Charge conjugation and spatial wave function

[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!

 

[ofirg]

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!

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 $(-1)^{L}$.

 

[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)