Lecture Note on Senior Laboratory Zeeman effect in Na, Cd, and Hg Masatsugu Sei Suzuki and Itsuko S. Suzuki Department of Physics, SUNY at Binghamton (Date March 13, 2011)

Abstract: In 1897, Pieter Zeeman observed the splitting of the atomic spectrum of cadmium (Cd) from one main line to three lines. Such a splitting of lines is called the normal Zeeman effect. According to the oscillation model by Hendrik Lorentz, the Zeeman splitting arises from the oscillation of charged particles in atoms. The discovery of the Zeeman effect indicates that the charged particles are electrons. In 1897 - 1899, J.J. (Joseph John) Thomson independently found the existence of electrons from his explorations on the properties of cathode rays. There are few atoms showing the normal Zeeman effect. In contrast, many atoms shows anomalous Zeeman effects. For the spectra of sodium (Na). for example, there are two D- lines (yellow) in the absence of the magnetic field. When the magnetic field is applied, each line is split into four and six lines, respectively. Although Zeeman himself observed the spectra of Na, he could not find the splitting of the D lines in the presence of the magnetic field because of the low resolution of his spectrometer. The electron configuration of Na is similar to that of hydrogen. There is one electron outside the closed shell. Instead, he chose Cd for his experiment and found the normal Zeeman effect (three lines). In the electron configuration of Cd, there are two electrons outside the closed shell. The three lines observed in Cd was successfully explained in terms of the Lorentz theory. It seems that the choice of Cd by Zeeman is fortunate to the development of atomic physics. If Zeeman found the anomalous Zeeman effect in Na by using spectrometer with much higher resolution, Lorentz might have some difficulty in explaining such a complicated phenomenon. In fact, only the quantum mechanics can explain the normal Zeeman effect, the anomalous Zeeman effect, and the Paschen-Back effect (Zeeman effect in an extremely large magnetic field). Here we note that the Fabry-Perot interferometer (which is used for the measurement of Zeeman effect in our laboratory ), designed in 1899 by C. Fabry and A. Perot, represents a significant improvement over the Michelson interferometer. In our Advanced laboratory [Senior Laboratory (Phys.427, Phys.429) and Graduate Laboratory (Phys.527)], students (both undergraduate and graduate students) are supposed to do the experiment for the Zeeman splitting of mercury (Hg), using an equipment consisting of magnetic field, Hg light source, polarizer, Fabry-Perot Etalon, CCD camera, and computer. The normal Zeeman splitting is observed in Hg. The electron configuration of Hg is similar to that of Cd, where two electrons are outside the closed shell. The introduction of such new techniques may lead to clear visualization of the Zeeman effect in the laboratory class. In this lecture note, we present both classical and quantum mechanical theories on the Zeeman effect. (1) Lorenz theory, (2) the Zeeman effect of Na using quantum mechanics, (3) the Zeeman effect of Cd and Hg. These notes will be helpful to understanding the Zeeman effect from a view point of quantum mechanics. "The atomic spectra are sort of voices which can be heard from the quantum world."

______Pieter Zeeman (25 May 1865 – 9 October 1943) was a Dutch physicist who shared the 1902 Nobel Prize in Physics with Hendrik Lorentz for his discovery of the Zeeman effect.

http://en.wikipedia.org/wiki/Pieter_Zeeman ______Hendrik Antoon Lorentz (18 July 1853 – 4 February 1928) was a Dutch physicist who shared the 1902 Nobel Prize in Physics with Pieter Zeeman for the discovery and theoretical explanation of the Zeeman effect. He also derived the transformation equations subsequently used by Albert Einstein to describe space and time. http://en.wikipedia.org/wiki/Hendrik_Lorentz

______Sir Joseph John "J. J." Thomson, OM, FRS (18 December 1856 – 30 August 1940) was a British physicist and Nobel laureate. He is credited for the discovery of the electron and of isotopes, and the invention of the mass spectrometer. Thomson was awarded the 1906 Nobel Prize in Physics for the discovery of the electron and for his work on the conduction of electricity in gases.

http://en.wikipedia.org/wiki/J._J._Thomson ______1. Normal Zeeman effect The explanation for the atomic spectra is due to the oscillation of charged particles inside atoms. These was no positive evidence that the particle should be an electron. The experimental evidence for the atomic spectra due to an electron was found by Pieter Zeeman in 1897. He observed splitting of Cd lines into three components (normal Zeeman effect) when an external magnetic field B is applied. He showed that the angular frequency of these lines are given by

   ,      ,      0 0 L 0 L ,

where 0 is the angular frequency in the absence of B. The angular frequency L is given by

eB 1     , L 2mc 2 c where - e is the charge of electron (we assume that e>0) and m is the mass of electron. In 1897, Hendrik Lorentz presented a theoretical interpretation for the observation by Zeeman. The Zeeman effect is the splitting of the energy levels of an atom when it is placed in an external magnetic field. The splitting occurs because of the interaction of the magnetic moment  of the atom with the magnetic field B slightly shifts the energy of the atomic levels by an amount,

E  μ  B .

The magnetic moment L of electron can be expressed by

B μ L   L  where L is the orbital angular momentum, B is the Bohr magneton of electron, and  is the Planck's constant,

e    B 2mc

z w0

w0 w0

Fig.1 Schematic diagram. B = 0.

z B

w0+wL B

w0-wL w0

Fig.2 Schematic diagram, where a magnetic field is applied along the z axis.

2. Oscillation model of Lorentz We consider a particular orbit, in which the particle is rotating around the z axis with angular frequency ±0, where +0 indicates the counterclockwise (CCW) rotation, and -0 indicates clockwise (CW) rotation. If we view the atoms in the z direction, we will obtain light that is circularly polarized with the electric vector rotating in the same direction as the electron.

CCW w0 CW -w0

Fig.3

If we view it normal to the z axis, the electric field that reaches us will depend only on the projection of the electron motion on the axis normal to the viewing. The projection of circular motions on such an axis is simple harmonic motion. So that the light viewed in a direction parallel to the plane of motion will be polarized in a direction normal to the direction of viewing and parallel to the plane of electron motion.

Viewed from the z axis B = 0

Circulary polarized 2 modes x-y plane  

  

w0

Fig4 Viewed from the z axis. B = 0. There are two modes (circularly polarized, xy plane).

Viewed from the x-y plane

Linearly polarized 1 mode z , 2 mode x-y plane

    

w0 Fig.5 Viewed from the x-y plane. B = 0. There are three modes at  = 0. 2 modes (linearly polalized, xy plane) and 1 mode (linearly polarized, z axis).

______(b) Magnetic field along the z axis: B ≠ 0. In the presence of an external magnetic field along the z axis, the components of the motion in the z direction are left unchanged, while the components of motion in the xy plane are altered. Those atoms with counterclockwise orbits will have their frequencies of rotation in the xy plane increased by L , while clockwise orbits will have their frequencies decreased by L , where

   c L 2 with

eB   (Larmor angular frequency). c mc

CCWw0+wL CW -w0+wL

Fig.6 Schematic diagram for the classical explanation of the normal Zeeman effect.

When viewed along the z axis (the direction of the magnetic field), the radiated line will therefore split into two lines, with opposite circular polarization. The electrons moving in the z direction cannot radiate in the z direction; as a result, there will be no undeviated lines in the light emitted in the z direction.

Viewed from the z axis

No linearly polarized z Circulary polarized x-y Circulary polarized x-y

     

w0-wL w0 w0+wL

Fig.7 Viewed from the z axis. B ≠ 0.

If the atom is viewed normal to the z direction, then there will be undeviated line, produced by electrons which move in the z direction. This will be polarized in the z direction. The components of electron motion in the xy plane will produce two deviated lines, each linearly polarized in a direction normal to z.

