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Finding Charge Conjugation Eigenvalues 
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#1
Feb2012, 10:04 PM

P: 49

I've just recently been introduced to charge conjugation while reading the introductory particle physics texts by Griffiths and Perkins, and neither one really seem to explain how you go about finding the values for C.
For example, if I wanted to find the value for the [itex]\rho^0[/itex] meson (which I believe should be 1), the only real example in Perkins simply says [itex]C_{\rho} = 1 [/itex] since [itex] \rho^0 \rightarrow e^+e^[/itex], from which I assume that we (somehow) know that the [itex]e^+e^[/itex] system has C=1 and by charge conjugation conservation so does the [itex]\rho^0[/itex]. But how does actually go about figuring out the [itex]C_{e^+e^}[/itex]? 


#2
Feb2112, 05:53 AM

P: 90

Both the [itex]\rho^{0}[/itex] and the [itex]e^{+}e^{}[/itex] system are fermionantifermion states. For these states, the eigenvalues of C depend on the spin and orbital angular momentum of the system. The reason is that the C operator transforms a particle to its antiparticle, but doesn't change the spin or momentum. Therefore, in order to return to the initial state one needs to exchange the momenta in the spatial wave function which gives a factor [itex](1)^{L}[/itex] and the spin in the spin wave function which gives a factor [itex](1)^{S+1}[/itex] (when adding two spin 1/2, s=0 antisymmetric, s=1 symmetric). in addition there is another (1) factor due the exchange of two fermions. therefore, in total you get that C=[itex](1)^{L+S}[/itex]. In the quark model [itex]\rho^{0}[/itex] is a quark antiquark bound state with S=1 and L=0 (J=1), and therefore C=1. Meaning that a [itex]e^{+}e^{}[/itex] state that it would decay to must have an odd L+S.



#3
Feb2112, 10:58 AM

P: 49

Ah I see, that makes sense. So charge conjugation in fermions is similar to parity then (when you say you need to exchange angular momentum/spin to return to the original state), or is this not a good way to think about it?



#4
Feb2112, 03:38 PM

P: 90

Finding Charge Conjugation Eigenvalues
They are both discrete symmetries, but they act differently.
In the case of parity, the spin doesn't change and the particle identity doesn't change, it is only the momentum which changes sign. In order to return to the initial state one has to change the momenta sign in the orbital wave function, which gives a factor [itex](1)^{L}[/itex]. In addition, the intrinsic parities of a fermion and its antiparticle are of opposite sign, therefore the a fermionantifermion pair has odd intrinsic parity. This give another factor (1), giving a value of P=[itex](1)^{L+1}[/itex]. For example the pion, which has L=0, has P=1 (pseudoscalar). 


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