# Tag Info

## New answers tagged spin-statistics

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

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 ...

0

Since there is no way in which the molecules can be labeled, the particles are indistinguishable. On the other hand, if the assembly is a crystal, the molecules can be labeled in accord with the positions they occupy in the crystal lattice and can be considered distinguishable.

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 $\psi$ ...

0

A potential well in the Schrodinger equation will produce energy levels similar to the energy levels of the hydrogen atom A nucleus with two protons to be neutral will need two electrons. These will be accommodated in the lowest two energy levels: because of the spin statistics of electrons they cannot occupy the exact same energy level. A nucleus with ...

1

Check out the section about Chapter 17 Identical particles in Ballentines, he not only points out why looking at the Permutation operators of two particles in a multi particle setting is misleading but also discusses some errors in previous claims.

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I think the best way to understand spin is to look at it from a complete abstract point of view: do not try to find classical analogs to it. What you find when you perform experiments with electrons is that they interact with a magnetic fields as if they had an intrinsic magnetic moment with two possible values: a positive one and a negative one (loosely ...

1

It is true, you have to "rotate twice" (or by $720^\circ$) to recover the original state. You can prove this in the following way. Let $$|a\rangle=|+\rangle\langle+|a\rangle+|-\rangle\langle -|a\rangle$$ be a general ket. Consider now a rotation by a finite angle $\theta$ around the $z$ axis. I remind here that if a ket of a spin $1/2$ system is ...

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