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					Quantum Mechanics

Chapter 6. Spin
         §6.1 The Spin Operators
           d2                2m              l (l  1) 2 
                [r R(r )]   2 E  V (r )               [rR (r )]   (8.1)
           dr 2                                2mr 2 


• Following the method of Section 2.3, we represent
  the eigenstates of the spin operator as two-
  dimensional vectors. In the case of electron spin,
  there are only two eigenfunctions instead of three or
  more.
• The Hilbert spin space is therefore two-dimensional
   and we can write the eigenfunctions as column
  vectors just as we did for orbital angular momentum;
• the unit vectors that are eigenfunctions of Sz are the
  two-component vectors




• Compare with Eq.(7.18) for the three eigenstates of
  Lz with l = 1. Following the reasoning given there,
  we say that any electron-spin function Fs can be a
  represented by a superposition of │+ >z and │- >z:
• Continuing this line of reasoning, we can represent
  the spin operators Sx , Sy, and Sz as the following 2
  × 2 matrices that operate on these vectors:



• These may be written



• where the dimensionless operators σx, σy, and σz
  (the Pauli spin matrices) are expressed by the
  matrices in Eq.(10.3).
• These matrices can be derived by using
• rules analogous to those of Eqs.(7.17):



• It is left as an exercise to verify that these equations
  are indeed satisfied by the matrices given. You can
  also verify the anticommutation relation σxσy + σyσx
  = 0, and the corresponding relations involving σz.
• Example Problem 10.1 Find values of the constants
  a and b that yield normalized eigenfunctions of Sx ,
  by applying this operator to the general expression
  for a spin state: a│+ >z + b│- >z.
    – Solution. We know that the eigenvalues of Sx ,
    must be the same as those of Sz .
                           a
•    The statement that  b  is an eigenfunction of Sx
                            
                            
  with eigenvalue  / 2 is expressed by the notation
        a  a
    S x     .
        b 2 b
             
•   Substituting the matrix for Sx and performing the
                                                      a    0 1  a    b
                                                   S   
  indicated matrix operation. we have  b  2  1 0  b   2  a .
                                                      
                                                             x
                                                             
                                                                    
                                                                    
                                                                             
                                                                             
               a b
• Therefore  b    a . or b = a. Normalization requires
  that a  b  1, so that a = b = l/(2)1/2.
              2      2


                          a
• Similarly, if is an eigenfunction with the
                           
                          b
                           

 eigenvalue -ћ/2 , then S  a     0 1  a     b .
                           
                                  
                                        x 
                                                   
                                                      
                                             b   2  1 0  b    2 a
                     a    b
• In this case           .
                     b    a
                           
• or b = -a = -1/(2)1/2.
                                        1 / 2 
                                              
• In conclusion, the eigenvectors are   1 / 2 
                                              
          1/ 2 
• and  1 / 2 
         
         
                
