# Weak Isospin, Chirality and Helicity

1. Apr 20, 2015

### agent009

Could someone please help me to understand the difference between the concepts of Weak Isospin, Chirality and Helicity. In particular, I have the following questions to which I was unable to find answers so far:

1. Since both spin direction and momentum are vectors, would not their apparent direction depend on the observer's reference frame and would not the same particle appear to have different Chirality to two different observers? E.g. Earth has both spin and momentum, but would not those vectors appear to have the same direction to an observer on Mars, while appearing to have different direction to an observer on Venus?

2. Would not knowing the direction of a particle's spin imply measuring its orientation in space? If so, would not determining directions of both its spin and momentum violate Heisenberg's Uncertainty Principle?

3. How comes that all common quarks and leptons are considered "left-handed" (i.e. have the same chirality), but up-type quarks and neutrino have positive weak isospin, while this value is negative for down-type quarks and e-/μ-/τ-?

2. Apr 20, 2015

### Orodruin

Staff Emeritus
Chirality is the projection of the spin on the momentum direction. As such, it is invariant under rotations (it is not invariant under boosts, but that is another matter).

The "orientation" in space is not related to the position in space. These are independent quantities. However, a generalisation of the HUP results in the fact that you cannot know a particles spin component in several directions at the same time.

All quarks and leptons are not left-handed. There are right-handed components of all quarks and charged leptons. Weak isospin is something else.

3. Apr 20, 2015

Staff Emeritus
I think that's helicity. Chirality is more complicated, or at least more abstract. I think the best way to think about it is in terms of chiral projection operators, defined so that it's an invariant that matches helicity for massive particles - i.e. it's an extension of the idea with nicer properties.

4. Apr 20, 2015

### agent009

Hmm... yes, I suppose if you consider an axis of spin that is anything but strictly perpendicular to the vector of momentum, the relation between those two vectors would be invariant as long as the direction of both remains unchanged. So, how do you define the axis of spin relative to the vector of momentum when you determine Chirality/Helicity and what direction of spin would be considered "the same" as direction of momentum?

Could you please elaborate on what exactly the difference between Chirality and Helicity is?

5. Apr 20, 2015

### Orodruin

Staff Emeritus
Yes, you are correct. I always write faster than I think with chirality vs helicity.

6. Apr 20, 2015

### ChrisVer

7. Apr 20, 2015

Staff Emeritus
Helicity was defined above. Chirality is the thing that gets projected out by (1+gamma5) or (1-gamma5). For a massless particle, it's the same as Helicity.

8. Apr 25, 2015

### vanhees71

It's a bit unfortunate that the Wikipedia article labels the positive (negative) helicity states as right (left) handed. The text is correct: Helicity is the projection of the total angular momentum (there's in general no unique and gauge independent splitting of total angular momentum in a spin and orbital part in relativistic QT!) to the direction of momentum. That's pretty intuitive, because it can be visualized (with some caveats) by a classically spinning moving object.

Chirality (or "handedness") is, as said, a bit more complicated. It is related to the representations of the Lorentz group which define the mathematical structure of the quantum fields used to describe the particles. A representation of the proper orthochronous Lorentz group (or more precisely its Lie algebra) is uniquely determined by the direct product of two "pseudorotation algebras", i.e., by two numbers that can be integer (including 0) or half-integer. In the case, where you have a representation (n,0), you can extend the representation to contain also the representation of spatial reflections, by looking at the direct sum $(n,0) \oplus (0,n)$ for $n$ integer or half-integer ($\neq 0$). The $(n,0)$ and $(0,n)$ are two kinds of spinor (or tensor) fields with $2n+1$ components, and in the direct sum the space-reflection transformation involves an exchange of the $(n,0)$ with the $(0,n)$ components. That's why you can call one of them, say the $(n,0)$, "left handed" and the other ones, then the $(0,n)$, "right handed".

In the standard model you have the case $n=1/2$ for the quarks and leptons. For QED and QCD there should be space-reflection symmetry, and thus you need the representation $(1/2,0) \oplus (0,1/2)$. This leads to the well-known Dirac spinors with 2+2=4 components. The above formalism leads to the socalled spin representation of the $\gamma^{\mu}$ matrices. The Dirac spinors with well-defined chirality are the eigenspinors of the $\gamma^5=\mathrm{i} \gamma^0 \gamma^1 \gamma^2 \gamma^3$ matrix.

