11
11.1
Spin
Historical
Spectroscopy of Alkali atoms First expt. to suggest need for
electron spin: observation of splitting of expected spectral
lines for alkali atoms: i.e. expect one line based on analogy
of alkali atom with Hydrogen, see two closely spaced ones
instead.
Naive argument–alkali atoms: Argue valence electron in alkali
atom feels mostly Coulomb force due to “extra” proton in
nucleus −→ analogous to hydrogen:
Z=3: Lithium=Helium + 1 proton + 1 electron
Z=11: Sodium= Neon + 1 proton + 1 electron
Sommerfeld (1920) noted energy levels of Li deduced from
spectroscopy looked like H, with slight adjustment of principal quantum number:
1
En '
1 Rydberg
(n + p`)2
(1)
due to the fact that once the valence electron gets inside the
He-like core electrons, it will feel a higher nuclear charge,
potential falls faster than 1/r—see figure. Correction term
p` < 0 gets small in magnitude for large `, however, since
these orbits are outside the core with high probability.
Expt’l. observations:
1. Some of the spectral lines for Li appear in close doublets
(pairs of closely spaced energies). Sommerfeld said must
imply influence of two possible nuclear or core states. Pauli
argued “two-valuedness” came from two possible quantum
states of valence electron.
2. Doublets observed to move farther apart in applied magnetic field, other lines where splitting too small to see in
B = 0 turn into doublets when B 6= 0. Zeeman effect.
To explain splitting in applied mag. field, Uhlenbeck and
Goudschmidt (1925) suggested electron has spin angular momentum s = 1/2, therefore two states |ms = ±1/2i.
Predicted observed splitting, with
E = −µ · B
e
µ=− S
m




energy of electron magnetic moment
(2)


 in applied magnetic field B




relation of magnetic moment


 to electron spin operator
2
(3)
S = (Sx, Sy , Sz ) = electron spin operator
(4)
S2
has e’value h̄2s(s + 1) = 3h̄2/4
(5)
Sz
has e’value h̄ms = (±1/2)h̄
(6)
Q: Should we have expected spin on classical grounds?
Electron might be spinning in analogy to planet revolving
around sun—total a.m. J is sum of orbital a.m. (L ≡ r ×
p) around sun plus its spin angular momentum S ≡ Iωn̂.
(Bogus argument: quantum mechanical spin is intrinsic and
permanent. Electron is as far as we know point particle.) It is
interesting to note that if we had a classical charge e orbiting
in a circle of radius a with angular velocity ω, the ratio of the
magnetic moment to the angular momentum (gyromagnetic
ratio) would be
µ
Ja
=
`
Iω
(e ω/(2π))(πa2)
=
(ma2)ω
e
=
,
2m
where J = eω/(2π) is the current carried by the orbiting particle. Note that the radius cancels, so we can in fact discuss
the classical gyromagnetic ratio of any spinning object using
this formula since it’s just made up of many such charge el3
ements orbiting around the same axis. Note also that the
gyromagnetic ratio for the quantum mechanical spin determined by fits to the field dependence of the doublet splitting
is a factor of two different from the classical expectation.
11.2
Free spin: Pauli Spin Matrices
Recall we showed that operators which obeyed the algebra
[L̂i, L̂j ] = ih̄²ijk L̂k
(7)
had eigenvalues of total L̂ ≡ iL̂2i restricted to h̄2`(` + 1),
with ` = 12 , 1, 32 , 2, 52 , . . .. (Orbital ang. mom. L̂2 only took
on values with ` =integer since we showed additionally that
L̂z = mh̄, −` ≤ m ≡ integer ≤ `.)
So let’s demand that the components of spin angular momentum obey same comm. rels.:
[Ŝi, Ŝj ] = ih̄²ijk Ŝk
(8)
X
2
because
1. by analogy with orbital angular momentum, looking for a
quantum operator which has exactly two discrete e’values
to explain spectral doublets, recall a.m. algebra allows this.
2. will use algebra (Eq. (8)) later to show S generates rotations of spins.
Spin algebra (8) implies [Ŝ 2, Ŝz ] = 0 as before, so we can
find simultaneous eigenstates of Ŝ 2 and Ŝz , with eigenvalues
h̄2s(s + 1) and msh̄, with s = 1/2 and ms = ±s as before.
4
Write two allowed spin states of a particle as | 21 i ≡ | ↑i (“spin
up”) and | − 12 i ≡ | ↓i, so we now need to specify the spin
index σ of a wave function, by ψσ , i.e.
ψ↑(r) = h↑ |ψi
ψ↓(r) = h↓ |ψi
(9)
(10)
Of course the operator S is completely specified by the matrix elements hms|Ŝi|m0si. Calculate:
• Ŝz :
hms|Ŝz |m0si = m0sh̄δmsm0s
(11)
or, in this basis,

