), we find what each pair of a does with |+ +>· The answer is:

Hence, (10.7) turns into


Table 10.2 spin operators FOR THE HYDROGEN ATOM

And since all our four basic states are orthogonal, this immediately leads to

Remembering that H| i>=<.i>|H|j>*, we can immediately write a differential equation for the amplitude FROM 1:

That's all! Only one member.

To now obtain the remaining Hamilton equations, we must patiently go through the same procedures with H^, acting on other states. First, practice checking that all sigma products in Table. 10.2 are spelled correctly. Then use them to get

And then, multiplying them all in order from the left by all other state vectors, we get the following Hamiltonian matrix H ij :

This, of course, means that the differential equations for the four amplitudes FROM i look like

But before moving on to solving them, it's hard not to tell you about one clever rule that Dirac derived. It will help you to feel how much you already know, although we will not need it in our work. From equations (10.9) and (10.12) we have

“Look,” said Dirac, “I can also write the first and last equations in the form

and then they will all be the same. Now I will come up with a new operator, which I will denote R spin. exchange and which definition, will have the following properties:

This operator, as you see, only exchanges the directions of the spin of two particles. Then I can write the whole system of equations (10.15) as one simple operator equation:

This is the Dirac formula. The spin exchange operator gives a handy rule to remember s e s p. (As you can see, you already know how to do everything. All doors are open for you.)

§ 3. Energy levels

We are now ready to calculate the energy levels of the ground state of hydrogen by solving the Hamiltonian equations (10.14). We want to find the energies of the stationary states. This means that we must find those special states |y> for which each of the amplitudes belonging to |y> C i=i|y> has the same time dependence, namely e - w t . Then the state will have energy E=hw. So we are looking for a set of amplitudes for which

where the quadruple of coefficients a i does not depend on time. To see if we can get these amplitudes, we substitute (10.17) into (10.14) and see what happens. Each ihdC i /dt in (10.14) goes into EU i . And after reducing by a common exponential factor, each FROM i will turn into a i; we get

This is what needs to be done to find a 1 , a 2 , a 3 and a four . Indeed, it is very nice of the first equation that it does not depend on the others, which means that one solution is immediately visible. If choose E=A, then

a 1=1, a 2 =a 3 =a 4 =0

will give a solution. (Of course, if we accept all a equal to zero, then this will also be a solution, but it will not give a state!) We will consider our first solution as a state | I>:

His energy

E I =A.

All this immediately gives a clue to the second solution, obtained from the last equation in (10.18):

a 1 =a 2 =a 3 =0, a 4 =1, E=A.

We will call this decision the state | II>:

|//> = |4> = |-->,(10.20)

E(a 2 + a 3 ) = A(a 2 + a 3 ). (10.21)

Subtracting, we will have

Looking around and remembering the ammonia we already know, we see that there are two solutions here:

These are mixtures of states | 2 > and | 3 >. Denoting them | III> and | IV> and inserting the factor 1/T2 for correct normalization, we have

E III =A(10.24)

We have found four stationary states and their energies. Note, by the way, that our four states are orthogonal to each other, so they can also be considered basic states if desired. Our problem is completely solved.

The three states have energy equal to BUT, and the last one has PER. The mean is zero, which means that when in (10.5) we have chosen E 0 = 0, thus, we decided to count all energies from their average value. The energy level diagram of the ground state of hydrogen will look like in Fig. 10.2.

Fig. 10.2. Energy level diagram of the ground state of atomic hydrogen.

The difference in energies between the state | IV> and any of the rest is 4 A. An atom that happens to be in states | I>, can fall from there to the state | IV>and emit light: not optical light, because the energy is very small, but a microwave quantum. Or, if we illuminate hydrogen gas with microwaves, we will notice the absorption of energy, because the atoms are in the state | IV>will intercept it and move to one of the higher states, but all this is only at a frequency of w=4 A/h. This frequency has been measured experimentally; the best result obtained relatively recently is as follows:

The error is only three hundred billionth! Probably none of the fundamental physical quantities is better measured than this; this is one of the most outstanding measurements in physics in terms of accuracy. The theorists were very happy when they were able to calculate the energy to within 3·10 -5 ; but by that time it had been measured with an accuracy of 2·10 -11, i.e. a million times more accurate than in theory. So the experimenters are far ahead of the theorists. In the theory of the ground state of the hydrogen atom and you, and we are in the same position. You can also take the value BUT from experience - and everyone, in the end, has to do the same.

