Hot answers tagged spin
28
Higher spin particles have to be coupled to conserved currents, and there are no conserved currents of high spin in quantum field theories. The only conserved currents are vector currents associated with internal symmetries, the stress-energy tensor current, the angular momentum tensor current, and the spin-3/2 supercurrent, for a supersymmetric theory.
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22
Note As David pointed out, it's better to distinguish between generic angular momentum and orbital angular momentum. The first concept is more general and includes spin while the second one is (as the name suggests) just about orbiting. There is also the concept of total angular momentum which is the quantity that is really conserved in systems with ...
21
I don't think that this is a physics restriction, but one of current engineering capability.
As you link points out, using 12 atoms allowed the information to be retained without effecting the information stored next to it.
You will also need enough data-mass to allow for the reading and writing of the information without affecting the data next to the one ...
20
Spin is a technical term specifically referring to intrinsic angular momentum of particles. It means a very specific thing in quantum/particle physics. (Physicists often borrow loosely related everyday words and give them a very precise physical/mathematical definition.)
Since truly fundamental particles (e.g. electrons) are point entities, i.e. have no ...
12
The expression you wrote for $\sigma_z^1 + \sigma_z^2$ is not quite right, but it's not surprising that you're unsure of how to proceed because the notation is somewhat obscuring the real math behind all of this. What's actually going on here is manipulations with tensor products of Hilbert spaces.
The spin state of a single spin-$\frac{1}{2}$ particle is ...
11
For any kind of magnetic data storage you need a magnetic state that is stable over time. The magnetic moment of an isolated single atom does not have any preferred direction, therefore the energy states are degenerated.
The 12 atoms used in this experiment is not a lower limit, in principle it can also work with 2 atoms given the right magnetic ...
9
How should one imagine a particle without dimensions - like an electron - to spin?
You don't. If you want to imagine, then you think classically and it is just a particle spinning... Thinking like that doesn't give you any other insight of what spin really is (an intrinsic angular momentum, behaving like an [orbital] angular momentum).
How should one ...
9
In the field of classical mechanics angular momentum is almost always derived form linear momemtum. This actually might be the problem because it is also possible to do it the other way around: linear momentum is a limiting case of angular momentum where the radius of rotation becomes infinite. In this view the split between rotational and linear vanishes - ...
9
For spin measurements the original experiment was the Stern-Gerlach experiment in which you will see that a prior unpolarized beam will split up in two (Spin up and down) orientations.
see: http://en.wikipedia.org/wiki/Stern%E2%80%93Gerlach_experiment
For helicity, a very ingenious and fascinating experiment is the famours Goldhaber experiment that uses a ...
9
Aram's answer seems perfect, but since you are also asking about the case for higher dimensional systems, let me add that there is a simply way to get somewhat non-trivial upper and lower bounds on $C(j_S,j_L)$. As a lower bound, you can simply synthesize an arbitrary gate which implements communication between the quantum systems (for an explicit algorithm ...
9
I recently was writing about this on wikipedia. The most intuitive way to see why an operator like $S_z$ has discrete values is based on its relation to rotation operators:
$R_{internal}(\hat{z},\phi) = \exp(-i\phi S_z / \hbar)$
where the left side means rotation of angle $\phi$ about the $z$-axis, but only rotating the "internal state" of particles not ...
8
Dear rubenb, yes, what your professor says is surely based on solid maths. The reason is that the 4-component Dirac spinor is actually composed of two separate 2-component pieces.
The elementary "spinors" for 3+1 dimensions have two complex components. That results from the isomorphism between groups
$$SL(2,C) \sim Spin (3,1).$$
Note that both groups have 6 ...
8
The proton ($uud$) turns into a neutron ($ddu$). Up and down quarks don't have equal charges; the up is $+\frac{2}{3}e$ and the down is $-\frac{1}{3}e$.
By the way, such an operation has a name - isospin symmetry transformation - corresponding to an approximate SU(2) symmetry that makes the proton and neutron have almost similar masses.
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One interesting fact is that there are some revolving structures in space that aren't mostly flat - they're known as elliptical galaxies. And the difference here is that elliptical galaxies usually don't have much gas or dust in them. Interestingly enough, the orbits of objects in the inner solar system also tend to be coplanar, whereas the orbits of the ...
8
For elementary particles one uses the angular distributions of the decay products in the center of mass and the spin properties of the decay products themselves.
In this particular case, because the resonance decays into two photons it has to be either spin 0 or spin 2. The angular distribution of the photons is distinctly different in the two cases.