                   respectively.
         §6.2 Fine Structure in The
           Hydrogen Spectrum
• Effect of Spin on Energy Levels
• The intrinsic magnetic moment of the electron,
  which results from its spin, leads to a splitting of the
  energy levels of hydrogen into sublevels, provided
  that there is a magnetic field in the atom, because in
  each of the states determined by the Schreodinger
  equation the electron would then have two possible
  va1ues for its magnetic potential energy, according
  to the value of the z component of its magnetic
  moment.
•   But why should there be a magnetic field in the
 hydrogen atom?
• The only other particle present is the proton. One
 might expect the proton, like the electron to have
 internal spin and magnetic moment, and it does, but
 it turns out that the proton's intrinsic magnetic
 moment produces a field which is too small to
 account for the observed fine structure in the
 hydrogen spectrum. (The proton's magnetic moment
 leads to hyperfine structure, to be discussed later in
 this section.)
• However, the proton's electric charge leads to the
 existence of a magnetic field of sufficient strength,
• because in the electron's rest frame the proton moves
  in an orbit around the electron. To put in another
  way, a body moving through an electric field also
  “sees” a magnetic field, according to the theory of
  relativity.
• Energy of the Spin-Orbit Interaction
• Let us use the preceding idea to estimate the
  magnitude of the splitting of the hydrogen levels.
  First we must know the magnetic moment of the
  electron, which may be deduced from the magnitude
  of the splitting of the atomic beam in the Stern-
  Gerlach experiment.
• This moment is very close to
• μ= -(e/me)S.        (or μ= -(e/mec)S in cgs units)
                                       (10.6)
•     where S is the spin angular momentum of the
  electron, and e is the magnitude of the electronic
  charge. Since Sz = ±ћ /2, the z component of μ is
  (in mks units)
•       μz = ±eћ /2me = ±1 Bohr magneton
•     [The Bohr magneton is often denoted by the
  symbol μb and simply called the magnetic moment
  of the electron. Strictly speaking, μb is not the
  electron's magnetic moment; it is only one
  component of that moment. However, one
  component is all that is observed in an experimental
  measurement of the magnetic moment of the
  electron.]
•    To determine the energy attributable to the
  existence of the electron's magnetic moment, we
  write the energy of a magnetic moment μ in a
  magnetic field B as W = -μ• B, and from Eq.(10.6)
  we have
•           W = -(e/me)S•B        (10.7)
•     B may be related to the electron's orbital angular
  momentum in the following way:
• If the velocity of the electron is v, the velocity of
  the proton relative to the electron is -v. The
  magnetic field B produced by the proton's charge e
  is therefore
• This may be written in terms of the electron's
    angular momentum L = r × mv as
                                 
                              eL
                     B                       (10.9)
                           4r 3m 0c 2

•      The interaction energy is then, in the electron's
    rest frame
                             e2       
          W    B                 S L       (10.10)
                           4r m 0c
                              3      2



• Expression (10.10) is not the correct one to use for
  W, however, because the energy splitting seen in the
  lab is not the same as that seen in the electron's rest
  frame.
• The reason for this is rather complicated, involving a
  phenomenon known as the Thomas precession,
• which is related to the fact that the time scale is not
  the same in the laboratory as in the rest frame, of the
  electron.
• The proper relativistic treatment shows that the
  energy shift seen in the laboratory is just one-half of
  the value given in Eq.(10.10), or



• We can now estimate the value of W. We set ∣L∣≌ћ .
  For S, we know that the component in any direction,
  such as the direction of L must be ћ /2 in magnitude.
  If we then let r ≌ a0, we find W to be of the order of
  10-4 eV,
• which is of the order of magnitude required by the
  observed fine structure. We shall see in Chapter 8
  that effects of this size may be calculated by the
  methods of perturbation theory.
• It should come as no surprise to find that as a first
  approximation this energy can be assumed to be the
  expectation value of S·L, computing this value by
  using the eigenfunctions already found for the
  hydrogen atom; that is

