In Chapters 34-36 of the Srednicki QFT book, 2-component spinors and their combinations in Dirac and Majorana spinors are carefully constructed. Specifically, in equations 36.14 and 36.15 the following left-handed spinors are defined:

$$ \begin{equation} \begin{split} &\chi = \frac{1}{\sqrt{2}}(\psi_1 + i\psi_2)\\ &\xi = \frac{1}{\sqrt{2}}(\psi_1 - i\psi_2) \end{split} \end{equation} $$

A couple of pages later, charge conjugation is then defined as "Charge conjugation simply exchanges $\chi$ and $\xi$".

Now, I have read in other books (e.g. Georgi) that if a particle is defined in one SU(2) representation of the Lorentz group (e.g. as left handed spinor), then the corresponding anti-particle is in the other representation (e.g. as right handed spinor).However, in the above definition in Srednicki, both fermion and anti-fermion are left-handed spinors.

Have I got this right? How can both of these statements be correct? Any help to clear up my confusion would be greatly appreciated.

As an aside, I have a more general issue understanding the intuition behind Dirac spinors. The state of a general fermion or anti-fermion is often given with a Dirac spinor $\Psi$. If this fermion has mass, then the fermion is a combination of a left-handed and right handed spinor, as in Srednicki: $$ \Psi = \begin{pmatrix} \chi_c \\ \xi^{\dagger\dot{c}} \end{pmatrix}. $$

As helicity and handedness are frame dependent, I can kind of understand how the Dirac fermion can be a combination of the two above. However, Dirac fermions can also by viewed as a combination of a fermion and anti-fermion in much the same way. I do not fully understand this. How can a fermion with mass be described by a field that is partly particle and partly anti-particle?

Again, any insights that the community could provide would be much appreciated.

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    $\begingroup$ I'm working on turning this into an answer, but I think this will be a useful read. (If someone else wants to post an answer before I get to it, go right ahead.) $\endgroup$
    – David Z
    Commented Jun 9, 2014 at 22:52
  • $\begingroup$ Somebody asked me the same question (in private) a month or so ago; she returned to me with this pdf which helped her immensely; I wonder if it helps you too, c.f., pages 199-201 physics.indiana.edu/~dermisek/QFT_09/qft-I-9-4p.pdf $\endgroup$
    – Autolatry
    Commented Nov 13, 2014 at 9:11

1 Answer 1


Fermion is left-handed spinor and anti-fermion is right-handed spinor.

Firstly, you can see that $\xi = \chi ^{\dagger}$. From Sredniski, "... that hermitian conjugation swaps the two SU(2) Lie algebras...the hermitian conjugate of a field in the (2,1) representation should be a field in the (1,2) representation"... Thus $\chi$ is a right-handed fermion while $\xi$ the opposite.

Maybe you should read Chap.34, especially between eqn. (34.10) to (34.13)...


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