Viewed from the xy plane

Linearly polarized xy Linearly polarized xy Linearly polarized z       w0-wL w0 w0+wL

Fig8 Viewed from the xy plane.

3. Rotating reference frame y

B P x2

e2 e2' C x1' D e1'

x2' q x1 Oxe1 A

Fig.9 x1 = m1. x2 = m2. x1' = m1'. x2 = m2'. Note that OP is fixed. The relation

between {e1, e2} and {e1', e2'}.

We consider the two coordinate systems.

xe1  ye2  Xe1'Ye2 ',

with

x  (Xe 'Ye ') e  X cos Y sin 1 2 1

y  (Xe1'Ye2 ') e2  X sin  Y cos

where  is the angle between e1 and e1'. In the matrix form, we get

 x cos  sin 0 X        y   sin cos 0 Y  ,       z   0 0 1 Z 

Now we consider the equation of motion for electron (mass m and charge -e),

2 e mr  m0 r  (r  B) c or

2 e 2 eB 2 r  0 r  (r  B)  0 r  (r  zˆ)  0 r  c (r  zˆ) mc mc or

2 x 0 x  c y 2 y  0 y  c x 2 z 0 z  0

This equation can be rewritten as

0  sin[(  2)X { 2  ( )}Y  Y X}] c 0 c 2  cos[Y(c  2) {0  (c )}X  X Y}]

0  cos[(  2)X { 2  ( )}Y  Y X}] c 0 c 2  sin[Y(c  2) {0  (c )}X  X Y}] in the new reference frame (X, Y). Here we assume that

  t , where  is constant,    , and   0 Then we have

0  sin[(  2)X { 2  (  )}Y  Y}] c 0 c  cos[Y(  2) { 2  (  )}X  X}]  c 0 c  , and

0  cos[(  2)X { 2  (  )}Y  Y}] c 0 c  sin[Y(  2) { 2  (  )}X  X}]  c 0 c  .

When we put

   c   2 L we get 1 0  sin[( 2   2 )Y  Y] 0 4 c 1  cos[( 2   2 )X  X] 0 4 c and

1 0  cos[( 2   2 )Y  Y] 0 4 c

2 1 2  sin[(0  c )X  X] 4 .

These lead to

X  2 X  0

2 Y   Y  0 , where

 2     2  c   . 0 4 0

 2 since  2  c . The solution of these equation, defined by 0 4

d 2 (X  iY )   2 (X  iY ) , dt 2 0 can be obtained as

X iY C ei0t C ei0t   1  2 .

The time dependence of z is given by

z C ei0t C ei0t  1  2 .

When

X iY ei0t     , we have

X  cos( t)  0 Y  sin( t)  0 .

When

X iY ei0t     , we have

X  cos( t)  cos( t)  0 0 Y  sin( t)  sin( t)  0 0 .

These results indicates that the electron motion of the (X, Y) plane is exactly the same as that in the absence of the magnetic field (This is called the Larmor's theorem).

((Larmor's theorem)) The theorem that for a system of charged particles, all having the same ratio of charge to mass, moving in a central field of force, the motion in a uniform magnetic field B is, to first order in B, the same as a possible motion in the absence of B except for the superposition of a common precession of angular frequency equal to the Larmor frequency. Y

C.C.W.

X

C.W.

Fig.10 ParametricPlot of {X-, Y-} and {X+, Y+}, where p is chosen appropriately.

4. Linear combination of modes

  x  X cos( c t)  Y sin( c t)   2  2    cos( t)cos( c t)  sin( t)sin( c t) 0 2 0 2   cos( t  c t)  cos( t   t) 0 2 0 L   y  X sin( c t)  Y cos( c t)   2  2    cos( t)sin( c t)  sin( t)cos( c t) 0 2 0 2

 sin(0t  Lt) and

  x  X cos( c t)  Y sin( c t)   2  2    cos( t)cos( c t)  sin( t)sin( c t) 0 2 0 2   cos( t  c t)  cos( t   t) 0 2 0 L

  y  X sin( c t)  Y cos( c t)   2  2    cos( t)sin( c t)  sin( t)cos( c t) 0 2 0 2   sin( t  c t)  sin( t   t) 0 2 0 L

Then we have

1 1 x  (x  x )  [cos( t   t)  cos( t   t)]  cos( t)cos( t) 2   2 0 L 0 L 0 L

1 1 y  (y  y )  [sin( t   t)  sin( t   t)]  sin( t)cos( t) 2   2 0 L 0 L 0 L

1 x  cos[(   )t]  cos[(1 ) ]  0 L p

1 x  cos[(   )t]  cos[(1 ) ]  0 L p 1 y  sin[(   )t]  sin[(1 ) ]  0 L p 1 y  sin[(   )t]  sin[(1 ) ]  0 L p

For simplicity we use a parameter p defined by

 p  0 . L

5. Simulation (Mathematica)

1.5

1.0

0.5

-1.5 -1.0 -0.5 0.5 1.0 1.5 2.0

-0.5

-1.0

-1.5

Fig.11 (a) Plot of {x, y} with p = 5. Moving transversely to the field, the x and y motions will take the form of rosettes. This is the same type motion as one encounters in the Faraday effect.

2

1

-2 -1 1 2

-1

-2

Fig.11 (b) Plot of {x,y} with p = 10.

2

1

-2 -1 1 2

-1

-2

Fig.11 (c) p = 20

2

1

-2 -1 1 2

-1

-2

Fig.11 (d) p = 50 2

1

-2 -1 1 2

-1

-2

Fig.11 (e) p = 100

6. Physics constants for the calculation of the Zeeman effect. cgs units (from NIST Fundamental Physical constants) http://physics.nist.gov/cuu/Constants/

-21 Bohr magneton: B = 9.27400915 x 10 emu; emu = erg/Oe.

Speed of light: c = 2.99792 x 1010 cm/s

-27 Planck's constant  = 1.054571628 x 10 erg s.

-28 mass of electron me = 9.10938215 x 10 g charge of electron e = 4.80321 x 10-10 esu eV eV = 1.60218 x 10-12 erg

______ B = 5.78838 x 10-9 (1/Oe) eV

 B  4.66865105 [1/(Oe cm)] 2c

7. Separation of 3p and 3s energy levels in Na Spin-orbit interaction 2 3 P3 2 2 n=3, { = 1, s = 1 2 xÑ { 2 3P  2 xÑ {+1 2 2  3 P1 2

  

n=3, { = 0, s = 1 2 3S Fig.12 Energy levels of Na with and without spin-orbit interaction. The 3P level is 2 slightly different from the 3S level. The 3P level is split into 3 P3/2 (4 2 degeneracies) and 3 P1/2 (3 degeneracies) due to the spin-orbit interaction.

The well known bright doublet which is responsible for the bright yellow light from a sodium lamp may be used to demonstrate several of the influences which cause splitting of the emission lines of atomic spectra. The transition which gives rise to the doublet is from the 3p to the 3s level, levels which would be the same in the hydrogen atom. The fact that the 3s (orbital quantum number l = 0) is lower than the 3p (l = 1) is a good example of the dependence of atomic energy levels on angular momentum. The 3s electron penetrates the 1s shell more and is less effectively shielded than the 3p electron, so the 3s level is lower (more tightly bound). The fact that there is a doublet shows the smaller dependence of the atomic energy levels on the total angular momentum . The 3p level is split into states with total angular momentum j = 3/2 and j = 1/2 by the magnetic energy of the electron spin in the presence of the internal magnetic field caused by the orbital motion. This effect is called the spin-orbit effect. In the presence of an additional externally applied magnetic field, these levels are further split by the magnetic interaction, showing dependence of the energies on the z-component of the total angular momentum. This splitting gives the Zeeman effect for sodium. When the wavefunctions for electrons with different orbital quantum numbers are examined, it is found that there is a different amount of penetration into the region occupied by the 1s electrons. This penetration of the shielding 1s electrons exposes them to more of the influence of the nucleus and causes them to be more tightly bound, lowering their associated energy states. In the case of Na with two filled shells, the 3s electron penetrates the inner shielding shells more than the 3p and is significantly lower in energy.

