Difficulty Finding the EigenX of an Operator

[Spurious J]

1. The operator is:
[ 0 -i]
[ i 0]


2. I'm trying to find the eigenvalues and vectors. By inspection they appear to be
|i> = (0)
...(1)
and |-i> = (1)
...(0)

(or, easier to type, <i| = ( 0 1) and <-i| = (1 0) )

But these values don't work when I plug them back into the general equation for finding eigenvalues, moreover, when I attempt to simply solve for the values/vectors by hand, I obtain the same values, but then get zero fro ever component of the related eigenvectors

eg.
(0 -i) (a) = -i (a)
( i 0) (b) (b), which is -bi= -ai and ai= -bi

Thanks

 

[collinsmark]

1. The operator is:
[ 0 -i]
[ i 0]2. I'm trying to find the eigenvalues and vectors. By inspection they appear to be
|i> = (0)
...(1)
and |-i> = (1)
...(0)

(or, easier to type, <i| = ( 0 1) and <-i| = (1 0) )

Your eigenvalues are incorrect, if I'm interpreting your post correctly. It seems you are concluding that the eigenvalues are i and -i. But that is not correct.

Solve for λ in the for characteristic polynomial det(A - λI) = 0,

Where A is your operator matrix. In this case,

$$\mathrm{A} = \left[ \begin{array}{cc}<br /> 0 & -i \\<br /> i & 0<br /> \end{array} \right]$$

And λI is unity matrix times λ,

$$\lambda \mathrm{I} = \left[ \begin{array}{cc}<br /> \lambda & 0 \\<br /> 0 & \lambda<br /> \end{array} \right]$$

So in other words, your characteristic polynomial for this problem is:

$$\left| \begin{array}{cc}<br /> 0 - \lambda & -i \\<br /> i & 0 - \lambda<br /> \end{array} \right| = 0$$

Solve for λ. You'll end up with two possible values for λ. These are your eigenvalues.

Then find your eigenvectors.