I have studied about exchange operators and rotation operators and I know that an exchange between 2 particles in a combined state is the same as rotating each particle 180 degrees (according to http://en.wikipedia.org/wiki/Exchange_symmetry). Could someone explain the mathematical equation behind this? I have tried to search but to no avail.
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An exchange between two particles is not the same as rotating anything, Feynman is wrong in the 1986 lectures, but he makes the statement anyway, probably because it clarified to him what Schwinger was doing way back when in the 1950s to do spin-statistics from time-reversals. Wikipedia is just copying Feynman's wrong statement. There is no relation between rotation and exchange outside of relativity, and the rotation argument is not conclusive without rotations involving time, i.e. relativistic Lorentz transformations analytically continued to Euclidean space. The intuition here is as Lubos says: if you have two particles at x and at y, you can exchange their positions by rotating around the point half-way between by 180 degrees, and this has the effect of rotating each particle by 180 and switching the two particles. But the idea doesn't lead to a spin-statistics relation, nor does it express any exchange operator (in formalisms where this operator makes sense) in terms of a rotation operator (which always makes sense in any formalism). An exchange operator exchanges the positions of two identical particles, so that acting on $\psi(x,y)$ it produces $\psi(y,x)$ (the same function of two variables, but now with the dependence of the two variables interchanged). the rotation operator rotates everything, including the two particles. You can't express a general exchange in terms of rotation. What you can do is relate the symmetry of the particles under exchange and under rotation. If you have a two-particle state $\alpha(x,y)\psi^\dagger(x)\psi^\dagger(y)$, you can express it in center of mass and relative coordinates, and in the relative coordinates, you have a rotational and exchange symmetry. So the relative wavefunction $\alpha(x-y)$ can be rotated by 180 degrees, and the sign you get from this wavefunction rotation must be correct for the statistics of the particles. That's all you can say in a nonrelativistic theory, barring additional assumptions. The nonrelativistic theory has no necessary relation between spin and statistics, as you can understand from the example of a nonrelativistic spinless fermion. The spinless fermionic Schrodinger field is completely consistent. This system can be realized as the low-energy limit of a physical system--- namely neutrons in a very strong magnetic field. In the low energy limit, the spin is decoupled from the dynamics, and the neutron magnetic moment aligns the spin with B, but the B doesn't affect the translational motion. The low energy limit is a 3d fermi gas with full rotational invariance and no spin. What happens with relativity is that the theory has to have a Euclidean continuation, or if you like the energies are always positive, or if you like, the particles can travel faster than light (these are closely related statements). So that in the relativistic theory (unlike the nonrelativistic theory), there is nonzero vacuum expectation value for fields: $$ \langle \phi(x) \phi(y) \rangle \ne 0 $$ Where $\phi$ is a real field, or a real component of a complex field. Then the Euclidean argument for spin-statistics is simply that you can rotate this correlation function by 180 degrees and it stays the same. This requires that the interchange of $\phi$ is the same as the rotation properties of $\phi$. The new thing here is that the field $\phi(x)\phi(y)$ are time ordered, and rotating them in relativity connects the different orders by the different ordering in time. This is essential--- in nonrelativistic physics, you don't have a nonzero expected value except into the future, and any rotation never switches the order of the operators in a path-integral. In short, don't listen to this Wikipedia page. Look at the spin-statistics page, which is the accurate one |
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"The nonrelativistic theory has no necessary relation between spin and statistics, as you can understand from the example of a nonrelativistic 2d spinless fermion (imagine 2d confined electrons with a huge magnetic field in the z-direction which forces the spin to always be along the z-axis). The electrons behave as spinless particles with respect to the 2d rotations, and yet they are still fermions." Ron I'm not sure this is correct. In a paper (That i tried to link but the site wont let me) called something like "Anomalous Quantum Hall effect, Fractionally charge excitations in incompressible quantum hall fluid" Laughlin uses the variational approach to model exactly the 2D electron system in the presence of a magnetic field which you are talking about. He derives that the excitations in this system have fractional charges. In this follow up-paper Arovas, Wilzcek and Schieffer show that these fractionally charged 2-dimensional excitations exhibit fractional exchange statistics, contrary to what you are saying about their fermionic behavior. That is, under the rotation, these charged particles accumulate a phase due to the aharnov-bohm effect. The phase is $e^{i\theta}$ where $\theta \ne \pi$ (as it would with fermions) or $2\pi$ (as it would with bosons) This is what gives rise to the entire study of anyons and some explanations of the fractional quantum hall effect. As mentioned in the first comment, you can imagine that the phase accumulated due to an aharonov-bohm interaction by each particle undergoing a rotation of $\pi$ radians about the other (the intuitive way to envision an exchange in 2D) is the same as the phase accumulated as one particle makes a $2\pi$ rotation about the other. Preskill explains this very well in parts 9.3 and 9.4 of this set of lecture notes The reason that you don't see fractional exchanges in 3 dimensions has to do with the ability to contract a circle to a trivial path around any 2D object. This is not possible if you draw a loop in 2D around say, an electron in the plane. (The actual explanation for this is some simple algebraic topology which I don't know currently.) Perhaps this is not the 2D spinless fermion model you had in mind though. Anyway, this is my first stack exchange post, so I hope it was helpful. |
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