S-1
Spin ½
Recall that in the H-atom solution, we showed that the fact that the wavefunction
Ψ
(r) is
single-valued requires that the angular momentum quantum nbr be integer:
l
= 0, 1, 2.
.
However, operator algebra allowed solutions
l
= 0, 1/2,
1,
3/2,
2…
Experiment shows that the electron possesses an intrinsic angular momentum called
spin
with
l
= ½.
By convention, we use the letter s instead of
l
for the spin angular
momentum quantum number : s = ½.
The existence of spin is not derivable from non-
relativistic QM.
It is not a form of orbital angular momentum; it cannot be derived from
.
(The electron is a point particle with radius r = 0.)
Lrp
=×
GG
G
Electrons, protons, neutrons, and quarks all possess spin s = ½.
Electrons and quarks are
elementary point particles (as far as we can tell) and have no internal structure.
However,
protons and neutrons are made of 3 quarks each.
The 3 half-spins of the quarks add to
produce a total spin of ½ for the composite particle (in a sense,
↑↑↓
makes a single
↑
).
Photons have spin 1, mesons have spin 0, the delta-particle has spin 3/2.
The graviton
has spin 2.
(Gravitons have not been detected experimentally, so this last statement is a
theoretical prediction.)
Spin and Magnetic Moment
We can detect and measure spin experimentally because the spin of a
charged particle is always associated with a magnetic moment.
Classically,
a magnetic moment is defined as a vector
µ
associated with a loop of
current.
The direction of
µ
is perpendicular to the plane of the current loop
(right-hand-rule), and the magnitude is
2
iA
i
r
µ
==
π
.
The connection
between orbital angular momentum (not spin) and magnetic moment can
be seen in the following classical model: Consider a particle with mass
m, charge q in circular orbit of radius r, speed v, period T.
µ
()
2
q2
r
q
v
q
v
i,v
i
i
A
r
TT
2
r
2
r
⎛⎞
π
⇒
=µ
=
= π
=
⎜⎟
ππ
⎝⎠
q
v
r
2
r
i
i
r
m, q
4/26/2008
Dubson, Phys3220