Charge conjugated Dirac equation

Question

I would very much like to understand the motivation behind the correlation between:

$(i\partial\!\!/-eA\!\!/-m)\psi=0$

and

$(i\partial\!\!/+eA\!\!/-m)\psi_c=0$

when dealing with the derivation of the charge conjugation operator.

1. Why is the first equation only for electrons (says wiki-section: ”charge conjugation for Dirac fields”), when the Dirac equation has negative & positive energy solutions (upper components for particles and lower components for antiparticles)?

2. Mathematically it’s not relevant, but physically: did we negate the potential, or did we negate the charge when going from the first to the second equation? And with what purpose? (I guess the potential, right? Negating the charge wouldn’t make much sense, since the spinor-components of the solution have both charges as solutions either way.)

3. Can one simply say that $\psi$ is the same vector as $\psi_c$, but upside down?

4. You are welcome to give other details or explanation on why we start with such an ansatz at all, because this is what I’m trying to understand.

Answer 1

1. Diracs classical first order 4momentum-4velocity equation for the states of a free electrons $$i \hbar \partial_{ct} \psi = \ \left( \vec \alpha \cdot \vec p + \beta m c \right) \ \psi $$ has solutions with complex phase $$\psi = e^{i\left( \omega t- \vec k \cdot \vec x\right)} \phi$$

$$\omega \ \phi(\omega, \vec k) = \left( \vec \alpha \cdot \vec k + \beta \frac{m c}{\hbar} \right) \ \phi(\omega.\vec k) $$

Its clear immediately, that negative values of $\omega$ for small $\vec k$ belong to the the negative eigenvalues of the matrix $\beta$.

2. Since masses and frequencies are positive by the very concept of gravity and direction of time, the representation of the particle-antiparticle map must be a product of commuting discrete operations, being their own inverses, applied to the full equation with an external, quasistatic em 4-potential $\Phi=A_0, \vec A$

Best expressed in its scalar form of positivity of the Compton wavelength $$\beta\left( i \partial_{ct} - \frac{e}{\hbar c}\ A_0(t,x) - \vec \alpha \cdot (-i \nabla -\frac{e}{\hbar c} \vec A(t,x) ) \right) \psi =\frac{mc}{\hbar} \ \psi$$

3. It's the mathematical conundrum of the Clifford algeba $\mathit {Cl}(\mathbb R, 3,1)$ that is can be written as a product of two commuting sets of commuting Pauli matrices $\sigma_k, \tau_l$ such that its basis is $$1, \sigma_{1,2,3}, \tau_{1,2,3},\ \text{with} \ (\sigma,\tau)_i (\sigma,\tau)_k + (\sigma,\tau)_k (\sigma,\tau)_i =2\delta_{ik},\quad \sigma_i \tau_k = \tau_k \sigma_i $$

$$\tau_3 \left( i \partial_{ct} - \frac{e}{\hbar c}\ A_0(t,x) - \tau_1 \vec \sigma \cdot (-i \nabla -\frac{e}{\hbar c} \vec A(t,x) ) \right)\ \psi =\frac{mc}{\hbar}\ \psi$$

Using the operations complex conjugate $*$, transpose $ ^t$, parity $y\to -y$ and multiplication on both sides with products of $\tau_k$, its indeed possible to change all signs twice, except for those of the electromagnetic terms $ \tau_1 A_0, \tau_3 \tau_1 \vec \sigma \vec A$ that change sign once.

The critical point: Physicists are used to represent parity by inversion of all three coordinates. Thats plain wrong in even dimensions. The volume element changes sign by a single coordinate reflection.

Here the interplay of reflexion and complex conjugate acts by the fact, that in the standard representation of Pauli, $\sigma_2 (-i\partial_y)$ is real, so the parity transform has to change $\tau_1 \sigma_2 \to -\tau_1 \sigma_2$ while the three other derivatives remain unchanged. The rest of the $\gamma$-gymnastics is left to the reader.

Why PAM Dirac was so lucky to find his equation with such a subtle algebra representing the full Poincaré group with its weak point of parity, and not Pauli?

A speciality of Cambridge (UK) as a place - rather retarded in 1900 modern mathematics in the tradtions of Newton - of geometric algebra in the succession of Hamiltons invention of quaternions as a extension of Gauss complex plane algebra to three dimensions.

By use quaternion algebra, already Maxwell was enabled to find the fully relativistic electromagnetic theory, while on the continent everybody in mthematical physics was concerned with analytic functions, partial differential equations, Riemannian geometry and understanding statistical thermodynamics.

Answer 2

If you make the choice that the Dirac field $\psi$ should be the "electron field", the corresponding field operator describes electrons $e^-$ (having electromagnetic charge $-e$) as "particles" and positrons $e^+$ (with electromagnetic charge $+e$) as "antiparticles". Calling electrons "particles" and positrons "antiparticles" is, of course, a pure convention. There is no physics in conventions. You could equally well call the positrons "particles" with the electrons as their associated "antiparticles". In this case, the "positron field" would be described by some Dirac field $\chi$, where the "particles" are positrons and the "antiparticles" are electrons. As both fields describe the same physics (namely electrons and positrons), the Dirac fields $\psi$ and $\chi$ must be related. Obviously, $\chi$ is nothing else than the charge conjugated field $\psi^c$ of $\psi$ (and vice versa).

The usual text-book strategy to find the mathematical relation between $\psi$ and $\psi^c$ is to consider the Dirac equations for $\psi$ and $\psi^c$ in the presence of an external electromagnetic field described by the $4$-potential $A_\mu(x)$. Starting with the equation for the electron field $\psi$ (charge $-e$) given by $(i \partial \!\!\!/-e A \!\!\!/-m) \psi=0$, it is obvious that the corresponding equation for the positron field $\psi^c$ (charge $+e$) must be of the same form with $\psi \to \psi^c$ and $-e \to e$. It is now an easy task to find $\psi^c = C \bar{\psi}^T$ and the properties of the charge conjugation matrix $C$.