• where u is an eigenfunction of Eq.(9.10). We
  should therefore be able to express this energy in
  terms of the quantum numbers n, l, and m for these
  eigenfunctions, if we can evaluate the S·L operator;
• that is, we need to know the result of the operation
  (S·L)opu.
• Combining Spin and Orbital Angular Momentum
• The S·L operator can be included in the operator
  for the total energy as it appears in the Schreodinger
  equation.
• The energy is then found by finding the
  eigenvalues of the Schreodinger equation in this
  form, with the variables S and L represented as
  operators. Thus
• where the function is labeled us to show that it is a
  function of the spin coordinate as well as the three
  space coordinates, because S·L must operate on it.
•    We may account for this spin dependence by
  assuming that us is some superposition of two wave
  functions ψ1 and ψ2 that differ in the value of the
  spin coordinate, as follows:
•       ψ1 = e-iωt u1(x, y, z)∣+ >s (10.14a)
•       ψ2 = e-iωt u1 (x, y, z)∣- >s (10.14b)
• where the state vectors ∣+ >s and ∣- >s are the two
  eigenfunctions introduced in Eqs.(10.1). There we
  saw that the spin operator Sz operates on these
  functions to yield the two possible eigenvalues of Sz:
•    Now we must examine the effect of all three
  components of S, as they appear in the S • L
  operator. To proceed, we must introduce the total
  angular momentum vector J = L + S.
•    Because there are no external torques on the
  system the total angular momentum is constant
  during the motion, but neither L nor S is necessarily
  constant.
• The B field caused by the electron's orbital
  motion exerts a torque on the spin magnetic moment;
• when this magnetic moment changes direction, L
  must also change direction, in order that L + S
  remain constant.
•     In quantum mechanical terms, this means that the
  wave function cannot be either of the functions of
  Eqs.(10.14), which have fixed values of S. Rather,
  the wave function must be a superposition of these
  two functions. The vector S, of course, remains
  constant in magnitude, because this is a fundamental
  property of the electron, but the direction of S is not
  determined.
•     Having introduced J, we now can express S • L
  in terms of J, L, and S, using the equation
• J2 ≡J·J = (L + S)·(L + S) = S2 + L2 + (S·L) + (L·S)
  (10.16)
•     But the L operator must commute with the S
  operator, because Lop and Sop operate on different
  coordinates. Therefore (S·L) = (L·S), and we can
  solve for S·L, obtaining
•         S·L = (J2 – S2 – L2)/2        (10.17)
•     The problem of evaluating (S·L) is now reduced
  to evaluating (J2 – S2 – L2).
•     We have already found [Eq.(6.58)] that the
  eigenvalues of L2 can be written in the form
•           L2 = l(l + 1)ћ2             (10.18)
• where l is defined as the maximum value of m, or of
  Lz / ћ (Section 2.3).
• By analog we might suspect that we could also
  write
•           S2 = s(s + 1)ћ2      (10.19)
•    and
•           J2 =j(j + l)ћ2       (10.20)
• where s is defined as the maximum value of Sz /ћ
  and j is defined as the maximum value of Jz /ћ.
• Because Sz =±ћ/2, the value of s must be l/2. This
  leads to a value for S2 that is consistent with what
  we know about orbital angular momentum:
• From the isotropy of space, Sx2 = Sy2 = Sz2 = ћ2/4,
  and by definition,
•        S2 = Sx2 + Sy2 + Sz2 = 3ћ2/4     (10.21)
  the same result that we find from Eq.(10.19).
• Notice the similarity of this discussion to the one
  that led to Eq.(6.58).
• The only difference is that Eq.(6.58) involved
  expectation values (e.g.,<Lx2>). In the case of spin,
  Sx2 has only one possible value: <Sx2> = Sx2 in all
  cases.
       §6.3 Stepping Operators and
           Eigenvalues of J2
•  The eigenvalues of J2 follow a similar pattern; we
 only have to determine the maximum value of Jz.
