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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 $s_z = \hbar /2$.


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At low energies, close to the Dirac points we can obtain the following effective Hamiltonians: \begin{align*} \begin{array}{ll} \mathcal{H}_{\mathbf{K}} = \hbar v_{F}\left(\Pi_{x}\sigma^{x} + \Pi_{y}\sigma^{y}\right) & \text{at the $K$ point} \\ \mathcal{H}_{\mathbf{K^{\prime}}} = -\hbar v_{F}\left(\Pi_{x}\sigma^{x} - \Pi_{y}\sigma^{y}\right) & ...


3

There's not a unique $|+_z\rangle$, because you can always choose a difference phase. This doesn't matter for the eigenstate, but once you take a superposition the relative phase makes a real difference. So your definitions of $|\psi\rangle$ and $|\psi'\rangle$ are ambiguous. Indeed, $|\psi\rangle$ could be $|+_z\rangle$ for one choice of phase, or could be ...


2

The fact that your integral is a surface term is indeed due to Stokes theorem: $$ \int d^3x\ (x^1\partial^2 -x^2\partial^1)(\overline\Psi(x) \gamma^0\ e^{ipx^3})\\ =\int d^3x\ \partial^2\left(x^1 \overline\Psi(x) \gamma^0\ e^{ipx^3}\right)-(1\leftrightarrow 2)\\ =\int dx_1 dx_3 \left(\int dx_2 \partial^2\left(x^1 \overline\Psi(x) \gamma^0\ ...


1

Quantum mechanics (QM; also known as quantum physics, or quantum theory) is a fundamental branch of physics which deals with physical phenomena at nanoscopic scales, where the action is on the order of the Planck constant The Planck constant is a very small number, 6.6*10^-34 Joulesecond Quantum mechanics was invented because the data showed that at ...


4

In theory, it is possible to measure chirality using the exact method that you mentioned. If you have an electron and you are unsure its chirality, you can send it through a dense bucket of $W$ bosons and see if it interacts with them. If it did, then you know that you had a left handed electron. There is, however, an important issue with this idea (and no ...


0

The Blandford-Znajek process and variants are leading contenders for explaining black hole jets. The driving energy for the jets is extracted from the black hole spin and transferred to the outgoing plasma by twisted magnetic fields. Along with energy, the jets must contain the lost angular momentum from the black hole, which then ultimately comes from ...


2

How does one explain the results that the spin of a particle is always measured with its full magnitude along the plane of measurement? In Bohmian mechanics, the spinor valued wave function (packet), under the influence of the apparatus, splits into two discrete packets only one of which guides the Bohmian particle; the other packet is 'empty'. Which ...


1

GENERALITIES : Let's precise some issue: the Bohm interpretation (BI) suggests that particles have from the beginning, i.e. even in the initial wave-function, not only positions but also velocities. The mathematics of the BI, in essence, predicts trajectories. In these calculi, the quantum potential tells through which regions the particle may pass, and ...


-4

Both, but they are not the same magnetic fields. As a clue you may consider rotation and revolution as 'not' two different phenomena, but even revolution is a rotation.


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The magnet has a finite moment of inertia. What would happen when the magnet with "wrong" orientation enters Stern-Gerlach apparatus? Of course, the magnetic field will exert torque on it. The magnet starts rotating. After it comes to the equilibrium orientation, i.e. is oriented along the field, the torque is zero, but angular velocity is at maximum, and ...


2

Suppose you shot a large number of small classical magnetic dipoles with magnetic moment $\vec{\mu}$ through the field. Imagine the dipoles to be small enough that they could be treated as the particles of an ideal gas, and they are "boiled" out of some source into the magnetic field. We would then expect each of the particle's components of velocity to ...


2

If $\phi=0$, the state lies in the $x$-$z$-plane, i.e., it is of the form $\alpha\lvert0\rangle+\beta\lvert1\rangle$ with $\alpha,\beta\in\mathbb R$.


2

An ordered SDW phase breaks both the continuous $SU(2)$ spin-rotation symmetry and the time-reversal symmetry (because the presence of either of these two symmetries would force the order parameter of SDW vanishing). It is the spontaneously broken of continuous spin-rotation symmetry that leads to the gapless Goldstone mode. Here is a related issue. The ...


2

There must be an error in the way you calculate the expectation values. The fact that your transformed operators have the same spectrum of eigenvalues and the same algebra comes from the fact that redefinitions of the phase lead to unitarily equivalent operators, i.e. they are related by $$\tilde j_+ = U^\dagger j_+ U,\quad \tilde j_- = U^\dagger j_- ...


3

The first step, probably you have already know, is to get the ground state. We may write the Hamiltonian on a set of orthogonal basis, e.g., $$\left| n_{1,\uparrow} n_{2,\uparrow} \cdots n_{N,\uparrow}, n_{1,\downarrow} n_{2,\downarrow} \cdots n_{N,\downarrow} \right\rangle =\prod_{i} (c_{i,\uparrow}^{\dagger})^{n_{i,\uparrow}} \prod_{i} ...


0

A really neat and intuitive way to deal with qubit systems like spin is to use the Bloch sphere. The Bloch sphere represents the two dimensional Hilbert space which the spin $\frac{1}{2}$ state vectors live in by a sphere in $\mathbb{R^3}$. Have a look at this wikipedia page for some more info http://en.wikipedia.org/wiki/Bloch_sphere. (Particularly the ...


3

The spin of the photon is intrinsic in its description as an elementary particle, +/-1 when measured. There is no zero component because the photon has mass=0., so the spin is always either parallel or antiparallel to the direction of motion of the photon. This wiki article describes the relation between light polarization and photon spin. I find this ...


