# Projection operator

1. Dec 1, 2013

### White_M

Hello,

Suppose P is a projection operator.
How can I show that I+P is inertible and find (I+P)^-1?
And is there a phisical meaning to a projection operator?

(Please be patient I have just started with QM).

Thanks.
Y.

2. Dec 1, 2013

### bhobba

Well I + |u><u| is diagonalizeable, hence invertable. If A is diagonalizeable it is of the form P(-1)DP where D is a diagonal matrix and hence easily and obviously invertable (eigenvalues non zero). Simply take the inverse to get P(-1)D(-1)P.

|u><u| is a projection operator and represents an operator that as an observable gives the expected value of observing a state to determine if its in state |u> - with 1 if it is in that state and zero otherwise ie the expected number of times it gives a true result.

This follows directly from the Born rule, which is the expected value of observing a system in state P with observable O is trace(OP).

Actually a projection operator in general form is Ʃ |ui><ui|, but I will leave you to do the grunt work of generalizing it - it's a good exercise.

Thanks
Bill

Last edited: Dec 1, 2013