You've probably heard of the "21 cm line" of hydrogen before. This is the wavelength of the spectral line at 1420 MHz between hyperfine states. Radiation with this wavelength is emitted or absorbed by atomic hydrogen gas in galaxies. This means that with the help of radio telescopes tuned to waves 21 cm(or about 1420 MHz), one can observe the velocities and arrangement of atomic hydrogen concentrations. By measuring the intensity, you can estimate its amount. By measuring the shift in frequency caused by the Doppler effect, the movement of gas in the galaxy can be determined. This is one of the great radio astronomy programs. So what we're talking about now is something very real, it's not some kind of artificial task at all.

§ 4. Zeeman splitting

Although we have coped with the task of finding the energy levels of the ground state of hydrogen, we will still continue our study of this interesting system. To say something else about it, for example, to calculate the speed with which a hydrogen atom absorbs or emits radio waves of length 21 cm, you need to know what happens to him when he is indignant. We need to do what we did with the ammonia molecule - after we found the energy levels, we went further and found out what happens when the molecule is in an electric field. And after that it was not difficult to imagine the influence of the electric field of the radio wave. In the case of a hydrogen atom, the electric field does nothing with the levels, except that it shifts them all by some constant value proportional to the square of the field, and we are not interested in this, because this does not change differences energies. This time it's important magnetic field. So the next step is to write the Hamiltonian for the more complicated case where the atom sits in an external magnetic field.

What is this Hamiltonian? We'll just tell you the answer, because we can't give you any "proof" other than to say that this is how the atom works.

The Hamiltonian has the form

Now it consists of three parts. First member BUT(s e s p) represents the magnetic interaction between an electron and a proton; it is the same as if there were no magnetic field. The influence of the external magnetic field is manifested in the other two terms. Second term (-m e s e · AT) is the energy that an electron would have in a magnetic field if it were alone there. Similarly, the last term (-m p s R · AT) would be the energy of a single proton. According to classical physics, the energy of both of them together would be the sum of their energies; according to quantum mechanics, this is also correct. The interaction energy arising due to the presence of a magnetic field is simply the sum of the interaction energies of an electron with a magnetic field and a proton with the same field, expressed in terms of sigma operators. In quantum mechanics, these terms are not really energies, but referring to the classical formulas for energy helps memorize the rules for writing the Hamiltonian. As if. however, (10.27) is the correct Hamiltonian.

Now you need to go back to the beginning and solve the whole problem again. But most of the work has already been done, we just need to add the effects called by the new members. We assume that the magnetic field B is constant and directed along z. Then to our old Hamiltonian operator H^ two new pieces must be added; let's denote them H^":

Using table. 10.1, we immediately get

See how convenient! Operator H", acting on each state, gives simply a number multiplied by the same state. In the matrix i|H"|j> there is therefore only diagonal elements, and one can simply add the coefficients from (10.28) to the corresponding diagonal terms in (10.13), so that the Hamiltonian equations (10.14) become

The form of the equations has not changed, only the coefficients have changed. And while AT does not change over time, you can do everything the same way as before.

Substituting

, we get

Fortunately, the first and fourth equations are still independent of the others, so the same technique will come into play again. One solution is the state | I> for which

Another solution

The other two equations require more work because the coefficients at a 2 and a 3 are no longer equal to each other. But on the other hand, they are very similar to the pair of equations that we wrote for the ammonia molecule. Looking back at equations (7.20) and (7.21), we can draw the following analogy (remember that indices 1 and 2 there correspond to indices 2 and 3 here):

Previously, the energies were given by formula (7.25), which had the form

Substituting (10.33) here, we obtain for the energy

In ch. 7 we used to call these energies E I and E II , now we will label them E III and E IV :

So, we have found the energies of the four stationary states of the hydrogen atom in a constant magnetic field. Let's check our calculations, for which we strive AT to zero and see if we get the same energies as in the previous paragraph. You can see that the weight is ok. At B= 0 energy E I , E II and E III turn to + BUT, a E IV - in - PER. Even our numbering of states is consistent with the previous one. But when we turn on the magnetic field, then each energy will begin to change in its own way. Let's see how it goes.

First, we recall that m e is negative for an electron and is almost 1000 times greater than m p, which is positive. Hence, both m e +m p and m e -m p are both negative and almost equal to each other. Let's denote them -m and -m":

(Both m and m" are positive and almost coincide in magnitude with m e, which is approximately equal to one Bohr magneton.) Our four energies will then turn into

Energy E I initially equal to BUT and increases linearly with AT with speed m. Energy E II is also equal to the beginning A, but with growth AT linearly decreases the slope of its curve is -m . Changing these levels from AT shown in FIG. 10.3. The figure also shows the energy graphs E III and E IV . Their dependence on AT different. At small AT they depend on AT quadratically; at first their inclination is equal to zero, and then they begin to bend, and at large B approach straight lines with a slope of ±m", close to the slope e i and E II

The shift in the energy levels of an atom caused by the action of a magnetic field is called the Zeeman effect. We say that the curves in Fig. 10.3 show Zeeman splitting the ground state of hydrogen.

