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I am having a bit of a trouble understanding how a parity transformation acts on Dirac spinors with a well-defined chirality and, in particular, the (intuitively correct, since chirality is related to helicity for massless particles) sentence parity transforms left-handed fields into right handed-fields. Take a general parity transformation to be:

$$ \hat{P} \psi(t,\mathbf{x}) \hat{P}^{-1} = \gamma^0 \psi(t, -\mathbf{x}) $$

Now suppose we have $\psi(t, \mathbf{x})$ a solution to the massless Dirac equation with, say, left-handed chirality ($\gamma^5 \psi = -\psi$). It seems correct then the previous sentence concerning the chirality of the P-transformed field, since:

$$ \gamma^{5} \hat{P} \psi(t,\mathbf{x}) \hat{P}^{-1} = - \gamma^0 \gamma^5 \psi(t, -\mathbf{x}) = \gamma^0 \psi(t, -\mathbf{x}) = \hat{P} \psi(t,\mathbf{x}) \hat{P}^{-1} $$

So this $\hat{P} \psi(t,\mathbf{x}) \hat{P}^{-1}$ is indeed right-handed. This also agrees with the following observation for a general (not necessarily massless) field with a non-defined chirality:

$$ P_L \hat{P} \psi(t,\mathbf{x}) \hat{P}^{-1} = \gamma^0 P_R \psi(t, -\mathbf{x}) = \gamma^0 \psi_R(t, -\mathbf{x}) $$ $$ P_R \hat{P} \psi(t,\mathbf{x}) \hat{P}^{-1} = \gamma^0 P_L \psi(t, -\mathbf{x}) = \gamma^0 \psi_L(t, -\mathbf{x}) $$

where $P_{R,L} = \frac{1}{2}(1\pm \gamma^5)$ are the usual chirality projectors. In the previous discussion with a left-handed massless field, since $\psi=\psi_L$ and $\psi_R=0$, we see that the transformed field gives zero when acted on with $P_L$, confirming that the transformed field is right-handed.

So everything seems to work fine until now. The problem comes when I try a different approach to the above calculations. Consider again the left-handed massless field $\psi=P_L \psi = \psi_L$. Since $\hat{P}$ is an operator on our Hilbert space, it commutes with gamma matrices (and therefore with projectors $P_{L,R}$). Then:

$$ \hat{P} \psi(t,\mathbf{x}) \hat{P}^{-1} = \hat{P} P_L \psi(t,\mathbf{x}) \hat{P}^{-1} = P_L \gamma^0 \psi(t, -\mathbf{x}) = \gamma^0 P_R \psi(t, -\mathbf{x}) = 0 $$

What is going on? It seems that our parity transformed field is zero now!

I am quite sure that this last argument is somehow wrong (mainly because all the previous results seemed more sensible, but also because I have tried with a particular example of a massless left-handed field and the result after parity transformation is not zero), so the question is, where is the mistake? I have been trying to find it for a while, but every time I look at the previous argument I find it perfectly valid. By the way, this problem also shows up when you try to discover the chirality of a field after charge conjugation or time reversal, but since I think the solution is the same in every case a good answer for the parity case should be more than enough!

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In most of your derivations, you have used the symbol $\psi$ for the genuine field operator (operator distribution). But your equation $$ \gamma^5 \psi = -\psi$$ clearly doesn't work for any Dirac field. This equation an operator equation equivalent to $$ (1+\gamma^5) \psi = 0$$ which says that one-half of the components of $\psi$ are zero as operators. But that's clearly impossible because all components of $\psi$ are operators in a dynamical theory – for example, regardless of the state of the system $|\phi\rangle$, there exist other operators that these operators don't commute (or anticommute) with, while the operator equation vanishing says that these two components are completely absent, zero as $c$-numbers.

In effect, this assumption of yours has "killed" one-half of the operators in a way that can't occur in QFT, and you saw this disappearance of one-half of the degrees of freedom at the end, too. You may only impose $(1+\gamma^5) u =0$ on particular classical solutions (which may be terms in the expansion of the operators), not on the whole operators.

In effect, what you assumed was that your theory only contained the chiral left-handed operator, and you just derived that in that case, the mirror image i.e. right-handed part of the operator was zero. It's common sense but your assumption that was shown to be problematic above was an assumption that the field was a Weyl field, not a Dirac field.

And for a Weyl field, i.e. with the field constrained by $(1+\gamma^5)\psi=0$, there is no natural definition of parity $P$ – the theory with a Weyl field and nothing else isn't invariant under parity.

Let me say the basic point in a slightly different way. The OP is mixing classical fields and quantum fields and thinks that the quantum field remembers the "state" just like the classical field did. But that's simply not true. It's the state vector (or density matrix) that remembers the state – or that is always needed to calculate any predictions from the operators. The operator itself is "always the same" up to a unitary transformation. Its evolution (if it evolves at all, i.e. in the Heisenberg picture) is given by a unitary transformation $\psi \to U \psi U^\dagger$ which doesn't change its spectrum at all. So if we learn that the operator is nonzero (it has nonzero eigenvalues), it just can't be equal to zero after any evolution or for any state: the evolution doesn't affect the spectrum of the operator at all!

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    $\begingroup$ +1, really dont see why you got downvoted $\endgroup$ – Wolpertinger May 13 '16 at 14:22
  • $\begingroup$ Thanks for the synergy, Numrok. But no prob. $\endgroup$ – Luboš Motl May 13 '16 at 15:58
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    $\begingroup$ Thank you, that was helpful! So, just to check I got it right, although you can split solutions to the massless Dirac equation (understood as a classical field equation) into positive chirality and negative chirality solutions (using the eigencalue of $\gamma^5$), this does not work when considering the Dirac field as a quantum field. In that case the only thing you can do is to isolate the left- and right-handed parts using the projector, and that is what you do e.g. when coupling the left-handed part of the electron field to the weak bosons but not the right-handed part. Is that right? $\endgroup$ – Alex V. May 20 '16 at 9:56
  • $\begingroup$ Dear @AlexV., if I understand well, I think it's right. But maybe I should have said it: you can have a counterpart of $\gamma_5$ at the level of the multi-particle Fock space, in QFT. But you must replace the simple $\gamma_5$ operation acting on components of the Dirac field by an operator that is bilinear in the Dirac field or exponential of bilinear. The operator "upgrading" $\gamma_5$ is $O=(-1)^N$ where $N$ is the $\int\bar\psi\psi$-style number operator counting electrons created in the $\gamma_5=-1$ state. On one-particle states, $O$ acts like $\gamma_5$ did in the one-particle theory. $\endgroup$ – Luboš Motl May 20 '16 at 10:03
  • $\begingroup$ This is quite a general way of the old-fashioned "second quantization" in which the operators from the one-particle quantum mechanical theory can be promoted to similar operators, sesquilinear in the field $\hat \psi$, acting on the multi-particle Hilbert space. All these operators have the form of $H=\psi^\dagger_j \psi_k h_{jk}$ when matrix elements $h_{jk}$ are those from the one-particle theory. If one has $[\psi_j,\psi^\dagger_k]=\delta_{jk}$, the operators $H$ obey the same algebra as the one-particle operators $h$. In this language, you have mixed up the 1st and 2nd quantized operators. $\endgroup$ – Luboš Motl May 20 '16 at 10:06

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