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4

If you measure one particle to be in $\left|\uparrow\right\rangle$, then the other will be in $\left|\downarrow\right\rangle$, just as you said. So your two eigenvalues are $1$ and $-1$. Multiplying (not averaging) the results together: $$1\times (-1) = -1.$$ You'll get the same answer if you measure the particles to be down/up rather than up/down. I think ...

3

Is there anyone knowledgable enough in this area who would be able to comment on some of the possible theoretical/ hypothetical implications of the existence of spin 3 particles? Is there any thought that their existence could imply additional fundamental forces? If you look at the presentation linked in the link you gave , in page five, you will see ...

0

As ACuriousMind said, the fact that the particle is composite makes it less earth-shattering. The interesting thing here (I think...not a high-energy guy) is that it's the first example of this kind of spin-3 particle, in particular flavored. Other composite spin-3 particles have been known before, the first one I found was a Boron nucleus.

0

To say I am skeptical would be an understatement. I briefly worked on this around 1997 and what I see here, although it matches his description of what he wanted, in no ways matches the effects he claimed for his "original" which exhibited massive positive feedback, ionization, extreme cooling and massive antigravity effects. The over-unity claims were just ...

0

For your first questions: Observables correspond to hermitian (or self-adjoint) operators. As such, the eigenvalues are real, and these values are the possible outcomes. Also because of the hermitian/self-adjointness of the operator, when you have eigenvectors with different eigenvalues, the eigenvectors are orthogonal. So you know the values, but what ...

0

Spin is the total angular momentum, or intrinsic angular momentum, of a body. The spins of elementary particles are analogous to the spins of macroscopic bodies. you can look at http://en.wikipedia.org/wiki/Spin_%28physics%29

1

In quantum mechanics an "observable" like the position of an atom, $\hat X$, is represented by a Hermitian operator. An "observable" is some physical quantity you can actually go out and measure, like the position. The eigenvalues of that operator are the possible measurement results. In quantum mechanics a "system" (some physical entity that has ...

1

since every sample has both protons and electrons, and all have magnetic spin, But the electron spins cancel out usually, because normal matter only has electrons in Pauli pairs. EPR is restricted to radicals in organic chemistry or transition metal complexes, or O2 gas :=)

0

So, as I now understand, it is (obviously) only in two dimensions that $h$ and $\bar{h}$ have the interpretation of the weights corresponding to left- and right- movers. And it is therefore clear that $h-\bar{h}$ is the spin. However, the spin-statistics theorem says that spin is either integer or half-integer, so it isn't quite clear how $(h-\bar{h})$ is ...

3

The propability $P$ that a system in state $|\Psi\rangle$ will eventually transit into state $|\Theta\rangle$ is given by: $$P = | \langle\Theta | \Psi \rangle |^2$$ So for example if you have a System in Spin-X-State "up" denoted by $|\Psi\rangle \hat{=} \frac{1}{\sqrt{2}} (1, 1)^T$, the propability of getting Spin-Z-State "up" (e.g. by measuring along ...

2

You are basically asking how to compute $$e^{i\gamma B \hat{S}_z t/\hbar} |+\rangle.$$ Now, $e^{i\gamma B \hat{S}_z t/\hbar} =\sum_{n=0}^{n=\infty} \frac{\left(i\gamma B \hat{S}_z t/\hbar\right)^n}{n!},$ so you can approximate it quite well by $$\sum_{n=0}^{n=N} \frac{\left(i\gamma B \hat{S}_z t/\hbar\right)^n}{n!}.$$ And you can figure out how to operate ...