Fig. 10.3. Ground state energy levels

hydrogen in a magnetic fieldAT .

Curves E III and E IV approach the dotted lines

A ± m "B.

When there is no magnetic field, one simply gets one spectral line from the hyperfine structure of hydrogen. State transitions | IV> and any of the other three occur with the absorption or emission of a photon whose frequency is 1420 MHz:1/h, multiplied by the energy difference 44. But when the atom is in a magnetic field B, then there are much more lines. Transitions between any two of the four states can occur. Hence, if we have atoms in all four states, then energy can be absorbed (or emitted) in any of the six transitions shown in Fig. 10.4 vertical arrows.

Fig. 10.4. Transitions between energy levels of the ground state of hydrogen in a certain magnetic fieldAT.

Many of these transitions can be observed using the Rabi molecular beam technique, which we described in Chap. 35, § 3 (issue 7).

What is the reason for the transitions? They arise if, along with a strong constant field B apply a small perturbing magnetic field that varies with time. We observed the same thing under the action of an alternating electric field on an ammonia molecule. Only here the culprit of the transitions is the magnetic field acting on the magnetic moments. But the theoretical calculations are the same as in the case of ammonia. They are most easily obtained if we take a perturbing magnetic field rotating in the plane hu, although the same will be from any oscillating horizontal field. If you insert this perturbing field as an additional term into the Hamiltonian, you will get solutions in which the amplitudes change with time, as was the case with the ammonia molecule. This means that you can easily and accurately calculate the probability of transition from one state to another. And you will find that it all agrees with experience.

§ 5. States in a magnetic field

Let us now deal with the shape of the curves in Fig. 10.3. First, if we talk about large fields, then the dependence of the energy on the field is quite interesting and easily explained. For sufficiently large AT(namely, when mB/A>>1) in formulas (10.37) one can be neglected. The four energies take the form

These are the equations of the four lines in Fig. 10.3. These formulas can be physically understood as follows. The nature of stationary states in zero the field is completely determined by the interaction of two magnetic moments. Mixing of basic states | + -> and | - +> in stationary states |III> and | IV>caused by this interaction. However, one can hardly expect that each of our particles (both proton and electron) in strong external fields will be influenced by the field of another particle; each will act as if it were alone in the external field. Then (as we have already seen many times) the spin of the electron will be directed along the external magnetic field (along or against it).

Let the spin of the electron be directed upward, i.e., along the field; its energy will be -m e B. At the same time, the proton can stand in different ways. If its spin is also directed upwards, then its energy is -m p b. Their sum is -(m e +m p) B=mB. And that's just what it is E I , and this is very nice, because we describe the state |+ +>=| I>. There is also a small additional member BUT(now (m B>>A), representing the interaction energy of a proton and an electron when their spins are parallel. (We thought from the beginning BUT positive, because it should have been so according to the theory in question; the same is obtained in experiment.) But the spin of the proton can also be directed downward. Then its energy in the external field will turn into +m Р B, and together with the electron their energy will be - (m e -m p) B= m AT. And the interaction energy turns into - BUT. Their sum will give energy E III , in (10.38). So state | III>in strong fields becomes the state |+ ->.

Now let the spin of the electron be directed downward. Its energy in the external field is equal to m e AT. If the proton also looks down, then their total energy is (m e + m p) B = - m Plus interaction energy BUT(the backs are now parallel). It leads just in time to energy E II in (10.38) and corresponds to the state |- ->=| II>, which is very nice. And finally, if the spin of an electron is directed downwards, and that of a proton is directed upwards, then we will get the energy (m e -m p )В-А (minus A because the spins are opposite), i.e. E IV . And the state answers |- +>.

