Spin-orbit interaction Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Due Date: March 14, 2015)
In quantum physics, the spin–orbit interaction is an interaction of a particle's spin with its motion. The first and best known example of this is that spin–orbit interaction causes shifts in an electron's atomic energy levels due to electromagnetic interaction between the electron's spin and the magnetic field generated by the electron's orbit around the nucleus.
((Llewellyn Hilleth Thomas))
http://www.aip.org/history/acap/biographies/bio.jsp?thomasl
Llewellyn Hilleth Thomas (21 October 1903 – 20 April 1992) was a British physicist and applied mathematician. He is best known for his contributions to atomic physics, in particular: Thomas precession, a correction to the spin-orbit interaction in quantum mechanics, which takes into account the relativistic time dilation between the electron and the nucleus of an atom. The Thomas–Fermi model, a statistical model of the atom subsequently developed by Dirac and Weizsäcker, which later formed the basis of density functional theory. Thomas collapse - effect in few-body physics, which corresponds to infinite value of the three body binding energy for zero-range potentials. http://en.wikipedia.org/wiki/Llewellyn_Thomas
1. Biot-Savart law 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.
The current I due to the movement of nucleus (charge Ze, e>0) is given by
Idl ZevN ,
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 r B 1 , v ve eff c 3 N r1 where v is the velocity of the electron. Then we have
Ze v r Ze ve r B N 1 1 . eff c 3 c 3 r1 r1
Since r1 r , Beff can be rewritten as
Ze ve r Zev r e Zev e Zem v B z e e eff c 3 c 3 c 2 2 z r r r mec r
where me is the mass of electron. The Coulomb potential energy is given by
Ze 2 dV (r) Ze 2 V (r) , c . c r dr r2
Thus we have
2 2 Ze rv 1 Ze 1 dVc (r) Beff 3 ez 2 mervez Lzez , cer mecer r mecer dr or
1 1 dVc (r) Beff Lzez , mece r dr
where Lz is the z-component of the orbital angular momentum, Lz mevr .
2. Derivation of the expression for the spin-orbit interaction
Fig. Electron in the proton frame, and proton in the electron frame. The direction of magnetic field B produced by the proton is the same as that of the orbital angular momentum L of the electron (the z axis in this figure).
We consider the circular motion of the nucleus (e) around an electron at the center. The nucleus rotates around the electron at the uniform velocity v. Note that the velocity of the proton in the electron frame is the same as the electron in the proton frame. The magnetic field (in cgs) at the center of the circle along with the current I flow.
2I B . cr according to the Biot-Savart law. The current I is given by
e e ev I , T 2r 2r v where T is the period. Then the magnetic field B at the site of the electron (the proton frame) can be expressed as
emevN r eL B 3 ez 3 , mecr mecr where L is the orbital angular momentum of the electron (the electron frame). Note that the direction of B is the same as that of L. The spin magnetic moment is given by
2B μs S , where S is the spin angular momentum and B is the Bohr magneton,
e B . 2mec
Then the Zeeman energy of the electron is given by
1 1 2 eL e2 H μ B ( B S) L S , s 3 2 2 3 2 2 mecr 2me c r where the factor (1/2) is the Thomas correction. In quantum mechanics we use the notation
2 ˆ e 1 ˆ ˆ H so 2 2 3 L S . 2me c r av
3. Thomas correction The spin magnetic moment is defined by
2B μs S , where S is the spin angular momentum. Then the Zeeman energy is given by
1 1 2 1 1 dV (r) B c H so μs Beff S L 2 2 mece r dr 1 1 dV (r) c 2 2 S L 2me c r dr (S L) where the factor 1/2 is the Thomas correction and the Bohr magneton B is given by
e B . 2mec
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 so S L , with
2 1 1 dV (r) 1 e 1 c . 2 2 3 2me c r dr 2 mec r av
When we use the formula
1 r 3 , 3 3 n aB l(l 1/ 2)(l 1) the spin-orbit interaction constant is described by
e2 m e8 e 2 2 3 3 2 3 6 2me c n aB l(l 1/ 2)(l 1) 2c n l(l 1/ 2)(l 1) where
2 aB 2 = 0.53 Å (Bohr radius). mee
The energy level (negative) is given by
2 e 0 En 2 2 , 2n aB n where
4 2 mee e 0 2 . 2 2aB
2 The ratio / En is
2 4 2 e 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.
((Summary)) 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).
