Tag Info

Hot answers tagged

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 ...


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, ...


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 ...


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 ...


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 ...


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. ...


3

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 ...


2

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 $s_z = \hbar /2$.


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 ...


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 ...


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 ...


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 ...


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 ...


1

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 ...


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 ...


1

"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 ...


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 ...


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 ...



Only top voted, non community-wiki answers of a minimum length are eligible