# I Charge conjugation and spatial wave function

1. Jul 14, 2017

### Josh1079

Hi,

I'm recently reading something which briefly introduces C symmetry. So the thing that confuses me is that how does the spatial wave function contribute the (-1)^L factor?

Thanks!

2. Jul 19, 2017

### PF_Help_Bot

Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

3. Jul 30, 2017

### ofirg

It would be beneficial to point to your reference.

But I would imagine you're referring to a particle-anti particle pair after a C symmetry transformation. The $(-1)^{L}$ factor comes from exchanging the particle coordinates in the spatial wave function after applying the C operator to return the state to its previous appearance. The parity of the spatial wave functions under that exchange is $(-1)^{L}$.

4. Jul 30, 2017

### ChrisVer

Suppose you have a particle $A$ and an antiparticle $\bar{A}$... the system of two together is an eigenstate of the charge conjugation operator $C$... that is because:
$C |A \bar{A}> = |\bar{A} A> =^{(?)} \lambda_C |A \bar{A}>$
See the questionmark.... again a reminder: a state $|a>$ is an eigenstate of an operator $O$ with eigenvalue $c_a$ if the following equation holds: $O |a> = c_a |a>$.

Now again you asked where does the $(-1)^L$ comes from. Well, L is the angular momentum of the system.... This becomes obvious if you drew the particle-antiparticle pair, but on maths it becomes obvious if you assign to them their position $x_{A}, \bar{x}_A$ for the particle and antiparticle respectively.
$C |A (x_A) \bar{A}(\bar{x}_A)> = |\bar{A}(x_A) A(\bar{x}_A)> = (-1)^L |A(x_A) \bar{A}(\bar{x}_A)>$
Since the middle step of the above equation is the parity applied on A which was on x_A and Abar which was on barx_A (got the positions exchanged)
(that's what happens in the spatial-coord space)