4. Commutation relations We introduce a new Hamiltonian given by
ˆ ˆ ˆ H H0 H so ,
The total angular momentum J is the addition of the orbital angular momentum and the spin angular momentum,
Jˆ Lˆ Sˆ ,
The spin-orbit interaction is defined by
1 Hˆ Lˆ Sˆ (Jˆ 2 Lˆ2 Sˆ 2 ) . so 2 where
ˆ ˆ ˆ ˆ ˆ ˆ L L iL , S S iS , and
ˆ ˆ ˆ ˆ ˆ ˆ [Lx ,S x ] 0 , [Lx ,S y ] 0 , [Lx ,Sz ] 0 ,
ˆ ˆ ˆ ˆ ˆ ˆ [Ly ,S x ] 0 , [Ly ,S y ] 0, [Ly ,Sz ] 0,
ˆ ˆ ˆ ˆ ˆ ˆ [Lz ,S x ] 0 , [Lz ,S y ] 0, [Lz ,Sz ] 0 .
ˆ (a) The unperturbed Hamiltonian H 0 ˆ ˆ ˆ The unperturbed Hamiltonian H0 commutes with all the components of L and S .
ˆ ˆ2 ˆ ˆ 2 ˆ ˆ [H0 , L ] [H0 , Lz ] [H0 , Lz ] 0 ,
ˆ ˆ 2 ˆ ˆ 2 ˆ ˆ [H0 ,S ] [H0 ,Sz ] [H0 ,Sz ] 0 , and
ˆ ˆ ˆ ˆ ˆ [H0, J z ] [H0, Lz Sz ] 0 .
We note that
ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ [H 0 , L S] [H 0 ,(Lx S x Ly S y Lz Sz )] 0 .
Then we have
ˆ ˆ 2 ˆ2 ˆ 2 [H0 ,J L S ] 0 , or
ˆ ˆ 2 [H 0 ,J ] 0 .
We also note that
[Jˆ 2 , Lˆ2 ] [Lˆ2 Sˆ 2 2Lˆ Sˆ, Lˆ2 ] 2[Lˆ Sˆ, Lˆ2 ] 0 ,
[Jˆ 2 , Sˆ 2 ] [Lˆ2 Sˆ 2 2Lˆ Sˆ,Sˆ 2 ] 2[Lˆ Sˆ,Sˆ 2 ] 0 ,
ˆ 2 ˆ ˆ2 ˆ 2 ˆ ˆ ˆ ˆ [J , J z ] [L S 2L S,Lz Sz ] ˆ ˆ ˆ ˆ 2[L S,Lz Sz ] 0 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2[Lz ,Lx ]S x 2[Ly ,Lz ]S y 2Lx [Sz ,S x ] 2Ly [S y ,Sz ] ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2iLy S x 2iLx S y 2iLx S y 2iLy S x 0
Thus we conclude that 0 is the simultaneous eigenket of the mutually commuting ˆ ˆ2 ˆ 2 ˆ 2 ˆ observables { H0 , L , S , J , and J z }.
0 j,m;l, s ,
ˆ (0) H0 0 En 0 ,
ˆ2 2 L 0 l(l 1) 0 ,
ˆ 2 2 S 0 s(s 1) 0 ,
ˆ 2 2 J 0 j( j 1) 0 ,
ˆ J z 0 m 0 .
ˆ (b) The perturbed Hamiltonian H so ˆ ˆ ˆ ˆ Here we note that L S does not commute with Lz or Sz .
ˆ ˆ ˆ [L S,Lz ] 0 ˆ ˆ ˆ ˆ ˆ ˆ [Lz ,Lx ]Sx [Ly ,Lz ]S y ˆ ˆ ˆ ˆ iLy S x iLx S y
ˆ ˆ ˆ [L S,Sz ] 0 ˆ ˆ ˆ ˆ ˆ ˆ Lx[Sz ,Sx ] Ly[S y ,Sz ] ˆ ˆ ˆ ˆ iLxS y iLy Sx and
ˆ ˆ ˆ [L S, J z ] 0
These commutation relations can be also derived from the invariant of the scalar product Lˆ Sˆ, under the rotation. The rotation operator around the z axis is given by ˆ ˆ i ˆ Rz () 1 J z . So we have
ˆ ˆ ˆ [L S, J x ] 0 from the rotation around the x axis.
ˆ ˆ ˆ [L S, J y ] 0 from the rotation around the y axis.
ˆ ˆ ˆ [L S, J z ] 0 from the rotation around the z axis.
Similarly we have
ˆ ˆ ˆ 2 ˆ ˆ ˆ 2 ˆ ˆ ˆ 2 [L S, J x ] 0 , [L S, J y ] 0 , [L S, J z ] 0 , and
[Lˆ Sˆ,Jˆ 2 ] 0
In summary, we have
ˆ ˆ ˆ 2 ˆ ˆ ˆ ˆ 2 ˆ [L S,J ] 0 , [L S, J z ] 0 , [J , J z ] 0 .
