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How are anyons possible? (another version)

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I know that this question has been submitted several times (especially see How are anyons possible?), even as a byproduct of other questions, since I did not find any completely satisfactory answers, here I submit another version of the question, stated into a very precise form using only very elementary general assumptions of quantum physics. In particular I will not use any operator (indicated by $P$ in other versions) representing the swap of particles.

Assume to deal with a system of a couple of identical particles, each moving in $R^2$. Neglecting for the moment the fact that the particles are indistinguishable, we start form the Hilbert space $L^2(R^2)\otimes L^2(R^2)$, that is isomorphic to $L^2(R^2\times R^2)$. Now I divide the rest of my issue into several elementary steps.

(1) Every element $\psi \in L^2(R^2\times R^2)$ with $||\psi||=1$ defines a state of the system, where $|| \cdot||$ is the $L^2$ norm.

(2) Each element of the class $\{e^{i\alpha}\psi\:|\; \psi\}$ for $\psi \in L^2(R^2\times R^2)$ with $||\psi||=1$ defines the same state, and a state is such a set of vectors.

(3) Each $\psi$ as above can be seen as a complex valued function defined, up to zero (Lebesgue) measure sets, on $R^2\times R^2$.

(4) Now consider the "swapped state" defined (due to (1)) by $\psi' \in L^2(R^2\times R^2)$ by the function (up to a zero measure set):

$$\psi'(x,y) := \psi(y,x)\:,\quad (x,y) \in R^2\times R^2$$

(5) The physical meaning of the state represented by $\psi'$ is that of a state obtained form $\psi$ with the role of the two particles interchanged.

(6) As the particles are identical, the state represented by $\psi'$ must be the same as that represented by $\psi$.

(7) In view of (1) and (2) it must be: $$\psi' = e^{i a} \psi\quad \mbox{for some constant $a\in R$.}$$

Here physics stops. I will use only mathematics henceforth.

(8) In view of (3) one can equivalently re-write the identity above as

$$\psi(y,x) = e^{ia}\psi(x,y) \quad \mbox{almost everywhere for $(x,y)\in R^2\times R^2$}\quad [1]\:.$$

(9) Since $(x,y)$ in [1] is every pair of points up to a zero-measure set, I am allowed to change their names obtaining

$$\psi(x,y) = e^{ia}\psi(y,x) \quad \mbox{almost everywhere for $(x,y)\in R^2\times R^2$}\quad [2]$$
(Notice the zero measure set where the identity fails remains a zero measure set under the reflexion $(x,y) \mapsto (y,x)$, since it is an isometry of $R^4$ and Lebesgues' measure is invariant under isometries.)

(10) Since, again, [2] holds almost everywhere for every pair $(x,y)$, I am allowed to use again [1] in the right-hand side of [2] obtaining:

$$\psi(x,y) = e^{ia}e^{ia}\psi(x,y) \quad \mbox{almost everywhere for $(x,y)\in R^2\times R^2$}\:.$$
(This certainly holds true outside the union of the zero measure set $A$ where [1] fails and that obtained by reflexion $(x,y) \mapsto (y,x)$ of $A$ itself.)

(11) Conclusion:

$$[e^{2ia} -1] \psi(x,y)=0 \qquad\mbox{almost everywhere for $(x,y)\in R^2\times R^2$}\quad [3]$$

Since $||\psi|| \neq 0$, $\psi$ cannot vanish everywhere on $R^2\times R^2$. If $\psi(x_0,y_0) \neq 0$, $[e^{2ia} -1] \psi(x_0,y_0)=0$ implies $e^{2ia} =1 $ and so:

$$e^{ia} = \pm 1\:.$$

And thus, apparently, anyons are not permitted.

Where is the mistake?

ADDED REMARK. (10) is a completely mathematical result. Here is another way to obtain it. (8) can be written down as $\psi(a,b) = e^{ic} \psi(b,a)$ for some fixed $c \in R$ and all $(a,b) \in R^2 \times R^2$ (I disregard the issue of negligible sets). Choosing first $(a,b)=(x,y)$ and then $(a,b)=(y,x)$ we obtain resp. $\psi(x,y) = e^{ic} \psi(y,x)$ and $\psi(y,x) = e^{ic} \psi(x,y)$. They immediately produce [3] $\psi(x,y) = e^{i2c} \psi(x,y)$.

So the physical argument (4)-(7) that we have permuted again the particles and thus a further new phase may appear does not apply here.

2nd ADDED REMARK. It is clear that as soon as one is allowed to write

$\psi(x,y) = \lambda \psi(y,x)$ for a constant $\lambda\in U(1)$ and all $(x,y) \in R^2\times R^2$

the game is over: $\lambda$ turns out to be $\pm 1$ and anyons are forbidden. This is just mathematics however. My guess for a way out is that the true configuration space is not $R^2\times R^2$ but some other space whose $R^2 \times R^2$ is the universal covering.

An idea (quite rough) could be the following. One should assume that particles are indistinguishable from scratch already defining the configuration space, that is something like $Q := R^2\times R^2/\sim$ where $(x',y')\sim (x,y)$ iff $x'=y$ and $y'=x$. Or perhaps subtracting the set $\{(z,z)\:|\: z \in R^2\}$ to $R^2\times R^2$ before taking the quotient to say that particles cannot stay at the same place. Assume the former case for the sake of simplicity. There is a (double?) covering map $\pi : R^2 \times R^2 \to Q$. My guess is the following. If one defines wavefunctions $\Psi$ on $R^2 \times R^2$, he automatically defines many-valued wavefunctions on $Q$. I mean $\psi:= \Psi \circ \pi^{-1}$. The problem of many values physically does not matter if the difference of the two values (assuming the covering is a double one) is just a phase and this could be written, in view of the identification $\sim$ used to construct $Q$ out of $R^2 \times R^2$: $$\psi(x,y)= e^{ia}\psi(y,x)\:.$$ Notice that the identity cannot be interpreted literally because $(x,y)$ and $(y,x)$ are the same point in $Q$, so my trick for proving $e^{ia}=\pm 1$ cannot be implemented. The situation is similar to that of $QM$ on $S^1$ inducing many-valued wavefunctions form its universal covering $R$. In that case one writes $\psi(\theta)= e^{ia}\psi(\theta + 2\pi)$.