1

Notice that $\hat{S_z}= \frac{\hbar}{2}\sigma_3$ and that $\sigma_3^2=\mathbb{1}$, where $\sigma_3$ is the Pauli matrix. I don't know if you know the identity $$\hat{O}^2= \mathbb{1} \implies e^{i \alpha \hat{O}}= \mathbb{1} \cos(\alpha) + i \hat{O} \sin(\alpha) \tag{1}$$ It is fairly easy to derive it via Taylor expansion. I set $\omega = \frac{\gamma ... 1 but I'm stuck to carry the computation forward: what does$e^{i\hat{S}_zt}$do to an eigenvector like$|+\rangle$? I'm stuck because the exponential contains an operator, and I want to get an expression without operator. Luckily, you only have to act the operator on its own eigenvector (|+> and |->), so you can replace the operator with the ... 3 The diagram is simply a way to remember two variables at the same time, one called the total spin$\hbar\sqrt{3}/2$, the other called the z-component which has two possibilities$\pm\hbar/2$. If you draw a picture of a vector of length$\hbar\sqrt{3}/2$that is at an angle so that the z-component of the vector is$\pm\hbar/2$(so it lies on two cones, one ... -1 It is true that for the electron the eigenvalue of the operator$ \ \hat {S^2} \ $is$ \ \frac {\hbar^2}{2} \left(1 + \frac {1}{2} \right)$s.t. the square root is as you say$\frac {\sqrt {3}}{2}$, however, about the angle of 60°. it is not clear where it comes from. In the spin-up state$|\uparrow\rangle$the spin projection is parallel with the axis$z$, ... 3 First of all you should not think of spin of an electron as a ball spinning on is own axis. At least this comparison should not go beyond your intuitive understanding of an electron spin. The quantity$s=\sqrt{s(s+1)}= \frac{\sqrt3}{2} \hbar$is an intrinsic property of an electron with spin one-half and has nothing to do with the axis of rotation. I don't ... 1 Actually I "believe" that this topic is not really "settled"... The expression 1/2 (σx⊗σx +σy⊗σy +σz⊗σz) is introduced by P. Dirac in his famous book "The Principles of Quantum Mechanics" IV ed. chap IX p. 221 where he uses this expression (actually he does not write explicitly the tensor or Kronecker product sign ⊗ but one "undersdands" that it corresponds ... 1 Actually I "believe" that this topic is not really "settled"... The expression $$\frac12 \left(\sigma_x\otimes\sigma_x+\sigma_y\otimes\sigma_y+\sigma_z\otimes\sigma_z\right)$$ is introduced by P. Dirac in his famous book The Principles of Quantum Mechanics IV ed. chap IX p. 221 where he uses this expression (actually he does not write explicitly the ... 1 Because you want the maximal spin particle allowed to be 2 (since there is no higher spin field theory interacting non trivially), thus in a supermultiplet starting with a particle with helicity$0$, say in$d=4$, the maximal number of susy you can apply is$8$, thus$N=8$in$d=4$is the maximal supersymmetry admitted, which means$32$supercharges. 4 The way you've written them, those are the spin operators in the$\hat{S}_z$eigenbasis for a spin-1/2 particle. The two$\hat{S}_z$eigenstates are spin up (written as$|+\rangle$or$\uparrow$) and spin down ($|-\rangle$or$\downarrow$), which can be written as$\begin{pmatrix}1\\0\end{pmatrix}$and$\begin{pmatrix}0\\1\end{pmatrix}$in the$\hat{S}_z$... 2 These are mainly conventions. Conventionally, the kets$|+\rangle$and$|-\rangle$are taken to be eigenkets of the z-spin operator with, respectively, z-spin of$+\hbar/2$and -$\hbar/2$. S_x and S_y are chosen such that they obey the canonical commutation relations for angular momenta $$[S_i,S_j]=i\epsilon_{ijk}S_k$$ E.g., $$[S_x,S_y]=iS_z$$ and so ... 0 It's made up of many particles. I would assume some of them "orbit" the center. I know many electron orbitals have angular momentum. The orbitals of nuclei are less well-understood, but I would expect the same is true. 0 The angular momentum of the nucleus is the combined contribution of the spin-orbit angular momenta of the constituent particles. In order for an entity to have orbital angular momentum of its own it must some conceptual orbit: electrons in the atom, protons and neutrons in the nucleus, atoms in a molecule. That's why the angular momentum of a nucleus is ... 4 There are a number of ways to represent a 720 degree spin of a particle, but the particle has to be a little more complex then a spinning ball. Imagine this wiki image (from http://en.wikipedia.org/wiki/Spin-1/2) as the inside of a larger ball and you can see an example of 720 degree spin. My favorite representation is the idea of breaking the electron into ... 6 You have fallen prey to a popular simplification of spinors. The statement "you have to turn electron by 720 degrees in order to get the same spin state" does not refer to an actual rotation of an actual electron. In quantum mechanics, we describe the states of objects as elements of a Hilbert space$\mathcal{H}\$. The crucial thing is that not all elements ...

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