axis, and the eigenvalues−^12 ,+^12 of this operator will be used to label the two
eigenstates, so
|+1
2
〉=
(
1
0
)
and |−1
2
〉=
(
0
1
)
Such eigenstates|+^12 〉and|−^12 〉provide a basis forC^2 , so an arbitrary
vector inHcan be written as
|ψ〉=α|+1
2
〉+β|−1
2
〉
forα,β∈C. Only ifαorβis 0 does the observableσ 3 correspond to a well-
defined number that characterizes the state and can be measured. This will be
either^12 (ifβ= 0 so the state is an eigenvector|+^12 〉), or−^12 (ifα= 0 so the
state is an eigenvector|−^12 〉).
An easy to check fact is that|+^12 〉and|−^12 〉are NOT eigenvectors for the
operatorsσ 1 andσ 2. One can also check that no pair of the threeσjcommute,
which implies that there are no vectors that are simultaneous eigenvectors for
more than oneσj. This non-commutativity of the operators is responsible for the
characteristic paradoxical property of quantum observables: there exist states
with a well defined number for the measured value of one observableσj, but
such states will not have a well-defined number for the measured value of the
other two non-commuting observables.
The physical description of this phenomenon in the realization of this system
as a spin^12 particle is that if one prepares states with a well-defined spin compo-
nent in thej-direction, the two other components of the spin can’t be assigned a
numerical value in such a state. Any attempt to prepare states that simultane-
ously have specific chosen numerical values for the 3 observables corresponding
to theσjis doomed to failure. So is any attempt to simultaneously measure
such values: if one measures the value for a particular observableσj, then going
on to measure one of the other two will ensure that the first measurement is no
longer valid (repeating it will not necessarily give the same thing). There are
many subtleties in the theory of measurement for quantum systems, but this
simple two-state example already shows some of the main features of how the
behavior of observables is quite different from that of classical physics.
While the basis vectors(
1
0
)
and(
0
1
)
are eigenvectors ofσ 3 ,σ 1 andσ 2take these basis vectors to non-trivial linear combinations of basis vectors. It
turns out that there are two specific linear combinations ofσ 1 andσ 2 that do
something very simple to the basis vectors. Since
(σ 1 +iσ 2 ) =(
0 2
0 0
)
and (σ 1 −iσ 2 ) =(
0 0
2 0
)
we have
(σ 1 +iσ 2 )(
0
1
)
= 2
(
1
0
)
(σ 1 +iσ 2 )