# Tag Info

18

Spin is determined from the representation of the Lorentz group the quantum field transforms in. The projective finite-dimensional representations of the Lorentz group are labeled by two half-integers $(s_1,s_2)$. The spin of a field is the sum $s = s_1+s_2$. For example, a scalar transforms in $(0,0)$, a vector field in $(\frac{1}{2},\frac{1}{2})$, a Dirac ...

17

The idea that there must be some reason that all terrestrial DNA has a right-handed twist (or D- vs. L-glucose, or whatever your favorite chiral biomolecule is), goes all the way back to the discovery of biomolecular chirality by Pasteur. The connection to the weak nuclear interaction is apparently known as the "Vester–Ulbricht hypothesis," after its first ...

14

The thing is that the operation "exchange of two particles" has to be defined properly. What is the meaning of $P$ ? We can imagine the operator $P$ is not physical (in the sense that is does not correspond to a physically possible operation). For instance, $P\psi(x_1,x_2)=\lambda\psi(x_2,x_1)$ in the sense that it only exchange the argument of the ...

13

I have a partial answer to my question, which I'm posting down here because the question was getting too long. After a good look online at a bunch of confusing (to me) papers from the nuclear physics literature, I came upon this review: Intrinsic reflection asymmetry in atomic nuclei. P. A. Butler and W. Nazarewicz. Rev. Mod. Phys. 68 no. 2, pp. 349-421 (...

12

This is a slightly counterintuitive approach, but very easy: take the harmonic oscillator eigenfunctions $\psi_n(x)$. It is known that the functions $\psi_{2k}(x)$ are even under parity and the functions $\psi_{2k+1}(x)$ are odd. Now the time evolution operator with the harmonic oscillator Hamiltonian is $$U(t) = \exp\left(\frac{-it}{2\hbar}\,\left(-\... 10 Now when we operate parity operator, does that mean we are taking any physical entity at x to −x. Or we are just reverting axes of the co-ordinate system? Well, either operation should adhere to the same rules, and you mention the correct term: it depends on whether we see the operation as active or passive. Either view has the same end result: we move "\... 10 Here is the particle table for exchange bosons. You will see that the massive intermediate bosons are not assigned a parity. Parity is an operator. To have a definite value the state must be an eigenvalue of this operator. In the case of the massive weak interaction mediating bosons no such eigenvalue exists because in the standard model they carry both ... 10 For massive spinors "right-handed" and "left-handed" chirality isn't tied so much to true rotations, as to the casting of Lorentz transformations as "space-time rotations". In this case, a very popular short answer to the conceptual question is that Lorentz transformations "rotate" (1/2, 0)-spinors one way in space-time, and (0, 1/2)-spinors the opposite ... 10 It is a pseudo-vector because it is the curl of a vector potential, or because its curl is a vector (\displaystyle{\vec{J}}, \displaystyle{\frac{d\vec{E}}{dt}}). Or: consider the Biot-Savart law, which expresses it as an integral over the cross product (pseudo-vector) of 2 vectors. One can also look at the Lorentz Law:$$\vec{F}= q\left(\vec{E} + \vec{...

9

Sorry I found David Z' answer a bit confused just when discussing the crucial point. Since the two functions ψ(x) and ψ(−x) satisfy the same equation, you should get the same solutions for them, except for an overall multiplicative constant; in other words, ψ(x)=aψ(−x) Normalizing ψ requires that |a|=1, which leaves two possibilities: a=+1 (even ...

9

I) Consider an arbitrary coordinate transformation $$x^{\mu}\longrightarrow x^{\prime \nu}~=~f^{\nu}(x).$$ Let $$J ~:=~\det(\frac{\partial x^{\prime \nu}}{\partial x^{\mu}})$$ denote the corresponding Jacobian. Traditionally in physics, a scalar $\sigma$ transforms as $$\sigma ~\longrightarrow~ \sigma^{\prime}~=~\sigma,$$ a pseudo-scalar $\... 9 The magnetic field is a "pseudovector" (more properly, a 2-form), as opposed to the electric field, which is a vector (or a 1-form). That is, under parity,$\mathbf B$is left unchanged. You can see this from the Lorentz force, $$\mathbf F = q(\mathbf E + \mathbf v\times \mathbf B)$$ where since force is a vector,$\mathbf E$must also be a vector. Since$\...