“Wait a minute,” you probably say. “States | Ill>and | IV>- is not states | + - > and | -+>; they are them mixtures." True, but the mixing here is barely noticeable. Indeed, at 5=0 they are mixtures, but we have not yet found out what happens at large AT. When we used the analogy between (10.33) and the formulas of Ch. 7, then at the same time it was possible to take the amplitudes from there. They will be obtained from (7.23):

Attitude a 2 /a 3 - of course this time C 2 /C 3 Inserting similar quantities from (10.33), we obtain

where instead of E you need to take the right energy (or E III , or E IV ). For example, for the state | III>we have

So, for large AT y state | ///> FROM 2 >>C 3 ;state almost completely becomes a state | 2>= |+ ->. Similarly, if in (10.39) we substitute e iv , then it turns out that (С 2 /С 3) IV> simply turns to the state |3> = |- +>. You see that the coefficients in the linear combinations of our basis states that make up the stationary states themselves depend on AT.

The state we call | III>, in very weak fields it is a mixture of |+ -> and |- +> in a ratio of 1:1, but in strong fields it shifts entirely to |+ ->. Similarly, the state | IV>, which in weak fields is also a mixture of |+ -> and |- +> in a ratio of 1:1 (with the opposite sign), goes into the state | - +), when the spins are no longer connected to each other due to a strong external field.

I would like to draw your attention, in particular, to what is happening in very weak magnetic fields. There is one energy ( -3A), which does not change when a weak magnetic field is turned on. And there is another energy +A), which, when a weak magnetic field is turned on, splits into three different energy levels. In weak energy fields with increasing AT change as shown in Fig. 10.5. Let's say that we have a somehow selected set of hydrogen atoms, all of which have energy equal to - 3A. If we pass them through a Stern-Gerlach device (with not very strong fields), then we will find that they simply pass completely through. (Because their energy does not depend on AT, then, according to the principle of virtual work, the gradient of the magnetic field does not create any force that would be felt by them.) Suppose, on the other hand, we would select a group of atoms with energy + BUT and passed them through the Stern-Gerlach device, say, through the device S.(Again, the fields in the apparatus should not be so strong as to destroy the interior of the atom; it is understood that the fields are small enough that the energies can be considered linearly dependent on AT.) We would have received three bundles. On states | I> and | II>opposite forces act, their energies change according to AT linearly with slope ±m, so that strength are similar to the forces acting on a dipole, in which m z = ±m , and state | III> goes right through. We are back to Chap. 3. A hydrogen atom with energy +A is a particle with spin 1. This energy state is a "particle" for which j=1, and can be described (with respect to some system of axes in space) in terms of the basis states |+ S>, | 0S> and |- S>, which we used in Chap. 3. On the other hand, when the hydrogen atom has an energy of -3 BUT, it is a particle with zero spin. (We remind you that, strictly speaking, everything that has been said is true only for infinitely small magnetic fields.) Thus, the states of hydrogen in a zero magnetic field can be grouped as follows:

In ch. 35 (Issue 7), we said that for any particle, the components of the angular momentum along any axis can take only certain values, which always differ by h. So, the z-component of the angular momentum J z may be equal jh,(j-1) h, (j- 2)h,..., (-j)h, where j is the spin of the particle (which can be integer or half-integer). Usually write

J z =mh,(10.43)

where t stands in place of any of the numbers j, j-1, j- 2, . . .,-j(at the time we did not mention this). Therefore, you will often find in books the numbering of the four basic states with the help of the so-called quantum numbers j and m[often referred to as the "quantum number of the total angular momentum" ( j) and "magnetic quantum number" (m)]. Instead of our status symbols | I>, |II> etc. many often write states in the form | j, m>. Our table of states for the zero field in (10.41) and (10.42) they would have depicted in the form of a table. 10.3. There is no new physics here, it's just a matter of notation.

Table 10.3 STATES OF THE HYDROGEN ATOM IN A ZERO FIELD

§ 6. Projection matrix for spin 1

Now we would like to apply our knowledge of the hydrogen atom to a special problem. In ch. 3 we talked about the particle with spin 1, located in one of the basic states (+, 0, -) with respect to the Stern-Gerlach device with some particular orientation (say, with respect to the device S), will have a certain amplitude of staying in one of three states in relation to the device T, differently oriented in space. There are nine such amplitudes jT|iS> , which together form a projection matrix. In ch. 3, § 7, without proof we have written out the elements of this matrix for various orientations T towards S. Now we want to show you one of the ways to display them.