 We know that Jz = Lz + Sz and that Sz = ±ћ/2.
• Thus it would seem clear that the maximum value
 of Jz must equal the maximum value of Lz plus the
 maximum value of Sz.
• Apparently, therefore, j = (lћ + ћ/2)/ћ = l + 1/2, by
 definition. This is a possible value for j, but there is
 more to it.
•     The value of j is related to the eigenvalue of J2.
    which in turn is related to the value of L·S.
•     For a given value of l, there must be two possible
    values of L·S, because the spin vector always has
    two possible directions relative to any other
    direction (including the direction of the L vector).
•    One of these directions yields a positive value for
    L·S, the other a negative value of equal magnitude.
    This tells us that there must be two possible values
    of j.
•   Stepping Operators for Angular Momentum Sates
•   To put the discussion on a firmer basis, we now
    generate the eigenvalues of J2 by using the stepping-
    operator method that was introduced in Section 4.2
• for harmonic-oscillator eigenfunctions. In doing this,
  it is convenient to introduce the symbol [Lx, Ly] to
  represent the so-called commutator of Lx and Ly: the
  operator LxLy – LyLx.
• It is easy to show (Exercise 6, Chapter 6) that the
  three operators of this type obey the commutation
  relations
• [Lx, Ly] = ihLz, [Ly, Lz] = ihLx, [Lz, Lx] = ihLy
                                             (10.22)
•     Each relation can be generated from the previous
  one by the replacement of x by y, y by z, and z by x.
• We can now derive the eigenvalues of J2 by making
  two reasonable assumptions:
• 1. Similar commutation relations apply to all forms
  of angular momentum, so that [Sx, Sy]= iћSz, [Jx, Jy]
  = iћ Jz, and so on.
• 2. We can find a function that is an eigenfunction of
  both J2 and Jz, with the eigenvalue aћ2 for J2 and
  eigenvalue bћ         for Jz, where a and b are
  dimensionless.
•    Denoting this function by the symbol uab, we
  write
•          J2uab = aћ2uab (10.23)
•          Jzuab = bћuab (10.24)
•    Let us now introduce the operator J+ = Jx + iJy
  and demonstrate that the function J+uab is also an
  eigenfunction of Jz.
• Applying Jz to J+uab gives us
   JzJ+uab = Jz(Jx + iJy)uab = (JzJx + iJzJy)uab (10.25)
•    We can apply Jz directly to uab by using the
  commutation relations in the form
      JzJx = JxJz + iћJy and JzJy = JyJz - iћJx
                                                 (10.26)
•    Substituting from Eq.(10.26) into Eq.(10.25)
  produces
      JzJ+uab = (JxJz + iћJy + iJyJz +ћJx)uabv (10.27)
•    The substitution of Jzuab = bћuab from Eq.(10.24)
  now yields
       JzJ+uab = (Jxbћ + iћJy + iJybћ + ћJx)uab (10.28)
•    which reduces to
• JzJ+uab = (Jx + iJy)(bћ + ћ)uab =(b + 1)ћJ+uab (10.29)
•    Thus we have proven that the function J+uab is an
  eigenfunction of Jz, with eigenvalue (b+l)ћ.
• Following the same procedure, we could prove
  that J-uab, where J- = Jx – Jy, is an eigenfunction of Jz
  with eigenvalue (b - l)ћ.
• Therefore we can generate from uab a set of
  eigenfunctions whose eigenvalues are ћ multiplied
  by b,b ± 1, b ± 2,…… Each of these functions is
  also an eigenfunction of J2, as you may verity by
  applying the J2 operator to the function J+uab, then
  applying the commutation relations and Eqs.(10.23)
  and (10.24).
•       This set of eigenfunctions cannot be infinite,
    because we know that no eigenvalue of Jz2 can be
    greater than the eigenvalue of J2, which equals aћ2
    for every eigenfunction in the set.
•      Thus there must be a largest eigenvalue of Jz,
    which we denote as jћ. We write the eigenfunction
    corresponding to this eigenvalue as uaj; that means
    that Jzuaj = jћuaj.
•   It is clear that
•                 J+uaj = 0                  (10.30)
•    for otherwise J+uaj would be an eigenfunction of Jz
    with eigenvalue (j + 1)ћ, in contradiction to the
