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I understand that the Dirac equation has negative and positive sets of solutions and this contributes to its quantization by a superposition of two Fourier modes represented as creation and annihilation operators. What about a complex Dirac field for representing antiparticle fields?

I don't understand why a real field can't describe the antiparticles alone since it was the negative solutions of the real Dirac field which first sparked the antiparticle debate.

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Um...what do you mean by "real" and "complex" Dirac fields? Fermion fields take values in a spin-1/2 representation of the (cover of the) Lorentz group, they are not real or complex in the usual sense. –  ACuriousMind Jul 27 at 15:02
    
the complex scalar field which describes particles with charge q and antiparticles with charge -q, how is it described from the solutions of the dirac field? –  pkjag Jul 27 at 15:07
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If you're asking why there are no charged real fields: Charge needs a $\mathrm{U}(1)$ symmetry, which scalar complex Lagrangians have, while real scalars do not (it's only a discrete $\mathbb{Z}_2$ there). If that's not what you're asking, please write down what "the Dirac field" is for you and why you think it has anything to do with charges of other fields. –  ACuriousMind Jul 27 at 15:11

3 Answers 3

up vote 4 down vote accepted

Here is some reasoning for representation both of particles and antiparticles by one (complex in general, not only in usual sense but also in a sense of the irreducible representation of the Lorentz group) field.

All of QFT processes are described by S-matrix, which can be written in a form $$ S_{\alpha \beta} = \langle \beta | \hat{S} | \alpha \rangle , \quad \hat{S} = \hat{T}e^{i\int \hat{H}_{int}(x)d^4 x}. $$ Here are some restrictions on a $\hat{H}_{int}(x)$ which is given by the primary properties of QFT:

  1. $\hat{S} $ must be Poincare-invariant operator (because S-matrix is poincare-covariant quantity, which is a consequence of its definition). It is Poincare-invariant operator only if $[\hat{H}_{int}(x), \hat {H}_{int}(y) ] = 0, (x - y)^{2} < 0$ (the zero commutator for spacelike intervals). It can be easily understood if use the expansion of S-operator: $$ \hat {S} = \sum_{n}\frac{i^{n}}{n!}\int d^{4}x_{1}...d^{4}x_{n}\hat {T}\left( \hat {H}_{int}(x_{1})...\hat {H}_{int}(x_{n})\right). $$ $H_{int}(x)$ is poincare-invariant operator due to its definition, $\int d^{4}x$ is also poincare-invariant quantity. But time-ordering operation of two points $x, y$ is poincare invariant only in a case of timelike interval $(x - y)^{2} > 0$. So we must get zero commutator $[\hat{H}_{int}(x), \hat {H}_{int}(y) ]$.

  2. Due to causality principle, for spacelike intervals we must have $[\hat {H}_{int}(x), \hat {H}_{int}(y)] =0 $ too: operators must commute because the information about measuring does not have time to travel from $x$ to $y$.

But $\hat {H}_{int}(x)$ as poincare-covariant object is constructed from the fields as $$ \hat {H}_{int}(x) = \sum_{all \Psi}\sum_{A, B}F^{A_{1}...A_{n}B_{1}...B_{m}}\hat {\Psi}_{A_{1}}(x)...\hat{\Psi}_{A_{n}}(x)\hat {\Psi}^{\dagger}_{B_{1}}(x)\hat {\Psi}^{\dagger}_{B_{m}}(x). $$ So we must have $$ [\hat {\Psi}(x), \hat {\Psi}^{\dagger}(y)]_{\pm} = 0, \quad (x - y)^{2} < 0 . $$ This equality is possible only if $\hat {\Psi}$-field is constructed from the operators $$ \hat {\Psi}(x) = \sum_{\sigma}\int \frac{d^{3}\mathbf p}{\sqrt{(2 \pi)^{3}2E_{\mathbf p}}}\left( \hat {a}_{\sigma}(\mathbf p)e^{-ipx}u_{\sigma}(\mathbf p) + e^{ipx}\hat{b}^{\dagger}_{\sigma}(\mathbf p)v_{\sigma}(\mathbf p)\right), $$ where $\hat {b}^{\dagger}$ creates the particles with the mass of $\hat {a}^{\dagger}$ particles and with charge which is opposite to $\hat {a}^{\dagger}$ particles.