In the hydrogen atom, we have found a system with spin 1, composed of two particles with spin 1/2. In ch. 4 we have already learned how to transform the amplitudes for spin 1/2. This knowledge can be applied to obtain a transformation for spin 1. Here is how it is done: there is a system (a hydrogen atom with energy + BUT) with spin 1. Let us pass it through the filter S Stern - Gerlach so that we now know that it is in one of the basic states with respect to S, say in |+ S). What is the amplitude of the fact that it will be in one of the basic states, say |+ T), relative to the device T? If you name the instrument's coordinate system S system x, y, z, then the state |+ S> - this is what was recently called the |+ +> state. But imagine that some buddy of yours spent his axis z along the axis T. He will attribute his states to some system x", y", z". His "up" and "down" states for the electron and proton would be different from yours. His"plus - plus" state, which can be written down | +"+">, noting the "hatching" of the system, there is a state |+ T> particles with spin 1. Are you interested in T|+ S> that there is just another way to write the amplitude.

The amplitude can be found as follows. AT your spin system electron out of state | + +> is directed upwards. This means that he has some amplitude e to be in your buddy's spin-up system and some amplitude e to be in this spin-down system. Likewise, proton able + + At has spin up in your frame and the amplitudes p and p turn out to be spin up or spin down in the "primed" frame. Since we are talking about two different particles, then the amplitude of that both particles together in his the system will turn out to be spins up, is equal to the product of the amplitudes

We put the signs e and p under the amplitudes so that it is clear what we are doing. But both are just transformation amplitudes for a spin 1/2 particle, so they're actually the same numbers. In fact, these are the same amplitudes that we in Chap. 4 called T|+ S quoted1 > > and which we have given in the table. 4.1 and 4.2.

But now, however, we are threatened by a confusion of notation. One must be able to distinguish the amplitude T|+ S) for the particle with spin 1/2 of what we are also called T|+ S> but for back 1-they have nothing in common! I hope it won't confuse you too much if we for a while we introduce other notation for the amplitudes for the spin 1 / 2. They are given in Table. 10.4. For spin-1 particle states, we will still use the notation | + S, | 0S> and |- S>.

Table 10.4 AMPLITUDES for SPIN 1 / 2

In our new notation, (10.44) simply becomes

This is just the amplitude T|+ S> for spin 1. Now let's say, for example, that your buddy has a coordinate system, i.e. a "hatched" fixture T, turned around your axes z at the angle j; then from Table. 4.2 turns out

Hence, from (10.44) the amplitude for spin 1 will be equal to

Now you understand how we will proceed further.

But it would be good to carry out calculations in the general case for all states. If a proton and an electron are in our system (system S) both look up, then the amplitudes of what is in the other system (system T) they will be in one of four possible states,

We can then write the |+ +> state as the following linear combination:

But now we notice that |+ "+"> is a state of |+ T>, that (| + "-">+|-"+">) is just C2, multiplied to state |0 T> [see (10.41)] and that | - "-">= |- T>. In other words, (10.47) is rewritten as

It is just as easy to show that

C |0 S> things are a bit more complicated, because

But each of the states | + - > and | - +> can be expressed in terms of "hatched" states and substituted into the sum:

Multiplying the sum (10.50) and (10.51) by 1/C2, we get

this implies

Now we have all the necessary amplitudes. The coefficients in (10.48), (10.49) and (10.52) are the matrix elements

jТ| iS>. Let's put them in one matrix:

We have expressed the spin 1 transformation in terms of the amplitudes a, b, with and d spin 1 / 2 transformations.

If, for example, the system T turned towards S at an angle a around the axis at(see Fig. 3.6, p. 64), then the amplitudes in Table. 10.4 are just matrix elements R y(a) in table. 4.2:

Substituting them into (10.53), we obtain formulas (3.38), which are given on page 80 without proof.

But what happened to the state | IV)?! This is a spin-zero system; it means that it has only one state - it in all coordinate systems same. You can check that everything works out like this if you take the difference between (10.50) and (10.51); we get

But (ad-bc)- is the determinant of the matrix for spin 1 / 2 , it just equals one. It turns out

|IV">=|IV> for any relative orientation of the two coordinate systems.

* To those who jumped over ch. 4, you will have to skip this paragraph.

* Recall that classically U= -m·B, so the energy is at its lowest when the moment is in the field. For positively charged particles, the magnetic moment is parallel to the spin, for negative ones, vice versa. Hence, in (10.27) m R is a positive number, and (m e - negative.

*Crampton, Kleppner, Ramsey, Physical Review Letters, 11, 338 (1963).

*Actually, the state is

but, as usual, we will identify the states with constant vectors, which at t=0 coincide with the real vectors.

* This operator is now called the spin exchange operator.

* For these operators, however, it turns out that nothing depends on their order.

