What is the intrinsic parity of anti-particles?

[dEdt]

I don't understand how it's possible for the intrinsic parity of any elementary particle to be anything other than one. The parity operator makes the transformation $\mathbf{x} \rightarrow -\mathbf{x}$, so the only thing about a state that can change after the parity operator is applied to it are the components of its wavefunction, right? ie for a spin 1/2 particle, $$P \left( \begin{array}{cc} \psi_{+}(\mathbf{x}) \\ \psi_{-}(\mathbf{x}) \end{array} \right)=\left( \begin{array}{cc} \psi_{+}(\mathbf{-x}) \\ \psi_{-}(\mathbf{-x}) \end{array} \right).$$ It's elementary to show that if a state has orbital angular momentum $l$, then its parity is $(-1)^l$.

But apparently some particles - specifically the antiparticles of fermions - have some sort of "intrinsic parity" equal to $-1$, which means that $$P \left( \begin{array}{cc} \psi_{+}(\mathbf{x}) \\ \psi_{-}(\mathbf{x}) \end{array} \right)=-\left( \begin{array}{cc} \psi_{+}(\mathbf{-x}) \\ \psi_{-}(\mathbf{-x}) \end{array} \right).$$
How is this possible? Where does that extra factor of $-1$ come from?

 

[The_Duck]

I don't understand how it's possible for the intrinsic parity of any elementary particle to be anything other than one. The parity operator makes the transformation $\mathbf{x} \rightarrow -\mathbf{x}$, so the only thing about a state that can change after the parity operator is applied to it are the components of its wavefunction, right?

This sort of reasoning doesn't fly in quantum mechanics. This kind of thinking would probably also lead you to also conclude that particles can only be bosons, because if you interchange two identical particles how can that possibly affect the wave function?

I think the proper way to think about such things is representation theory. In QM, symmetries correspond to groups of operators. Particles live in representations of these groups. In general you should expect that all possible representations will be realized. For example, the parity operator generates a simple group with only two operators: P and I, with the rule that $P^2 = I$. There are two possible representations of this group: P = 1 (positive parity) and P = -1 (negative parity). In both cases we have $P^2 = 1 = I$, as required. We should expect both possibilities to occur, because why not?

You object, "but when I defined parity I defined it to only act on spatial stuff, I didn't want it to notice any sort of internal properties of elementary particles!" But you don't actually define parity that way. If you want parity to be a good symmetry of your theory then you need an operator P such that $[P, H] = 0$ (parity is a good symmetry) and $P\vec{x}P = -\vec{x}$ and $P\vec{p}P = -\vec{p}$ (parity reverses positions and momenta). Nature can satisfy these commutation relations any way She likes: the only requirement is that $P$ have a representation on the Hilbert space. But that leaves open the possibility of the negative-parity representation, if Nature chooses to take advantage of it.

As to why specifically why the anti-particles of spin-1/2 particles have negative parity: Dirac fermions have a component with left-handed chirality and a component with right-handed chirality. Parity flips chirality (the mirror image of a left hand is a right hand). So in the "chiral basis" the parity operation on a Dirac spinor looks like

$$P \left( \begin{array}{cc} \psi_L \\ \psi_R \end{array} \right)=\left( \begin{array}{cc} \psi_R \\ \psi_L \end{array} \right)$$

This sort of interchange operation has two eigenvalues: +1 and -1, corresponding to the eigenvectors

$$\left( \begin{array}{cc} 1 \\ 1 \end{array} \right) \mbox{ and } \left( \begin{array}{cc} 1 \\ -1 \end{array} \right)$$

One of these eigenvectors corresponds to a fermion and the other to an antifermion, though I think it's a matter of convention which one you take to be the positive parity eigenvector.

 

[andrien]

One can take the quantized dirac field to show that positive energy and negative energy spinors behave differently under space inversion.As is known to everyone,that dirac field transforms under parity just as the wavefunction transforms and it is related by
ψ'(x',t)=γ₀ψ(x,t),where x' is -x.We can also write it as ψ'(x,t)=γ₀ψ(-x,t).Now one can use the plane wave expansion of the field operator and in that one will get terms like b^s_pγ₀u^s(p) and d^s_p⁺γ₀v^s(p),where these b and d appear as creation or annihilation operator in the expansion with their anticommutation rule.Now one can show that γ₀u^s(p)=u^s(-p) but γ₀v^s(p)=-v^s(-p).Note the minus sign here.It shows that positive energy spinor and negative energy spinor behave oppositely under parity.One can also deduce the transformation of creation and annihilation operator from here.The point is that as a particle with designation (p,s) goes into (p,-s),the negative energy part(positron here) will acquire a minus sign.As a consequence of this one can show the famous result that parity of a s-state electron positron system is odd.

 

[dauto]

That's why it is called intrinsic.