2 2 2 8. The states of 3 S1/2 , 3 P3/2, and 3 P1/2 in Na The electron configuration of Na is given by

Na: (1s)2(2s)2(2p)6(3s)

The inner 10 electrons can be visualized to form a spherically symmetrical electron cloud. We are interested in the excitation of the 11-th electron from 3s to a possible higher state.

(a) The 3s state For the electron with 3s state (l = 0, s = 1/2),

D0 x D1/2 = D1/2

2 2 Thus we have j = 1/2 (the degeneracy 2). The state is described by 3 S1/2 (or simply S1/2). 2S+1 Here we use the notation of n Lj where n is the principal quantum number, s is the spin number, l is the orbital angular momentum, and j is the resultant angular momentum.

j 1/ 2,m 1/ 2  ml  0,ms 1/ 2

j 1/ 2,m  1/ 2  ml  0,ms  1/ 2

ms 1 j= 2

1 1 1 1 1 -1, > j= ,m= >= 0, > 1, > 2 2 2 2 2

   

ml

-1 2 1 1 1 1 1 -1,- > j= ,m=- >= 0,- > j=-1,- > 2 2 2 2 2     

1 Fig.13 j  ,m . (l = 0). m = 1/2. m = -1/2. The recursion relation to obtain the 2 Clebsch-Gordan coefficients.

(b) The 3p state For the electron with 3p state (l = 1, s = 1/2), we have

D1 x D1/2 = D3/2 + D1/2

Thus we have j = 3/2 (the degeneracy 4) and j = 1/2 (degeneracy 2). These states are 2 2 described by 3 P3/2 (j = 3/2) and 3 P1/2 (j = 1/2).

Note that ms 3 j= 2 1 1 1 -1, > 0, > 1, > 2 2 2 3 3 j= ,m= > 2 2   

 3 1 3 1 ml j= ,m=- > j= ,m= > 2 2 2 2 3 3 j= ,m=- > 2 2  -1 2  1 1 1 -1,- > 0,- > 1,- > 2 2 2     

3 Fig.14 j  ,m . (l = 1). The recursion relation to obtain the Clebsch-Gordan 2 coefficients.

The Clebsch-Gordon coefficients can be calculated using the Mathematica.

2 (i) For j = 3/2 (3 P3/2),

j  3/ 2,m  3/ 2  ml  1,ms  1/ 2

2 1 j  3/ 2,m  1/ 2  m  0,m  1/ 2  m  1,m 1/ 2 3 l s 3 l s

1 2 j  3/ 2,m 1/ 2  m 1,m  1/ 2  m  0,m 1/ 2 3 l s 3 l s

j  3/2,m  3/2  ml  1, ms  1/2

______

ms 1 j= 2

1 1 1 -1, > 0, > 1, > 2 2 2

  

1 1 1 1 ml j= ,m=- > j= ,m= > 2 2 2 2

  -1 2 1 1 1 -1,- > 0,- > -1,- > 2 2 2    

1 Fig.15 j  ,m . l = 1. The recursion relation to obtain the Clebsch-Gordon 2 coefficients.

2 (ii) For j = 1/2 (3 P1/2),

1 2 j 1/ 2,m  1/ 2  m  0,m  1/ 2  m  1,m 1/ 2 3 l s 3 l s

2 1 j 1/ 2,m 1/ 2  m 1,m  1/ 2  m  0,m 1/ 2 3 l s 3 l s

(c) g factors

The Lande g-factor is defined by

3 s(s 1)  l(l 1) g   J 2 2 j( j 1)

(see the Appendix)

Table

Term j l S gJ

2 3 P3/2 3/2 1 1/2 4/3

2 3 P1/2 1/2 1 1/2 2/3

2 3 S1/2 1/2 0 1/2 2

______

m=3 2 2 m=1 2 3 P3 2 m=-1 2 m=-3 2    2 m=1 2 3 P1 2 m=-12

  

Dm=-1001 -1-10011

2 m=1 2 3 S1 2 m=-1 2

Fig.16 Schematic diagram of energy levels in Na (n = 3) with and without magnetic  field B. The splitting of energy levels occurs due to the spin-orbit interaction and the Zeeman effect.

The wavelength of sodium D lines is given by

3 3  = 589.6 nm (3 P1/2 - 3 S1/2).

3 3  = 589.0 nm (3 P3/2 - 3 S1/2).

The sodium D lines correspond to the 3p  3s transition. In the absence of a magnetic field 2 2 B, the spin orbit interaction splits the upper 3p state into 3 P3/2 and 3 P1/2 terms separated -1 2 by 17 cm . The lower 3 S1/2 has no spin-orbit interaction.

9. Quantum mechanics for system with one electron (a) Spin orbit interaction The spin-orbit interaction serves to remove the l degeneracy of the eigenenergies of hydrogen atom. If the spin-orbit interaction is neglected, energies are dependent only on n (principal quantum number). In the presence of spin-orbit interaction (n, l, s = 1/2; j, m) are good quantum numbers. Energies are dependent only on (n, l, j). We introduce a new Hamiltonian given by

Hˆ Hˆ Hˆ  0  LS ,

Jˆ  Lˆ  Sˆ ,

1 Hˆ  Lˆ  Sˆ   (Jˆ 2  Lˆ2  Sˆ 2 ) , LS 2

ˆ ˆ ˆ The Hamiltonian H0 commutes with all the components of L and S .

[Hˆ , Lˆ 2 ] [Hˆ ,Lˆ 2 ] [Hˆ , Lˆ ] 0  0 z  0 z ,

ˆ ˆ 2 ˆ ˆ 2 ˆ ˆ [H 0, S ]  [H 0, S z ]  [H 0,S z ],

ˆ ˆ ˆ [H 0, H LS ]  0 , or

ˆ ˆ ˆ [H , H 0 ]  0 ,

ˆ ˆ ˆ [H,Sz ]  0 , and

ˆ ˆ ˆ [H, Lz ]  0 .

Thus we conclude that

ˆ ˆ2 ˆ 2 ˆ2  is the simultaneous eigenket of the mutually commuting observables{ H0 , L , S , J , ˆ and J z }.

ˆ (0) H0   En  ,

ˆ2 2 L    l(l 1) ,

ˆ 2 2 S    s(s 1) ,

ˆ 2 J    j( j 1) ,

ˆ J z   m .

We consider the hydrogen atom problem with the spin orbit interaction. The eigenket  is expressed by n,l,s; j,m . The angular part of this eigenket is

j,m  l,s; j,m , with

s = 1/2, j = l ±1/2, where

Dl  D1/ 2  D 1  D 1 . j l  j l  2 2

The spin-orbit interaction is given by

 H  Lˆ  Sˆ  [Jˆ 2  Lˆ2  Sˆ 2 ], LS 2

ˆ ˆ H LS j,m  L  S j,m   [Jˆ 2  Lˆ2  Sˆ 2 ] j,m 2   [Jˆ 2  Lˆ2  Sˆ 2 ] j,m 2   2[ j( j 1)  l(l 1)  3/ 4] j,m 2 

(a) For j = l +1/2,

J- m-1 2,1 2 m+1 2,1 2

       

m

 

m-1 2,-1 2 m+1 2,-1 2

1      Fig.17 j  l  ,m . The recursion relation to obtain the Clebsch-Gordan 2 coefficients.

ˆ H LS j  l 1/ 2,m  ELS j  l 1/ 2,m , with

  E  2[(l 1/ 2)(l  3/ 2)  l(l 1)  3/ 4]  2l . LS 2  2 

We note that

l  m 1/ 2 j  l 1/ 2,m  m  m 1/ 2,m 1/ 2 2l 1 l s

l  m 1/ 2  m  m 1/ 2,m  1/ 2 2l 1 l s

ˆ ˆ The expectation values of Lz and S z are obtained as follows.

