PDA

View Full Version : QM: Sum of projection operators = identity operator?


Simfish
Jun19-10, 01:41 PM
1. The problem statement, all variables and given/known data

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
Jun19-10, 02:16 PM
Hi Simfish! :wink:
… 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 ? :smile:

Simfish
Jun19-10, 02:18 PM
Hi. :) Okay, good idea. Thanks!