    definition of jћ as the largest eigenvalue.
• Let us now apply the operator (Jx - iJy) to this
  function to obtain
      (Jx - iJy)(Jx + iJy)uaj = 0      (10.31)
•    or
      [Jx2 + Jy2 +i(JxJy – JyJx)]uaj = 0 (10.32)
•     The commutation rule for Jx and Jy shows that
  this can be written as
       (Jx2 + Jy2 - ћJz)uaj = 0 (10.33)
•    or, since Jzuaj = jћuaj
       (Jx2 + Jy2 - jћ2)Uaj = 0 (10.34)
•    Adding (Jz2 +jћ2)uaj to both sides, we have
•      (Jx2 + Jy2 + Jz2)uaj = (Jz2 +jћ2)uaj (10.35)
•     which reduces to
       J2uaj = j(j + 1)ћ2uaj           (10.36)
• In summary, we have shown directly from the
  assumed commutation relations that the eigenvalues
  of J2 are j(j + 1)ћ2 and that the states with a given
  value of j form a set for which the eigenvalue of Jz is
  one of the sequence +jћ, +(j - 1)ћ,..., -jћ.
• Thus we may write Jz = mjћ, where mj has a
  maximum value of j and a minimum value of -j.
• Since the difference between any two values of mj
  must be an integer and the difference between j and
  -j is 2j, the value of j must be either an integer or a
  half-integer.
• This derivation is based solely upon the
  commutation relations; nothing in the derivation
  depends upon the nature of the particle involved. For
  a single electron, the fact that s = 1/2 makes j a half-
  integer; for a particle whose spin is an integer, j
  would also be an integer.
•     Now we must determine the general relation
  between the value of j and the value of s. We do this
  by referring to the z components of the vectors J, L,
  and S. Since J = L + S we know that Jz = Lz + Sz.
• The eigenvalue relation is mjћ = mlћ + msћ, so that
  mj = ml + ms, or
•        mj = ml ± 1/2 (10.37)
• The maximum value of mj is j. so
       mj ≤ l + 1/2                          (10.38)
•     and there must be a set of states for which mj
  takes on the values
   mj = l + 1/2, l – 1/2, l - 3/2,...,-(l + 1/2) (j = l + ½)
                                                  (10.39)
• and for every state in this set the value of j is l +
  1/2, This means that the magnitude of J is greater
  than the magnitude of L for each of these states, and
  we deduce that S·L is positive for each of these
  states.
• There must also be states for which S·L is negative;
  these form a set for which j equals l – 1/2,
• and in this set
•      mj = l – 1/2,l – 3/2,...,-(l – 1/2)    (j = l – 1/2)
                                                 (10.40)
• In both of these sets we find values of mj in the
  range l – 1/2 ≥ mj ≥ -(l – 1/2). Thus there is only one
  state for which mj = l + 1/2, only one state for which
  mj = -(l + ½), and two states for each of the other
  possible values of mj.
• Let us now examine in detail the distinction
  between two states with different j and the same mj.
• Consider the case mj = l – 1/2. This value of mj can
  be produced in two ways; either ml = l - 1 and the
  spin is positive, or ml = l and the spin is negative.
•  In Dirac notation (as presented in Section 2.4) we
 would write these states as ∣l,l – 1>∣+ >s and ∣l,l>∣->s,
 respectively.
• These two states are of course orthogonal to each
 other. Neither of them is an eigenstate of the J2
 operator, but we can form two linear combinations
 of them, such that one combination is an eigenstate
 of the J2 with eigenvalue j = l + 1/2, and the other
 combination is an eigenstate of the J2 with
 eigenvalue j = l – 1/2.
• As illustrated in the following example, S·L is
 positive when j = l + 1/2, and S·L is negative when j
 = l – 1/2.
• Example Problem 10.2 Find the eigenvalue of J2
  for each of the two independent linear combinations
  of ∣1,0>∣+ >s and ∣1,1>∣- >s that are eigenfunctions
  of J2.
• Solution. Applying the S·L operator to the general
  expression
       F = a∣1,0>∣+ >s + b∣1,1>∣- >s
  gives us the expression
∣SxLx + SyLy + SzLz[a∣1,0>∣+ >s + b∣1,1>∣- >s].

