My bra-ket calcs seem to be going wrong - help

  • Context: Graduate 
  • Thread starter Thread starter andrewkirk
  • Start date Start date
  • Tags Tags
    Bra-ket
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 2K views
Messages
4,140
Reaction score
1,741
What am I doing wrong here?

Let [itex]\psi[/itex] be a ket whose representation in the X basis is given by
[itex]\psi(x)\ =\ \langle x|\psi\rangle\ =\ e^{-x^{2}/2}[/itex]

Then
[itex]\psi(-x)\ =\ \langle -x|\psi\rangle\ =\ e^{-x^{2}/2}\ = \psi(x)[/itex] (1)

But we also have:
[itex]\psi(-x)\ =\ \langle -x|\psi\rangle[/itex] (2)
[itex]\ =\ \langle (-1)\times x)|\psi\rangle[/itex] (3), by the linearity of the inner product
[itex]\ =\ (-1)^*\times\langle x|\psi\rangle[/itex] (4)
[itex]\ =\ -\langle x|\psi\rangle[/itex] (5)
[itex]\ = -\psi(x)[/itex] (6)

and this contradicts (1).

I must have gone wrong here somewhere. I think it might be in (2) or (3). But I can't see the problem.

Thank you very much for any help.
 
Physics news on Phys.org
Going from (3) to (4) is wrong. The ket |x> is the eigenket of the position operator with eigenvalue x. The notation generally |-x> means the eigenket of the position operator with eigenvalue -x. It is NOT "the operator -1 acting on the eigenket |x>." This confusion is understandable since we sometimes write "the ket obtained by acting on the ket |ψ> with the operator A" as |Aψ>; however that is not the meaning intended in this case.

So you need to be clear on the distinction between -|x> and |-x>. For instance, while both are eigenstates of the position operator, the first has eigenvalue x, while the second has eigenvalue -x. The first one is a multiple of the ket |x>, while the second one is completely orthogonal to |x>.
 
I agree with what The Duck said. I will take this a bit further for your curiosity's sake.

Define the Parity operator P by

P|x> = |-x>.

The eigenvalues of P can be either 1 or -1 (try and prove this or tell me if you can't. Hint for proof: consider P^2 and it's eigenvalues).

These are the ONLY eigenvalues P can take.

In the -1 (odd) case, we have P|x> = -|x>, and since P|x> = |-x>, we get |-x> = -|x>.

This is the case you have described here.

In the other case, (eigenvalue of P is +1), you would get

-|-x> = |x>.

Edit: I'm not sure if what I've described is exactly the same as what you have in the OP because you are doing the inner product whereas I am acting an operator on |x>. Food for thought...