18.10.12Eigenvectors ofSu
As an example, lets take theudirection to be in thexzplane, between the positivexandzaxes,
30 degrees from the x axis. The unit vector is then ˆu= (cos(30), 0 ,sin(30)) = (
√
3
4 ,^0 ,1
2 ). We may
simply calculate the matrixSu= ˆu·S~.
Su=√
3
4
Sx+1
2
Sz=̄h
2
1
2√
3
√^4
3
4 −1
2
We expect the eigenvalues to be± ̄h 2 as for all axes.
Factoring out theh ̄ 2 , the equation for the eigenvectors is.
1
2√
3
√^4
3
4 −1
2
(
a
b)
=±
(
a
b)
1
2 a+√
3
√^4 b
3
4 a−1
2 b
=±
(
a
b)
For the positive eigenvalue, we havea=
√
3 b, giving the eigenvectorχ(+u)=(√
3
4
1
2)
. For the
negative eigenvalue, we havea=−
√
1
3 b, giving the eigenvectorχ(u)
− =(
√−^12
3
4)
. Of course each of
these could be multiplied by an arbitrary phase factor.
There is an alternate way to solve the problem using rotation matrices. We take the statesχ(±z)and
rotate the axes so that theuaxis is where thezaxis was. We must think carefully about exacty
what rotation to do. Clearly we need a rotation about theyaxis. Thinking about the signs carefully,
we see that a rotation of -60 degrees moves theuaxis to the oldzaxis.
Ry=(
cosθ 2 sinθ 2
−sinθ 2 cosθ 2)
Ry(−60) =(
cos(30) −sin(30)
sin(30) cos(30))
=
√
3
4 −1
2
1
2√
3
4
χ(+u)=
√
3
4 −1
2
1
2√
3
4
(
1
0
)
=
(√
3
4
1
2)
χ(−u)=
√
3
4 −1
2
1
2√
3
4
(
0
1
)
=
(
√−^21
3
4)
This gives the same answer. By using the rotation operator, the phase of the eigenvectors is consistent
with the choice made forχ(±z). For most problems, this is not important but it is for some.