# Tag Info

## Hot answers tagged chirality

15

For spin measurements the original experiment was the Stern-Gerlach experiment in which you will see that a prior unpolarized beam will split up in two (Spin up and down) orientations. see: http://en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment For helicity, a very ingenious and fascinating experiment is the famours Goldhaber experiment that uses a ...

11

Related answer: http://math.stackexchange.com/a/532746/24293 Looking at the comments, you seem to be asking why there are chiral 'pairs' and not chiral multiplets. Looking at the tag, it looks like you want an analysis of higher dimensions as well. TL;DR Short answer: In any number of dimensions, chiral objects come in pairs. This is because numbers ...

11

No, it really is arbitrary. The reason we use the right hand rule today (although it may have been chosen for different reasons of convenience in the past) is simply that our coordinate system of choice is right-handed. Mathematically, this means that we define the directions of the axes so that you have to use the right-hand rule to evaluate this cross ...

8

At first glance, chirality and helicity seem to have no relationship to each other. Helicity, as you said, is whether the spin is aligned or anti aligned with the momentum. Chirality is like your left hand versus your right hand. Its just a property that makes them different than each other, but in a way that is reversed through a mirror imaging - your left ...

8

There is no chiral anomaly/gauge anomaly if the spacetime dimension $2\ell+1$ is odd, partly because $SO(2\ell+1)$ has real or pseudo-real representations, but no complex representations. There may instead be parity anomalies in odd spacetime dimensions. In fact, there is a dimensional ladder of related anomalies $$\text{Abelian chiral anomaly in}~ ... 8 The clockwise direction is normally defined by the right hand grip rule. When your thumb is pointing away from you, your fingers are curled clockwise. So when you look at a clock the axis of rotation is away from you through the clock. I'd guess the downvotes are because people believe your question is not physics related, but in fact this rule is how ... 7 Standard model doesn't predict that neutrinos are massless. It's a "Model", where initially neutrinos are considered massless, because no mass was observed. The way we know, now, that neutrinos have masses, is through the mixing between the different neutrino types, through a matrix called the PMNS matrix (similar to the CKM matrix for quarks). This mixing ... 7 Presumably you are asking about the communication ambiguity in physics: can we unambiguously specify what we mean by "a right handed coordinate system" to a correspondent far away without a pre-arrnage communications channel (i.e. using SETI)? For a long time the answer seemed to be "no", but the discovery of parity violation in 1957 changed the answer to ... 7 There isn't a good definition of chirality in (2+1)D or any other odd dimension. This is because the \gamma_5 matrix can't be defined usefully in a Clifford algebra with an odd number of generators. For instance try to define \gamma_5 = \gamma^0\gamma^1\gamma^2. This commutes (not anti-commutes) with \gamma^0,\gamma^1,\gamma^2 and thus commutes with ... 6 There is a very simple and enlightening explanation due to N.V.Gribov given in his following conference article and also beautifully explained by Dmitri Kharzeev in the following arxiv article(section 1). Gribov's argument doesn't involve the heavy machinery of quantum field theory. He actually proves that in the case of colinear electric and magnetic ... 6 I think its really important to differentiate between helicity and chirality. Helicity is the spin angular momentum of a particle projected onto its direction of motion. For a massive particle this quantity is frame dependent. Furthermore, since angular momentum is conserved, as a particle propagates helicity is conserved. On the other hand, chirality is an ... 5 You are correct that for a massive spinor, helicity is not Lorentz invariant. For a massless spinor, helicity is Lorentz invariant and coincides with chirality. Chirality is always Lorentz invariant. Helicity defined$$ \hat h = \vec\Sigma \cdot \hat p, $$commutes with the Hamiltonian,$$ [\hat h, H] = 0, $$but is clearly not Lorentz invariant, because ... 5 Simply think of a Weyl Spinor as a Dirac spinor where the other two components are set to zero. Equivalently, a Weyl spinor (of chirality +1) belongs to the two-dimensional subspace with eigenvalue 1 under the action of the projection operator P_+=\frac{(1-\gamma^5)}2. For negative chirality, use the other projection operator, i.e., ... 5 Your confusion arises because the term (anti)clockwise, when used by itself, is ambiguous, and should always be used with a statement like "as seen from the top" (unless that is absolutely obvious^1). The reason for this is that "clockwise" defines a direction of rotation within a plane, but does not specify which side the plane is observed from. (In more ... 