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48 Physics Formulary by ir. J.C.A. Wevers

10.10 Spin

For the spin operators are defined by their commutation relations: [S x, S y] = ihS z . Because the spin operators

do not act in the physical space (x,y,z) the uniqueness of the wavefunction is not a criterium here: also half odd-integer values are allowed for the spin. Because [L, S ] = 0 spin and angular momentum operators do not

have a common set of eigenfunctions. The spin operators are given by  S  = 1

2h σ, with

 σx =

0 11 0

,  σy =

0 −ii 0

,  σz =

1 00 −1

The eigenstates of  S z are called spinors: χ = α+χ+ + α−χ−, where χ+ = (1, 0) represents the state with

spin up (S z = 1

2h) and χ− = (0, 1) represents the state with spin down (S z = −1

2h). Then the probability

to find spin up after a measurement is given by |α+|2 and the chance to find spin down is given by |α−|

2. Of 

course holds |α+|2 + |α−|2 = 1.

The electron will have an intrinsic magnetic dipole moment  M  due to its spin, given by  M  = −egS  S/2m,

with gS  = 2(1 + α/2π + · · ·) the gyromagnetic ratio. In the presence of an external magnetic field this gives

a potential energy U  = −  M  ·  B. The Schrodinger equation then becomes (because ∂χ/∂xi ≡ 0):

ih∂χ(t)

∂t=

egS h

4mσ ·  Bχ(t)

with σ = ( σx,  σy,  σz). If   B = Bez there are two eigenvalues for this problem: χ± for E  = ±egS hB/4m =±hω. So the general solution is given by χ = (ae−iωt, beiωt). From this can be derived: S x = 1

2h cos(2ωt)

and S y = 1

2h sin(2ωt). Thus the spin precesses about the z-axis with frequency 2ω. This causes the normal

Zeeman splitting of spectral lines.

The potential operator for two particles with spin ± 1

2h is given by:

V (r) = V 1(r) +1

h2( S 1 ·  S 2)V 2(r) = V 1(r) + 1

2V 2(r)[S (S + 1) − 3

2]

This makes it possible for two states to exist: S  = 1 (triplet) or S  = 0 (Singlet).

10.11 The Dirac formalism

If the operators for p and E  are substituted in the relativistic equation E 2 = m20c4 + p2c2, the Klein-Gordon

equation is found: ∇2 −

1

c2∂ 2

∂t2−

m20c2

h2

ψ(x, t) = 0

The operator2− m20c2/h2 can be separated:

∇2 − 1c2 ∂ 

2

∂t2 − m

2

0c

2

h2 =

γ λ ∂ ∂xλ− m

2

0c

2

h2

γ µ ∂ ∂xµ+ m

2

0c

2

h2

where the Dirac matrices γ  are given by: γ λγ µ + γ µγ λ = 2δ λµ. From this it can be derived that the γ  are

hermitian 4 × 4 matrices given by:

γ k =

0 −iσk

iσk 0

, γ 4 =

I  00 −I 

With this, the Dirac equation becomes:

γ λ

∂ 

∂xλ

+m2

0c2

h2

ψ(x, t) = 0

where ψ(x) = (ψ1(x), ψ2(x), ψ3(x), ψ4(x)) is a spinor.