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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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No, if the spin were classical, as any classical angular momentum it weren't quantized. It's easier to take as an example the orbital angular momentum. It is quantized because of the wave-nature of the quantum particle. To put it in an intuitive form, not every function can represent the wave-function of the electron in the atom. Imagine an oscillating ...


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Let's take a simple example, a free electron in a magnetic field $H$, and let's take for simplicity the direction $z$ in the direction of the field. The electron has a magnetic dipole, $\vec {\mu}$. In Stern-Gerlach experiments we find this dipole either oriented in the direction of the field, or oppositely. We say, accordingly, that the projection of the ...


1

You are correct that spin states live in complex space, however, the expectation value $\left\langle\psi|S_z|\psi\right\rangle$ lives in real space. It is simply a real number, which represents the expectation value of spin if a series of measurements is made in the $z$ direction. You can see that the expression must be a scalar because ...


3

The spin operator $\vec S = \left(\begin{matrix} S_x \\ S_y \\S_z \end{matrix}\right)$ is just like the (orbital) angular momentum operator. $\langle \psi \rvert S_i \lvert \psi \rangle$ gives you the expectation value for the component of the spin angular momentum. $\langle \psi \rvert \vec S \lvert \psi \rangle$ is the expectation for the full spin vector. ...


1

Unless I'm missing something, your expression is just an expectation value of $\hat{S}_z$ when in the state $| \psi \rangle$. This is an actual measurement you could make on an ensemble of atoms, by running them through a Stern-Gerlach apparatus and counting +$\hbar/2$ for every one that hit the "top" detector and -$-\hbar/2$ for every one that hit the ...


1

Nothing prevents the electron's spin from being measured along a particular axis, and then subsequently measured along an axis perpendicular to the first. In this situation, however, the spins along the perpendicular axes would not be known simultaneously, so the uncertainty principle would not be violated. As an example, say that we perform our own ...


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"By the uncertainty principle" is the answer. In more detail, let's say we're talking about x and y axes. The first measurement puts the electron into an eigenstate of the spin X observable (the question of how it does this is the quantum measurement problem). Whichever of the two X eigenstates this "collapse" ends up in, it not an eigenstate of the spin Y ...


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This may not be exactly what you want, but it's something. The definition $$ \vec{L}=\vec{r}\times\vec{p} $$ combined with $\vec{p}=-i\hbar\vec{\nabla}$ implies that the the $z$ projection may be written as $$ L_{z} = -i\hbar\frac{\partial}{\partial \phi}.$$ Where $\phi$ is the azimuthal angle. Since $L_{z}|\ell m\rangle = m\hbar |\ell m\rangle$, the $\phi$ ...


1

The identity you used, $$ \exp(i\theta \, \hat s)=\cos(\theta)+i\sin(\theta)\,\hat s, \tag{$\ast$} $$ is crucially dependent on the operator $\hat s$ being idempotent, and particularly on the fact that $\hat{s}^2=\mathbb1$. This is generally not the case for angular momenta other than spin-1/2. In general, the total angular momentum is a scalar within the ...


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The point is that the spin operator is defined to be (1/2) times SU(2) generator while the orbital angular momentum is defined to be only SU(2)(or SO(3), is the same) generator. The proof is the same, and is " representation independent", in the sense that the structure of identity multiplied by something plus a linear combination of sigma matrices ...


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You can use similar derivation, but with momentum operator J for integer spin instead of Pauli matrix $1/2\sigma$. Then you get integer coefficient in from of $\alpha$ instead of half-integer in the final formula... and the same state after $2\pi$ rotation.


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Say you had didn't know about quantum mechanics and no idea what a spinor is. You're given an "operator" that rotates things. One of the most basic assumptions you will make is that a rotation of $2\pi$ changes nothing. This is really quite reasonable. As @Phoenix87 said, we can identify $SU(2)$ has having a Lie algebra isomorphic to that of $SO(3)$. We ...


