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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 ...


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

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 ...


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 $\downarrow_P= -1$. ...


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 ...


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 ...


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

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 ...


0

The $|00\rangle$ and $|1,M_S\rangle$ represent the spin singlet and triplet states. The overall wavefunction must contain both the 'space' part and the 'spin' part. We can schematically express this as follows: $$\psi \sim \psi(\mathbf{r}_1, \mathbf{r}_2) |s\rangle$$ Now, Pauli's exclusion principle demands the antisymmetry of the overall wavefunction. For ...


0

"The Barnett effect is the magnetization of an uncharged body when spun on its axis. It was discovered by American physicist Samuel Barnett in 1915... The magnetization occurs parallel to the axis of spin. Barnett was motivated by a prediction by Owen Richardson in 1908, later named the Einstein–de Haas effect, that magnetizing a ferromagnet can induce a ...


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

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.


0

Let $$|s\rangle = \alpha|\uparrow\rangle + \beta|\downarrow\rangle$$ We assume that $s$ is normalized i.e. $\langle s | s\rangle = 1 \implies |\alpha|^2+|\beta|^2 = 1$. Then the expectation value of $\hat{\mu}_e$ is: $$\langle\hat{\mu}_e\rangle = \langle s|\hat{\mu}_e|s\rangle$$ $$\implies \langle\hat{\mu}_e\rangle = |\alpha|^2\langle\uparrow| \hat{\mu}_e ...


0

$\newcommand{\ket}[1]{| #1 \rangle}$ An arbitrary spin state $\ket{s}$ can be broken down as a sum of the $\ket{+}$ and $\ket{-}$ eigenstates: $$ \ket{s} = \alpha \ket{+} + \beta \ket{-} $$ Where $\alpha$ and $\beta$ are complex numbers. We'll write the overall vector as: $$ \ket{s} = \begin{pmatrix} \alpha \\ \beta \end{pmatrix} $$ Where we remember that ...


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 ...


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 ...


0

Spin is an intrinsic quantity of a particle similar to a particle's mass or charge, and has units of angular momentum. It is an observable, and the corresponding operator has a matrix representation with respect to a particular basis.


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 \\ ...


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 ...


0

I hear that analogy too. Spin 0: any rotation left the "object" invariant, like a circle who rotates. Spin 1/2: half rotation to het the initial state of the object, and here we are: any figure of the playing card "has spin 1/2". Spin 1: any non figure card, like who knows, the ace of clubs. One integer rotation to get is as it was initially. Spin 2: no ...


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) ...


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

"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

\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

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.


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 $\mathrm{U}(1)\times\mathrm{SU}(2)$.


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

But why is there no such interaction between the electrons spin $\vec{s}$ and photons? Actually, maybe there is. See Compton scattering. The incident photon is partially absorbed and decelerated in the vector sense, whilst the free electron moves. IMHO you can visualize this by drawing repeated circles on a piece of paper without lifting your pencil. Now ...


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

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 ...


1

I would re-state the problem slightly to conform to conventional notation. In any discussion of magnetic resonance -- of spin 1/2 particles, be they electrons, or say, protons-- the direction of the magnetic (polarizing) field is always chosen as +z in a laboratory coordinate frame. Then the two stationary states, which (for convenience) I write as |+> ...


-2

No, there are just spin-0, spin-1 and spin-1/2 particles, even the Higgs-Boson is 0.


0

To say that the spin "is pointed along the $x$-axis [and will] be reoriented into the $\pm z$-direction" isn't quite right to begin with. A quantum-mechanical spin doesn't exactly point in a specific direction; it's just that it only has a definite projection onto (at most) a single axis. Indistinguishable particle questions aside, you can say that an ...



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