So we have the consequence: we must describe the particles as well as antiparticles by the one field $\hat {\Psi}$ because we need the poincare-covariant and causal QFT.

This result, of course, can be applied to your particular case.

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For fermions we have anticommutation relations $\{\psi(x),\psi(y)\}=0, x\neq y$ and these are postulated in the quantization to find out that $a,a^\dagger$ can be viewed as creation and annihilation operators. –  Void Jul 30 at 16:39
    
@void : Your statement is only the particular case of the Pauli theorem which is the result of the requirement of lorentz-covariance and causal QFT. For the details, read Weinberg QFT vol.1. –  Andrew McAddams Jul 30 at 17:14

As I showed in my article http://akhmeteli.org/wp-content/uploads/2011/08/JMAPAQ528082303_1.pdf (published in J. Math. Phys.), in a general case you can use one real function instead of the four complex components of the Dirac spinor, as 3 out of four components can be algebraically eliminated from the Dirac equation, and the remaining component can be made real by a gauge transform.

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Let us consider for a moment just a classical scalar field before quantization. For this field we have a Fourier decomposition: $$\phi = \int \frac{d^3 \!p}{...} (a_{\vec{p}}e^{ipx} + b^*_{\vec{p}} e^{-ipx})$$ Where $a_{\vec{p}}, b^*_{\vec{p}}$ are just numbers and the complex conjugation of $ b_{\vec{p}}$ is just a convention. In this description, the field can be real only if $a_{\vec{p}} = b_{\vec{p}}$ - such a condition will always give us $\phi^* = \phi$. Under quantization, this leads to the fact that the scalar particles are identical to their antiparticles.

On the other hand, if we did not impose the real-field condition, $b_{\vec{p}}^*, b_{\vec{p}}$ would lead to distinct creation and annihilation operators of antiparticles. I.e., for every field the "negative energy solutions" always correspond to antiparticles and for a scalar field, we can switch between particles/antiparticles by complex conjugation of the field.

For a Dirac field $\psi$ the structure of the Fourier decomposition is more complicated, we have to sum over spins and as we have a column vector with four components, we have to express the decomposition in a certain basis: $$\psi = \sum_s \int \frac{d^3 \! p}{...} (a^s_{\vec{p}} u^s_p e^{-ipx} + (b^{s}_{\vec{p}})^* v^s_p e^{ipx})$$ Where again $a,b$ are just numerical coefficients but $u^s_p, v^s_p$ are four-component column "vectors".

How do you even define if such a field is real or complex? The closest analogy of the complex conjugate for multi-component objects is the Hermitian conjugation which makes a complex conjugation of every component and reverts the column "vector" into a row. We denote this operation $\psi \to \psi^\dagger$. However, we find that the condition $\psi^\dagger = \psi$ is not covariant under Lorentz transforms and furthermore does not restrict antiparticles of the field. (Or more accurately, gives us a weird restricted mix of particles and antiparticles)

The correct condition for the particle-antiparticle equivalence is the "Charge conjugation" which requires $$\psi = \psi_C = i \gamma^2 \psi^\dagger$$ This condition actually gives us $a_{\vec{p}}^s = b_{\vec{p}}^s$ after some algebra and thus identifies the particles and antiparticles. Such a description can be used for neutrinos and does not in any way violate causality or covariance.

Nevertheless, as the name already suggests, a charge-self-conjugate spinor (giving rise to a so-called Majorana fermion) is not considered to be charged. A canonical answer why this is so would tell us that charge conservation would be broken.

I believe this would be true only for certain interaction terms in the Lagrangian. For a Dirac field $\psi$, we should be able to construct a Lagrangian from Majorana fields $\xi$ and an oppositely charged "anti-field" $\chi$ with the substitution $\psi \to \xi + \chi$ so that e.g. the electromagnetic interaction term would be $$ (\bar{\xi} + \bar{\chi})\gamma^\mu (\xi +\chi) A_\mu$$ By this process it should be possible to gain an equivalent Lagrangian to the one using the unrestricted Dirac spinor. The only argument why we don't do this is the fact that it is plainly unpractical and unelegant. Containing both the particle and anti-particle in one field just makes much more sense.

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