Although we have coped with the task of finding the energy levels of the ground state of hydrogen, we will still continue our study of this interesting system. To say something else about it, for example, to calculate the speed with which a hydrogen atom absorbs or emits radio waves of length 21 cm, you need to know what happens to him when he is indignant. We need to do what we did with the ammonia molecule - after we found the energy levels, we went further and found out what happens when the molecule is in an electric field. And after that it was not difficult to imagine the influence of the electric field of the radio wave. In the case of a hydrogen atom, the electric field does nothing with the levels, except that it shifts them all by some constant value proportional to the square of the field, and we are not interested in this, because this does not change differences energies. This time it's important magnetnoe field. So the next step is to write the Hamiltonian for the more complicated case where the atom sits in an external magnetic field.

What is this Hamiltonian? We'll just tell you the answer, because we can't give you any "proof" other than to say that this is how the atom works.

The Hamiltonian has the form

Now it consists of three parts. First member BUT(σ e ·σ p) represents the magnetic interaction between an electron and a proton; it is the same as if there were no magnetic field. The influence of the external magnetic field is manifested in the other two terms. The second term (- μ e σ e B) is the energy that an electron would have in a magnetic field if it were alone there. In exactly the same way, the last term (- μ r σ r ·B) would be the energy of a single proton. According to classical physics, the energy of both of them together would be the sum of their energies; according to quantum mechanics, this is also correct. The interaction energy arising due to the presence of a magnetic field is simply the sum of the interaction energies of an electron with a magnetic field and a proton with the same field, expressed in terms of sigma operators. In quantum mechanics, these terms are not really energies, but referring to the classical formulas for energy helps memorize the rules for writing the Hamiltonian. Be that as it may, (10.27) is the correct Hamiltonian.

Now you need to go back to the beginning and solve the whole problem again. But most of the work has already been done, we just need to add the effects called by the new members. We assume that the magnetic field B is constant and directed along z. Then to our old Hamiltonian operator H two new pieces must be added; let's denote them H′:

See how convenient! The operator H , acting on each state, gives simply a number multiplied by the same state. In the matrix<¡|H′| j>exist therefore only diagonal elements, and one can simply add the coefficients from (10.28) to the corresponding diagonal terms in (10.13), so that the Hamiltonian equations (10.14) become

The form of the equations has not changed, only the coefficients have changed. And while AT does not change over time, you can do everything the same way as before.
Substituting FROM= a l e-(¡/h)Et, we get

Fortunately, the first and fourth equations are still independent of the others, so the same technique will come into play again. One solution is the |/> state, for which

The other two equations require more work because the coefficients for a 2 and a 3 are no longer equal to each other. But on the other hand, they are very similar to the pair of equations that we wrote for the ammonia molecule. Looking back at equations (7.20) and (7.21), we can draw the following analogy (remember that indices 1 and 2 there correspond to indices 2 and 3 here):

Previously, the energies were given by formula (7.25), which had the form

In Chapter 7 we used to call these energies E I and E II, we will now label them E III and EIV

So, we have found the energies of the four stationary states of the hydrogen atom in a constant magnetic field. Let's check our calculations, for which we strive AT to zero and see if we get the same energies as in the previous paragraph. You see that everything is in order. At B=0 energy E I , E II and E III apply to +A, a EIV - in - 3A. Even our numbering of states is consistent with the previous one. But when we turn on the magnetic field, then each energy will begin to change in its own way. Let's see how it goes.

First, we recall that the electron μ e negative and almost 1000 times more μ p, which is positive. Hence, both μ e +μ r and μ e -μ r are both negative and almost equal to each other. Let's denote them -μ and -μ′:

(AND μ , and μ′ are positive and almost coincide in magnitude with μ e, which is approximately equal to one Bohr magneton.) Our four energies will then turn into

Energy E I initially equal to BUT and increases linearly with AT with speed μ. Energy E II is also equal to the beginning BUT, but with growth AT linearly decreases the slope of its curve is - μ . Changing these levels from AT shown in Fig.10.3. The figure also shows the energy graphs E III and EIV. Their dependence on AT different. At small AT they depend on AT quadratically; at first their inclination is equal to zero, and then they begin to bend, and at large B approach straight lines with a slope of ± μ ′ close to the slope E I and EII.

The shift in the energy levels of an atom caused by the action of a magnetic field is called the Zeeman effect. We say that the curves in Fig. 10.3 show Zeeman splitting the ground state of hydrogen. When there is no magnetic field, one simply gets one spectral line from the hyperfine structure of hydrogen. State transitions | IV> and any of the other three occur with the absorption or emission of a photon whose frequency is 1420 MHz:1/h, multiplied by the energy difference 4A. But when the atom is in a magnetic field B, then there are much more lines. Transitions between any two of the four states can occur. Hence, if we have atoms in all four states, then energy can be absorbed (or emitted) in any of the six transitions shown in Fig. 10.4 vertical arrows. Many of these transitions can be observed using the Rabi molecular beam technique, which we described in Chap. 35, § 3 (issue 7).