l  m 1/ 2 Lˆ j  l 1/ 2,m  (m 1/ 2) m  m 1/ 2,m  1/ 2 z  2l 1 l s

l  m 1/ 2  (m 1/ 2) m  m 1/ 2,m  1/ 2  2l 1 l s

 l  m 1/ 2   (m 1/ 2)  l  m 1/ 2 l  m 1/ 2  ˆ   2l 1  j  l 1/ 2,m Lz j  l 1/ 2,m    2l 1 2l 1  l  m 1/ 2    (m 1/ 2)  2l 1  l  m 1/ 2 l  m 1/ 2    (m 1/ 2)  (m 1/ 2)  2l 1 2l 1  2lm   2l 1

Similarly,

l  m 1/ 2 Sˆ j  l 1/ 2,m  (1/ 2) m  m 1/ 2,m 1/ 2 z  2l 1 l s

l  m 1/ 2   (1/ 2) m  m 1/ 2,m  1/ 2  2l 1 l s

l  m 1/ 2 l  m 1/ 2  j  l 1/ 2,m Sˆ j  l 1/ 2,m  (1/ 2)  (1/ 2) z  2l 1 2l 1    m   2l 1

Then we have

1 j  l 1/ 2,m Lˆ  2Sˆ j  l 1/ 2,m  m(1 ) z z  2l 1

______(b) For j = l -1/2

J- m-1 2,1 2 m+1 2,1 2

       

m

 

m-1 2,-1 2 m+1 2,-1 2

1       Fig.18 J  l  . The recursion relation to obtain the Clebsch-Gordan coefficients. 2

ˆ H LS j  l 1/ 2,m  ELS j  l 1/ 2,m , with

 2 1 1 3  2 ELS   [(l  )(l  )  l(l 1)  ]   (l 1),  2 2 2 4 2

l  m 1/ 2 l  m 1/ 2 j  l 1/ 2,m   m  m 1/ 2,m  1/ 2  m  m 1/ 2,m  1/ 2 2l 1 l s 2l 1 l s or

 Hˆ j  l 1/ 2,m   2 (l 1) j  l 1/ 2,m . LS 2 

ˆ ˆ The expectation value of Lz and S z

l  m 1/ 2 l  m 1/ 2  j  i 1/ 2,m Lˆ j  i 1/ 2,m  (m 1/ 2)  (m 1/ 2) z  2l 1 2l 1    2 m(l 1)   2l 1 l  m 1/ 2 l  m 1/ 2  j  l 1/ 2,m Sˆ j  l 1/ 2,m  (1/ 2)  (1/ 2) z  2l 1 2l 1    m    2l 1

Then we have

1 j  l 1/ 2,m Lˆ  2Sˆ j  l 1/ 2,m  m(1 ) . z z  2l 1

In summary, the energy shift due to the spin-orbit interaction is given by

 2 E  ELS   l , for j = l+1/2,  2

 2 E  ELS   (l  1) , for j = l-1/2,  2

 2    l 0  Hˆ  2  LS     0  2(l 1)  2    under the basis of j  l 1/ 2,m and j  l 1/ 2,m , where

2 e4 Z2 (Z)2 1    , E c2n 2 2l(l 1/2)(l 1) n2 l(l 1/2)(l 1)  n  or

Z 4 1 2  R 2 .  n 4 l(l 1/2)(l  1) 

(b) Zeeman effect in a system with a single electron such as Na The magnetic moment is given by a sum of the orbital magnetic moment and spin magnetic moment as

 μˆ   B (Lˆ  2Sˆ) , 

e where    (>0) is the Bohr magneton. The Zeeman energy is given by B 2m c  e

ˆ B ˆ ˆ B ˆ ˆ B B ˆ ˆ H B  μˆ B  (L  2S) B  (J  S)  B  (J z  Sz ) ,    for B//z. We now calculate

ˆ B B ˆ ˆ l  m 1/ 2 H B j  l 1/ 2,m  (Lz  2Sz ) ml  m 1/ 2,ms  1/ 2  2l 1 B B ˆ ˆ l  m 1/ 2  (Lz  2Sz ) ml  m 1/ 2,ms  1/ 2  2l 1 or

ˆ 1 l  m 1/ 2 H B j  l 1/ 2, m  B B(m  ) ml  m 1/ 2, ms  1/ 2 2 2l 1 1 l  m 1/ 2   B(m  ) m  m 1/ 2, m  1/ 2 B 2 2l 1 l s

Noting that

Hˆ j  l 1/ 2,m  E j  l 1/ 2,m LS LS , we have

ˆ ˆ  2 1 l  m 1/ 2 (H LS  H B ) j  l 1/ 2, m  [ l  B B(m  )] ] ml  m 1/ 2, ms  1/ 2 2 2 2l 1  1 l  m 1/ 2  [ l 2   B(m  )] m  m 1/ 2, m  1/ 2 2  B 2 2l 1 l s

Similarly,

ˆ 1 B B ˆ ˆ l  m 1/ 2 H B j  l  ,m   (Lz  2S z ) ml  m 1/ 2,ms  1/ 2 2  2l 1

B B ˆ ˆ l  m 1/ 2  (Lz  2Sz ) ml  m 1/ 2,ms  1/ 2  2l 1 or

ˆ 1 1 l  m 1/ 2 H B j  l  ,m  B B(m  ) ml  m 1/ 2,ms  1/ 2 2 2 2l 1 1 l  m 1/ 2   B(m  ) m  m 1/ 2,m  1/ 2 B 2 2l 1 l s or

ˆ ˆ  2 1 l  m 1/ 2 (H LS  H B ) j  l 1/ 2,m  [ (l 1)  B B(m  )] ml  m 1/ 2,ms  1/ 2 2 2 2l 1  1 l  m 1/ 2 [ (l 1) 2   B(m  )] m  m 1/ 2,m  1/ 2 2  B 2 2l 1 l s

Note that

1 l  m 1/ 2 l  m 1/ 2 j  l  ,m  m  m 1/ 2,m  1/ 2  m  m 1/ 2,m  1/ 2 2 2l 1 l s 2l 1 l s

1 l  m 1/ 2 l  m 1/ 2 j  l  ,m   m  m 1/ 2,m  1/ 2  m  m 1/ 2,m  1/ 2 2 2l 1 l s 2l 1 l s

ˆ ˆ Then the matrix elements of H LS  H B in the basis { j  l 1/ 2,m , j  l 1/ 2,m } can be obtained as

j  l 1/ 2,m j  l 1/ 2,m 1  1  B (l 1/ 2)2  m2 j  l  ,m l 2   Bm(1 )  B 2 2  B 2l 1 2l 1 1  B (l 1/ 2)2  m2  1 j  l  ,m  B  (l 1) 2   Bm(1 ) 2 2l 1 2  B 2l 1 for the same l and m. The eigenvalues of this matrix are given by

 2 1    Bm    4 2B2  8 Bm 2   2 (2l 1)2 4 , 1 B 4 4 B B  

 2 1    Bm    4 2B2  8 Bm 2   2 (2l 1)2 4 . 2 B 4 4 B B  

______10. Zeeman effect in Na We now consider the D lines of Na.

1 m=3 2 2 m=1 2 3 P3 2 m=-1 2 1 m=-3 2 22    2 m=1 2 3 P1 2 m=-12

  

2 1 m=1 2 3 S1 2 1 m=-1 2

  Fig.19 Energy levels for Na in the presence of weak magnetic field. In the presence 2 of a strong magnetic field, the states j  3/ 2,m  1/ 2 in 3 P3/ 2 2 j  1/ 2,m  1/ 2 in 3 P1/ 2 are no loner eigenstate. The appropriate linear 2 combination of j  3/ 2,m  1/ 2 in 3 P3/ 2 and j  1/ 2,m  1/ 2 in 3 2 ˆ ˆ P1/ 2 for the same m (= ±1/2) becomes eigenstates of H LS  H B .