4

The spin of a vector boson in any dimension is spin 1. What changes with the number of dimensions is the number of degrees of freedom associated with a given spin. A massless vector in four dimensions has two independent degrees of freedom, which can be seen from the rank of what's called the "little group" in the literature. It is the subgroup of the ...


0

Different Lie groups can have the same (up to isomorphism) Lie algebra. This is the case of, say, $SO(3)$ and $SU(2)$, the latter being the universal 2-cover of the former. When you are given a Lie algebra $\mathfrak g$ and you want to integrate it to a Lie group $G$ having $\mathfrak g$ as a Lie algebra, you will end up with a simply connected group. Hence, ...


1

The classic example here is the Thirring model, which describes fermions in 1+1 dimensions. While this is a very special two dimensional model (so the conclusions cannot be generalized), Sidney Coleman found that it is equivalent to the Sine-Gordon model, a theory of bosons. The demonstration is rather technical (Coleman essentially proved that the ...


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Some of my recent results may be relevant, but they use electromagnetic field as input, and electromagnetic field is associated with spin 1. Three out of four components of the Dirac spinor function can be algebraically eliminated from the Dirac equation in an arbitrary electromagnetic field. The resulting equation for one complex function (which can be ...


7

The answer is Yes. See A physical understanding of fractionalization http://arxiv.org/abs/hep-th/0302201 Quantum order from string-net condensations and origin of light and massless fermions, Xiao-Gang Wen; Spin-1/2 and Fermi statistics from qubits http://arxiv.org/abs/hep-th/0507118 Quantum ether: photons and electrons from a rotor model, ...


1

The time evolution of the two spins can be separated if they are independent, i.e. if they don't interact. Under this assumption the time operator splits in the tensor product $$U_1\otimes U_2=(U_1\otimes I)(I\otimes U_2)$$ and therefore it is clear how to define the time evolution for the single spin: for the $j$th particle one simply needs to take the ...


7

First, the electron is not a point particle. The abstraction you are thinking of is what we would call a naked electron. In an experiment, you do not see the naked particle ever. It is always surrounded by virtual pairs. Hence, what you measure as the electron is really a many-body system. Second, you might want to read this. The take-home message is "the ...


2

Just to clarify to Robin Ekman's answer, superpositions of the Pauli matrices exponentiate to $SU(2)$, not $SO(3)$, but both these Lie groups have $\mathfrak{so}(3)$ as their Lie algebra - but I am sure you already know this. Also, there is another way to look at the problem that you might find helpful, even though it is a mathematical insight rather than a ...


4

Yes. This commutation relation is that of the Lie algebra $\mathfrak{so}(3)$ corresponding to the rotation group in three dimensions. Thus the commutation relation states that the Pauli matrices generate rotations. To understand why this is the commutation relation of $\mathfrak{so}(3)$, one can draw a diagram showing that the commutator of two ...


1

The wave function only contains all the information about the system im so far as you consider it. Meaning each qualitatively different physical system needs its a modified Hilbert space to fit what can happen with the system. In case you have something like spin on its own in $ H_{Spin}$ and you want to look at a freely moving particle in $H_{free}$ that ...


4

The general question is quite hard to tackle I think, because a rigorous motivation of Hilbert space would end up in the theory of operator algebras (see e.g. this answer) and the OP is probably not interested in these aspects at the moment. As for the example of spin, the Hilbert space in this case is still an $L^2$ space, but the functions are no longer ...


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Understanding spin both conceptually and mathematically...what are we measuring when we measure spin, and what justification do we have for calling it spin? I think the confusion comes from the name. Suppose energy was named for what was applied originally to kenetic energy: maybe “speed” to pick something, that also has strong classical ...


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The maximum $s_z$ eigenvalue an electron can have is $\hbar/2$. Therefore, the only way that a quantum state can have $\langle s_z \rangle = \hbar/2$ is if the state lies completely within the $s_z = \hbar/2$ eigenspace. Thus your answer may only include pure states with $s_z = \hbar /2$.



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