What is the reason for the transitions? They arise if, along with a strong constant field AT apply a small perturbing magnetic field that varies with time. We observed the same thing under the action of an alternating electric field on an ammonia molecule. Only here the culprit of the transitions is the magnetic field acting on the magnetic moments. But the theoretical calculations are the same as in the case of ammonia. They are most easily obtained if we take a perturbing magnetic field rotating in the plane hu, although the same will be from any oscillating horizontal field. If you insert this perturbing field as an additional term into the Hamiltonian, you will get solutions in which the amplitudes change with time, as was the case with the ammonia molecule. This means that you can easily and accurately calculate the probability of transition from one state to another. And you will find that it all agrees with experience.

Until now, we have been talking about the features of the structure of the spectra, which are explained by the properties of the electron cloud of the atom.

However, details in the structure of the spectra that cannot be explained from this point of view have long been noted. This includes the complex structure of individual lines of mercury and the double structure of each of the two yellow lines of sodium discovered in 1928 by L. N. Dobretsov and A. N. Terenin. In the latter case, the distance between the components was only 0.02 A, which is 25 times less than the radius of the hydrogen atom. These details of the structure of the spectrum are called hyperfine structure (Fig. 266).

Rice. 266. Hyperfine structure of the sodium line.

For its study, the Fabry-Perot standard and other devices with high resolution are usually used. The slightest expansion of spectral lines, caused by the interaction of atoms with each other or by their thermal motion, leads to the merging of the components of the hyperfine structure. Therefore, the method of molecular beams, first proposed by L. N. Dobretsov and A. N. Terenin, is widely used at present. With this method, the glow or absorption of a beam of atoms flying in a vacuum is observed.

In 1924, the Japanese physicist Nagaoka made the first attempt to relate hyperfine structure to the role of the atomic nucleus in spectra. This attempt was made in a very unconvincing form and caused completely mocking criticism from the well-known

spectroscopist I. Runge. He assigned to each letter of the Nagaoka surname its ordinal number in the alphabet and showed that an arbitrary combination of these numbers among themselves gives the same good agreement with the experimental data as Nagaoka's theory.

However, Pauli soon established that there was a grain of truth in Nagaoka's ideas and that the hyperfine structure was indeed directly related to the properties of the atomic nucleus.

Two types of hyperfine structure should be distinguished. The first type corresponds to a hyperfine structure, the same number of components for all lines of the spectrum of a given element. The appearance of this hyperfine structure is associated with the presence of isotopes. When studying the spectrum of one isolated isotope, only one component of the hyperfine structure of this type remains. For light elements, the appearance of such a hyperfine structure is explained by simple mechanical considerations. In § 58, considering the hydrogen atom, we considered the nucleus to be motionless. In fact, the nucleus and electron revolve around a common center of mass (Fig. 267). The distance from the nucleus to the center of mass is very small, it is approximately equal to where the distance to the electron, the mass of the electron, the mass of the nucleus.

Rice. 267. Rotation of the nucleus and electron around a common center of mass.

As a result, the energy of the atom acquires a slightly different value, which leads to a change in the Rydberg constant

where the value of the Rydberg constant corresponding to the fixed nucleus

Thus, depends on and, consequently, the frequency of the lines must depend on The latter circumstance served as the basis for the spectroscopic discovery of heavy hydrogen. In 1932, Urey, Maffey, and Brickwid discovered weak companions of the Balmer series line in the hydrogen spectrum.

Assuming that these satellites correspond to lines of a heavy hydrogen isotope with an atomic weight of two, they calculated, using (1), the wavelengths and compared them with experimental data.

According to formula (1), for elements with medium and large atomic weights, the isotope effect should be vanishingly small.

This conclusion is confirmed experimentally for elements with medium weights, but, oddly enough, is in sharp contradiction with the data for heavy elements. Heavy elements clearly exhibit an isotopic hyperfine structure. According to the available theory, in this case, it is no longer the mass that plays a role, but the finite dimensions of the nucleus.

The definition of the meter in the SI system (GOST 9867-61) takes into account the role of the hyperfine structure by indicating the isotope of krypton: "The meter is a length equal to 1650763.73 wavelengths in the vacuum of radiation corresponding to the transition between the levels of the krypton atom 86".