(a) l = 1

2 2 The mixed state of 3 P3 / 2 (j = 3/2, l = 1, s = 1/2), and3 P1/ 2 (j = 1/2, l = 1, s = 1/2) is the eigenstate.

3 1 j  ,m j  ,m 2 2 9  B  m2 3  4 B j  ,m 2   Bm  4 2 2  B 3 3 9  B  m2 1 B 2 j  ,m  4  2   Bm 2 3  B 3

with m = ±1/2, where the factors 4/3 and 2/3 of the Zeeman terms correspond to the g 2 2 factors for P3 / 2 and P1/ 2 , respectively. The eigenvalue can be obtained from the eigenvalue problem for the (2 x 2) matrix. We get

 2 1 E (m)  E(3p)   Bm    4 2B2  8 Bm 2  9 2 4 , 2 B 4 4 B B   for m = ±1/2, and

 2 1 E (m)  E(3p)   Bm    4 2B2  8 Bm 2  9 2 4 , 4 B 4 4 B B  

3 3 for m = ±1/2. We note that j  ,m   is the eigenket of Hˆ  Hˆ , where 2 2 LS B

3 3  3 3 (Hˆ  Hˆ ) j  ,m   ( 2  2 B) j  ,m  , LS B 2 2 2  B 2 2 with

 E  E(3p)  2  2 B , 5 2  B for m = 3/2, and

ˆ ˆ 3 3  2 3 3 (H LS  H B ) j  ,m    (   2B B) j  ,m   2 2 2 2 2 , with

 E  E(3p)  2  2 B , 3 2  B for m = -3/2.

(b) j = 1/2, l = 0, s = 1/2

1 j  1/ 2,m   is the eigenket of Hˆ  Hˆ , 2 LS B with the eigenvalue

E (m)  E(3s)  2 Bm 1 B ,

1 for m   . 2

11. Paschen-Back effect in Na The Paschen-Back effect is the splitting of atomic energy levels in the presence of a strong magnetic field. This effect is the strong-field limit of the Zeeman effect. The effect was named after the German physicists Friedrich Paschen and Ernst E. A. Back. In the limit of strong magnetic field B, the energy levels of Na are strongly dependent on the magnetic field, and are given by

1 E (m  )  E(3s)   B , 1 2 B

1 E (m   )  E(3s)   B , 1 2 B

1  2 E (m  )  E(3p)   , 2 2 4

1  2 E (m   )  E(3p)     B , 2 2 4 B

 E  E(3p)  2  2 B , 3 2  B

2 1  E4 (m  )  E(3p)   B B , 2 4

1  2 E (m   )  E(3p)   , 4 2 4

 E  E(3p)  2  2 B . 5 2  B

((Mathematica)) We use B = 1,   1 E(3p) - E(3s)= -7.  = 1. for the calculation using the Mathematica.

Energy levels

5 E5

E4 1 2

E 3p E4 -1 2 B 0.5 1.0 1.5 E 2.01 2 2.5 3.0 spin-orbit 2   E2 -1 2   E3   -5    E 1 2 1   E 3s

E1 -1 2    -10  

Fig.20 Splitting of energy levels of Na in a magnetic   field B (anomalous Zeeman effect). Energy levels 20

10

B 2 4 6 8 10

-10

-20

Fig.21

Energy levels 200

E5

100 E4 1 2 E1 1 2

E4 -1 2 B 20 40 60 E2 1 2 80  100    E2 -1 2 -100  E  -1 2   1    E  3   -200

Fig.22 Paschen-Back effect. Zeeman splitting in the very large magnetic field for Na. 12. Paschen-Back effect in Na; quantum mechanical treatment Suppose that an extremely strong magnetic field is applied for the case of Na. The Zeeman term of the Hamiltonian is much significant compared to the spin-orbit interaction. in this case, the Hamiltonian H is simply given by

ˆ B ˆ ˆ H B  (Lz  2S z )B  in the presence of the magnetic field along the z axis.

(1) 3p states

l = 1, s = 1/2.

Hˆ m ,m   B(m  2m ) m ,m B l s 3 p B l s l s 3 p

m ,m is the eigenket of Hˆ with the eigenvalue  B(m  2m ) . l s 3 p B B l s

m ,m (m = 1, 0, -1, m = 1/2, -1/2). l s 3 p l s

(2) 3s states

l = 0, s = 1/2.

Hˆ m ,m   B(m  2m ) m ,m B l s 3s B l s l s 3s

m ,m is the eigenket of Hˆ with the eigenvalue  B(m  2m ) . l s 3s B B l s

m ,m (m = 0, m = 1/2, -1/2). l s 3s l s

In the extremely high magnetic fields, the energy levels of 3s and 3p states split into five levels. The difference between adjacent energy levels is the same and is equal to BB. ______Table

Energy (BB) Eigenkets 2 1,1/ 2 3 p 1 0,1/ 2 0,1/ 2 3 p 3s 0 1,1/ 2 1,1/ 2 3 p 3 p -1 0,1/ 2 0,1/ 2 3 p 3s -2 1,1/ 2 3 p

1 1, >3 p 2 2mBB

1 1 0, >3 p, 0, >3 s 2 2 mBB

1 1 1, - >3 p, -1, >3 p 2 2 0

1 1 0, - >3 p, 0, - >3 s, 2 2 -mBB

1  -1, -  > 2 3 p -2mBB

Fig. The level splitting of Na for the Pashen-Back effect. m ,m is the  l s 3 p eigenket for 3p states (6, states; m = 1, 0, -1, m = 1/2, -1/2). m ,m is the l s l s 3s

eigenket for 3s states (2 states; ml = 0, ms = 1/2, -1/2).

______13. Zeeman splitting in Cd The electron configuration of Cd is given by (Kr) 4d10 5s2. This is similar to the outer electron configuration of He but also of Hg.

(a) 5s5s

D0 x D0 = D0 l = 0

D1/2 x D1/2 = D1 + D0 s = 1, s = 0

l = 0 and s = 1

D0 x D1 = D1 j = 1 3 5S1

l = 0 and s = 0

D0 x D0 = D0 j = 0 1 5 S0

(b) 5s5p

D1 x D0 = D1 l = 1 D1/2 x D1/2 = D1 + D0 s = 1, s = 0

l = 1 and s = 1

D1 x D1 = D2 + D1 + D0

3 j = 2 5P2 3 j=1 5P1 3 j = 0 5P0

l = 1 and s = 0

D1 x D0 = D1 j = 1 1 5 P1 (g = 1)

(c) 5s5d

D2 x D0 = D2 l = 2 D1/2 x D1/2 = D1 + D0 s = 1, s = 0

l = 2 and s = 1

D2 x D1 = D3 + D2 + D1

3 j = 3 5D3 3 j = 2 5D2 3 j = 1 5D1

l = 2 and s = 0

D2 x D0 = D2 j = 2 1 5 D2 (g = 1)

m=2 m=1 1 5 D2 m=0 m=-1 m=-2

643.8 nm Dm=-1 Dm=0 Dm=1

m=1 1 5 P1 m=0 m=-1

1 Fig.23 Schematic diagram for the Zeeman splitting in Cd. 643.8 nm. g = 1 for 5 D2 1 (j = 2, l = 2, s = 0) and 5 P1 (j = 1, l = 1, s = 0).

We can observe the normal Zeeman effect in the red spectral line of Cd (643.8 nm). It corresponds to the transition

1 1 5 D2 (j = 2, l = 2, s = 0) → 5 P1 (j = 1, l = 1, s = 0).

