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6 votes
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A puzzle about relativistic spin

With all thanks to Valter Moretti, I believe we now have a complete solution to my original puzzle. In the identities for $\mathbf{K}$ and $\mathbf{J}$ that disagreed with the celebrated ones due to ...
ac2357's user avatar
  • 141
3 votes

Angular momentum quantum number $l$ either integer or half integer

The book is correct in using “either - or”. But the reason relies upon more fundamental instances of QM than the structure of angular momentum operator. More precisely a certain superselection rule. ...
Valter Moretti's user avatar
0 votes

Angular momentum quantum number $l$ either integer or half integer

The other answers and comments are all great; I will solely focus on the word either as emphasized in the question. The book is answering the question "what are the possible values of $l$?" ...
Quantum Mechanic's user avatar
2 votes
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The meaning of a representation in one-dimensional quantum mechanics

In abstract terms the meaning of representation in these cases is the ordinary meaning of representation in the sense of representation theory. Concretely, you are representing the Heisenberg algebra ...
ACuriousMind's user avatar
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2 votes
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Orthogonal singlet states?

Is there an issue in my reasoning or in my understanding of spin states? In both. You are abusing Dirac notation, meant to defocus from side issues and omit other labels (and bases of tensor calculus ...
Cosmas Zachos's user avatar
3 votes
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Given a representation $(n, m)$ of the Lorentz group, is the little group representation just the tensor product $n \otimes m$?

Consider the restricted Lorentz group $SO^+(3,1;\mathbb{R})$ and its complexification $$SO(3,1;\mathbb{C}).\tag{1}$$ Picking the COM frame the massive little group becomes the 3D rotation group $SO(3,\...
Qmechanic's user avatar
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2 votes

Given a representation $(n, m)$ of the Lorentz group, is the little group representation just the tensor product $n \otimes m$?

I am not going to read Weinberg's book with you to your satisfaction, nor should I. Indeed, the little group of a massive particle (go to its rest frame) is the rotation group SO(3), sharing a Lie ...
Cosmas Zachos's user avatar
0 votes

Why is half-integer spin not observed classically?

Electron spin has been interpreted classically in the context of the theory of stochastic electrodynamics (classical electromagnetism + zero-point field - the field causing the Casimir force): Cetto, ...
Andrei's user avatar
  • 805
1 vote

Why is half-integer spin not observed classically?

Ferromagnets are a classical manifestation of electron spin. MRI is a classical manifestation of nuclear, mostly proton, spin. Both are spin half. What you are looking for is a -1 factor upon 360$^\...
my2cts's user avatar
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5 votes

Why is half-integer spin not observed classically?

The other answers are great, and it's very much true that you can totally write down a classical field theory with half-integer-spin fields. That notion still feels kind of odd, though, because all ...
Rokas Veitas's user avatar
2 votes

Why is half-integer spin not observed classically?

Why is half-integer spin not observed classically? Why do you single out half-integer spin? Are you implying that integer spin can be observed classically? If neither integer spin nor half-integer ...
MadMax's user avatar
  • 4,452
0 votes

What is the importance of $SU(2)$ being the double cover of $SO(3)$?

It's actually Spin(n) which is the universal and double cover of SO(n) for n>=3. However, it turns out that Spin(3) is isomorphic to SU(2). This is important because SU(2) is defined via complex 2 ×...
Mozibur Ullah's user avatar
3 votes
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Generators of rotations: $[J_i, J_j] = \epsilon_{ijk} J_k$ and $(J_i)_{jk} = -\epsilon_{ijk}$. Is this a coincidence?

This is not a coincidence. The three matrices you have chosen as generators for your rotations live in the adjoint representation of $\mathfrak{so}(3)$. This explicit matrix elements of the adjoint ...
CStarAlgebra's user avatar
  • 2,772
0 votes

How to prove $e^{+i(\theta/2)(\hat{n}\cdot \sigma)}\sigma e^{-i(\theta/2)(\hat{n}\cdot \sigma)} = e^{\theta \hat{n}\cdot J}\sigma$?

$\newcommand{\bl}[1]{\boldsymbol{#1}} \newcommand{\e}{\bl=} \newcommand{\p}{\bl+} \newcommand{\m}{\bl-} \newcommand{\mb}[1]{\mathbf {#1}} \newcommand{\mc}[1]{\mathcal {#1}} \newcommand{\mr}[1]{\...
Frobenius's user avatar
  • 16k
2 votes
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$Ad\circ\exp=\exp\circ ad$ and $e^{i(\theta/2)\hat{n}\cdot\sigma}\sigma e^{-i(\theta/2)\hat{n}\cdot\sigma}=e^{\theta\hat{n}\cdot J}\sigma$

The original question is really almost there. The answer is that we can almost prove $(1)$ directly from $(2)$. All we need in addition is a few facts about the pauli matrices $\sigma$ and the ...
Jagerber48's user avatar
  • 14.4k
1 vote

How to prove $e^{+i(\theta/2)(\hat{n}\cdot \sigma)}\sigma e^{-i(\theta/2)(\hat{n}\cdot \sigma)} = e^{\theta \hat{n}\cdot J}\sigma$?

You appear to have two unrelated gaps, the final one being the rotation formula for the vector (triplet) representation, the celebrated Rodrigues rotation formula. $\vec \sigma$ rotates like a vector. ...
Cosmas Zachos's user avatar
0 votes

How to prove $e^{+i(\theta/2)(\hat{n}\cdot \sigma)}\sigma e^{-i(\theta/2)(\hat{n}\cdot \sigma)} = e^{\theta \hat{n}\cdot J}\sigma$?

Your equation $(*)$ is a bit hand-wavy since it doesn't hold exactly (on the r.h.s. you have a vector in $\mathbb{R}^3$ and on the l.h.s. a vector of Pauli matrices), but it's clear what you mean. I ...
fulis's user avatar
  • 835
2 votes
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How to prove $e^{+i(\theta/2)(\hat{n}\cdot \sigma)}\sigma e^{-i(\theta/2)(\hat{n}\cdot \sigma)} = e^{\theta \hat{n}\cdot J}\sigma$?

Physically, you can solve the Heisenberg equations of motion. It is equivalent to all the general arguments of the adjoint action, but at least it cuts to the chase and gives your formula explicitly. ...
LPZ's user avatar
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1 vote
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Is the Dirac adjoint in the representation dual to Dirac spinor?

In order for $\overline{\psi} \psi$ to be a Lorentz scalar, we should have $$\lambda^{\dagger}\gamma^0\lambda = \gamma^0$$ or equivalently (assuming $(\gamma^0)^2 = 1$) $$\lambda^{-1} = \gamma^0\...
MadMax's user avatar
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