### 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