1 In the presence of the magnetic field, the 5 D2 level splits into 5 Zeeman components and 1 the 5 P1 level splits into 3 Zeeman component. The optical transitions between these levels are only possible in the form of electrical dipole radiation. The following selection rules apply for the magnetic quantum number m of the states involved;

m  1 for  components,

m  0 for  components,

Thus we observe three spectral lines

______14. Energy levels in Hg: system with two electrons The neutral mercury (Hg) atom in its ground state has 80 electrons in the configuration 1s22s22p63s23p63d104s24p64d104f145s25p65d106s2 in which the n = 1, 2, 3, 4, and 5 electrons form an inert core for two 6s valence electrons. The optical emission spectrum of Hg results from transitions of the two valence electrons between various excited two-electron configurations. The Hg spectrum therefore has many features in common with the two- electron helium system.

2S 1 LJ

Orbital angular momentum

Dl1 x D1/2 = Dl1+l2 +..... + D|l1-l2|

Spin:

D1/2 x D1/2 = D1 + D0

(a) 6s6s

l = 0 and l = 0→ D0 x D0 = D0

l = 0 and s = 1

D0 x D1 = D1 3 j = 1 6 S1

l = 0 and s = 0

D0 x D0 = D0 1 j = 0 6 S0

(b) 6s6p

l = 0 and l = 1→ D0 x D1 = D1

l = 1 and s = 1

D1 x D1 = D2 + D1 + D0

3 j = 2 6 P2 (g = 3/2). 3 j = 1 6 P1 3 j = 0 6 P0

l = 1 and s = 0

D1 x D0 = D1 1 j = 1 6 P1

(c) 6s6d

l = 0 and l = 2→ D0 x D2 = D2

l = 2 and s = 1

D2 x D1 = D3 + D2 + D1

3 j = 3 6 D3 3 j = 2 6 D2 3 j = 1 6 D1

l = 2 and s = 0

D2 x D0 = D2 1 j = 2 6 D2

(d) 6s7s

l = 0 and l = 0→ D0 x D0 = D0

l = 0 and s = 1

D0 x D1 = D1

3 j = 1 7 S1 (g = 2)

l = 0 and s = 0

D0 x D0 = D0

1 j = 0 7 S0

(e) 6s7p

l = 0 and l = 1→ D0 x D1 = D1

l = 1 and s = 1

D1 x D1 = D2 + D1 + D0

3 j= 2 7 P2 3 j = 1 7 P1 3 j = 0 7 P0

l = 1 and s = 0

D1 x D0 = D1 1 j = 1 7 P1

(f) 6s7d

l = 0 and l = 2→ D0 x D2 = D2

l = 2 and s = 1

D2 x D1 = D3 + D2 + D2

3 j = 3 7 D3 3 j = 2 7 D2 3 j = 1 7 D1

l = 2 and s = 0

D2 x D0 = D2 1 j = 2 7 D1

______15. Zeeman splitting in Hg

m=1 3 7 S1 m=0 g=2 m=-1

Dm=-1 Dm=0 Dm=1

m=2 m=1 3 3 6 P2 m=0 g= m=-1 2 m=-2

Fig.24 Schematic diagram for the Zeeman splitting in Hg. 546.07 nm (Green line). 3 3 3 3 7 S1 (6s7s). 6 P2 (6s6p). 7 S1 (j = 1, l = 0, s = 1) and 6 P2 (j = 2, l = 1, s = 1).

3 3 The Hg green line corresponds to the transition from 7 S1 to 6 P2.

3 The state of the 7 S1 level is described by

j2  1,m2 (j2= 1, s2= 1, l2 = 0, g2= 2)

3 with m2 = 1, 0, and -1. The state of 6 P2 level is described by

j1  2,m1 (j1 = 2, s1= 1, l1= 1, g1= 3/2)

with m1 = 2, 1, 0, -1, and -2. According to the selection rule (m = 1, 0, and -1), there are 9 lines.

m = ±1 (6 lines):  lines.

m = 0 (3 lines):  lines.

16. Evaluation of observed wavelenghts

3 E2  E20  g2Bm2 B , for 7 S1 level

3 E1  E10  g2Bm1B for 6 P2 level

The energy separation is given by

0 E12  E2  E1  E12  B B(m2 g2  m1g1)

with

m  m2  m1 = -1, 0, or 1.

m2 =1, 0, -1. m1 = 2, 1, 0, -1, -2.

g2=2 g1 = 3/2

Here we note that

c 0 c E12  h , E12  h 0 12 12

Then we have

0 1 1 12  12 1 (  0 )  0  B B(m2 g2  m1g1) 12 12 1212 2c or

 0     1 12 12  12   B(m g  m g )   0 0 2 2c B 2 2 1 1 12 12 12  or

 1 3 12   B(2m  m ) 0 2 2c B 2 1 2 12  where

1 5 -1 -1 B  4.6686510 (Oe cm ) 2c and

0 12  = 546.07 nm (Green)

We note that

3 f(m1, m2) = (2m  m ) 2 1 2 takes discrete values of 3/2, 1, 1/2, 0, -1/2, -1, and -3/2. ______REFERENCES E.U. Condon and G.H. Shortley, The Theory of Atomic Spectra (Cambridge University Press, Cambridge, 1991). H.E. White, Introduction to Atomic Spectra, McGraw-Hill, New York, 1934). G. Herzberg, Atomic Spectra & Atomic Structure (Dover Publications, New York, 1944). D. Bohm, Quantum Theory (Englewood Cliffs, NJ, Prentice Hall, 1951). J.J. Sakurai, Modern Quantum Mechanics, Revised Edition (Addison-Wesley, Reading, MA, 1994). D.H. Goldstein and E. Collett, Polarized light, 2nd-edition (Marcel Dekker, Inc., 1993, New York). p.353.

J.F. Mulligan, :Who are Fabry and Perot?",Am. J. Phys. 66, 797 (1998). ______APPENDIX A. Fundamentals of quantum mechanics A1. Orbital angular momentum and its magnetic moment of one electron

Fig.A1 Orbital (circular) motion of electron with mass m and a charge –e. The direction of orbital angular momentum L is perpendicular to the plane of the motion (x-y plane).

The orbital angular momentum of an electron (charge –e and mass m) L is defined by

L  r  p  r  (mv) , or Lz  mvr .

According to the de Broglie relation, we have

h p(2r)  2r  nh , 

h where p (= mv) is the momentum ( p  ), n is integer, h is the Planck constant, and  is  the wavelength.

Fig.A2 Acceptable wave on the ring (circular orbit). The circumference should be equal to the integer n (=1, 2, 3,…) times the de Broglie wavelength . The picture of fitting the de Broglie waves onto a circle makes clear the reason why the orbital angular momentum is quantized. Only integral numbers of wavelengths can be fitted. Otherwise, there would be destructive interference between waves on successive cycles of the ring.

Then the angular momentum Lz is described by

nh L  pr  mvr   n . z 2 

The magnetic moment of the electron is given by

1   I A, z c 

2 where c is the velocity of light, A = r is the area of the electron orbit, and I is the current due to the circular motion of the electron. Note that the direction of the current is opposite to that of the velocity because of the negative charge of the electron. The current I is given by

e e ev I       ,  T (2r / v) 2r where T is the period of the circular motion. Then the magnetic moment is derived as

1 evr e e Lz B z  I A     Lz     Lz (e > 0), c 2c 2mc 2mc  

e -21 where B (= ) is the Bohr magneton. B = 9.27400915 x 10 emu. emu=erg/Oe. Since 2mc

Lz  n , the magnitude of orbital magnetic moment is nB.

