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1

Ok, here is my proof: Firstly, the $\gamma^\mu$ matrices, whatever the basis, must be traceless. Indeed, $\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}$ implies that 1) $\gamma^\mu\gamma^\nu$=$-\gamma^\nu\gamma^\mu$ when $\mu \neq \nu$, and 2) $\gamma^\nu\gamma^\nu=\pm I_4$ (no summation over $\nu$), and therefore $$2 tr(\gamma^\mu)=\pm 2 tr(\gamma^\nu\gamma^\nu\... 0 The bar notation is necessary in order to make all of the quantities above Lorentz invariant. In short the relationship between the bar'ed quantitites and charge conjugates of fermions have to do with taking the hermitian conjugate of the creation / annihilation opeartors. Let us focus for now on a free theory of Fermions. A Dirac Fermion can be written down ... -1 The Hamiltonian associated with the Dirac equation is not really the Hamiltonian for any physical system. The reason is that the "relativistic Hamiltonian" you speak of assumes that there is a fixed, finite number of fermions—and this is never really true. There is always the presence of the Dirac sea of negative-energy electrons (and muons, and ... 1 In QFT, when making a field operator, negative frequency solutions correspond to annihilation operators while positive frequency solutions correspond to creation operators. Not worrying about normalizations and things like that,$$ \hat \phi(x) \propto \int d^3 k \left( \hat a(k) e^{- i k \cdot x} + \hat a^\dagger(k) e^{ i k \cdot x} \right). $$Furthermore, ... 1 It might be up to you to recast the extra, nonminimal-coupling, unrenormalizable, gauge-invariant, hermitian, Pauli moment term, stuck in by hand, to accommodate arbitrary magnetic moments,$$ -{e\over 2M} \left (\frac{1}{2} F^{\mu\nu} \bar \psi \sigma _{\mu\nu}\psi\right )=i{e\over 2M} \left ( \partial^\nu A^\mu \bar \psi {[\gamma_\mu,\gamma_\nu]\over 4}\...

1

While I'm not sure why you would evaluate the commutator of a fermionic field, the process is the same for the anticommutator so the next reasoning is general. If you take the form of the field you gave, then the commutator (taken at two different spacetime points) \left[\psi(x),\psi^\dagger(y)\right] = \left[\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}}e^...

2

The order of $a_p^{s\dagger}$ and $u^{s\dagger}$ doesn't matter because $a_p^{s\dagger}$ doesn't act on $u^{s\dagger}$. $u^{s\dagger}$ is a finite dimensional vector, whereas $a_p^{s\dagger}$ is an operator that acts on the infinite dimensional space of field states.

1

The severity of this problem is exaggerated far too often. These actions (which have nothing wrong with them) look different because the first defines the Dirac bar by $\bar{\psi} = \psi^\dagger i \gamma^0$ while the second defines it by $\bar{\psi} = \psi^\dagger \gamma^0$. This is enough to show that both of them are S = \int dV \, \psi^\...

4

There is a different field for every type of particle. There is a $\psi_\tau$, $\psi_e$, $\psi_{\nu_e}$, etc. You describe a complex field just like you describe a complex number. You can treat it as a complex number $z$, or write it as $z=x+iy$ or as $z=re^{i\theta}$ with $x,y,r,\theta \in {\mathbb R}$. Whatever you find convenient! Real fields are ...

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