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When I first started to study quantum mechanics, my physics text book told that particles have spin of either 1/2 or -1/2. That's wrong. Particles can have any integer or half-integer spin. (There are some deeply technical reasons that fundamental particles are expected to have spin ranging from -2 to 2, but if you include composite particles, any ...


4

No. The commutation relation merely means that the $T_i$ form the Lie algebra $\mathfrak{su}(2)$. There are $\mathrm{SU}(2)$s (and consequently $\mathfrak{su}(2)$s) which have nothing to do with angular momentum, e.g. the $\mathrm{SU}(2)$ in the electroweak symmetry group $\mathrm{U}(1)\times\mathrm{SU}(2)$.


4

The Schrödinger equation is only correct in the non-relativistic limit $v << c$, for particles without spin. The correct equation for spinless (=spin $0$) particles is the Klein-Gordon equation, which reduces in the non-relativistic limit to the Schrödinger equation. If we want to talk about spin $\frac{1}{2}$, the correct, relativistic equation is ...


3

$\vert+\rangle$ and $\vert-\rangle$ are really just shorthand notations for the two eigenvectors of the diagonal spin operator $\sigma_z$. This means concretely: $$\vert+\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} $$ $$\vert-\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $$ Therefore the action of the sigma operator gives you simply the corresponding ...


2

A magnetic field consists of photons. Photons are spin $1$ particles, which means for a given $z$ axis a measurement of spin can yield $+1$ or $-1$. If a photon collides with an electron, we know that spin must be conserved. Let's assume the electron is in a spin up state $\uparrow_e= + \frac{1}{2}$ and the photon in a spin down state $\downarrow_P= -1$. ...


2

Application 1: It's the $z$-component of the vector valued angular momentum observable for a spin $\frac{1}{2}$ particle, when the basis states are the $z$-component angular momentum eigenstates. If this sounds a bit circular and tautological, it is the reason why $\sigma_z$ is diagonal. So the $n^{th}$ moment of the probability distribution of an angular ...


2

The answer is yes and no, but first, let me point out that you cannot "prove Bell's inequality", the whole point is that you violate the inequality in quantum mechanics. Now, let me come to the yes/no part: It's "no, you cannot violate Bell's inequality with this state", if you refer to what according to wikipedia is "the" Bell inequality: $$ \rho(a, c) ...


2

I haven't thought about this one before, so here is an approach that will work if you work hard enough at it. Before I begin banging on, point number 1: Should I assume a spin 3/2 system (4x4 Matrix) or an entangled Hilbert space with spin 1/2 and spin 1 (6x6 Matrix)? Unquestionably the latter. It is a bipartite system and its state space is the ...


2

Quantum mechanic predicts, that the allowed directions of the spins are quantized. This is one of the main findings of the Stern–Gerlach experiment. In a thermal beam I suppose the the spins to be equally in up and down. (There is no reason why they should not.) But "up" and "down" only correspond to a specific direction in space if there is an external ...


2

The angle is the same as long as you consider a free electron. Then they are parallel: $\vec{\mu}_\mathrm{elec}=-g_\mathrm{elec}\mu_\mathrm{Bohr}\frac{\vec{S}}{\hbar}$ with $g_\mathrm{elec}\approx 2$ (neclecting effects from quantum electro dynamics). But when dealing with bound electrons (e.g. in an atom), where the electron also has some orbital angular ...


2

When you have only one electron then $\Delta S=0$ makes intuitive sense: you can change the angular momentum $l$ of the atom by changing it's internal structure (by pushing the electron in "another orbit" if you will), while you certainly can't change the internal structure of the electron to change $s$. Would it be possible to change $s$ then you could ...


2

You're probably used to the convention where a hat is used to denote that something is an operator. But that convention is not universal. In many cases, when it's clear from the context whether something is an operator or not, we just write it without a hat either way. For this case in particular, $\boldsymbol{J}$ is defined to be an operator. The fact that ...


1

The reason is not really spin - we get circulating magnetic fields due to currents in straight wires, where spin is effectively absent. Many would argue that the most fundamental quantity in electromagnetism is not the fields themselves, but what is called the electromagnetic vector potential, $\mathbf{A}$. This potential is directly 'sourced' from the ...


1

When someone says that spin measured about different axis can't both be known, they mean that whatever state you pick will have variability in at least one of the possible spin measurements you can do. So that is what you will get when measure the spin, you will get variable results. This happens even with entanglement with even just one particle. With ...