•      Using the spin operators [Eqs.(10.3)] combined
    with the operators for the components of L
    [Eqs.(7.19)] we find that the individual terms are
•    SxLx[a∣1,0>∣+ >s + b∣1,1>∣- >s]
  = ћ2/2(2)1/2[a∣1,1>∣- >s + a∣1,-1>∣- >s
    + b∣1,0>∣+ >s]
•    SyLy[a∣1,0>∣+ >s + b∣1,1>∣- >s]
  = ћ2/2(2)1/2[a∣1,1>∣- >s - a∣1,-1>∣- >s
    + b∣1,0>∣+ >s]
•    SzLz[a∣1,0>∣+ >s + b∣1,1>∣- >s
  = ћ2/2[-b∣1,1>∣- >s]
•    This becomes, after adding and grouping terms:
  S·LF = ћ2/(2)1/2[a∣1,1>∣- >s + b∣1,0>∣+ >s]
  –ћ2/2[b∣1,1>∣- >s]
        = Aћ2[a∣1,0>∣+ >s + b∣1,1>∣- >s]
• where Ah2 is the assumed eigenvalue, A being a
  dimensionless number.
•    Equating corresponding coefficients we have
     a/(2)1/2 – b/2 = Ab and b/(2)1/2 = Aa (10.41)
•    We can solve these to find two roots for A: A = +
  1/2 or A = -l.
•    Therefore the eigenvalues of S·L are + ћ2/2 and -
  ћ2. Since L2 = 2ћ2 and S2 = 3ћ2/4, J2 = L2 + S2 +
  2S·L= 15ћ2/4 (when S·L = + ћ2/2), and J2 = 3ћ2/4
  (when S·L = - ћ2).
• Notice that these values agree with the formula J2
  = j(j + 1)ћ2 with j = 3/2 in the first case and j = l/2 in
  the second case. To summarize:
• 1. There are two sets of states for a given value of l,
  when s = 1/2; these have the values of ml given in
  Eqs.(10.39) and (10.40).
•     2. There are two independent states for each
  value of mj from mj = l - (l/2) down to mj = -[l -
  (l/2)].
•     3. If one of these states is an eigenstate of J2, that
  state is not an eigenstate of either Lz or Sz, but it is a
  linear combination of the two eigenstates ∣l,ml>∣- >s
  and ∣l,ml – 1>∣+ >s (as illustrated in the preceding
  example).
•     4. In general. the quantities Lz, Sz, and J2 are not
  simultaneously observable, because the J2 operator
  does not commute with the Lz and Sz operators.
•    FIGURE 10.l (a) State in which Lz and Sz are known J2 is unknown.
    because the direction of L relative to S is unknown. (b) State in which
    J2 is known. the values of Lz and Sz are not known. because S and L
    rotate around J.
•  Figure 10.1 illustrates this point, Figure 10.1a
 shows the case in which L2, Lz, S2, and Sz are all
 known. Each of the vectors S and L lies on a
 cone( whose axis is the z axis, but the position of the
 vector on its cone is completely undetermined.
• Therefore the direction of S relative to L is
 unknown, S·L is unknown, and J2 must be unknown.
• In contrast, Figure 10.1b shows the situation in
 which J2, L2, S2, and Jz are known. In this case S·L
 must be known; thus we know that the angle
 between S and L must be fixed, making S lie on a
 cone whose axis is the direction of L.
• Observe that the triangle formed by S, L, and J is
  completely determined in size and shape, but this
  triangle rotates about the direction of J. Thus Lz and
  Sz are undetermined, but J2 and Jz are known.
• Evaluation of the Spin-Orbit Energy
• We now have the necessary information to
  evaluate the spin-orbit energy. Let us generalize
  Eq.(10.12) to the case of a nuclear charge of Ze in a
  one-electron ion. Then we have



