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Eigenvectors of
for Spin

First the quick solution. Since there is no difference between x and z,
we know the eigenvalues of
must be
.
So, factoring out the constant, we have

These are the eigenvectors of
. We see that if we are in an
eigenstate of
the spin measured in the z direction is equally
likely to be up and down since the absolute square of either amplitude is
.
The remainder of this section goes into more detail on this calculation but
is currently notationally challenged.

Recall the standard method of finding eigenvectors and eigenvalues:

For spin
system we have, in matrix notation,

For a matrix times a nonzero vector to give zero, the determinant of the
matrix must be zero.
This gives the ``characteristic
equation'' which for spin
systems will be a quadratic equation in
the eigenvalue
:

whose solution is

To find the eigenvectors, we simply replace (one at a time) each of the
eigenvalues above into the equation

and solve for
and
.
Now specifically, for the operator
, the eigenvalue equation
becomes, in
matrix notation,

The characteristic equation is
, or

These are the two eigenvalues (we knew this, of course). Now, substituting
back into the eigenvalue equation, we obtain

The last equality is satisfied only if
(just write out the
two component equations to see this). Hence the normalized eigenvector
corresponding to the eigenvalue
is

Similarly, we find for the eigenvalue
,

Jim Branson
2013-04-22