# Tag Info

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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 ...

8

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 ...

6

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 ...

6

The classic place to start would be the book "PCT, Spin & Statistics, and All That", by R.F.Streater and A.S.Wightman. The spin statistics theorem can be proved as rigorously as you likely can want in the context of the Wightman axioms. The difficulty with this statement relative to your question is that we cannot prove that interacting fields satisfy ...

5

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 ...

5

I will firstly point out some apparent misconceptions in the question and subsequently I will explain what goes wrong when quantizing a theory of integer spin fields or particles with anticommutators, and vice versa. First, if we quantize a real Klein-Gordon field using anticommutators, the Hamiltonian is going to vanish (or be a field independent ...

5

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 ...

5

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 ...

5

Let me start from the beginning, also making explicit what I understand from user35388's answer. Consider a system of a couple of identical particles. Their states are pictured by normalized wavefunctions $\psi(x,y)$. However this representation is not one-to-one (this holds for every quantum system): states are wavefunctions up to phases. That is $\psi$ ...

4

The way Shankar addresses the problem (pg. 278) is by introducing an "Exchange Operator" $P_{1,2}$, which would swap your two particles as follows: $P_{1,2} |\xi_1, \xi_2 \rangle = |\xi_2, \xi_1 \rangle$ I like the operator notation because it makes it clear (to me, at least) that applying the operator twice is just the identity operator, since swapping ...

4

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 ...

3

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 ...

3

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 ...

3

Suppose we assume that an object's statistics depend only on its spin and not on whether the object is composite or fundamental. This assumption seems natural, since if it failed, it would be too good to be true -- it would give us a way of finding out about the internal structure of any particle, at all scales, without having to build particle accelerators. ...

3

How to obtain this braiding matrix from Non-Abelian Chern-Simon theory? To obtain braiding matrix $U^{ab}$ for particle $a$ and $b$, we first need to know the dimension of the matrix. However, the dimension of the matrix for Non-Abelian Chern-Simon theory is NOT determined by $a$ and $b$ alone. Say if we put four particles $a,b,c,d$ on a sphere, the ...

2

Assuming the question is about why anticommutators rather than commutators,as per David's comment: If I denote the Dirac particle properties by a general index $\alpha$ - these can include spins, momenta - then if I create a two particle state, with particle 1 having $\alpha_1$ and particle 2 $\alpha_2$ then the state is given by applying creation operators ...

2

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 ...

2

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 ...

2

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 ...

2

You are right. In a space-time with one time dimension and $D$ spatial dimensions, finding possible different statistics is equilalent to look at the fundamental group (first homotopy group) of $SO(D)$ For $D=1$, the fundamental group is trivial. For $D=2$, the fundamental group is $\mathbb{Z}$. For $D>=3$, the fundamental group is $\mathbb{Z}_{2}$ ...

2

1) The Pauli exclusion principle says that either 0 or 1 identical fermion, but not more, may occupy the same quantum state. It's been originally induced from the properties of atoms (periodic table). Today, we may derive it from the first principle out of the antisymmetric wave functions for many fermions, those obeying $$\psi(\vec x_1,\vec x_2) = ... 2 An SE user was kind enough to send me a PDF of the paper, so I'll try to give a summary of my own understanding of it. The historical context is interesting. This was before the neutron was discovered experimentally, so they thought nuclei were made out of protons and electrons. What we would today describe as a nucleus with mass number A and atomic ... 2 Yes, by introducing a homogenous field in the z direction you break full SU(2) rotational symmetry. Otherwise, you would have more degeneracies, but since the Hamiltonian conserves magnetization you can easily add or remove a homogenous field. If you do remove the field then any spin axis is fine and the ground state will be further degenerate. By ... 2 It depends on the non-trivial topological solution of the equations of motion under consideration. The most famous example is the 't Hooft–Polyakov monopole (see http://en.wikipedia.org/wiki/%27t_Hooft%E2%80%93Polyakov_monopole) which has the following solution: $$\phi = h \frac{x^a} {r} t^a$$ A_0=0 ... 2 The (unitary) "phase" factor for non-Abelian anyons satisfies the (non-Abelian) Knizhnik-Zamolodchikov equation:$$\big (\frac{\partial}{\partial z_{\alpha}} + \frac{1}{2\pi k} \sum_{\beta \neq \alpha} \frac{Q^a_{\alpha}Q^a_{\beta}}{z_{\alpha} - z_{\beta}}\big )U(z_1, ....,z_N) = 0  Where $z_{\alpha}$ is the complex plane coordinate of the particle ...

1

As you noted, the Ising model has spins that are $\pm 1$ whereas in a full quantum model such as the Heisenberg model, the spins are represented by Pauli matrices. This means that they are not interchangeable. The biggest difference is that at zero temperature, there are no spin fluctuations in an Ising model, whereas there are fluctuations in the ...

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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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