# Charge Conjugation Operation

by maverick280857
Tags: charge, conjugation, operation
 P: 1,770 Hi, According to Perkins (4th edition, pg 73 section 3.6) the operation of charge conjugation reverses the sign of the charge and the magnetic moment of a particle. Does this mean the spin also flips? But according to Griffiths, the spin is untouched by charge conjugation. What operation flips a particle to its antiparticle? I'm a bit confused, because I wrote in my class notes that spin flips under charge conjugation. But I don't see how it should. Thanks in advance.
 P: 527 A particle and it's anti-particle have the same spin, so it doesn't change under charge conjugation. A particle's spin state is described by a state such as $|s,m\rangle$. The quantum number s is the spin, which is unaffected by charge conjugation, time reversal or parity. The quantum number m is the spin component along some axis and changes sign under time reversal alone.
P: 1,770
 Quote by xepma The quantum number s is the spin, which is unaffected by charge conjugation, time reversal or parity.
That is not correct. Time reversal flips the spin (Table 3.2, Perkins, page 82). Parity does not. Perhaps you meant something else?

P: 527

## Charge Conjugation Operation

It flips the spin component $n$, not the total spin $s$ of the particle. A negative, total spin doesn't exist. It's the magnitude of the spin. A negative spin-component does exist, and this indeed flips under time reversal.

Just for the record, the total spin is the eigenvalue of the spin operator squared, $S^2$. The spin component is the eigenvalue of the spin operator along some particular axis, $S_z$
P: 1,770

 Quote by xepma It flips the spin component $n$, not the total spin $s$ of the particle. A negative, total spin doesn't exist. It's the magnitude of the spin. A negative spin-component does exist, and this indeed flips under time reversal. Just for the record, the total spin is the eigenvalue of the spin operator squared, $S^2$. The spin component is the eigenvalue of the spin operator along some particular axis, $S_z$
Ok I think I see why I'm so confused. When you said total spin $s$, did you mean $s^2$?

Also, what does the notation $^{x}S_{y}$ mean? I know it means a singlet spin state, but what do x and y denote? So many holes in my atomic physics :-( [never did a course on atomic or nuclear physics. Did two courses on QM, never really encountered this notation.]

I have another question, which I think is related: http://www.physicsforums.com/showthread.php?t=375609.

EDIT: I think its just a matter of notation. Correct me if I'm wrong: you're saying $S_{z}$ flips sign under time reversal. The eigenvalue of $S_z$, denoted by $m_s$ therefore flips sign. The total spin angular momentum squared is [tex]S^2 = \boldsymbol{S}\cdot\boldsymbol{S}[/itex] and its eigenvalue is $s(s+1)$.

PS - Please have a look at the other question too.
 P: 527 Yea, by total spin I meant the $s$ in $s(s+1)$ which is the eigenvalue of the total spin angular momentum squared, $S^2$. I agree with you that it can be a little confusing, because "spin" can really refer to the total spin, but also the spin component along some axis. These are, ofcourse, not really interchangeable. I haven't seen the notation $^{x}S_{y}$ before.. do you have a reference for it? We all have gaps in our knowledge. No shame in that :)
P: 1,770
 Quote by xepma I haven't seen the notation $^{x}S_{y}$ before.. do you have a reference for it?
I'm sorry I think this is called S-state (?). I came across something like this with a P instead of an S, in Perkins. Its supposed to be a favorite thing to put in PGRE

 We all have gaps in our knowledge. No shame in that :)
Particle Physics exam tomorrow morning