QM: Sum of projection operators = identity operator?

[Simfish]

Homework Statement

So we have an observable K = $$\begin{bmatrix} 0 & -i \\ -i & 0 \end{bmatrix}$$

and its eigenvectors are v1 = (-i, 1)^T and v2 = (i, 1)^T corresponding to eigenvalues 1 and -1, respectively.

Now if we take the outer products, we get these...

|1><1| = (-i, 1)^T*(i, 1) = $$\begin{bmatrix} 1 & -i \\ i & 1 \end{bmatrix}$$

|-1><-1| = (i, 1)^T*(-i, 1) = $$\begin{bmatrix} 1 & i \\ -i & 1 \end{bmatrix}$$

Then we add them and they sum up to form 2*(identity operator).

But isn't it supposed to sum up to the identity operator? What's wrong, or what happened?

 

[tiny-tim]

Hi Simfish!

… and its eigenvectors are v1 = (-i, 1)^T and v2 = (i, 1)^T …

Don't you need to normalise them (to unit vectors), by multiplying by 1/√2 ?

 

[Simfish]

Hi. :) Okay, good idea. Thanks!