• Using Eq.(1.17) for the operator, we have
                             Ze2
    W  S  L     u*                     ( J 2  S 2  L2 ) op udxdydz
                           16 0 r 3m 2 c 2
                                                                (10.43)

• If u is an eigenfunction of J2, S2, and L2, we may
  write
•    (J2 - S2 - L2)opu = ћ2[j(j + 1) - l(l + l) - s(s + 1)]u
                                               (10.44)
• And after setting s equal to 1/2, we have




• The angular part of this integral is equal to l,
• because of the normalization of the spherical
  harmonics.
• It is a simple matter to evaluate the radial part for
  any particular state, by plugging in the appropriate
  radial function (see Table 9.1 or Appendix G).
• Evaluating the integral for the general case is
  more difficult; the result (for l ≠ 0) is


• where a’0 is defined as a’0=a0(me+M)/M.
• It is instructive to write W in terms of the energy
  of the unperturbed state [En = -mrc2Z2α2/2n2, where
  α= e2/4πε0hc ≌ 1/137 (the finestructure constant)]:
•     Since j equals either l + 1/2 or l - 1/2, we can write
    this as two expressions:




•   Thus ∣W∣is proportional to ∣En∣α2, while ∣En∣ itself
 is proportional to mc2α2.
• Thus we could say that we have the first three
 terms in an expansion of the total energy in a power
 series in α2.
•   However, there is another effect, a relativistic
 contribution to the energy, that is comparable to the
 effect just computed.
• This effect is a consequence of the fact that the
 kinetic energy of a particle of mass m and
 momentum p is not accurately given by the
 expression p2/2m, as had been assumed in
 constructing the Schreodinger equation.
• With both effects included, the energy shift ΔE is
 given by a single formula
•   Notice that the total energy depends on n and j
  only, and not on l. (The calculation involves an
  approximation that we will discuss in Chapter 7.)
• For this reason, the labeling of states includes a
  subscript giving the value of j. For example, 2s1/2 is
  the state with n = 2, l = 0, and j = 1/2, and 2p1/2 is
  the state with n = 2, l = 1, and j = 1/2.
• Hyperfine Structure
• The proton and neutron also have intrinsic spin and
  magnetic moment. These are observable because of
  the energy shift resulting from the interaction of this
  magnetic moment and the orbital and spin magnetic
  moments of the electron.
• Although the energy involved is tiny, compared to
  the S·L effect, it has clearly been observed. The
  magnitude of this “hyperfine” energy splitting can
  be estimated theoretically by an extension of the
  methods used above.
• In general, an atom has a total angular momentum
  that can be defined as the vector F = J + I, where I
  is the internal angular momentum of the nucleus
  (analogous to S for the electron).
• In the 1H atom, I is simply the spin angular
  momentum of the proton, which is equal to the spin
  angular momentum of the electron, with two
  possible values, ±ћ/2,
• of the component along any specified direction.
• In all other nuclei, I is the vector sum of the
  angular moment of all of the constituent protons
  and neutrons. These vectors include spin angular
  momentum as well as angular momentum resulting
  from orbital motions of these particles within the
  nucleus.
• The vectors F and I obey the same rules that apply
  to all angular momentum vectors.
• The z component of the vector I is Iћ , and the
  possible values of I2 are I(I + 1)ћ2. The z component
  of the vector F is Fћ . and the possible values of F2
  are F(F + 1)ћ2.
•   The quantum number F takes on integrally spaced
  values between J + I and │J - I│. The magnetic
  energy resulting from the magnetic moment of the
  nucleus is proportional to I·J, and this energy has as
  many possible values as there are values of F—
  namely, either 2I + 1 or 2J + 1, whichever is smaller.
  (You may verify this for yourself.)
• The energy splitting is so tiny (or hyperfine)
  because the magnetic moment μp of the proton is
  very small compared to that of the electron; we find
  that
•      μp = 2.79 eћ /2Mp (whereas μB = eћ /2me >> eћ
  /2Mp)                            (10.50)
• The quantity eћ/2Mp is called the nuclear magneton
  (μN).
•     [The magnetic moment of the neutron is smaller
  than the proton's; its magnitude is 1.93 μN. You
  might wonder why a neutron, being neutral, would
  have any magnetic moment at all.
• But the neutron is not uniformly neutral; it is
  composed of other particles (three quarks) that are
  charged-two of them with a charge of +e/3, one with
  a charge of -2e/3. It is the motion and spin of these
  charges that produces the neutron's magnetic
  moment.]
• Because the nuclear magneton is so small relative
  to the Bohr magneton, hyperfine structure is much
  more difficult to resolve than fine structure
•   However, the nuclear magnetic moment is
 responsible for the splitting of the 1s level of
 hydrogen into two levels which differ in energy by
 about 6×10-6eV (considerably smaller than the fine-
 structure splitting, which is of order 10-4eV).
• The wavelength of protons of this energy is 21 cm;
 this wavelength is quite prominent in radiation from
 outer space, and it had made it possible to map the
 distribution of atomic hydrogen in our galaxy.
• The agreement between theory and experiment
 over the energy range from 10 eV to less than
  10-5eV is impressive, but even more impressive
 results are obtained
• by including effects of relativity and quantization
  of the energy levels of the electromagnetic field.
• Calculations have shown that the 2s1/2 and 2p1/2
  levels are not equal, as Eq.(10.48) implies. Instead,
  they should be split by 4.3743×10-6eV; this result
  (the Lamb shift) has been confirmed by
  experiment..

				
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