1

Well you can get Weyl spinors, it's exactly the same description of electron in terms of light helicity - 'left' and 'right'-handed. But if you will go deeper you have to remember, that in Standard Model left-handed electrons are paired with neutrinos, but right-handed electrons are single (there is no right-handed neutrino for couple). Ah, one more ...


1

Suppose I have two different spins, spin $A$ is in state $\left|\uparrow \right\rangle$ and sping $B$ is in state $\left|\downarrow \right\rangle$. I give you one of these spins randomly, so you have a 1/2 probability of spin up and 1/2 probability of spin down. There's no coherence between up and down because you either have a spin which is definitely up or ...


1

$A^\mu$ is a quantum field, i.e. an operator-valued distribution, and not a wavefunction. In QFT, there are no wavefunctions, since we have no naive position basis for our theory (the theory of the relativistic position representation is called Newton-Wigner theory). The analogue to the QM wavefunction is a wavefunctional that is a functional of time and ...


2

The photon wave function is a solution of a quantized Maxwell's equation , in the potential form. There the variables exist which will allow to build up the classical electromagnetic wave from an enormous number of photons, as shown in this link : The photons also have polarizations so the wave function has many components, too. I don't want to scare ...


3

As in the comments, fulfilment of the Klein Gordon equation is only a necessary condition for a field and it is fulfilled by all fields. For example, the Dirac equation for an electron implies the Klein Gordon equation, but not conversely. If you've never seen this, try working out the following. Begin with the Dirac equation $(i{\partial\!\!\!\big /} - m) ...


2

Recall that two ket vectors which differ by a global phase represent identical physical states. In this case, $|\psi_2\rangle = i|\psi_1\rangle$, and the states are therefore equivalent. One can always use this global phase freedom to make the coefficient of $\lvert+z\rangle$ real, and then a parametrisation in terms of the angles $\theta$ and $\phi$ becomes ...


0

(I understood your question as "how is it possible to change the electron spin at all in presence of a magnetic field", hope that is alright.) I expand a bit on the electron paramagnetic resonance. In the presence of an externally applied magnetic field the normally degenerate energy levels split up due to the zeeman effect. So let's say that you have one ...


2

The Pauli equation, which describe non-relativistic, spin-1/2 electrons and is the non-relativistic limit of the Dirac equation, can be obtained from the Schrödinger equation throught minimal coupling in a rather similar way of its relativistic counterpart. Consider the Schrödinger equation for a particle with charge $q$ in an external electrostatic ...


0

It's certainly true that a classical charged object that spins generates a magnetic field; the moving charges form current loops. If you're trying to understand NMR this is probably a good-enough model for the nucleon magnetic moments. Quantum-mechanical spin is a different creature, though, and thinking about it classically can get you only so far.


0

The equation of motion is $$\partial^2 A_\mu = 0$$ which in fact is a set of four KG equations, one for each field $A_\mu$, $(\mu = 0,\dots,4)$. This gives us four degrees of freedom for the $A_\mu$ field. But this cannot describe the electromagnetism, since the light has only two degrees of freedom (polarization). We must not forget that we have set a ...


0

yes, we have massless particle but not spin zero because A has 4 index.so it is basically four fields. but for EM we have only two states of polarization so it is cut down to two only not four. for refrence see peskin book.


0

Why planets, stars and other extended masses have rotation. Firstly, a few points: Most bodies in the universe are unhinged (i.e there is no physical 'hinge' holding the body in place) and move in space. To move any body, you have to give it a momentum. The momentum need not be oriented in any particular direction and need not be transferred at any ...


1

In order to deal with fermion fields in general relativity, that are representations of the Lorentz group, you want to work in a local inertial frame (LIF). To go back and forth between space-time indices $\mu=0,1,2,3$ and LIF indices $a=0,1,2,3$ you introduce the formalism of frame fields: $$g_{\mu \nu}=e^a_\mu (x) \eta_{ab} e^b_\nu (x)$$ The ...


0

There are a couple of mistakes in what you say. It may ne that Sakurai is not so clear. This is a book mentioned by other people too. So, let's fix the things. 1) You say "in this discrete case, we can see quantum behavior more clearly than with wave functions." No, wave-function is not something opposite to the continuous case. Both in the discrete and in ...


3

The momentum decomposition you wrote is valid only for a scalar (spinless), real field, satisfying the Klein-Gordon equation. When considering a field with spin, like a spin-$1/2$ field satisfying the Dirac equation, you must include the polarization vectors, obtaining something of the form $$ \psi_\alpha(x) = \sum_{\textbf{p},s} N_{s,\textbf{p}}[ ...


2

The short answer is that the spin states of a photon come in two kinds, based on helicity, how the circular polarization tracks with the direction of the photons momentum. You can think of them as circularly polarized in the sense that we can define the relative relationship between the different polarizations the same way we do for classical ...


0

I'm not exactly sure whether this applies for the angular momentum but I know it is true for the spin.. When things condense they begin to rotate or their existing rotation is accelerated. This can easily be seen in the form of neutron stars, and as an example from here on earth, a tornado. In the case of the neutron star, it spins rapidly, but before its ...


1

Photon's spin is different from spin for other particles. If we talk about a massive particle with spin 1, it'll have three possibilities for, lets say, $S_z$ which are $−ℏ,0,ℏ$. The fact that photon is massless, causes some mathematical peculiarities that rules out the 0 case. Then for photon, we don't talk about $S_z$, instead we say its helicity is either ...



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