ˆ ˆ ˆ 2 ˆ Thus the state vector is the simultaneous eigenkets of L S , J , and J z . In other words,
Lˆ Sˆ ,
ˆ 2 2 J j( j 1) ,
ˆ J z m .
5. Application of the perturbation theory (degenerate case)) We need to choose the unperturbed states that diagonalize the perturbation (Cardinal rule). So the best way we can do is to choose the state
j,m , where j and m are the good quantum numbers. The perturbation Lˆ Sˆ mixes states of 2 same Jz and J but different Lz and Sz. Here we use the following notation.
2 J j( j 1) , J z m , Lz ml , Sz ms ; with
m ml ms where
ml = l, l - 1, l -2, …. ,-l
ms = 1/2, -1/2.
We note that the spin-orbit interaction occurs only for l 1. In order to get the eigenvalues and eigenkets of Lˆ Sˆ , we choose two states:
1 1 1 l,m m S ,m , 1 l 2 2 s 2
1 1 1 l,m m S ,m 2 l 2 2 s 2
ˆ ˆ where m ml ms . Next we calculate the matrix element of L S under the basis of the kets, 1 and 2 ;
ˆ ˆ ˆ ˆ L S 1 , L S 2 . or
Lˆ Sˆ Lˆ Sˆ Lˆ Sˆ 1 1 1 2 ˆ ˆ ˆ ˆ 2 L S 1 2 L S 2
We can solve the eigenvalue problem of this matrix (2x2), leading to the two eigenvalues. The corresponding eigenstates are denoted by
1 1 j l ,m , for j l 1 2 2
1 1 j l ,m , for j l 2 2 2
1 We will check the value of j l later. In fact the value of j can be obtained the 2 triangular law as
Dl Ds1/ 2 Dl 1/ 2 Dl 1/ 2 leading to
1 1 j l , and j l 2 2
Then the energy eigenvalues are obtained as
1 1 1 Hˆ j l ,m (Jˆ 2 Lˆ2 Sˆ 2 ) j l ,m so 2 2 2 2 3 1 [ j( j 1) l(l 1) )] j l ,m 2 4 2
2 1 3 3 1 [(l )(l ) l(l 1) )] j l ,m 2 2 2 4 2 l 2 1 [ j l ,m 2 2
1 1 1 Hˆ j l ,m (Jˆ 2 Lˆ2 Sˆ 2 ) j l ,m so 2 2 2 2 3 1 [ j( j 1) l(l 1) )] j l ,m 2 4 2
2 1 1 3 1 [(l )(l ) l(l 1) )] j l ,m 2 2 2 4 2 (l 1) 2 1 [ j l ,m 2 2
In summary 1 l 2 (i) j l ,m is the eigenstate of Hˆ with the energy . 2 so 2 1 (l 1) 2 (ii) j l ,m is the eigenstate of Hˆ with the energy . 2 so 2
6. The choice of 1 and 2 with the same j and m
6.1 l = 1, s=1/2, leading to j = 3/2 and j = ½
(a) j =1/2 (m = 1/2 and m = 1/2)
ms
1 2
m 1 2 m 1 2 ml 10 1
1 2
Fig. (ml, ms) plane for j = 1/2. l = 1 and s = 1/2. m ml ms (denoted by blue lines).
For m = 1/2, the superposition of ml 0,m3 1/ 2 and ml 1,m3 1/ 2
For m = -1/2, the superposition of ml 1,m3 1/ 2 and ml 0,m3 1/ 2
(b) j =3/2 (m = 3/2, 1/2, -1/2, and -3/2)
ms
1 2 m 3 2
m 1 2 m 1 2 ml 10 1
m 3 2 1 2
Fig. (ml, ms) plane for j = 3/2. l = 1 and s = 1/2. m ml ms (denoted by blue lines).
For m = 3/2, the superposition of ml 1,m3 1/ 2
For m = 1/2, the superposition of ml 0,m3 1/ 2 and ml 1,m3 1/ 2
For m = -1/2, the superposition of ml 1,m3 1/ 2 and ml 0,m3 1/ 2
For m = -3/2, the superposition of ml 1,m3 1/ 2
6.2 l = 2, s=1/2, leading to j = 5/2 and j = 3/2
(a) j = 5/2 (m = 5/2, 3/2, 1/2, -1/2, -3/2, -5/2) ms
1 2 m 5 2
m 3 2 m 1 2 m 1 2 m 3 2 ml 2 10 1 2
m 5 2 1 2
Fig. (ml, ms) plane for j = 5/2. l = 2 and s = 1/2. m ml ms (denoted by blue lines).