For massless chiral fields it turns out that chirality and helicity are the same, but that's not the case for massive ones.

9. Apr 25, 2015

### ChrisVer

Is there a proof for the massless-same helicity/chirality?
I think this should have something to do with axial currents? or am I wrong?

10. Apr 26, 2015

### vanhees71

For Dirac fields it's most easy to see in the chiral representation, where the Dirac matrices have the form
$$\gamma^0=\begin{pmatrix} 0 & \mathbb{1}_2 \\ \mathbb{1}_2 & 0 \end{pmatrix}, \gamma^{j}=\begin{pmatrix} 0 & \sigma^j \\ -\sigma^j & 0 \end{pmatrix},$$
where the $\sigma^j$ are the usual Pauli matrices (with $\sigma^3=\mathrm{diag}(1,-1)$ in the conventional representation). For massless particles the standard Dirac spinors can be constructed from the four linearly independent canonical basis spinors
$$u'(0,+1/2)=\begin{pmatrix} 1 \\ 0 \\ 1 \\0 \end{pmatrix}, u'(0,-1/2)=\begin{pmatrix} 0 \\ 1 \\ 0 \\1 \end{pmatrix},$$
$$v'(0,+1/2)=\begin{pmatrix} 1 \\ 0 \\ -1 \\0 \end{pmatrix}, u'(0,-1/2)=\begin{pmatrix} 0 \\ 1 \\ 0 \\-1 \end{pmatrix},$$
The spin operator is given by
$$\Sigma^j=\epsilon^{jkl} \frac{\mathrm{i}}{4} \gamma^k \gamma^l =\frac{1}{2} \begin{pmatrix} \sigma^j & 0 \\ 0 & \sigma^j \end{pmatrix}$$
Obviously the four spinors $u'(0,\pm 1/2)$ and $v'(0,\pm 1/2)$ are eigenspinors of $\Sigma^3$ with the eigenvalues $\pm 1/2$.

For a massless particles the spinors in the momentum eigenmode decomposition of the Dirac field are given by
$$u(\vec{k},\sigma)=\frac{1}{\sqrt{2E}} \gamma^{\mu} k_{\mu} u(\sigma), \quad v(\vec{k},\sigma)=-\frac{1}{\sqrt{2E}} \gamma^{\mu} k_{\mu} u(\sigma),$$
with $k^{\mu}$ fulfilling the on-shell condition $k\dot k=0$, i.e., $k^0=E=|\vec{k}|$.

Now the choice of the Pauli matrices in their conventional form implies that the standard momentum in the representation theory of the proper orthochronous Poincare group is along the $3$-axis. Thus, the spin has its usual meaning in a reference frame, where the momentum of the particle is along the $3$ axis. Since the bispinor spin matrices $\vec{\Sigma}$ commute with the $\gamma^{\mu}$, for $\vec{k} =k \vec{e}_3$ the above spinors are helicity eigenspinors:
$\hat{h} u(\vec{k},\sigma)=\sigma u(\vec{k},\sigma), \quad \hat{h} v(\vec{k},\sigma)=\sigma v(\vec{k},\sigma),$
where the helicity operator is defined by
$\hat{h}=\frac{1}{E} \vec{k} \cdot \vec{\Sigma}, \quad E=|\vec{k}|.$
Now in this representation of the Dirac matrices we have
$$\gamma^5=\mathrm{i} \gamma^0 \gamma^1 \gamma^2 \gamma^3=\begin{pmatrix} -\mathbb{1}_2 & 0 \\ 0 & \mathbb{1}_2 \end{pmatrix}.$$
On the other hand the explicit form of the above given mode spinors is precisely such that for $\vec{k}=E \vec{e}_3$
$\gamma^5 u(\vec{k},\sigma)=2 \sigma u(\vec{k},\sigma), \quad \gamma^5 v(\vec{k},\sigma)=2 \sigma v(\vec{k},\sigma).$
That proves that in the frame where the momentum is in standard direction in sense of the Wigner basis, the helicity and chirality have the same sign.

It is also very easy to verify that the massive standard Dirac spinors are helicity but not chirality eigenstates (again in the frame where the momentum has the standard direction, i.e., 3 direction in our standard representation).