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## Hot answers tagged quantum-spin

16

By definition of spin $S$ it is a positive integer number or zero. Not to confuse with the spin projection possible values $S_z$, which may run from $-S$ to $S$.

12

Photons, as each massless particles, are characterized not by spin (which is defined as total angular momentum at rest, and mathematically corresponds to irreducible representation of the little group of representation), but by helicity $\lambda$ - the projection of total angular momentum on the direction of motion. Actually, the Casimir operator, which ...

11

Massless particles with spin do not have a "$S_z = 0$" state because they actually do not have spin like massive particles do. They have helicity, which is the value of the projection of the spin operator onto the momentum operator. The reason for this is the representation theory of the group of spacetime symmetry, the Poincaré group. To understand this, ...

7

When we say that the electron has "spin half," we mean half of the quantum of angular momentum, $\hbar$. A good quantum mechanics text or other reference will help you derive that the Laplacian operator transforms into spherical coordinates like \begin{align} \nabla^2 &= \left(\frac\partial{\partial x}\right)^2 + \left(\frac\partial{\partial y}\right)^...

5

Spin 1 just means that the spin in any direction can assume values out of {-1,0,1}. The 0 is only possible for massive particles, so the photon can have spin -1 or +1. That's like clockwise and anticlockwise circular polarization

4

The short story is that the Hyperphysics article you link to is using classical and semiclassical heuristics to justify the numbers it present. As I'm sure you're aware, the hydrogen atom cannot be described in any rigorous detail using classical mechanics, and instead requires quantum mechanics for any appropriate treatment, particularly where the ...

4

In Quantum Field Theory the one particle states are defined as the states of an irreducible unitary representation of the Poincare Group. If this was not true, there would be states of a reducible representation that would not be connected by a Poincare transformation. These states are rather different particles. The Casimirs If we have an irreducible ...

4

Given an angular momentum operator with components $S_1, S_2, S_3$ and commutation relations $[S_i, S_j] = \sum_k \epsilon_{ijk}S_k$, where $\epsilon_{ijk}$ are structure constant of the $\mathfrak{su}(2)$ algebra, the Casimir operator $S^2 = S_1^2+S_2^2+S_3^2$ can be diagonalised simultaneously with any of the original components $S_j$ onto their ...

4

Because it doesn't have total spin $s=0$ - it has total spin $s=1$, with the spin component parallel to the $z$-axis being zero. If you looked at that state in a different basis (e.g. the $x-$ or $y-$ basis) it would very clearly not have spin 0.

3

I will address the question from the current body of the question, asking for $\operatorname{Tr}(S^2\rho)$ on a system of $N$ spin-1/2 particles in the thermal state $\rho = \frac1Ze^{-\beta S_z}$. This has little to do with the photons in the question title. The total spin operator splits into single-particle terms and two-particle cross terms. $$S^2=\... 2 Your initial state is entangled, which means it exhibits a correlation between the spin of the two particles. Each particle is spin up or spin down with equal probability, but the spins are (anti)correlated so that they cannot have the same spin. The states \left|\uparrow \downarrow\right> and \left|\downarrow \uparrow\right> are in agreement ... 2 Spin operator of the total 2 electron system is tricky: the statistical requirement reduces the Hilbert space to a 3-d rather than 4-d version. Like a spin-1 system, the S_z, as represents in basis |S_z=1,0,-1\rangle is (see http://quantummechanics.ucsd.edu/ph130a/130_notes/node247.html, for example)$$S_z=\hbar\left(\begin{matrix}1&0&0\\0&0&...

2

As a matter of fact, there is a much easier way to derive those matrices. What you are really after is the matrix representation of the $\mathfrak{su}(2)$ Lie algebra generators $J_x$, $J_y$ and $J_z$ in the irreducible $\mathrm{SU}(2)$ representation with spin $j$, given by $$J_z|j,m\rangle = m |j,m\rangle,\quad J_\pm|j,m\rangle = C_\pm(j,m) |j,m\pm 1\... 2 According to quantum electrodynamics, the most accurately verified theory in physics, a photon is a single-particle excitation of the free quantum electromagnetic field. More formally, it is a state of the free electromagnetic field which is an eigenstate of the photon number operator with eigenvalue 1. The single-particle Hilbert space of the photon ... 2 The two papers talk about very different things. In Kapustin's paper, he considered non-orientable space-time manifold to classify SPT (i.e. the partition function of the phase on these manifolds). To do that, one has to first Wick rotate to Euclidean space-time, where time-reversal becomes a mirror reflection, but with a sign change. In Watanabe et. al. ... 2 A PSPACE machine can simulate any BQP machine by using path integrals. Since quantum computers are like BQP machines, and BQP is a subset of PSPACE, quantum computers can't efficiently solve problems outside of PSPACE. 2 I think the argument is that the proton as well as the neutron and electron are spin half particles. There is no way to add two spin half such that you get a spin half as a result. This is also a short argument why two fermion behave as a boson. 1 Sometimes a picture can help. Its not hugely rigorous (especially for a physics site) but using the vector model, pictures can be constructed to illustrate the states that two spins \alpha and \beta can form. 1 Mutually non-commuting operators cannot have simultaneous eigenstates, namely the eigenstates of the former must by all means be expressed as a linear combinations of (all) the eigenstates of the latter. In the case at hand, given {|+\rangle}_z as eigenstate of the operator S_z, the following must hold:$$ {|+\rangle}_z = c_1 {|+\rangle}_x + c_2 {|-\...

1

I want to give an answer for electrons which have both an intrinsic spin and a magnetic dipole moment. The key for the understanding why electrons in rest in relation to an external magnetic field get aligned and do not precess while moving non-paralle to the external magnetic field electrons undergo a precession is their kinetic energy in relation to the ...

1

I had the same doubt as I was figuring out if a certain coupling of angular momenta in the 3$\gamma$ could be found such that parity would be conserved. Turns out (if I am correct) you cannot have three photons with coupled total angular momentum $J=0$ in the first place. For the coupled photon angular momentum $L(3\gamma)$ and spin $S(3\gamma)$ to be able ...

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