# Lorentz invariance of chirality

1. Jun 12, 2012

### Eruvaer

Hi folks,

I've been reading into the concepts of chirality & helicity and often I find a statement that chirality is Lorentz invariant in contrast to helicity (which of course depends on the frame). BUT I don't see in which way chirality IS Lorentz invariant.

For massless particles things are easy, of course. In Weyl notation for a left chiral particle, you have some Weyl bispinor $\left( \begin{array}{c} \Phi \\ 0 \end{array} \right)$ with a Weyl spinor $\Phi$ depending on the energy and spin orientation of the particle. Since boosting to another frame means multiplying the bispinor with a diagonal matrix, it stays in this form: a left chiral particle is left chiral in every frame. Since the Dirac representation of the Lorentz group is reducible into left and right chiral states, it is not surprising that they don't mix.

But for a massive particle things seems strange to me. Suspect we have an electron. Then the positive frequence solution of the dirac eq. for momentum $\vec{p}$=0 and spin up in z-direction is $u\left(p=0\right) = \sqrt{m}\left( \begin{array}{c} \xi \\ \xi \end{array} \right)$ with the spinor $\xi = \left( \begin{array}{c} 1 \\ 0 \end{array} \right)$. Obviously $\frac{\left(1-\gamma^{5}\right)}{2}u$ and $\frac{\left(1+\gamma^{5}\right)}{2}u$ are of the same size; the mixing angle between the left and right chiral amount of the electron is 45 degrees. But boosting this bispinor in z-direction gives $u\left(p\right) = \left( \begin{array}{c} \sqrt{E-p^{3}}\xi \\ \sqrt{E+p^{3}}\xi \end{array} \right)$. In the limit $p^{3}$→∞ this becomes $u\left(p\right) = \left( \begin{array}{c} 0 \\ \sqrt{2E}\xi \end{array} \right)$. So an electron which has equal amount of left and right chirality in one frame is fully right chiral in another!

How can we then state that chirality is a Lorentz invariant concept?

2. Jun 13, 2012

### USeptim

Hello Eruvaer,

I don't have experience on this stuff but I would like to suggest and idea

Dirac spinors have four components. not two, and both solutions for a free electron (positive and negative) have always two zero components, one in the "big wave component" and the other in the "small wave component".

So, maybe you must take only the terms in one of the wave components, that way the chiliarity will be kept with any boost.

3. Jun 13, 2012

### vanhees71

States with proper chirality are represented by the eigenspinors of $\gamma_5=\mathrm{i} \gamma^0 \gamma^1 \gamma^2 \gamma^3.$

It's easy to see that $\gamma^5 \gamma^{\mu}=-\gamma^{\mu} \gamma^5$ for $\mu \in \{0,1,2,3 \}$. The generators of the proper orthochronous Lorentz transformations in the Dirac representation is given by

$$\sigma^{\mu \nu}=\frac{\mathrm{i}}{4} [\gamma^{\mu},\gamma^{\nu}].$$

Now, since $\sigma^{\mu \nu}$ has two $\gamma$-matrices obviously $\gamma^5$ commutes with it and thus an eigenspinor of $\gamma_5$ with a given chirality ($\pm 1$ since $\gamma_5^2=1$) remains an eigenspinor of this same chirality under proper orthochronous Lorentz transformations. This holds true for both massive and massless particles.

You can also show that chirality flips under spatial inversion (parity). That's why it's called chirality.

For massive particles you have to distinguish between chirality and helicity. While chirality is a good quantum number under proper orthochronous Lorentz transformations that's not true for helicity, which is the projection of the total angular momentum to the direction of momentum. You can easily imagine that you can overtake a massive particle since it's moving with a velocity less than the speed of light, and thus you can flip the direction of helicity with an appropriate Lorentz boost. Only for massless particles helicity and chirality coincide, and helicity becomes also a good quantum number under proper orthochronous Lorentz transformations.

4. Jun 13, 2012

### geoduck

Hi,

It's entirely possible to lose all your chirality under a boost. The boost operator is not unitary, unlike rotations. So as you say, the left chiral Weyl spinor will become zero as you boost to infinity. This does not mean however that the left and right chiral spinors mix with each other! They don't. The left Weyl spinor transforms among itself to produce zero.

Hope that helps.

5. Jun 13, 2012

### Eruvaer

I think this corresponds to what I've written about massless particles in my first post. If you start with an eigenspinor of $\gamma^5$ it stays an eigenspinor under proper orthochronous Lorentz transformations and therefore also under simple spatial boosts. So far everything is clear.

This doesn't fully answer my question regarding massive particles, because as in the example I gave above a positive frequency solution of the Dirac equation for a massive particle is not an eigenspinor of $\gamma^5$. That means that although a spatial boost doesn't mix the left chiral and right chiral parts of the Weyl bispinor, they of course transform differently under the transformation. Therefore it seems to me that the relative amount of left and right chirality of an electron depends on the reference frame. Or is this not a sensible thing to say?

That's clear to me, thanks anyway