Projector Operator: P^2, Eigenvalues & Eigenfunctions

[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?

 

[LAHLH]

Try acting again with the operator P

 

[CompuChip]

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?

 

[Fredrik]

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

 

[alphaneutrino]

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

 

[sgd37]

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

 

[Matterwave]

|-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).

 

[Fredrik]

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?

 

[dextercioby]

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