1

For a famous example of a nucleus with internal orbital angular momentum, consider the deuteron. Considerations of exchange symmetry, spin, and isospin demand that the deuteron have unit spin, rather than zero spin. However the pion-nucleon interaction, gleaned from neutron-proton scattering and deuteron formation, suggests that about 4% of the deuteron ...


1

Yes, the statement can be explicitly verified from the matrix representation of the spin operators acting on different spins. Acting on the spin-1/2 object, the spin operators read $$S^x=\left( \begin{array}{cc} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \\ \end{array} \right), S^y=\left( \begin{array}{cc} 0 & \frac{i}{2} \\ -\frac{i}{2} & 0 \\ ...


1

You chose the $\lvert \pm \rangle$ to be an eigenvector of $S_z$ with eigenvalue $\pm\frac{1}{2}$ - that's what the $m_s$ is: The eigenvalue of the state w.r.t. the $z$-spin. Since $S_x$ and $S_y$ do not commute with $S_z$, $\lvert \pm \rangle$ is not an eigenvector of them, hence the state cannot stay the same after they are applied to it. That the spin ...


1

Not necessarily. Some operators representing other physical quantities can be transformed so that they have the same algebraic structure of the angular momentum operators. For example, the inverse of "Jordan-Wigner transformation". Of course, you can think of them as effective angular momentum.


1

\begin{align*} -i\gamma^5 \gamma^i \gamma^j \partial_j &= \gamma^0 \gamma^1 \gamma^2 \gamma^3 \gamma^i \gamma^j \partial_j \\ &= \tfrac{1}{3!}\gamma^0 \epsilon_{klm}\gamma^k\gamma^l\gamma^m\gamma^i\gamma^j\partial_j\\ &= \tfrac{1}{2}\gamma^0 \epsilon_{kli}\gamma^k\gamma^l\gamma^j\partial_j\\ &= \gamma^0 \epsilon_{kji}\gamma^k\partial_j\\ ...


1

"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 ...


1

Hmmm, an old question without a satisfactory answer. I'll have a go. The spins of the two $B$ may combine as \begin{align} \text{singlet}\quad|s_1s_2\rangle &= \frac{\left|\uparrow\downarrow\right> - \left|\uparrow\downarrow\right\rangle}{\sqrt2}, & \text{or triplet}\quad|s_1s_2\rangle &= \frac{\left|\uparrow\downarrow\right> + ...


1

In that case, $\hat{\sigma}$ here refers to a vector formed by $\hat{\sigma}_x$, $\hat{\sigma}_y$ and $\hat{\sigma}_z$ as its Cartesian components. The individual components of the expectation value of the magnetic moment vector would then be obtained using the corresponding components of the Pauli spin operators.


1

I'm not sure what you mean by macroscopic? The experiment was first done with silver atoms which, due to having 1 unpaired electron of spin 1/2, split into 2 distinct beams. This was a quantum mechanical event that was visible macroscopically. If by macroscopic you mean on a large object instead of a beam of atoms then I don't think it would be feasible or ...


1

I am not supposed to place an answer here, as I am no more active in this site: however, a comment doesn't offer enough place. So, you ask: 1) "spin is a property of the wave function, and not of the particle?" Please pay attention to the following differences between the standard quantum theory (SQT) and the Bohmian interpretation (BI): SQT ...


1

The Dirac notation is simply an alternative to vector notation. Certainly there are PDEs describing the quantum state of a lone particle with spin and they are: The Pauli equation (see Wiki page of this name) was historically the first, and here the quantum state is two $\mathcal{L}^2(\mathbb{R}^3, \mathbb{R})$ functions of space and time. The two ...


1

The Dirac equation is mentioned in other answers as PDE describing spin. As you ask "what would Schrodingers equation (or some kind of generalization that allows for you to include spin) look like?", the following may be relevant. Yes, the Dirac equation adequately describes spin. However, it is actually a system of four partial differential equations for ...


1

The energy difference along with the larger thermodynamical likelihood for occupation of the lower level is real. There is an application, nuclear magnetic resonance (NMR) spectroscopy/imagining/quantum computing. But due to the very small energy difference for technically achievable magnetic fields, the effect is usually negligible at roomtemperature. NMR ...


1

If the particle source is "unpolarized", that literally means it is equally likely to find particles from this source in either energy eigenstate - that's the definition of "unpolarized", so you shouldn't be surprised about that. When the spin of a particle is "perpendicular to the magnetic field", that's another way of saying that the particle is in an ...



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