Hot answers tagged spin-statistics
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Indistinguishableness of particles is formulated in QM in terms of the total wave function symmetry. Then the wave functions can be symmetric or antisymmetric on their arguments. In case of antisymmetry ($\psi(x_1,x_2) = -\psi(x_2,x_1))$ the particles are called fermions and $\psi(x_1,x_2=x_1) = 0$. They say "the particles cannot occupy the same state". It's ...
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The exchange interaction is, on the contrary, one of the approximate low-energy consequences of the Pauli exclusion principle and the identical character of the particles.
The fundamental justification of these facts is offered by quantum field theory. Relativity requires observables to be linked to regions of spacetime - so that they don't communicate over ...
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Majorana fermions are by definition fermions which are their own antiparticles, i.e. the do have spin and it's 1/2. An introduction to these fermions can be for example found here: http://arxiv.org/abs/0806.1690. In contrast bosons are their own antiparticles, e.g. photons, i.e. one does not need a "Majorana-boson" definition.
Now, one has to say that these ...
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The spin-statistics thing isn't a problem, it is a theorem (a demonstrably valid proposition), and it shouldn't be addressed, it should be understood and celebrated.
The Higgs field gives us interactions between chiral fermions and the Higgs, $yh\cdot \chi_\alpha\eta^\alpha$ which produces mass terms $m \chi_\alpha\eta^\alpha$ if the Higgs field has a ...
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BebopButUnsteady answer is obviously correct. Actually most wave functions cannot be written as single Slater determinants and certainly no energy eigenstates of N-fermion systems (atom, molecules or solids) interacting by two-body potentials (like the Coulomb potential). The Slater determinant is just a nice simple approximation to ground state ...
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For the partition sum, you have so sum $e^{-E}$ ($T=1$) over all possible eigenstates of the system where $E$ is the energy of the corresponding state.
Two bosons can be in the states 10
$|kl\rangle$, with $1\leq k \leq l \leq 4$ where we accounted for the degeneracy by introducing an additional state with $E_4 =2E$. The corresponding partition sum reads ...
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On the deepest level, particles are indistinguishable if and only if they have the same quantum numbers (mass, spin, and charges).
However, in statistical mechanics one ofte studies effective theories where there are additional means of distinguishing particles. Two important examples:
In modeling molecular fluids, two atoms on the same molecule are ...
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Both fractional/non-Abelian statistics and fractional charges
come from the same origin: long-range entanglements.
This is why fractional/non-Abelian statistics common for fractional charges.
One way to realize long-range entanglements
is through the string-net liquid phase of a pure bosonic model.
The ends of strings in string-net liquid are non-local and ...
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Antisymmetric wave functions instantly imply the Pauli exclusion principle – essentially because $\psi(x,x)=-\psi(x,x)=0$, to write the concept schematically – which implies that the occupation numbers are $N=0,1$ and statistical physics is therefore inevitably governed by the Fermi-Dirac statistics which may be derived from Boltzmann/statistical physics for ...
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The CPT theorem is a consequence of the statement that the Euclidean theory is invariant under 180 degree rotations in a plane involving (the analytic continuation of) the time axis. The result of this rotation flips the sense of Euclidean time, and flips one space axis. The analytic continuation that defines the Minkowski theory is a PT transformation (it ...
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There are two possible answers to why $T^2=-1$:
a) Why not. The total phase of a quantum state is unphysical. So a symmetry
may be realized as a projective representation. Here T may be viewed as a projective representation of time reversal $T_{phy}$ which satisfy $T^2_{phy}=1$.
b) If we define the time reversal symmetry to be realized as a regular ...
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I wrote the Wikipedia page in question, so I feel bad. I thought it was clear.
There is a recent textbook by Banks which covers the spin/statistics theorem pretty good. I hope it is ok. The main difficulty is that there is no quantum field theory book that covers analytic continuation to Euclidean space, and this is the essential thing.
This is worked out ...
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If you think anti-commuting field is too arbitrary and do not like anti-commuting field, you may ask do we really need to use anti-commuting field
to describe fermions? Can a theory with only bosons have fermionic excitations
emerging at low energies? The answer is yes! So we do not need to
use anti-commuting fields to describe fermions, and this is true in ...
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First of all, the Standard Model doesn't treat bosonic fields as classical. They're quantum mechanical i.e. non-classical, they're just not anticommuting or Grassmann-odd. Second, a consistent theory just requires the relationship between spin and statistics, see e.g. the
http://en.wikipedia.org/wiki/Spin-statistics_theorem
Combining integer spin with ...
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Okay, my comments are getting too much, so I will answer.
If I understand your question correctly it says this:
Papers show that the non-planar Ising model (finding its ground state) is NP complete
On the other hand, finding the eigenvalues of a matrix is polynomial.
So how do these points reconcile?
The important point here is in the size of the input. ...
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Half-integer spin particles obey Fermi-Dirac statistics and integer spin fields obey the Bose-Einstein statistics – it's true because of Pauli's spin-statistics theorem.
Concerning the second question, I suppose you meant fractional spin, not fractional charges. In the case of 2 spatial dimensions, the trajectory of one particle around another is ...
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It seems that getting two electrons too close should annihilate both of them.
Your assumption is wrong. Only antimatter cancels out matter. In case of Electron, only a Positron can cancel it out. Mind it, two electrons repel each other due to identical charges (in classical sense).
In true sense, you shouldn't talk about distance. To share same quantum ...
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Feynman in multiple writings suggested thinking about "exchanging particles" in terms of exchanging them as they move through time. That is, they can either move in two parallel paths as they move forward, or they can cross paths (exchange roles).
The antisymmetric cancellation applies to the latter, but not to the former. Now if you think that through, it ...
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While writing the question, I realized there is an example of a system of this sort. Consider a gas of free neutrons at very low energies, where the spin and orbit are decoupled, in a strong constant magnetic field. The low energy dynamics is for the low-energy spin configuration, and it is the ordinary Schrodinger dynamics. So the resulting low-energy ...
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The answer for CPT transformations is obvious. A CPT transformation is a 180 degree rotation in the Euclidean theory, so CPT followed by CPT is a 360 degree rotation, which gives you a minus sign on fermionic states, and a plus sign on bosonic states. This is true for all Lorentz invariant theories, or even theories that spontaneously violate Lorentz ...
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It's a good question. The answer is that the bound on the density is given by the requirement that the interactions between the bosons have to remain weak for the Bose-Einstein condensate to exist. In practice, the helium-4 atoms have to be further away from each other than their radius.
Why it is so? Well, if you're talking about the bosons occupying the ...
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There's a repulsive van der Waals force between the atoms. This shows up in the second quantized formalism as
$$H = \int d^3x \left[ \frac{\hbar^2}{2m} \nabla \Psi^\dagger \cdot \nabla \Psi - \mu\Psi^\dagger \Psi \right] + \int d^3x\, d^3y\, \frac{1}{2}V\left( \vec{y} - \vec{x} \right) \Psi^\dagger\left( \vec{x} \right) \Psi^\dagger\left( \vec{y} \right) ...
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