# Charge Conjugate of the Vacuum

1. ### ChrisVer

What is the result of the charge conjugation acting on the state of vacuum?
$C|0>=...$
I have two intuitive problems.... If I see the vacuum as something which has no particles, then the charge conjugate would have to lead in the vacuum itself...

$C|0>=|0>$

However, if I think of the vacuum as the state on which creation operators act to create states, the above result doesn't seem correct. For example I would guess that the C acting on the vacuum would create a new vacuum:

$C|0>=|\bar{0}>$
On which now the creation operators will act destructively (because they would have to create particle in the antiparticle vacuum- the exact opposite of destruction operator acting on the initial vacuum).... On the other hand the destruction operators will bring out antiparticle states....

I am seeing the creation operators as particles and destruction operators as antiparticles, since they propagate (in momentum space) to opposite directions...

Which is correct (I already know it's the 1st) and more importantly, why?

Last edited: May 15, 2014
2. ### The_Duck

963
C|0> = |0> in a theory with charge conjugation invariance.

I think maybe you are a little confused about creation/annihilation operators and particles/antiparticles. There are separate creation/annihilation operators for particles and antiparticles. C swaps the particle creation operators with the antiparticle creation operators, and similarly for the annihilation operators. |0> is annihilated by both particle and antiparticle annihilation operators.

For example, let ##a^\dagger## create particles and ##b^\dagger## create antiparticles. The charge conjugation operator obeys ##C a^\dagger C = b^\dagger## and ##C b^\dagger C = a^\dagger## and ##CC = 1##. Then if we act on a state of a single particle with ##C##, we get a state of one antiparticle, as expected:

##C a^\dagger | 0 \rangle = C a^\dagger CC | 0 \rangle = b^\dagger C | 0 \rangle = b^\dagger |0\rangle##.

Note that we used the fact that ##C|0\rangle = |0\rangle##.

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3. ### Bill_K

4,159
Yes, bearing this in mind...