# Tag Info

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An arbitrary normalized quantum state on two dimensions can always be written as $$|\psi⟩=e^{i\alpha}\left(\cos\theta|\uparrow⟩+e^{i\phi}\sin\theta|\downarrow⟩\right)$$ without loss of generality. The phase factor $e^{i\alpha}$ has no bearing on experiment, as all measurements will be proportional to $⟨\psi|\propto e^{-i\alpha}$. This means that you can ...

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The fact that the overall phase factor does not matter means that we can choose it to be whatever we like. This gives us an extra constraint (even if is one we choose arbitrary) and so reduces the number of degrees of freedom by 1. For example, given that $|a_u|^2 + |a_d|^2 = 1$ we can write our coefficients as \begin{align} a_u = &\cos\theta ...

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Whether forces can be repulsive or not depends on the spin of their mediating field. A scalar (spin-0) force is universally attractive, as is a spin-2 force, while a spin-1 is attractive for different charges and repulsive for like charges. So the electromagnetic, the weak and the strong force can be repulsive, while gravity cannot.

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Almost. Whereas I guess the E-de-H effect does require the magnetic moment to be along the spin direction, what the effect is really used to show is that an electron spin flip imparts $\hbar$ of macroscopic angular momentum to the cylinder. Just because we know that the electron is a two state object transforming under rotation by the s=1/2 representation ...

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There are many forms of angular momentum. Spin angular momentum for this particle or that particle, orbital angular momentum of this particle about that point, totals of various angular momentum. None of it straight forward. And none of it just involves saying a non inertial frame is totally the same as an inertial one. And none of it involves confusing ...

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The spin indicates the length $(=2s+1)$ of the vector that a real world particle rotates like. They do not all rotate like pencils (3-vectors). Your questions are not silly! Part of Quantum mechanics involves 1) making a correspondence between a symbol (a |ket>) that you write on piece of paper and an object in the real world, and 2) making a ...

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Under your assumption of simultaneously well-defined x and z value, you reach predictions which are inconsistent with quantum theory. This is exactly what leads to Bell's inequality which is (experimentally!) violated by quantum theory, see https://en.wikipedia.org/wiki/Bell%27s_theorem or the explanation in Preskill's lecture notes ...

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The spin of the electromagnetic field tensor $F_{\mu\nu}$ is best understood by writing it as a spinor. A spin 1 field is a represented by a symmetric spinor $\xi^{AB}$ or by a dotted symmetric spinor $\eta_{\dot{A}\dot{B}}$. In order to get the field transforming correctly under parity, the electromagnetic field has to be a direct sum using the symmetric ...

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So the state is clearly $$\left(\begin{matrix} 1 \\ 0\end{matrix}\right)_1\otimes\left(\begin{matrix} 1 \\ 0\end{matrix}\right)_2.$$ Or $\left|s_1 s_{z1} s_2 s_{z2}\right\rangle=\left|\frac{1}{2} \frac{+1}{2} \frac{1}{2} \frac{+1}{2}\right\rangle.$ So really you are just trying to write it in the total angular momentum basis. This is what Clebsch-Gordan ...

1

The experiment rests on the assumption that if the particles contained hidden variables that determine whether they are measured with spin-up or spin-down, the measurements of spin would be the same (both spin-up or both spin-down) only 4/9ths of the time. The experiment doesn't rest on assumptions, the experiment is an experimental setup that produces ...

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You can transform to the rotating frame as follows: $$\psi_{\mathrm{rot}} (t) = \hat{U}(t)\psi(t),$$ where the time-dependent unitary transformation $U(t)$ is defined by U(t) \equiv \exp\left(i\omega t S_{z}/\hbar\right) = \begin{pmatrix} e^{i\omega t}&0&0\\ 0&1&0\\0&0&e^{-i\omega ...

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I) The main point is that when we apply Noether's theorem for a field theory, the total angular momentum Noether current $$J^{\mu,\nu\lambda}~=~L^{\mu,\nu\lambda}+S^{\mu,\nu\lambda}$$ splits in an orbital angular momentum current $$L^{\mu,\nu\lambda}=x^{\nu}T^{\mu,\lambda}-(\nu\leftrightarrow \lambda)$$ and an internal spin angular momentum Noether ...

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First, it is worth mentioning that while the word measurement is used, the experimental process involved actually changes the state rather than revealing some preexisting property. This can be known by considering an interaction for the z component followed by the z component again followed by the x component as compared to the interaction for z component ...

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Once the stable particles , electron and proton have had their spin determined by the stern gerlach method as discussed in the other answer, one can start building up the spins of the elementary particles and the resonances. The spins of the particles have been determined by the angular distributions of decay products. Example: the Higgs has been declared ...

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Finally I solved it using the Cayley–Hamilton theorem. Since the characteristic polynomial of $M$ is $$\lambda^3=\Omega^2\lambda,$$ where $\Omega=\sqrt{\delta^2+\omega_{\perp}}$, we have that $$M^3=\Omega^2M.$$ Then ($M^4=\Omega^2M^2$, $M^5=\Omega^4M$, $\dots$)  ...

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