A.2 Spin angular momentum and its magnetic moment The spin magnetic moment is given by

2B μS   S ,  where S is the spin angular momentum. In quantum mechanics, the above equation is described by

2 ˆ   B Sˆ ,  using operators (Dirac). When Sˆ   ˆ , we have ˆ   ˆ . The spin angular momentum 2 B is described by the Pauli matrices (operators)

Sˆ   ˆ , Sˆ   ˆ Sˆ   ˆ . x 2 x y 2 y z 2 z

Using the basis,

1 0    ,     . 0 1 we have

0 1 0  i 1 0  ˆ ˆ ˆ  x    , y    , z   . 1 0  i 0  0 1

The commutation relations are valid;

[ˆ x ,ˆ y ]  2iˆ z , [ˆ y ,ˆ z ]  2iˆ x , [ˆ z ,ˆ x ]  2iˆ y .

The resultant magnetic moment of an electron is given by

 μ   B (L  2S) . 

A.3. Magnetic moment of atom We consider an isolated atom with incomplete shell of electrons. The orbital angular momentum L and spin angular momentum S are given by

L  L1  L2  L3  ... , S  S1  S2  S3  ...

The total angular momentum J is defined by

J  L  S .

The total magnetic moment  is given by

 μ   B (L  2S) . 

The Landé g-factor is defined by

g J B μ J   J ,  where

Fig.A3 Basic classical vector model of orbital angular momentum (L), spin

angular momentum (S), orbital magnetic moment (L), and spin magnetic

moment (S). J (= L + S) is the total angular momentum. J is the

component of the total magnetic moment (L + S) along the direction (-J).

Suppose that

L  aJ  L  and S  bJ  S   where a and b are constants, and the vectors S and L are perpendicular to J.

Here we have the relation a  b 1, and L  S  0. The values of a and b are determined as follows.

J  L J S a  , b  . J 2 J 2

Here we note that J 2  L2  S2 J 2  L2  S2 J S  (L  S) S  S2  L S  S2   , 2 2 or

J 2  L2  S2 2 J S    [J (J 1)  L(L 1)  S(S 1)] , 2 2 using the average in quantum mechanics. The total magnetic moment  is

B B μ   (L  2S)   [(a  2b)J  (L  2S )] .   Thus we have

B B g J B μ J   (a  2b)J   (1 b)J   J ,    with

J S 3 S(S 1)  L(L 1) g  1 b  1   . J J 2 2 2J (J 1)

((Note)) The spin component is given by

S  bJ  S  (g J 1)J  S ,

with b  g J 1 . The de Gennes factor is defined by

2 2 (gJ 1) J 2 2  (gJ 1) J (J 1) .  In ions with strong spin-orbit coupling the spin is not a good quantum number, but rather the total angular momentum , J = L+S. The spin operator is described by

S  (g J 1)J .

A.4 Spin-orbit interaction in an electron around the nucleus The electron has an orbital motion around the nucleus. This also implies that the nucleus has an orbital motion around the electron. The motion of nucleus produces an orbital current. From the Biot-Savart’s law, it generates a magnetic field on the electron.

Fig.A4 Simple model for the spin-orbit interaction. The orbital current due to the

circular motion of the nucleus (with velocity vN and charge Ze) produces an magnetic field at the center where the electron is located.

The current I due to the movement of nucleus (charge Ze, e>0) is given by

Idl  Zev N ,

dl where v is the velocity of the nucleus and  v . Note that N dt N

q dl Idl  dl  q  Zev . t dt N

The effective magnetic field at the electron at the origin is

I dl  r1 Beff  3 , v N  ve , c r1 where v is the velocity of the electron. Then we have

Ze v N  r1 Ze ve  r1 Beff  3  3 . c r1 c r1

Since r1  r , Beff can be rewritten as

Ze v N r1 Ze ve  r1 Beff  3  3 , c r1 c r1 or

Ze ve  r Zev e  r Zev e Zev e Zemv B       z  z  e . eff c r 3 c r 2 c r 2 c r 2 mcr 2 z

The Coulomb potential energy is given by

Ze2 dV (r) Ze2 V (r)  , c  . c r dr r2

Thus we have

Zemrv Ze 2 mrv Ze 2 m rv B  e  e  0 e eff mcr 3 z mcer 3 z m cerr 2 z . 1 Ze 2 1 dV (r)  m rve  c L e mcer r 2 0 z mcer dr z z or

1 1 dV (r) B  c L e , eff mce r dr z z

where Lz is the z-component of the orbital angular momentum, Lz  mvr . The spin magnetic moment is given by

2B μs   S . 

The Zeeman energy is given by

1 1  2   1 1 dV (r)  H   μ  B    B S  c L LS 2 s eff 2 mce r dr      1 1 dV (r)  c S  L   (S  L) 2m2c2 r dr

((Thomas correction)) Thomas factor 1/2, which represents an additional relativistic effect due to the acceleration of the electron. The electron spin, magnetic moment, and spin-orbit interaction can be derived directly from the Dirac relativistic electron theory. The Thomas factor is built in the expression.

H LS  S  L , with

2 1 1 dVc (r) 1  e  1   2 2  Z  3 . 2m c r dr 2  mc  r av

When we use the formula

3 3 Z r  4 3 , n a0 l(l 1/ 2)(l 1) the spin-orbit interaction constant  is described by

e2Z 4 me8Z 4   2 2 4 3  2 4 6 , 2m c n a0 l(l 1/ 2)(l 1) 2c n  l(l 1/ 2)(l 1) where 2 a   = 0.52917720859 Å. 0 me2 (Bohr radius) (from NIST physics constants)

The energy level (negative) is given by

Z 2 me4 Z 2e2 | En | 2 2  2 . n 2 2n a0

2 The ratio   / En is

2 4 2 2   e Z (Z) 1  2 2 2  2 , En c n  l(l 1/ 2)(l 1) n l(l 1/ 2)(l 1) with

e2 1    .  c 137.037

((Note)) For l = 0 the spin-orbit interaction vanishes and therefore  = 0 in this case.

______B. Paschen-Back effect in Na Mathematica calculation Calculation of Matrix element for the Zeeman effect

Eigenvalue problem for the Zeeman effect

Clear "Global`" ;

 —2 1 BB l  1 2 2  m2 M1   l  BBm 1  ,  , 2 2 l  1 2 l  1

BB l  1 2 2  m2  —2    1  ,  l  1  BBm 1   ; 2 l  1 2 2 l  1

eq1  Eigensystem M1 Simplify;     The eigenvalues;   1  eq1 1, 1

 —2 1 BmB   4B2 B2  8BmB  —2  2l 2 —4 4 4

2  eq1 1, 2   1 4BmB —2  4B2 B2  8BmB  —2  2l 2 —4 4  

The eigenvectors:  

1  eq1 2, 1

2B 1  4l 4l2  4m2 B ,1   4BmB  1  2l 1  2l  —2  4B2 B2  8BmB  —2  2l 2 —4

  1  eq1 2, 2  FullSimplify   

2B 1  2l 2  4m2 B ,1    4BmB  1  2l 1  2l  —2  4B2 B2  8BmB  —2  2l 2 —4

    C. Zeeman splitting for Na    Here we discuss the eigenvalue problem in more detail. j = 3/2, l = 1, s = 1/2 Clebsch-Gordan coefficient

3 3 j  ,m   l  1,m 1  2 2 l

3 1 1 2 j  ,m   l 1, m 1   l 1,m  0  2 2 3 l 3 l

3 1 2 1 j  ,m    l 1,m  0   l 1,m  1  2 2 3 l 3 l

3 3 j  ,m    l 1,m  1  2 2 l

______j = 1/2, l = 1, s = 1/2 Clebsch-Gordan coefficient;

1 1 2 1 j  , m   l 1, m 1   l 1, m  0  2 2 3 l 3 l

1 1 1 2 j  ,m    l 1,m  0   l 1,m  1  2 2 3 l 3 l

Note that

 2 1  1 1      j  , m   l 1, ml 1  2 2     3 3    l 1, m  0    1 2  3 1   l    j  , m    3 3  2 2 

 1 2  1 1      j  , m    l 1, ml  0  2 2     3 3    l 1, m  1    2 1  3 1   l    j  , m     3 3  2 2 

The Hamiltonian is expressed by

 B  2 2 2  B H  L S  (L  2S)B  (J  L  S )  (Lz  2S z )B .  2 

We need to calculate H j,m  with j = 1/2 ( m  1/ 2 ) and j = 3/2 (m = ±3/2, ±1/2).