Sz =

h̄  1 0 


2 0 −1
(12)
(Now for Sx and Sy we’ll need ladder operators:
Ŝ+ = Ŝx + iŜy
Ŝ− = Ŝx − iŜy
(13)
(14)
with the usual conditions Ŝ+| ↑i = 0 and Ŝ−| ↓i = 0
to terminate the “ladder”. The proportionality constant
given as usual (i.e., as for the orbital a.m. ladder ops):
r
Ŝ±|msi = h̄ s(s + 1) − ms(ms ± 1)|ms ± 1i
i.e. for s = 1 and ms = ± 12
5
(15)
Ŝ+| ↓i = h̄| ↑i and Ŝ−| ↑i = h̄| ↓i
(16)
So, using Ŝx = (Ŝ+ + Ŝ−)/2 and Ŝy = (Ŝ+ − Ŝ−)/(2i) can
now calculate (check it!)
• Ŝx:

hms|Ŝx|m0si
=
v
u
u
u
u
t

 Ŝ+ + Ŝ− 
h̄
0

 |m i
= hms| 

s
2
2
h̄ 3
− m0s(m0s + 1)δms,m0s+1 +
2 4v
u
u3
h̄ u
u
t
− m0s(m0s − 1)δms,m0s−1
2 4
So

Sx = Sx =
(17)
(18)

h̄  0 1 


2 1 0
(19)
and similarly
• Ŝy :


h̄  0 −i 
Sy = Sy = 

2 i 0
Notation: 3 dimensionless
matrices





0 
 0 1 
 0 −i 
 1


; σy ≡ 
; σz ≡ 
σx ≡ 



1 0
i 0
0 −1
(20)
(21)
are called Pauli spin matrices, sometimes written in vector
form, ~σ = (σx, σy , σz ), so that S = (h̄/2)~σ .
6
Mathematical aside: Rank 2 unitary matrices of determinant 1 form a group in the mathematical sense, called SU (2).
The Pauli matrices are a representation of the generators of
SU (2), meaning any 2 × 2 matrix of determinant 1 can be
written as
i
exp n̂ · ~σ .
(22)
2
11.3
Two spin-1/2 particles
Consider two electrons, and suppose that only dynamical variables of interest for moment are spins, Ŝ1 and Ŝ2. 1 and 2 refer
to two different particles, so Ŝ1 and Ŝ2 commute. So we have
[Ŝ1i, Ŝ1j ] = ih̄²ijk Ŝ1k
[Ŝ2i, Ŝ2j ] = ih̄²ijk Ŝ2k
Ŝ1i, Ŝ2j ] = 0.
(23)
(24)
(25)
S = S1 + S2
Ŝ 2 = Ŝ12 + Ŝ22 + 2S1 · S2
(26)
(27)
Total spin operator
As in Prob. we did in homework, can specify two distinct
sets of commuting observables:
1)
2)
Ŝ12, Ŝ1z , Ŝ22, Ŝ2z
Ŝ 2, Ŝz , Ŝ12, Ŝ22
7
(28)
(29)
Note, e.g. Ŝ 2 can’t be added to set 1) because of Eq. (27):
S1 · S2 won’t commute with Ŝ1z , e.g.
Assume both particles spin 1/2, then all state vectors are
eigenstates of Ŝ12 and Ŝ22 with eigenvalues, e.g. h̄2S1(S1 +
1) = 3h̄2/4, same for spin 2. Drop S labels, complete set of
eigenstates for this problem is then
1
1
|m1, m2i, m1 = ± , m2 = ±
(30)
2
2
or , could choose as our complete set
|S, mi , |m| ≤ S with
Ŝ 2|S, mi = h̄2S(S + 1)|S, mi
Ŝz |S, mi = h̄m|S, mi
Completeness =⇒
|S, mi =
X
m1 ,m2
and
(31)
(32)
(33)
|m1, m2ihm1, m2|S, mi
(34)
Find expansion coefficients |m1, m2ihm1, m2|S, mi:
Use spin version of favorite identities:
Ŝ 2 = Ŝ−Ŝ+ + Sz2 + h̄Ŝz
Ŝ 2 = Ŝ+Ŝ− + Sz2 − h̄Ŝz
As usual start with state with max. ang. momentum,
|m1 = 1/2, m2 = 1/2i
Check:
8
(35)
(36)
(37)
11
2 11
Ŝ | i = 2h̄ | i,
22
22
11
11
Ŝz | i = h̄| i
22
22
2
(38)
(39)
where good use was made of
Ŝz = Ŝ1z + Ŝ2z
(40)
So clear we can write
|S = 1, m = 1i = |m1 = 1/2, m2 = 1/2i
(41)
Now play usual game and apply lowering op. Ŝ− = Ŝ1− + Ŝ2−
to find (check!):