For m = 5/2, the superposition of ml 2,m3 1/ 2
For m = 3/2, the superposition of ml 1,m3 1/ 2 and ml 2,m3 1/ 2
For m = 1/2, the superposition of ml 0,m3 1/ 2 and ml 1,m3 1/ 2
For m = -1/2, the superposition of ml 1,m3 1/ 2 and ml 0,m3 1/ 2
For m = -3/2, the superposition of ml 1,m3 1/ 2 and ml 2,m3 1/ 2
For m = 5/2, the superposition of ml 2,m3 1/ 2
(b) j = 3/2, (m = 3/2, 1/2, -1/2, -3/2)
ms
1 2
m 3 2 m 1 2 m 1 2 m 3 2 ml 2 10 1 2
1 2
Fig. (ml, ms) plane for j = 3/2. l = 2 and s = 1/2. m ml ms (denoted by blue lines).
For m = 3/2, the superposition of ml 1,m3 1/ 2 and ml 2,m3 1/ 2
For m = 1/2, the superposition of ml 0,m3 1/ 2 and ml 1,m3 1/ 2
For m = -1/2, the superposition of ml 1,m3 1/ 2 and ml 0,m3 1/ 2
For m = -3/2, the superposition of ml 1,m3 1/ 2 and ml 2,m3 1/ 2
6. Solving the eigenvalue problem We calculate the matrix element of the spin-orbit interaction under the basis of
1 1 1 1 1 m m ,m l,m m S ,m , 1 l 2 s 2 l 2 2 s 2
1 1 1 1 1 m m ,m l,m m S ,m , 2 l 2 s 2 l 2 2 s 2 where m is fixed. Here we we use the formula
ˆ J j,m ( j m)( j m 1) j,m 1 ,
ˆ J j,m ( j m)( j m 1) j,m 1 ,
ˆ S ,
ˆ S ,
ˆ 2 2 J j,m j( j 1)( j,m , where
Jˆ Lˆ Sˆ ,
ˆ 2 ˆ2 ˆ 2 ˆ ˆ ˆ2 ˆ 2 ˆ ˆ ˆ ˆ ˆ ˆ J L S 2L S L S 2Lz Sz (L S L S ) ,
ˆ ˆ ˆ ˆ 1 ˆ ˆ ˆ ˆ L S L S (L S L S ) . z z 2
Then we get
1 1 Lˆ Sˆ Lˆ Sˆ m m ,m 1 l 2 s 2 ˆ ˆ 1 ˆ ˆ ˆ ˆ 1 1 [L S (L S L S )] m m ,m z z 2 l 2 s 2 2 1 1 ˆ ˆ 1 1 [ (m ) L S ] m m ,m 2 2 2 l 2 s 2
2 1 1 1 (m ) m m ,m 2 2 l 2 s 2 2 1 1 1 1 (l m )(l m ) m m ,m 2 2 2 l 2 s 2 2 1 1 1 [(m ) (l m )(l m ) ] 2 2 1 2 2 2
1 1 Lˆ Sˆ Lˆ Sˆ m m ,m 2 l 2 s 2 ˆ ˆ 1 ˆ ˆ ˆ ˆ 1 1 [L S (L S L S )] m m ,m z z 2 l 2 s 2 2 1 ˆ ˆ 1 1 [ (m ) L S ] m m ,m 2 2 l 2 s 2
2 1 1 1 1 (l m )(l m ) m m ,m 2 2 2 l 2 s 2 2 1 1 1 (m ) m m ,m 2 2 l 2 s 2 2 1 1 1 [ (l m )(l m ) (m ) ] 2 2 2 1 2 2
ˆ ˆ Thus we obtain the matrix of L S under the basis of 1 and 2 , as
1 1 1 (m ) (l m )(l m ) 2 ˆ ˆ 2 2 2 2 L S . 1 1 1 (l m )(l m ) (m ) 2 2 2
We solve the eigenvalue problem. The eigenvalues are obtained as
1 l , 2 (l 1) .
