# Projector Operator: P^2, Eigenvalues & Eigenfunctions

• alphaneutrino

#### alphaneutrino

An Operator is defined as P|x> = |-x>,
1. Is P^2 a Projector operator?
2. What are the eigen value and eigen function of P?

Try acting again with the operator P

Be careful, P is not a projector operator (P² is not equal to P); the question asks if P² is a projection operator.
alpha, What is the definition of such an operator?

Alphaneutrino, you should state the definition of "projection operator" that you intend to use, and tell us where you get stuck.

CompuChip said:
Be careful, P is not a projector operator (P² is not equal to P); the question asks if P² is a projection operator.
alpha, What is the definition of such an operator?

Thank you Compuchip!
I am asking about P. Yes, I know that P^2|x> = |x> which is not equal to p|x>. So it is not projection operator. My next confusion is can I write
|-x> = -|x> ?

How can we calculate the eigen value and eigen function of P

alphaneutrino said:
Thank you Compuchip!
I am asking about P. Yes, I know that P^2|x> = |x> which is not equal to p|x>. So it is not projection operator. My next confusion is can I write
|-x> = -|x> ?

How can we calculate the eigen value and eigen function of P

what you have there is a parity operator and they have eigenvalues $$\pm1$$ and the most generic eigenfunctions I can think of are $$A(e^{kx} \pm e^{-kx}) ; k \in C$$ respectively for the +1 and -1 eigenvalues

|-x>=-|x> only if the states have odd parity in x. Maybe it'd be clearer if you just used say f(-x) and -f(x).

Is |x> supposed to be a position "eigenstate" or an arbitrary state? If it's a position "eigenstate", then -|x> represents a particle located at x, and |-x> a particle located at -x, so these kets can't be the same unless x=0. If |x> is an arbitrary state, then what does |-x> mean?

Well, $P^2$ is a projector. It can be shown using the definition of a projector operator on a Hilbert space.