5 General explanation Chirality and helicity are in general different quantities. Chirality is connected with the representation of the Lorentz group (left or right) while helicity is connected with projection of spin on momentum direction and becomes to characterize representation only in massless case. I have meaned following. Particularly (!), ... 5 Roughly sketched, for the quantized Dirac field one has: $$\hat\psi(x)\sim \int d\mathbf{p}\, \sum_r \bigg[ u^{r}(p)\, \hat a^r_\mathbf{p}\,e^{-ipx}+v^{r}(p)\, {\hat b^r_\mathbf{p}}^\dagger e^{ipx}\bigg],$$ where r=\pm1 denotes helicity. The {\hat a^r_\mathbf{p}}^\dagger operator creates a helicity-r particle state when it ... 4 Helicity (the correlation of spin and momentum) and chirality (whether a particle couples to the left-handed or right-handed part of the weak interaction) are strongly correlated in the relativistic limit v\to c, but not strongly coupled at low speeds. The electron, following the decay, is happy to believe that it's in its own rest frame and to evolve ... 4 In theory, it is possible to measure chirality using the exact method that you mentioned. If you have an electron and you are unsure its chirality, you can send it through a dense bucket of W bosons and see if it interacts with them. If it did, then you know that you had a left handed electron. There is, however, an important issue with this idea (and no ... 4 I'm not very strong in group theory, so could someone please explain in simple terms what it means for the left handed parts to transform as triplets and right handed parts to transform as singlets? How would you go about writing down such terms in a Lagrangian? Groups are abstract. They have elements that can be "multiplied" and they have other ... 4 Consider a one-particle state of a relativistic quantum field theory, and let this state be an eigenstate of the 4-momentum operator, \hat P^\mu |p^\mu\rangle = p^\mu|p^\mu\rangle. Other than the 4-momentum, what other quantum numbers can the state have, and how should they transform? That is, if there are any quantum numbers collectively labeled by s, ... 4 I think you are sort of reversing the logic of chirality and helicity in the massless limit. Chirality defines which representation of the lorentz group your Weyl spinors transform in. It doesn't 'become' helicity, helicity 'becomes' chirality in the massless limit. That is, chirality is what it is, and it defines a representation of a group and that can't ... 4 Dear lurscher, the quote is the kind of C-physics described by the C-word which is a favorite word of mine but is discouraged on this server, so I won't use it - but you have used it. You don't misunderstand anything - quite on the contrary, you're right on the money. These comments about a non-existent test of parity in the equivalence principle are due to ... 4 I agree with the answer of Quantum physicist , that zero mass for neutrinos was an input to the standard model , not a prediction, because measurements showed a mass compatible with zero. But I will add that the discovery that neutrinos must have mass does not destroy the Standard Model, just different Lagrangian for the neutrinos has to be included. ... 4 You seem want to introduce gauge invariance into a theory that doesn't appear to have the global symmetry need in the first place. One way to think of gauge invariance is that you 'gauge' the global symmetry, then you just change your derivative terms to covariant derivatives like you mentioned. In other words, we can only concern ourselves with the global ... 4 The importance of chiral symmetry breaking is a point that e.g. David Gross, a co-father of Quantum Chromodynamics, likes to make whenever some people suggest that the mass is entirely due to the Higgs field. In fact, most of the mass of visible matter is due to the QCD, especially chiral symmetry breaking, and it has nothing to do with the Higgs field. The ... 3 The Majorana bound state inside a vortex of a topological superconductor is, indeed, not a chiral edge state. It does not follow that the topological superconductor does not have a chiral edge state. It does! Solve, for example, the BdG equations for a p+ip superconductor with open boundary conditions, and you'll see it. The existence of edge modes is ... 3 No, it's not true. Suppose I'm floating in outer space (presumably in a space suit or something else to keep me alive). I'm still me, and I still know that, for example, my left hand is the one on the left, and my right hand is the one I can write with. Even on Earth, we don't need environmental clues to distinguish left from right; it's more a matter of ... 3 Tarek (OP) e-mailed me to contribute to this thread. Here's the response that I gave him (slightly edited for clarity). I see why this was confusing, my apologies! I was perhaps too glib in the post. Iwas implicitly talking about a chiral rotation but wanted to present it somewhat more intuitively. Let me try to spell it out more carefully, and hopefully ... 3 The charged current part of the Lagrangian of the electoweak interaction, for the first generation of leptons, is :$$L_c = \frac{g}{\sqrt{2}}(\bar \nu_L \gamma^\mu e_L W^+_\mu + \bar e_L \gamma^\mu \nu_L W^-_\mu ) The first part corresponds to different versions of the same vertex : $e_L + W^+ \leftrightarrow \nu_L \tag{1a}$ \$(\bar\nu)_R + W^+ ...

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