For 1 l , the eigenket is obtained as
l m 1/ 2 m m 1/ 2,m 1/ 2 1 2l 1 l s
l m 1/ 2 m m 1/ 2,m 1/ 2 2l 1 l s
Using 1 l , we get
2 2 2 3 2 J j( j 1) [l(l 1) l] (l 1/ 2)(l 1/ 2 1) . 4
1 which means that j l 2
For 2 (l 1) , we have
l m 1/ 2 m m 1/ 2,m 1/ 2 2 2l 1 l s
l m 1/ 2 m m 1/ 2,m 1/ 2 2l 1 l s
Using 2 (l 1) , we get
2 2 2 3 2 J j( j 1) [l(l 1) (l 1)] (l 1/ 2)(l 1/ 2 1) . 4
1 which means that j l . 2
Then we can obtain
(i)
1 j l ,m 1 2 l m 1/ 2 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
l 2 Hˆ Lˆ Sˆ (which is independent of m) so 1 1 2 1
(ii)
1 j l ,m 2 2 l m 1/ 2 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
(l 1) 2 Hˆ Lˆ Sˆ (which is independent of m. so 2 2 2 2
1 j 1 2 E 1 1 l, s 2
E 1 2 j 1 2
l 2 Fig. Splitting of the energy level due to the spin-orbit interaction. E . 1 2 (l 1) 2 E . This splitting is independent of the quantum number m. 2 2
7. Mathematica: the eigenvalue problem
1 Clear "Global`" ; rule1 k2 m2 L ; 2 m 1 k 1 1 2 rule2 k L m L m ;A1 ; 2 2 k m 1 2 eq1 Eigensystem A1 ; 1 eq1 1, 2 . rule1 Simplify , L 0 & L 2 eq1 1, 1 . rule1 Simplify , L 0 &
1 L f1 eq1 2, 2 . rule1 Simplify; f2 eq1 2, 1 . rule1 Simplify;
Normalize f1 . rule2 FullSimplify , L m 1 2 0, L m 1 2 0 & 1 m 1 m , 2 1 2L 2 1 2L
Normalize f2 . rule2 FullSimplify , L m 1 2 0, L m 1 2 0 &
1 m 1 m , 2 1 2L 2 1 2L
8. Another method (Sakurai) We start from the assumption that
1 j l ,m 2 1 2
1 j l ,m 2 1 2 where we need to determine the values of , , , and . The normalization condition:
2 2 1
2 2 1
The condition of the orthogonality:
0
There are four unknown parameters and three equations. So we need one more equation. Later, we show that
1 l m 2 . 1 l m 2
We want to determine the values of , , and . To this end, we assume that
cos, sin cos( / 2) sin, sin( / 2) cos
y
1 B g,d A a, b
p 2 q 1 x O
Since
2 2 1, we have
1 1 l m l m 1 2 2 . 2l 1 2l 1
Then we get
l m 1/ 2 l m 1/ 2 , . 2l 1 2l 1 where 0 , 0 , 0 , and 0 (assumption). The final result is the same as one derived from the Clebsch-Gordan coefficient (addition of angular momentum).
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
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 or
l m 1/ 2 j l 1/ 2,m 2l 1 , l m 1/ 2 2l 1
l m 1/ 2 j l 1/ 2,m 2l 1 . l m 1/ 2 2l 1
((Note))
The ratio / can be determined as follows. We demand that
ˆ 2 1 2 1 2 1 3 1 J j l ,m j( j 1) j l ,m (l )(l ) j l ,m , 2 2 2 2 2 where
1 1 1 1 1 j l ,m m m ,m m m ,m . 2 l 2 s 2 l 2 s 2
Here we note that
ˆ 2 ˆ2 ˆ 2 ˆ ˆ ˆ2 ˆ 2 ˆ ˆ ˆ ˆ ˆ ˆ J L S 2L S L S 2Lz Sz (L S L S )
Then we get
ˆ 2 2 3 1 ˆ ˆ J [l(l 1) (m ) L S ] 1 4 2 1
3 1 1 1 2[l(l 1) (m ) 2 (l m )(l m ) 4 2 1 2 2 2
ˆ 2 2 3 1 ˆ ˆ J [l(l 1) (m ) L S ] 2 4 2 2 3 1 2[l(l 1) (m ) 4 2 2 1 1 2 (l m )(l m ) 2 2 1
Since
1 3 1 Jˆ 2 j l ,m Jˆ 2 ( ) 2 (l )(l )( ) 2 1 2 2 2 1 2 we have
1 3 2 (l )(l )[ ] 2 2 1 2 3 1 2[l(l 1) (m )] 4 2 1 1 1 2 (l m )(l m ) 2 2 2 3 1 2[l(l 1) (m )] 4 2 2 1 1 2 (l m )(l m ) 2 2 1 or
1 1 1 [ (l m )(l m ) (l m )] 2 2 2 1 1 1 1 [ (l m )(l m ) (l m )] 2 2 2 2 0
Then we have
1 l m 2 . 1 l m 2
9. Energy splitting due to the spin-orbit interaction The eigenket can be described by
j,m;l,s
Note that the expression of the state can be formulated using the Clebsch-Gordan coefficient. When the spin orbit interaction is the perturbation Hamiltonian, we can apply the degenerate theory for the perturbation theory,
