3
$\begingroup$

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?

$\endgroup$
2
6
$\begingroup$

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.

$\endgroup$
10
  • $\begingroup$ Thanks. I think antiparticles are the wrong term. My question comes from gauge theory. And what we call positive and negative charge is just arbitrary. That is the symmetry I am looking for. $\endgroup$
    – NicAG
    Jul 9 at 20:00
  • 1
    $\begingroup$ @NicAG It's true the labels positive and negative (or particle or antiparticle) are arbitrary. But there is a physical difference between positive and negative, or particle and anti-particle. They are different objects (they are even different in gauge theory :)). And they are not related by a rotation of the phase of the wavefunction. Anyway I'd be happy to look again at the question if you are able to more precisely formulate what you want to know. $\endgroup$
    – Andrew
    Jul 9 at 20:03
  • $\begingroup$ Thanks. I have edited the question $\endgroup$
    – NicAG
    Jul 9 at 20:17
  • $\begingroup$ @NicAG I don't understand the premise of your question. I agree that the wavefunction of an electron is a complex valued function, and I agree that you can split any complex number into real and imaginary parts (I take this to be your "canonical way of splitting a complex number into a vector in $R^2$). But I don't understand how you get from there to identifying the first component of the "vector" (ie the real part of the complex number) with an electron and the second component with the positron. My answer shows that this identification is not a consistent thing to do. $\endgroup$
    – Andrew
    Jul 9 at 20:20
  • $\begingroup$ Maybe you are confused because in field theory one writes a four component Dirac spinor and loosely speaking one can think of 2 components as representing the electron and 2 as the positron. But that is a very different construction from your question; for example each component in the spinor is a complex number $\endgroup$
    – Andrew
    Jul 9 at 20:21
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Thanks for the answer. Can you maybe elaborate a bit further on your first paragraph. My idea is that the complex dimension of my representation space is the number of charges. And the real dimension the number of charges and anti-charges. Why is $\varphi^*$ not the anti-particle, when we deal with just spin-0 and the Schrödinger or Klein-Gordon theory, in a first quantizes theory $\endgroup$
    – NicAG
    Jul 9 at 20:47
  • $\begingroup$ $U(1)\times SU(3)=U(3)$, so I'm not sure it's correct to say that using complex fields means we work with $U(1)\times SU(3)$. For example, $U(3)$ doesn't work as a gauge theory for the strong interactions because it implies there are 9 gluons instead of 8. I suppose if by the $U(1)$ factor here, you mean the $U(1)$ hypercharge gauge symmetry, then yes part of the gauge group is $U(1)_Y \times SU(3)_c$, but it is not true that using complex fields implies there must be a $U(1)_Y$ gauge symmetry, nor is the converse true if you are willing to work with real representations. $\endgroup$
    – Andrew
    Jul 9 at 23:25

Not the answer you're looking for? Browse other questions tagged or ask your own question.