1
1 
1 
1 1
√

i + | − i
|1, 0i =
|−
2 2
2
2
2
and
(42)
1 1
(43)
|1, −1i = | − − i
2 2
(Note I dropped S =, m = and m1 =, m2 labels since the distinction is clear: S, m are always integers for this problem.)
Finally we need remaining state for combination of two spins
with total S = 0. Must be orthogonal to all 3 states. Could
construct formally, but easy to guess, and verify orthonormality:


1 
11
1
1 
|0, 0i = √ | − i − | − i
(44)
22
2
2
2
9
singlets and triplets. The three S = 1 states (41-43) are referred to together as triplet states, whereas the S = 0 combination Eq. (44) is called a singlet.
11.4
Pauli principle
Pauli: two identical particles in quantum mechanics have symmetric (antisymmetric) wavefunctions under interchange of all
particle labels if they have integer (half-integer) spin. Particles
with integer (half-integer) spin are called bosons (fermions).
So
bosons
ψ(1, 2) = ±ψ(2, 1)
(45)
fermions
Remarks on Pauli principle:
1. Immediate consequence for atomic physics: no two electrons can be in same single-particle quantum state. (Then
wave function would be φ(1)φ(2), manifestly symmetric–
no good.)
2. Two electrons are in 2-particle wavefctn. which consists
of product of spatial part (dependence on e− coordinates
r1 and r2) and spin part. Could write
ψ(1, 2) = R(r1, r2)χ(1, 2)
(46)
Note if spatial part has definite sign under interchange,
spin part must have opposite sign to make total wavefctn
antisymmetric!
10
• Example 1. Suppose R = R(|r1 − r2|) — symmetric
under 1 ↔ 2, so must have
χ(1, 2) = −χ(2,¶1). System is
µ
in spin singlet state! ( √12 | 12 − 12 i − | − 12 21 i
• Example 2. Suppose R ∝ (r1 − r2) · ẑ (relative p-wave
state). Then mustµhave χ(1, 2) = +χ(2,
1)—lin. comb. of
¶
triplet states. ( √12 | 12 − 12 i + | − 21 12 i , | 12 21 i, or | − 12 − 12 i)
11.5
Spin 1/2 in external magnetic field
Spin was assumed by Goudschmidt and Uhlenbeck to give rise
to magnetic moment just as orbiting charged particle does.
Small difference: orbiting charge −e has moment (check!)
e
~µ = − L
(47)
2m
whereas Goudschmidt and Uhlenbeck found they needed to
assume for the electron
e
~µ = − S
(48)
m
in order to explain experiment. They assumed then the coupling to the external field was just as in classical electromagnetism:
e
HZeeman = −~µ · B = S · B
(49)
m
≡ µ0~σ · B
(50)
where µ0 ≡ eh̄/(2m) is called the Bohr magneton.
Time evolution of spin states. Ignore all parts of Hamiltonian
except HZeeman for now, and take B = B0ẑ. Time evolution
11
operator is exp −iHt/h̄, so amplitude for spin state |mi at
time t is
χm(t) = hm|ψ(t)i = hm|e−iHt/h̄|ψi = e−iµ0mB0t/h̄χm (51)
Note if, at t = 0, |ψi is in an eigenstate of Ŝz , it stays that
way, as must be since [Ŝz , H] = 0! (Anything which commutes
with the Hamiltonian is conserved...) On other hand, if it is
in another state, spin precesses in general.
Example: start spin off in eigenstate of Ŝx:


or,



h̄  0 1   χ↑ 
h̄  χ↑ 


 =


1
0
χ
χ
2 {z } | {z↓ }
2 ↓
|
Sx
|ψi

=

h̄  χ↓ 


2 χ↑
(52)
(53)
(54)
In other words, χ↑ = χ↓ for this state at t = 0. Using normalization and (49) we have




−iµ
B
t/h̄
0
0
1 1
1 e

|ψi = √   ; |ψ(t)i = √  iµ0B0t/h̄ 
(55)
2 1
2 e
? Note |ψ(t)i generally not eigenstate of Ŝx anymore. However at t = T = πh̄/(µ0B),