ˆ ˆ ˆ H LS L S j,m;l,s 1 (Jˆ 2 Lˆ2 Sˆ 2 ) j,m;l,s 2 2 [ j( j 1) l(l 1) s(s 1)] j,m;l,s 2
ELS j,m;l,s
1 l 2 For j l , E E 2 LS 1 2
1 (l 1) 2 For j l , E E 2 LS 2 2
Thus the energy eigenvalue due to the spin-orbit interaction is
2 E [ j( j 1) l(l 1) s(s 1)] so 2 2 2 e [ j( j 1) l(l 1) s(s 1)] 2 2 3 3 4me c n aB l(l 1/ 2)(l 1) 2 En n[ j( j 1) l(l 1) s(s 1)] 2 mec l(l 1/ 2)(l 1) and the eigenket is j,m;l,s , where
2 2 2 e 1 nEn 3 3 me 4n aB
e2 2 =7.24524 x 10-4 eV 2 2 3 2me c aB
We note that
1 m c2 4 Z 4 l j l E e 2 so 1 1 4 n3l(l )(l 1) (l 1) j l 2 2 with
m c2 4 e = 362.263 eV. 4
We consider the hydrogen atom (1 electron) [spin 1/2, and l] problem with the spin orbit interaction, where Z = 1. The eigenket is expressed by n,l,s; j,m . The angular part of this eigenket is
j,m j,m;l, s; with
s = 1/2, j = l ±1/2
Hˆ Lˆ Sˆ (Jˆ 2 Lˆ2 Sˆ 2 ) so 2
ˆ ˆ ˆ H so j,m L S j,m (Jˆ 2 Lˆ2 Sˆ 2 ) j,m 2 2[ j( j 1) l(l 1) 3/ 4] j,m 2
((Note)) The energy levels with degenerate states are split to the state with j = l + 1/2 and the state with j = l - 1/2, due to the spin-orbit interaction. However, these states are still degenerate. When the magnetic field is applied to the system, these states are finally separated into the non-degenerate states.
j l 1/ 2,m , where m+ = l + 1/2, l - 1/2,….., - (l + 1/2) and
j l 1/ 2,m , where m- = l - 1/2, l - 3/2,….., - (l - 1/2)
(a) For j = l +1/2,
E 2[(l 1/ 2)(l 3/ 2) l(l 1) 3/ 4] 2l so 2 2
Then we have
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
In summary, the energy shift due to the spin-orbit interaction is given by
E 2l , for j = l+1/2 so 2
E 2 (l 1) , for j = l-1/2 so 2
2 l 0 Hˆ 2 so 0 2 (l 1) 2 under the basis of j l 1/ 2,m and j l 1/ 2,m , where
2 4 2 e 1 2 2 2 2 En c n l(l 1/ 2)(l 1) n l(l 1/ 2)(l 1) or
2 m c2 4 1 e 2 2 n3 l(l 1)(2l 1)
10. The Zeeman splitting In the presence of an external magnetic field along the z axis, the Zeeman energy is given by
ˆ B ˆ 2B ˆ B B ˆ ˆ H B Lz Sz B (Lz 2Sz ) .
We now consider the influence of the magnetic field on the eigenstates j l 1/ 2,m and j l 1/ 2,m
(a) j l 1/ 2,m
Hˆ j l 1/ 2,m 2l j l 1/ 2,m so 2
ˆ ˆ The expectation value of Lz and S z
ˆ l m 1/ 2 Lz j l 1/ 2,m (m 1/ 2) ml m 1/ 2,ms 1/ 2 2l 1
l m 1/ 2 (m 1/ 2) ml m 1/ 2,ms 1/ 2 2l 1
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,
ˆ l m 1/ 2 Sz j l 1/ 2,m (1/ 2) ml m 1/ 2,ms 1/ 2 2l 1
l m 1/ 2 (1/ 2) ml m 1/ 2,ms 1/ 2 2l 1
ˆ l m 1/ 2 l m 1/ 2 j l 1/ 2,m Sz j l 1/ 2,m (1/ 2) (1/ 2) 2l 1 2l 1 m 2l 1
ˆ Then we have the expectation value of H B as
ˆ EB1 j l 1/ 2, m H B j l 1/ 2 ˆ ˆ j l 1/ 2, m Lz 2S z j l 1/ 2
B B 1 m(1 ) 2l 1 1 m B(1 ) B 2l 1 l 1 2mB B 2l 1
(b) For j = l -1/2,
2 1 1 3 2 ELS [(l )(l ) l(l 1) ] (l 1) 2 2 2 4 2
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 or
Hˆ j l 1/ 2,m 2 (l 1) j l 1/ 2,m so 2
ˆ ˆ The expectation value of Lz and S z
l m 1/ 2 l m 1/ 2 j l 1/ 2,m Lˆ j l 1/ 2,m (m 1/ 2) (m 1/ 2) z 2l 1 2l 1 2 m(l 1) 2l 1
ˆ l m 1/ 2 l m 1/ 2 j l 1/ 2,m Sz j l 1/ 2,m (1/ 2) (1/ 2) 2l 1 2l 1 m 2l 1
ˆ ˆ 1 j l 1/ 2,m Lz 2Sz j l 1/ 2,m m(1 ) 2l 1
ˆ Then we have the expectation value of H B as
ˆ EB2 j l 1/ 2, m H B j l 1/ 2 ˆ ˆ j l 1/ 2, m Lz 2S z j l 1/ 2
B B 1 m(1 ) 2l 1 1 2m B( ) B 2l 1
Eso1 EB1 Eso1 j l 1 2,m
Eso2 EB2 j l 1 2,m
Eso2
H0 Hso Hso HB
Fig. Energy level splitting due to the spin orbit interaction and the Zeeman effect