9

The QCD and QED themselves conserve parity. The conclusion of this statement is that all corresponding effective vertices must conserve the parity. The only coupling of $\pi^{0}$ to $\gamma$ conserving the parity is $$L_{\pi^{0}} \simeq \frac{\pi^{0}}{\Lambda}\epsilon^{\mu\nu\alpha\beta}F_{\mu\nu}F_{\alpha\beta},$$ which doesn't allow your decay process $\... 9 Mathematical physicists will tell you the question you're asking has no answer: only CPT as a whole has a rigorous definition. That means that practicing physicists, who consider concrete problems, are free to define it however they want! So while I don't know the mathematical niceties, let me lay out what I think particle physicists usually mean when they ... 9 The parity operator does not have a generator in the way that the translation or rotation operators do. Notice how you gave the translation operators and the rotation operators a parameter, like$\hat{T}(\lambda)$. There isn't just one translation operator, but a whole family of translation operators that form a group. What's more, the operator family is ... 8 We don't observe the right handed neutrinos directly because, to good approximation, only left handed neutrinos interact with the weak force, and the weak force is the only mechanism we have observed neutrinos to interact with at all. As far as I know, directly observed right handed neutrinos would require observing it interacting via gravity. We can barely ... 7 So, the issue with your argument is that you assume the Hilbert space of two particles is the tensor product of two one-particle Hilbert spaces, and then indistinguishability gives you a quotient of the two-particle Hilbert space. This is not the most correct way of thinking about it. The correct way, going back to the (beautiful) original paper of Leinaas ... 7 You are not applying the transformations correctly. Your transformation,$P$, is linear map that changes a vector into another vector. Well,$F^{\mu\nu}$is a rank (2,0) tensor, not a vector (rank (1,0) tensor). This all becomes much clearer if you use index notation, rather than writing matricies. I will work in Cartesian basis. So, let us denote your$P$... 7 I gather you want the "seat of the pants" beastie: $$\bbox[yellow]{\hat P= \exp \left ( \frac{-\pi}{2\hbar}(\hat {x}\hat p+\hat {p} \hat {x}) \right )}.$$ This is clearly hermitean,$\hat P ^\dagger = \hat P$, but also unitary,$\hat P ^{-1}=\hat P ^\dagger =\hat P$: compose the exponentials in$\hat P ^2= \exp \left ( \frac{-\pi}{\hbar}(\hat {x}\hat p +...

6

I assume OP means an even potential $V(x)~=~V(-x)$, e.g., a finite square well potential $V(x) ~\propto~ \theta(|x|-a)$. Then the answer to the question(v1) is No. Sketched proof: Under the assumption that $V$ is even, the Hamiltonian $$H= \frac{p^2}{2m}+V(x)$$ then commutes with the parity operator $P$. So the operators $H$ and $P$ can be ...

6

"Spin parity" isn't a thing. It's saying the xi baryon has spin $\frac{1}{2}$ and positive parity; they're separate properties whose names tend to be run together for some reason. As for why we use the word spin even though some of the angular momentum may be orbital: it allows you to imagine the $\Xi^-$ as an elementary particle which has the same amount ...

6

About your questions: "It is known that the mass of the neutrino is not 0, therefore there are chances (depending on the mass) to find right-handed neutrinos and left-handed anti-neutrinos. But we don't." Actually no. The way you wrote it down assumes that the mass is a so-called Dirac mass. In that case the left-handed and right-handed neutrinos ...

6

Like many pseudovectors, $\mathbf{B}$ is more correctly (or more completely) thought of as a two-form. It's the lower right $3\times 3$, skew-symmetric block (the spatial part) of the Faraday tensor, so it actually represents a directed plane. Alternatively, as the exterior derivative of the vector potential one-form, it's a two form. In 3 dimensions, we can ...

6

The reason that the weak nuclear force is called 'weak' is that it has a minimal role compared to electromagnetism and the strong nuclear force. Generally speaking it only really appears in nuclear decay, and while it can be responsible for the appearance of chiral ground states of nuclei (see e.g. Why are pear-shaped nuclei possible?) its influence on ...

6

The full Lorentz group is $$O(1,3) = SO^+(1,3)\rtimes D_4\,,$$ where $D_4$ is generated by $P$ and $T$, the discrete symmetries of parity and time reversal. Its representations can be labeled by representations of $SO^+(1,3)$, hence of $\mathfrak{so}(1,3)$, and of $D_4$. In physics' language this means that a particle is defined by the spin quantum ...

5

Ronak M. Soni's answer is correct. Your mistake is in the very first step, in assuming that the many-body Hilbert space is the tensor product of single-particle Hilbert spaces. When this is true, then your logic is completely correct and anyons are impossible in any dimension. Now it's true that in a real solid with anyonic excitations, the exact Hilbert ...

5

You are right concerning the parity transformation, it implies the degeneracy of all states with finite momentum. The effect of the translation symmetry does not imply more than what is known from the conservation of momentum, as both operators are closely related. Indeed, the translation operator is given by $$\hat T(a)=e^{i \hat P a}$$ (up to a sign, ...

5

No, the parity symmetry is not the reason free particle states are doubly degenerate, because the group $\mathbb{Z}_2$ is an abelian group, and hence has no irreducible representations of dimension greater than one. The translational symmetry isn't responsible either, for the same reason. You need both. Suppose that we instead are working with a symmetric ...

5

spin 1/2 fermions (electron, proton, neutron, muon, tau, quarks) have +1 parity (by convention as pointed out in Anna's comment). The corresponding anti-fermions have -1 parity. Bosons and their anti-particles have the same parity. See this and this lecture for more information on parity.

5

If you're referring to elementary fermions (described by the Dirac equation), there is no ambiguity: fermions have an intrinsic parity +1 while antifermions have -1. This is easy to establish by starting from the Dirac equation, you deduce that the parity operator is $\gamma^0$ and using the general solutions for spinors $u$ (particles) and $v$ (anti-...

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