The second type of hyperfine structure is not associated with the presence of a mixture of isotopes; in particular, a hyperfine structure of this type is observed in bismuth, which has only one isotope.

The second type of hyperfine structure has a different shape for different spectral lines of the same element. The second type of hyperfine structure was explained by Pauli, who attributed to the nucleus its own mechanical torque (spin), a multiple of

Rice. 268. Origin of the hyperfine structure of the yellow lines of sodium.

The total rotational moment of an atom is equal to the vector sum of the nuclear moment and the moment of the electron shell. The total torque must be quantized, as are all atomic moments. Therefore, spatial quantization again arises - only certain orientations of the nuclear torque with respect to the electron shell torque are allowed. Each orientation corresponds to a certain sublevel of atomic energy. As in multiplets, here different sublevels correspond to a different amount of magnetic energy of the atom. But the mass of the nucleus is thousands of times greater than the mass of the electron, and therefore the magnetic moment of the nucleus is approximately the same number of times less than the magnetic moment of the electron. Thus, changes in the orientation of the nuclear moment should cause only very small changes in energy, which show up in the hyperfine structure of the lines. On fig. 268 shows diagrams of the hyperfine structure of sodium. To the right of each energy level is a number characterizing the total torque. The spin of the atomic nucleus of sodium turned out to be equal to

As can be seen from the figure, each of the yellow sodium lines consists of a large number of components, which, with insufficient resolution, look like two narrow doublets. The rotational moments of nuclei determined from the analysis of the hyperfine structure (in particular, for nitrogen) turned out to be in conflict with the hypothesis of the existence of electrons in the composition of the nucleus, which was used by D. D. Ivanenko to assert that the nuclei consist of protons and neutrons (§ 86).

Subsequently (since 1939), the much more accurate Rabi radiospectrographic method began to be used to determine nuclear moments.

Rabi's radiospectroscopic scheme for determining nuclear magnetic moments is, as it were, two Stern-Gerlach facilities (p. 317) arranged in series with mutually opposite directions of inhomogeneous magnetic fields. The molecular beam penetrates both installations in succession. If in the first setup the molecular beam is deflected, for example, to the right, then in the second setup it is deflected to the left. The effect of one setting compensates for the effect of another. Between these two settings is a device that violates the compensation. It consists of an electromagnet that creates a uniform magnetic field, and electrodes connected to a generator of high-frequency oscillations. The uniform magnetic field is directed parallel to the magnetic field in the first Stern-Gerlach installation.

A particle with a magnetic moment directed at an angle to the direction of the field has potential energy (vol. II, § 58). The same angle determines the amount of beam deflection in the first Stern-Gerlach setup. Under the action of a high-frequency field, the orientation of the magnetic moment can change and the magnetic energy becomes equal. This change in magnetic energy must be equal to the energy of the photon that caused the transition (absorption or forced transition, § 73):

Possible values ​​are determined by the spatial quantization law. Beam deflection in the second setup depends on the angle Since the angle is not equal to the angle, this deflection will not be equal to the deflection in the first setup and the compensation will be violated. Violation of the compensation of deviations is observed only at frequencies that satisfy the specified ratio; in other words, the observed effect is a resonance effect, which greatly improves the accuracy of the method. From the measured frequencies, the magnetic moments of the nuclei are calculated with great accuracy.

However, conventional optical spectroscopy retains its full value for the study of isotopic effects, where radiospectroscopy is fundamentally inapplicable. Isotopic effects are of particular interest for the theory of nuclear forces and intranuclear processes.

In recent years, spectroscopists have again returned to a thorough study of the spectrum of hydrogen. The spectrum of hydrogen turned out to be a literally inexhaustible source of new discoveries.

In § 59 it has already been said that, when examined with high resolution equipment, each line of the hydrogen spectrum turns out to be double. It has long been believed that the theory of these subtle details of the hydrogen spectrum is in excellent agreement with experimental data. But, beginning in 1934, spectroscopists began to carefully point out the existence of small discrepancies between theory and experience. The discrepancies were within the measurement accuracy. The smallness of the effects can be judged by the following figures: according to the theory, the line should basically consist of two lines with the following wave numbers: 15233.423 and The theoretical difference of the wave numbers is only a thousandth of a percent of each wave number. The experiment gave a value for this difference, about 2% less Michelson once said that "we should look for our future discoveries in the sixth decimal place." Here we are talking about the discrepancy in the eighth decimal place. In 1947, Lamb and Riserford returned to the same problem, but using the latest advances in physical experiment technology. The old theory led to a scheme of lower energy levels for the line shown in fig. 269.