2 2 l 1 Eso1 l , Eso2 (l 1) , EB1 2mB B , and 2 2 2l 1 1 E 2m B( ) B2 B 2l 1
11. Effect of the spin-orbit interaction on the spectrum in hydrogen atom For hydrogen, Z = 1.
2 E (J 2 L2 S 2 ) [ j( j 1) l(l 1) s(s 1)] SO 2 2
m c2 4 [ j( j 1) l(l 1) s(s 1)] e 2 n3 l(l 1)(2l 1) where is the fine structure
e2 1 =7.2973525698 x 10-3 c 137.036
(a) The energy level of the 2S state does not change since L = 0. (b) The 2P state is split into two states due to the spin-orbit coupling.
2 2 P3 2
D 2P
2D
2 2 P1 2
Spin-orbit interaction
2 2 Fig. Splitting of 2 P level into two states (2 P3/2 and 2 P1/2).
l = 1 (2p) and spin s = 1/2 →j = 3/2 and j = 1/2
(D1 x D1/2 = D3/2 + D1/2).
2 2 P3/2 (j = 3/2, l = 1, s = 1/2)
2 2 P1/2 (j = 1/2, l = 1, s = 1/2)
When s = 1/2,
2 ESO = 15.0943 eV for the 2 P3/2 state
2 ESO 2 = -30.1886 eV for the 2 P1/2 state
3 45.2829 eV.
The energy difference is 3 = 45.2829 eV.
Note that the Landè g-factor is defined by.
3 s(s 1) l(l 1) g . J 2 2 j( j 1)
2 For P3/2 (j = 3/2, l = 1, s = 1/2)
3 3 1(11) 4 g 4 . 1 3 5 2 2 3 2 2
2 For P1/2 (j = 1/2, l = 1, s = 1/2)
3 1(11) 3 2 g 4 . 2 1 3 2 2 3 2 2
(c) The 3p state is split into two states due to the spin-orbit coupling.
l = 1 (3p) and spin s = 1/2 →j = 3/2 and j = 1/2
(D1 x D1/2 = D3/2 + D1/2).
2 3 P3/2 (j = 3/2, l = 1, s = 1/2)
2 3 P1/2 (j = 1/2, l = 1, s = 1/2)
Then we have
2 ESO = 4.47239 eV for the 3 P3/2 state
2 ESO 2 = -8,94478 eV for the 3 P1/2 state
3 13.4172 eV.
2 3 P3 2
D 3P
2D
2 3 P1 2
Spin-orbit interaction
2 2 Fig. Splitting of 3 P level into two states (3 P3/2 and 3 P1/2)
2 ESO = 4.47239 eV for the 3 P3/2 state
2 ESO 2 = -8.94478 eV for the 3 P1/2 state
3 13.4172 eV.
(d) We note that the energy level of the 3S state does not change since L = 0.
______12. Sodium D lines The electron configuration of Na is (1s)2(2s)2(2p)6(3s)1. The atomic number is Z = 11. The famous Na doublet arises from the spin-orbit splitting of Na 3p level, and consists of the closely spaced pair of spectral lines at wavelength of D1 line (589.592 nm) 2 2 2 2 for the transition 3 P1/2 S1/2, and D2 line (588.995 nm) P3/2 S1/2. (Serway).
2 P3 2
2 P12
l=589.0 nm l=589.6 nm
2 S1 2
2 2 2 Fig. 3 P3/2 (j = 3/2, l = 1, s = 1/2). 3 P1/2 (j = 1/2, l = 1, s = 1/2). 3 S1/2 (j = 1/2, l = 0, s 2 2 = 1/2). The splitting of the energy levels (3 P3/2 and the 3 P1/2) is due to the spin- orbit interaction. The D1 line (denoted by blue line). The D2 line (denoted by red line).
