New answers tagged

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 ...


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 ...


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\...


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=\...


0

Why has the $S_z$ operator suddenly become part of the generator of translation? Because the authors want their $U$ operation to make the spin and the position interact. The process that moved the $S_z$ into the generator wasn't anything physical or simulated, it was just the authors defining a useful operation. Also where did that factor of 2 come ...


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.


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.


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.


0

This might be easier to understand in terms of the Pauli matrices: $\sigma_z=\pmatrix{0 &1\\1 &0}$, $\sigma_y=\pmatrix{0 &-i\\i &0}$, $\sigma_x=\pmatrix{1 &0\\0 &-1}$, where the operators for $z, y$, and $x$, are given by $\hat{S_z}=\tfrac{\hbar}{2}\sigma_z$, $\hat{S_y}=\tfrac{\hbar}{2}\sigma_y$, and $\hat{S_x}=\tfrac{\hbar}{2}\...


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)^...


-4

Below is in response to the title of your post: what exactly is spin 1/2 The long answer is that spin is evidently an internal symmetry of a particle. There is no good way to think of spin in intuitive terms. Spin 1/2 particles are those objects which transform under the spinor representation of the poincare group (a spinor is a weird thing that has been ...


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 ...


0

In general you are getting it right: Non-commuting operators do not share eigenstates, thus measuring $S_x$ on an eigenstate of $S_z$ will result in a state that is not an eigenstate of $S_z$ anymore. The spin operators do not commute because they are defined via the Lie-algebra relation $[S_i, S_j] = i \hbar \varepsilon_{ijk} S_k$. Next, if the particle ...


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 ...


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 ...


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 ...


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, ...


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 ...


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 $S$.


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


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. ...


0

It seems that I've derived an answer. The spin of one photon in some sense is determined as the quantity, which characterizes the non-coordinate transformation of 4-potential $A_{\mu}$ under spatial rotation (i.e., mixing components without changing the coordinates, and hence without change the momentum of photon). Let's fix $A_{0} = 0$ (therefore we ...



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