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Just conserve angular momentum. If I have two photons on a collision course, their spin can either be aligned or anti-aligned, since photons must have spins lying on the same plane as their motion by virtue of their masslessness.
Then, you can either add one to one to get two, or you can subtract one from one to get zero. If you have a decay to two ...
8
It's because 4-vector fields in the $(1/2,1/2)$ representation don't produce any spin-0 excitations, at least not in consistent theories.
Electromagnetism is the canonical example. The vector field $A_\mu$ would create both positive-norm ($A_i$) and negative-norm ($A_0$) polarizations. The latter is time-like. However, probabilities can't be negative, so ...
7
I believe there is such a representation, as follows:
You need only consider operators which can be written as $\Omega = \sum_k \alpha_k \gamma_k$, where $\gamma_k = \sum_\ell P_\ell \big( \sigma_{k_1} \otimes \cdots \otimes \sigma_{k_N} \big)P_\ell^\dagger$ with $P_\ell$ being the operator corresponding to the $\ell^{\text{th}}$ permutation of qubits. Note ...
6
I think that you are confused. When you rotate something by 360 degrees, you won't change the direction in space of anything. You will only change the wave function to minus itself - if there is an odd number of fermions in the object (which is usually hard to count for large objects).
If you have electrons with spins pointing up and you rotate them around ...
6
About this:
How are some useful ways to imagine a particle with spin 1/2 to make a 360° turn without returning to it's original position (the wave function transforms as: Ψ→−Ψ).
There is a nice example of such an objects -- "the Dirac scissors":
The picture is from the book by Penrose and Rindler "Spinors and space-time." I suggest to read it.
6
Conservation of angular momentum is invoked for the neutrinos because beams of neutrinos cannot be collimated for an experimental measurement. Neutron spin can be measured in a Stern Gerlach setup.
The interactions and decays were carefully examined in various experiments and the only consistent spin values are the ones assigned.
Edit: I see that the ...
6
It's easy to prove the formula if you just look at the individual basis vectors of the tensor product. Let's use $(2j+1)$ and $(2s+1)$ eigenvectors of $j^2, j_z$ and $s^2, s_z$ called $|j,j_z\rangle$ and so on.
Now let's ask about the multiplicity of basis vectors of the tensor product with a given eigenvalue of $J_z = j_z+s_z$. The maximum eigenvalue of ...
6
Yes, what you are suggesting is exactly what is happening, but that is if you have an expression which transforms like an axial vector which you can identify with the the spin of the photon. The inherent spin property of photons ($1\hbar$) and electrons ($\tfrac12\hbar$) is of course reference frame independent.
Maybe without realizing you brought forward ...
6
Starting with the Lagrangian for a massive $U(1)$ vector boson $A_\mu$ which like you said has 3 DOF:
$\mathcal{L} = - \frac{1}{4 e^2} F^{\mu \nu} F_{\mu \nu} - m^2 A^\mu A_\mu$
now if we change variables to $A_\mu\rightarrow A_\mu - \partial_\mu \theta$ and we have (Note that $ F^{\mu \nu} $ and hence $F_{\mu \nu}F^{\mu \nu}$ is invariant under this ...
6
This is often confusing to people getting acquainted to QM and you need to stare at it for a while and convince yourself about how it works.
Firstly $\sigma^{x,y,z}$ are the Pauli spin matrices and $\vec{\sigma}_1 \cdot \vec{\sigma}_2 \equiv \sigma_1^x \otimes \sigma_2^x + \sigma_1^y \otimes \sigma_2^y + \sigma_1^z \otimes \sigma_2^z$
The $\sigma_z$ ...
5
Consider two independent systems, two distant atoms, for example. Each atom has is own angular momentum $J_1$ and $J_2$. If these atoms are different and distant (non interacting), then the QM variables are separated, the total wave function becomes a product of the two atomic wave functions, and the total system angular momentum is a sum of the two. The ...
5
Imagine going to the rest frame of a massive particle. In this frame, there is rotational symmetry, which means that the Lie algebra of rotations acts on the wave function. So the wave function is a vector in a representation of Lie(SO(3)) = Lie(SU(2)). "Spin" is the label of precisely which representation this is. Note that while SO(3) and SU(2) share a ...
5
whether you call a similar concept "fundamental" is a matter of taste - and the proposition is just a meaningless emotional slogan. The angular momentum is surely an important quantity that is, in a very well-defined sense, as important as the normal momentum. Incidentally, both of them are conserved if the physical laws are symmetric with respect to ...
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