The sodium D lines correspond to the 3p 3s transition. In the absence of a 2 magnetic field B, the spin orbit interaction splits the upper 3p state into 3 P3/2 and 3 2 -1 2 P1/2 terms separated by 17 cm . The lower 3 S1/2 has no spin-orbit interaction. For l = 1, the energy shift due to the spin-orbit interaction is given by
2 Es0 , for j = 3/2 (3 P3/2)
2 Eso 2 , for j = 1/2 (3 P1/2)
For the Na doublet, the observed wavelength difference is
2 1 0.597 nm, since
2 589.592 nm, 1 588.995 nm.
2 2 Then the energy difference between the 3 P3/2 state and 3 P1/2 state is derived as follows.
hc E 3 = 2.1314 x 10-3 eV 12
______13. Clebsch-Gordan co-efficient in Na
(a) For the electron with 3s state (l = 0, s = 1/2)
D0 x D1/2 = D1/2
2 Thus we have j = 1/2. The state is described by S1/2.
j 1/ 2,m 1/ 2 ml 0,ms 1/ 2
j 1/ 2,m 1/ 2 ml 0,ms 1/ 2
(b) For the electron with 3p state (l = 1, s = 1/2)
D1 x D1/2 = D3/2 + D1/2
2 2 Thus we have j = 3/2 and j = 1/2. The state is described by P3/2 and P1/2
(i) j = 3/2
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
(ii) j = 1/2
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
ms
j= 3 2
-1,1 2 0,1 2 1,1 2 m=3 2 m=-1 2 m=1 2
-1,-1 2 0,-1 2 1,-1 2 m=-3 2
Fig. Clebsh-Gordan diagram for j,m with j = 3/2. m = 3/2, 1/2, -1/2, and -3/2.
ms
j= 1 2
-1,1 2 0,1 2 ml m=-1 2 m=1 2
0,-1 2 1,-1 2
Fig. Clebsh-Gordan diagram for j,m with j = 1/2. m = 1/2 and -1/2.
14. Total angular momentum and the total magnetic moment 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 μJ J , where
Fig. 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 S 2 J 2 L2 S 2 J S (L S) S S 2 L S S 2 , 2 2 or
J 2 L2 S 2 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 μJ (a 2b)J (1 b)J J , with the Landé g-factor
J S 3 S(S 1) L(L 1) g 1 b 1 . J J 2 2 2J (J 1)
15. Lande g-factor in Na The Lande g-factor is given by
3 s(s 1) l(l 1) g J 2 2 j( j 1)
Table
Term j l S g
2 2 P3/2 3/2 1 1/2 4/3 2 2 P1/2 1/2 1 1/2 2/3
______REFERENCES J.J. Sakurai Modern Quantum Mechanics, Revised Edition (Addison-Wesley, Reading Massachusetts, 1994). D.J. Griffiths Introduction to Quantum Mechanics (Prentice Hall, Upper Saddle River, NJ, 1995). E. Fermi Notes on Quantum Mechanics (University of Chicago, 1961). R. Serway, C.J. Moses, and C.A. Moyer, Modern Physics, 3rd edition (Thomson Brooks/Cole, 2005). ______APPENDIX Mathematica program for the Clebsch-Gordan coefficients Clear "Global`" ;
CCGG j1_, m1_ , j2_, m2_ , j_, m_ : Module s1 , s1 If Abs m1 b j1 && Abs m2 b j2 && Abs m b j, ClebschGordan j1, m1 , j2, m2 , j, m ,0 CG j_, m_ , j1_, j2_ : Sum CCGG j1, m1 , j2, m m1 , j,m a j1, m1 b j2, m m1 , m1, j1, j1 j1 = 1 and j2 = 1/2; j1 1; j2 1 2;
CG 3 2, 3 2 , j1, j2 1 1 a 1, 1 b , 2 2
CG 3 2, 1 2 , j1, j2 1 1 a 1, 1 b , 2 1 1 2 2 a 1, 0 b , 3 3 2 2 CG 3 2, 1 2 , j1, j2 1 1 2 1 1 a 1, 1 b , a 1, 0 b , 2 2 3 2 2 3 CG 3 2, 3 2 , j1,j2 1 1 a 1, 1 b , 2 2
CG 1 2, 1 2 , j1 , j2 1 1 2 1 1 a 1, 0 b , a 1, 1 b , 2 2 3 2 2 3 CG 1 2, 1 2 , j1,j2 1 1 a 1, 0 b , 2 1 1 2 2 a 1, 1 b , 3 3 2 2 j1 = 0 and j2 = 1/2
j1 0; j2 1 2;
CG 1 2, 1 2 , j1, j2 1 1 a 0, 0 b , 2 2
CG 1 2,1 2, j1, j2 1 1 a 0, 0 b , 2 2