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

1. Feb 17, 2012

### andrewkirk

What am I doing wrong here?

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

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

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

2. Feb 17, 2012

### The_Duck

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>.

3. Feb 17, 2012

### andrewkirk

Thank you Duck. It all makes sense now.

I always thought using numbers to label kets was a bit dicey, and now I see why.
From now on I'll remind myself that
$|x\rangle$ is really $|\delta_x\rangle$

4. Feb 17, 2012

### jewbinson

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...