Download Spin The evidence of intrinsic angular momentum or spin and its

Spin
The evidence of intrinsic angular momentum or spin and its associated magnetic moment
came through experiments by Stern and Gerlach and works of Goudsmit and Uhlenbeck.
The spin is called intrinsic since, unlike orbital angular momentum which is extrinsic, it is
carried by point particle in addition to its orbital angular momentum and has nothing to
do with motion in space.
The magnetic moment µ
~ of silver atom was measured in 1922 experiment by Stern and
Gerlach and its projection µz in the direction of magnetic field ẑ B (through which the
silver beam was passed) was found to be just −|~
µ| and +|~
µ| instead of continuously varying
between these two as limits. Classically, magnetic moment is proportional to the angular
momentum (12) and, assuming this proportionality to survive in quantum mechanics, quantization of magnetic moment leads to quantization of corresponding angular momentum S,
which we called spin,
q
q~
µz = g
Sz ⇒ µz = g
ms
(57)
2m
2m
as is the case with orbital angular momentum, where the ratio q/2m is called Bohr magneton, µb and g is known as Lande-g factor or gyromagnetic factor. The g-factor is 1 for
orbital angular momentum and hence corresponding magnetic moment is l µb (where l is
integer orbital angular momentum quantum number). A surprising feature here is that
spin magnetic moment is also µb instead of µb /2 as one would naively expect. So it turns
out g = 2 for spin – an effect that can only be understood if linearization of Schrödinger
equation is attempted i.e. make it consistent with special theory of relativity.
The algebra for spin angular momentum is considered very similar to the orbital angular
momentum algebra,
S 2 |s, ms i = s(s + 1)~2 |s, ms i
Sz |s, ms i = ms ~ |s, ms i,
−s ≤ ms ≤ +s
p
S± |s, ms i =
(s ∓ ms )(s ± ms + 1) ~ |s, ms ± 1i,
(58)
where S± = Sx ± iSy . Any component of spin angular momentum S has 2s + 1 eigenvalues
and therefore we expect ms to have 2s + 1 disctinct values between −s ≤ ms ≤ s. However,
unlike orbital angular momentum quantum number l, the spin angular momentum quantum
number can take both integer and half-integer values, s = 0, 1/2, 1, 3/2, 2, . . .. In fact, the
experimental result of Stern and Gerlach showing only 2 distinct µz can explained if s = 1/2
is considered bacause ms = ±1/2. Every elementary particle has specific value of spin s: π
has spin 0, electron spin is 1/2, photons have spin 1, deltas’ spins are 3/2 and so on. Matrix
representation for spin S can be derived using the equations in (58), the same way as L,
and they are,
3
1
1 0
1 0
S 2 = ~2
,
Sz = ~
,
0 1
0 −1
4
2
1
1
0 1
0 −i
Sx = ~
,
Sy = ~
,
1 0
i 0
2
2
0 1
0 0
S+ = ~
,
S− = ~
.
(59)
0 0
0 1
We may write this representation as
S=
1
~σ
2
1
(60)
where,
σx =
0 1
1 0
,
σy =
0 −i
i 0
,
σz =
1 0
0 −1
(61)
are the Pauli spin matrices. The commutation and anti-commutation relations obeyed by
Pauli matrices are,
[σi , σj ] = 2i ijk σk {σi , σj } = 2 δij .
(62)
A few other properties of Pauli spin matrices are,
σx2 = σy2 = σz2 = 1,
σx σy = i σz etc.,
σj† = σj
Tr (σj ) = 0, det (σj ) = −1.
(63)
These relations are peculiar to spin 1/2 representations and do not hold for l = 1 matrices.
The eigenstates of S for spin 1/2 particles will be represented by a two-component column
matrix, called spinor:

1


spin − up : χ+ =

0
a
, and |a|2 + |b|2 = 1. (64)
χ = a χ+ + b χ− =
0
b


spin − dn : χ− =

1
As we have discussed before, there is nothing special about the z-component of angular
momentum operator and just because the σz is diagonal the spin-states ±~/2 are easy to
see, otherwise any other directions are just fine. To illustrate this, letus try to figure out
the eigenvalues and eigenspinors of Sx . To find the eigenvalues, diagonalize the Sx ,
⇒
Sx |χ(x) i = λ |χ(x) i → (Sx − λ) |χ(x) i = 0
−λ ~/2 2
2
~/2 −λ = 0 ⇒ λ = (~/2) → λ = ±~/2
(65)
which says the spin eigenvalues are the same as that for Sz as it should be. However, the
space in which Sx is diagonal will obviously be different and they are obtained as,
~ 0 1
~
α
α
α
α
β
=
Sx
= ±
⇒
= ±
,
(66)
β
β
β
β
α
2 1 0
2
i.e. β = ±α. Hence the eigen spinors of Sx are,
1
~
α
1
(x)
χ+ =
=√
e − val :
α
2
2 1
1
~
α
1
(x)
χ− =
=√
e − val : − .
−α
2
2 −1
2
(67)