Can a complex wave function just be seen as two real functions describing a particle and antiparticle state?

Question

Let's assume electromagnetism. There are two charges. The wave function is complex but can be seen canonically as a vector in $\mathbb{R}^2$. Can we see one of the components as the electron and the other one as the positron component?

But since the assignment \begin{equation} electron \rightarrow \begin{pmatrix} { \varphi }_{ 1 } \\ 0 \end{pmatrix} \end{equation} \begin{equation} positron \rightarrow \begin{pmatrix} 0 \\ { \varphi }_{ 2 } \end{pmatrix} \end{equation}

is arbitrary, we just say \begin{equation} electron \rightarrow \begin{pmatrix} { \varphi }_{ 1 } \\ { \varphi }_{ 2 } \end{pmatrix} \end{equation} is an electron and the complex conjugate \begin{equation} positron \rightarrow \begin{pmatrix} { \varphi }_{ 1 } \\ {- \varphi }_{ 2 } \end{pmatrix} \end{equation} is a positron. I am asking because I guess the same pattern holds for quarks and $SU(3)$. There are 3 color charges and 3 anti-charges and a quark is a vector \begin{equation} \begin{pmatrix} { \varphi }_{ r } \\ { \varphi }_{ g } \\ { \varphi }_{ b } \end{pmatrix} \end{equation} where $r,g,b$ are the colors. But it could be seen as a vector in $\mathbb{R}^6$ and the pattern above holds. I.e. is the complex structure in quantum mechanics just connected with the fact that there are particles and antiparticles? Are particles and antiparticles just the two (real) basis vectors for the complex plane?

Edit: Let's say we have a wave function $\varphi(x)$. We can always decompose it into

\begin{equation} \varphi= \begin{pmatrix} { \varphi }_{ 1 } \\ { \varphi }_{ 2 } \end{pmatrix} \end{equation} Charge is just the name for a basis vector in a space $\mathbb{C}^n$. E.g. for color charge we use $\mathbb{C}^3$, for electromagnetism only $\mathbb{C}$. Can we see $\varphi$ and $\varphi^*$ as the basis for $\mathbb{C}$ when seen as a real vector space? And why should $\varphi^*$ not be the anti-particle?

Answer 1

No. For example if you take a free electron moving with momentum $\vec{k}$, then its wavefunction is complex $\psi \sim e^{i \vec{k}\cdot \vec{x}}$. If for some reason you had a prejudice that caused you to prefer working with real and imaginary parts instead of complex numbers, you could write $\psi=\psi_R+i\psi_I$, with $\psi_R\sim \cos(\vec{k}\cdot\vec{x})$ and $\psi_I \sim \sin(\vec{k}\cdot\vec{x})$ both real. Let's say you somehow transformed the wavefunction so you projected out either the real or imaginary part, so $\psi \rightarrow \psi_R$ or $\psi\rightarrow \psi_I$. Either way, if you measured the electric charge you would always find it was $-e$. Since positrons have charge $+e$, not $-e$, neither $\psi_R$ nor $\psi_I$ can represent a positron.

Answer 2

Identifying the positron and electron component cannot be done as straightforwardly. In fact, there is no way to do this by a local operation on the field, the positron and electron component are hidden in the field global mode by global mode, as already stated by Andrew.

However, it is true that for a complex field $\phi$, it is often instructive to understand $\phi$ and its complex conjugate $\phi^*$ as independent fields. This is done for instance when deriving the equations of motion by varying the action functional. In principle, you can also write exactly as you suggest: $\phi \leftrightarrow (\varphi_1, \varphi_2)$ where $\varphi_1,\varphi_2$ are real and imaginary parts. Then you have operations such as $$\phi \to i \phi\; \leftrightarrow\; (\varphi_1,\varphi_2) \to (\varphi_1,\varphi_2)\cdot \begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}$$ or $$\phi \phi^* \leftrightarrow (\varphi_1,\varphi_2) \cdot \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\cdot (\varphi_1,\varphi_2)^T$$ The U(1) gauge transform then acts on this vector as $$e^{i\theta} \phi \; \leftrightarrow\; (\varphi_1,\varphi_2) \to (\varphi_1,\varphi_2)\cdot \begin{pmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos \theta\end{pmatrix} $$ I find this useful to show the similarities between U(1) gauge theory and SU(3) gauge theory - the gauge transform can be understood as always acting on vectors. And yes, again a correct observation that by using complex fields we are in fact implicitly working in $U(1)\times SU(3)$! As you say, we could in principle write this theory using $\mathbb{R}^2\times \mathbb{R}^3$ vectors, but it is almost never done.