# What is really interacting in weak interactions?

Only particles with chirality $-1$ do interact weakly. The corresponding eigenstate in the Dirac basis is $\Psi_L = \begin{pmatrix}f \\ -f \end{pmatrix} = \begin{pmatrix}u_r {\mathrm{e}}^{-imt} \\ -u_r {\mathrm{e}}^{-imt} \end{pmatrix}$, whereas a pure particle state, say for a muon is $\Psi_{\mu^-}= \begin{pmatrix}f \\ 0 \end{pmatrix} = \frac{1}{2} \Big( (\Psi_{\mu^-})_L + (\Psi_{\mu^-})_R \Big)= \frac{1}{2} \Big ( \begin{pmatrix}f \\ -f \end{pmatrix} + \begin{pmatrix}f \\ f \end{pmatrix} \Big )$. The lower two component object inside a Dirac spinor in this basis can be related to antiparticles, and my problem is understanding the lower component that takes part in the weak interaction of a pure $\mu^-$, because it cannot represent some $\mu^+$ mixture that always goes with $\mu^-$ (because of charge conservation.)

Some background

We can solve the Dirac equation in the rest frame in the Dirac basis, which yields four independent solutions

$$U_r= \begin{pmatrix} u_r \\ 0 \end{pmatrix} {\mathrm{e}}^{-imt} \qquad \text{and} \qquad V_r= \begin{pmatrix} 0 \\ v_r \end{pmatrix} {\mathrm{e}}^{+imt}$$

with $u_1 = v_1= \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $u_2 = v_2= \begin{pmatrix} 0 \\ 1 \end{pmatrix}$.

The $U_r$ spinors are directly connected to particle states with different spin configuration, whereas $V_r$ correspond to antiparticle states. This can be seen for example in QFT where a general solution of the Dirac equation is used

\begin{align} \Psi &= \sum_r \sqrt{\frac{m}{(2\pi)^3}} \int \frac{ d^3p}{\sqrt{E_p}} \left(c_r(p) u_r(p) {\mathrm{e }}^{-ipx}+ d^\dagger_r(p) v_r(p) {\mathrm{e }}^{+ipx} \right) \end{align}

and the fourier coefficents $c_r$, and $d^\dagger$ are interpreted as operators destroying particles and creating antiparticles respectively.

This shows that the upper two component object inside a Dirac spinor is in the Dirac basis directly related to particles and the lower two component object to antiparticles.

In this basis the chiral eigenstates are $\Psi_L = \begin{pmatrix}f \\ -f \end{pmatrix}$ and $\Psi_R = \begin{pmatrix}f \\ f \end{pmatrix}$.

A pure particle, say muon, state is for example: $\Psi_{\mu^-}= \begin{pmatrix}f \\ 0 \end{pmatrix}$. Such a pure particle state is always a mixture of both chiralities:

$\Psi_{\mu^-}= \frac{1}{2} \Big( (\Psi_{\mu^-})_L + (\Psi_{\mu^-})_R \Big)= \frac{1}{2} \Big ( \begin{pmatrix}f \\ -f \end{pmatrix} + \begin{pmatrix}f \\ f \end{pmatrix} \Big)$. Only the first part does interact weakly and we can see that we have a lower component there. With the discussion above this corresponds to some antiparticle state and my problem is understanding what we are dealing with here. How can this lower component be interpreted? Clearly it does not represent some mixture of $\mu^-$ and $\mu^+$, because this would mean that weak interactions could change a $\mu^-$ into a $\mu^+$ which would violate charge conservation.

We can see this, because $(\Psi_{\mu^-})_L = \begin{pmatrix} u_r {\mathrm{e}}^{-imt} \\ - u_r {\mathrm{e}}^{-imt} \end{pmatrix} \neq U_r + V_r = \begin{pmatrix} u_r {\mathrm{e}}^{-imt} \\ v_r {\mathrm{e}}^{imt} \end{pmatrix}$, because of the different sign in the exponent.

The $\mu^+$ is related to $V_r= \begin{pmatrix} 0 \\ v_r {\mathrm{e}}^{imt} \end{pmatrix}$, which we can also see by charge conjugation the $\Psi_{\mu^-}$ state $$\Psi_{\mu^-}^c= i \gamma_2\Psi_{\mu^-}^\star$$

and I want to understand how an Dirac spinor with lower component and exponent factor ${\mathrm{e}}^{-imt}$ is interpreted.

• Do not forget that the weakly interacting part of the $\mu^+$ also contains a "particle part" by this interpretation (it has an upper component!). Since this is pretty much symmetric, doesn't the "charge conservation problem" you have go away? – ACuriousMind Nov 21 '14 at 13:43
• Thanks for your comment. I'm not sure what you have in mind regarding how a particle part of the weakly interacting part of $\mu^+$, which in this interpretation certainly exists, solves the charge conjugation problem. Let's say we have a $\mu^-$ in the rest frame, which then decays via the weak interaction. The pure $\mu^-$ at rest is described by $\Psi_{\mu^-}= \begin{pmatrix}f \\ 0 \end{pmatrix}$, and after the weak inteaction, which couples only to the left-chiral component, the new object has a lower "antiparticle" component. – jak Nov 21 '14 at 14:05
• Although this newly appearing lower component must be (at least in my understanding) somehow related to antiparticles, it can't be related to $\mu^+$, because of charge conservation. In this interpretation a $\mu^+$ is of the form $\psi = \begin{pmatrix} 0 \\ \tilde f \end{pmatrix}$, which we can again decompose in a left-chiral and a right-chiral component. The weakly interacting right-chiral component is $\psi_R = \frac{1}{2} \begin{pmatrix} \tilde f \\ \tilde f \end{pmatrix}$ – jak Nov 21 '14 at 14:08
• Hm. Now that I think of it, I think the premise of interpreting weak interactions like this is flawed - why do you think that weak interaction would mean that "a new lower component appears"? The weak interaction either does nothing to the left-chiral part, so the muon stays that same, or it changes it completely. Recall that the scattering-out states do not live in literally the same Hilbert space as scattering in-states. – ACuriousMind Nov 21 '14 at 14:10
• My goal is to understand the components inside a Dirac spinor a little better. The "new lower component" idea comes from the obersveration that a pure particle state, consists always of left- and right-chiral parts: $\Psi_{\mu^-}= \frac{1}{2} \Big( (\Psi_{\mu^-})_L + (\Psi_{\mu^-})_R \Big)= \frac{1}{2} \Big ( \begin{pmatrix}f \\ -f \end{pmatrix} + \begin{pmatrix}f \\ f \end{pmatrix} \Big)$. – jak Nov 21 '14 at 14:27