1  1 
√
|ψ(T )i = −
(56)


2 1
again an eigenstate of Ŝx with eigenvalue h̄/2! (“spin points
along x̂”).
12
Quick calculation shows (check) that at T /4,


1  1 − i 
|ψ(T /4)i = − 

2 1+i
(57)
which is an eigenstate of Ŝy now with eigenvalue h̄/2 (“spin
points along ŷ axis). At T /2,


i  1 
(58)
|ψ(T /2)i = − √ 

−1
2
which is an eigenstate of Ŝx again but with eigenvalue −h̄/2
(“spin points along −x̂”). And so on... see that spin is rotating in xy plane with period T !
Alternative viewpoint on same problem:
Q: System is in state |ψ(t)i. What is probability you will
obtain a result h̄/2 when you measure Ŝx?
A: E’state of Ŝx with e’value h̄/2 is


1  1 
|φi = √  
(59)
1
2
as we saw.
Therefore probability of obtaining h̄/2 when measuring Ŝx
is
P = |hφ|ψ(t)i|2
¯

¯2
¯ 1
¯
−iµ
B
t/h̄
0
0
¯ √ [1 1]
¯
1
 e
¯
¯
2


¯
= ¯¯
· √  iµ0B0t/h̄ ¯¯¯ = cos2 πt/T
¯
¯
2 e
consistent with previous picture of a rotating spin!
13
(60)
(61)
Or, calculate expectation value of Ŝx in |ψ(t)i:
hψ(t)|Ŝx|ψ(t)i =
√1 [e−iµ0 B0 t/h̄
2
= cos 2πt/T
11.6
iµ0 B0 t/h̄ ∗
e
]

−iµ0 B0 t/h̄

1 e

· √  iµ0B0t/h̄ 
2 e
(62)
Magnetic resonance
Put electron in large static applied field B0ẑ, assume spins
initially in eigenstate of Ŝz corresponding to −h̄/2 ( | ↓ i
= “spin down”). (Note this is equilibrium config. since µ
likes to align with field, but µ ∝ −S for an electron!) Apply
oscillating field B1 cos ωt x̂ ⊥ to it as shown.
Hamiltonian is then
H = µ0~σ · B




0 
 0 1 
 1



= µ0B0 
 + µ0 B1 cos ωt 

1 0
0 −1
14
(63)
(64)
If we once again put


|ψi = 

χ↑ 
,
χ↓
(65)
Then Schrödinger eqn. has two components:
∂χ↑
ih̄
= µ0B0χ↑ + µ0B1 cos ωtχ↓
(66)
∂t
∂χ↓
ih̄
= −µ0B0χ↓ + µ0B1 cos ωtχ↑
(67)
∂t
Exact soln. possible but not enlightening. Suppose B1 ¿
B0, assume initial conditions χ↑ = 0, χ↓ = 1 at t = 0. So for
small t, can assume χ↑ is small, and S.-eqn is approx.
∂χ↓
ih̄
' −µ0Bχ↓, so
∂t
χ↓ ' eiµ0B0t/h̄
and substituting into (64) we get
∂χ↑
ih̄
' µ0B0χ↑ + µ0B1 cos ωt eiµ0B0t/h̄
∂t
Rewrite as
∂ Ã iµ0B0t/h̄!
ih̄
χ↑e
= µ0B1 cos ωte2iµ0B0t/h̄
∂t
15
(68)
(69)
(70)
Integrate, use cos x = (eix+e−ix)/2, and define ω0 = 2µ0B0/h̄:
iµ0 B0 t/h̄
ih̄χ↑e
Z
0
t
= µ0B1 0 dt0 cos ωt0 e2iµ0B0t /h̄


i(ω+ω0 )t
i(ω0 −ω)t
µ0B1  e
−1 e
− 1 

 (71)
=
+

2
i(ω + ω0)
i(ω0 − ω) 
From (69) we find by 1) noting that for ω ' ω0 2nd term is
much bigger, then 2) taking complex magnitude

2
µ
B
sin2(ω − ω0)t/2
0
1
2



|χ↑| ' 
(72)
2
h̄
(ω − ω0)
Recap: At t = 0, spin is down. At time t, χ↑ is probability
that spin is up. Clearly prob. is max. when ω = ω0, “resonant
frequency”. Note resonance occurs when h̄ω = 2µ0B0, i.e.
difference between two magnetic field–split levels. Consistent
with picture that photon of energy just equal to splitting is
causing spin flip. (Not quite kosher since we are treating Bfield classically!)
16