1 1 B 1 B 2 H Zeeman j  ,m    B(Lz  2Sz ) l 1, ml  0   B(Lz  2Sz ) l 1, ml  1  ) 2 2  3  3 1   B l 1, m  0  B 3 l  B 1 1 3 1   B ( j  , m    2 j  , m   ) 3 2 2 2 2

1 1  2 2 2 1 1 2 1 1 H spinorbit j  , m    (J  L  S ) j  , m     j  , m   2 2 2 2 2 2 2

______

1 1 B 2 B 1 H Zeeman j  ,m   B(Lz  2Sz ) l 1, ml 1   B(Lz  2Sz ) l 1, ml  0  2 2  3  3 1   B l 1,m  0  B 3 l 1 1 1 1 2 3 1   B ( j  , m   j  ,m  B 3 3 2 2 3 2 2  B 1 1 3 1   B ( j  , m   2 j  , m  3 2 2 2 2

1 1  2 2 2 1 1 2 1 1 H spinorbit j  , m   (J  L  S ) j  , m    j  , m  2 2 2 2 2 2 2

______3 3 B H Zeeman j  , m    B(Lz  2Sz ) l 1, ml  1  2 2  3 3  2 B l 1, m  1   2 B j  , m   B l B 2 2

3 3  2 2 2 3 3 1 2 3 3 H spinorbit j  , m    (J  L  S )B j  , m     j  , m   2 2 2 2 2 2 2 2 ______

3 1 B 2 B 1 H zeeman j  ,m    B(Lz  2Sz )( l  1,ml  0   B(Lz  2Sz ) l  1,ml  1  ) 2 2  3  3 2   B l  1,m  0  B 3 l 2 1 1 1 2 3 1   B ( j  ,m    j  ,m   ) B 3 3 2 2 3 2 2  B 1 1 3 1   B ( 2 j  ,m    2 j  ,m   ) 3 2 2 2 2 3 1  3 1 1 3 1 H j  ,m    (J 2  L2  S2 ) j  ,m     2 j  ,m   spinorbit 2 2 2 B 2 2 2  2 2

______

3 1 B 1 B 2 H zeeman j  ,m   B(Lz  2Sz )( l 1,ml 1   B(Lz  2Sz ) l 1, ml  0  ) 2 2  3  3 2   B l 1,m  0  B 3 l 2 1 1 1 2 3 1   B ( j  , m   j  ,m  B 3 3 2 2 3 2 2  B 1 1 3 1  B ( 2 j  , m   2 j  , m  ) 3 2 2 2 2

3 1  2 2 2 3 1 1 2 3 1 H spinorbit j  ,m   (J  L  S )B j  ,m    j  , m  2 2 2 2 2 2 2 2 ______3 3 B H Zeeman j  , m   B(Lz  2Sz ) l 1, ml 1  2 2  3 3  2 B l 1, m 1   2 B j  , m  B l B 2 2

3 3  2 2 2 3 3 1 2 3 3 H spinorbit j  , m   (J  L  S )B j  , m    j  , m  2 2 2 2 2 2 2 2 ______Then we have

3 3 1 3 3 H j  ,m   (2 B   2 ) j  ,m  2 2 B 2  2 2

3 1 2 1 2 3 1 2 1 1 H j  ,m   ( B B   ) j  ,m   B B j  ,m  2 2 3 2 2 2 3 2 2

3 1 2 1 3 1 2 1 1 H j  ,m    (  B   2 ) j  ,m     B j  ,m   2 2 3 B 2  2 2 3 B 2 2

3 3 1 2 3 3 H j  ,m    (2B B   ) j  ,m   2 2 2 2 2

1 1 2 1 1 1 2 3 1 H j  , m   (  B B)( j  , m   B B j  , m  2 2 3 2 2 3 2 2

1 1 2 B B 1 1 2 3 1 H j  , m    (  ) j  , m    B B j  , m   2 2 3 2 2 3 2 2

From this, we find that

3 3 1 2 j  ,m  is the eigenket of H with the energy (2B B   ) . 2 2 2

3 3 1 2 j  , m   is the eigenket of H with the energy (2B B   ) . 2 2 2

3 1 1 1 There are two subspaces of the matrices for the basis of { j  , m  , j  , m  , 2 2 2 2 3 1 1 1 and { j  , m   , j  , m   } 2 2 2 2

3 1 1 1 (i) For the basis of { j  , m  , j  , m  2 2 2 2 ,

 2 1 2   2  B B    B B  3 2 3  H sub1   2 1 2    B B B B    3 3 

((Mathematica)) Clear "Global`" ;

m  1 2;   2  —  BB 4 m  BB 9  m2  2 3 3 4 M1  ;  BB 9  m2  —2  BB 4 m 3 4 3

eq1  Eigensystem M1 Simplify 1 8BB  3  —2  32 B2 B2  81 2 —4 , 12   1 8BB  3  —2  32 B2 B2  81 2 —4 ,  12   9  —2  32 B2 B2  81 2 —4 9  —2  32 B2 B2  81 2 —4  ,1 ,   ,1 4 2BB 4 2BB

E1 eq1 1, 1 Series  , B, 0, 3 & Simplify  ,   0, —  0& 2 BB 4 B2 B2  —2    O B 4 3  —2  27        

E2  eq1 1, 2 Series , B, 0, 3 & Simplify  ,   0, —  0 &    —2 2 BB 4 B2 B2    O B 4 2 3 27  —2          

1  eq1 2, 1 Series , B, 0, 3 & Simplify  ,   0, —  0 & 2 2 BB 16 2 B3 B3   4 2 3 6 O B ,1 9  — 729   —          2  eq1 2, 2 Series  , B, 0, 3 & Simplify  ,   0, —  0 &   9  —2 2 2 B B 16 2 B3 B3     4 2 3 6 O B ,1 2 2B B 9  —   729  —    

     3 1 1 1    (ii) For the basis of { j  ,m  , j  , m   } 2 2 2 2

 2 1 2   2   B B    B B  3 2 3  H sub2   2 1 2    B B  B B    3 3 

((Mathematica)) Clear "Global`" ; m 1 2;   2  —  BB 4 m  BB 9  m2  2 3 3 4 M1  ;  BB 9  m2  —2  BB 4 m 3 4 3 eq1  Eigensystem M1 Simplify 1 8BB  3  —2  32 B2 B2  81 2 —4 , 12   1 8BB  3  —2  32 B2 B2  81 2 —4 ,  12   9  —2  32 B2 B2  81 2 —4 9  —2  32 B2 B2  81 2 —4  ,1 ,   ,1 4 2BB 4 2BB

E1 eq1 1, 1 Series  , B, 0, 3 & Simplify  ,   0, —  0& 2 BB 4 B2 B2  —2    O B 4 3  —2  27        

E2  eq1 1, 2 Series , B, 0, 3 & Simplify  ,   0, —  0 &    —2 2 BB 4 B2 B2    O B 4 2 3 27  —2          

1  eq1 2, 1 Series , B, 0, 3 & Simplify  ,   0, —  0 & 2 2 BB 16 2 B3 B3   4 2 3 6 O B ,1 9  — 729   —          2  eq1 2, 2 Series  , B, 0, 3 & Simplify  ,   0, —  0 &   9  —2 2 2 B B 16 2 B3 B3     4 2 3 6 O B ,1 2 2B B 9  —   729  —    

           