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Projector Operator

  • #1
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?
 

Answers and Replies

  • #2
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Try acting again with the operator P
 
  • #3
CompuChip
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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?
 
  • #4
Fredrik
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Alphaneutrino, you should state the definition of "projection operator" that you intend to use, and tell us where you get stuck.
 
  • #5
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
 
  • #6
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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 [tex]\pm1[/tex] and the most generic eigenfunctions I can think of are [tex] A(e^{kx} \pm e^{-kx}) ; k \in C [/tex] respectively for the +1 and -1 eigenvalues
 
  • #7
Matterwave
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|-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).
 
  • #8
Fredrik
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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?
 
  • #9
dextercioby
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Well, [itex] P^2 [/itex] is a projector. It can be shown using the definition of a projector operator on a Hilbert space.
 

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