3rd ADDED REMARK I think I solved the problem I posted focusing on the model of a couple of anyons discussed on p.225 of this paper matwbn.icm.edu.pl/ksiazki/bcp/bcp42/bcp42116.pdf suggested by Trimok. The model is simply this one: $$\psi(x,y):= e^{i\alpha \theta(x,y)} \varphi(x,y)$$
where $\alpha \in R$ is a constant, $\varphi(x,y)= \varphi(y,x)$, $(x,y) \in R^2 \times R^2$ and $\theta(x,y)$ is the angle with respect to some fixed axis of the segment $xy$. One can pass to coordinates $(X,r)$, where $X$ describes the center of mass and $r:= y-x$. Swapping the particles means $r\to -r$. Without paying attention to mathematical details, one sees that, in fact: $$\psi(X,-r)= e^{i \alpha \pi} \psi(X,r)\quad \mbox{i.e.,}\quad \psi(x,y)= e^{i \alpha \pi} \psi(y,x)\quad (A)$$ for an anti clock wise rotation. (For clock wise rotations a sign $-$ appears in the phase, describing the other element of the braid group $Z_2$. Also notice that, for $\alpha \pi \neq 0, 2\pi$ the function vanishes for $r=0$, namely $x=y$, and this corresponds to the fact that we removed the set $C$ of coincidence points $x=y$ from the space of configurations.)

However a closer scrutiny shows that the situation is more complicated: The angle $\theta(r)$ is not well defined without fixing a reference axis where $\theta =0$. Afterwards one may assume, for instance, $\theta \in (0,2\pi)$, otherwise $\psi$ must be considered multi-valued. With the choice $\theta(r) \in (0,2\pi)$, (A) does not hold everywhere. Consider an anti clockwise rotation of $r$. If $\theta(r) \in (0,\pi)$ then (A) holds in the form $$\psi(X,-r)= e^{+ i \alpha \pi} \psi(X,r)\quad \mbox{i.e.,}\quad \psi(x,y)= e^{+ i \alpha \pi} \psi(y,x)\quad (A1)$$ but for $\theta(r) \in (\pi, 2\pi)$, and always for a anti clockwise rotation one finds $$\psi(X,-r)= e^{-i \alpha \pi} \psi(X,r)\quad \mbox{i.e.,}\quad \psi(x,y)= e^{- i \alpha \pi} \psi(y,x)\quad (A2)\:.$$ Different results arise with different conventions. In any cases it is evident that the phase due to the swap process is a function of $(x,y)$ (even if locally constant) and not a constant. This invalidate my "no-go proof", but also proves that the notion of anyon statistics is deeply different from the standard one based on the groups of permutations, where the phases due to the swap of particles is constant in $(x,y)$. As a consequence the swapped state is different from the initial one, differently form what happens for bosons or fermions and against the idea that anyons are indistinguishable particles. [Notice also that, in the considered model, swapping the initial pair of bosons means $\varphi(x,y) \to \varphi(y,x)= \varphi(x,y)$ that is $\psi(x,y)\to \psi(x,y)$. That is, swapping anyons does not mean swapping the associated bosons, and it is correct, as it is another physical operation on different physical subjects.]

Alternatively one may think of the anyon wavefunction $\psi(x,y)$ as a multi-valued one, again differently from what I assumed in my "no-go proof" and differently from the standard assumptions in QM. This produces a truly constant phase in (A). However, it is not clear to me if, with this interpretation the swapped state of anyons is the same as the initial one, since I never seriously considered things like (if any) Hilbert spaces of multi-valued functions and I do not understand what happens to the ray-representation of states. This picture is physically convenient, however, since it leads to a tenable interpretation of (A) and the action of the braid group turns out to be explicit and natural.

Actually a last possibility appears. One could deal with (standard complex valued) wavefunctions defined on $(R^2 \times R^2 - C)/\sim$ as we know (see above, $C$ is the set of pairs $(x,y)$ with $x=y$) and we define the swap operation in terms of phases only (so that my "no-go proof" cannot be applied and the transformations do not change the states):

$$\psi([(x,y)]) \to e^{g i\alpha \pi}\psi([(x,y)])$$

where $g \in Z_2$. This can be extended to many particles passing to the braid group of many particles. Maybe it is convenient mathematically but is not very physically expressive.

In the model discussed in the paper I mentioned, it is however evident that, up to an unitary transformation, the Hilbert space of the theory is nothing but a standard bosonic Hilbert space, since the considered wavefunctions are obtained from those of that space by means of a unitary map associated with a singular gauge transformation, and just that singularity gives rise to all the interesting structure! However, in the initial bosonic system the singularity was pre-existent: the magnetic field was a sum of Dirac's delta. I do not know if it makes sense to think of anyons independently from their dynamics. And I do not know if this result is general. I guess that moving the singularity form the statistics to the interaction and vice versa is just what happens in path integral formulation when moving the external phase to the internal action, see Tengen's answer.

This post imported from StackExchange Physics at 2014-04-11 15:20 (UCT), posted by SE-user V. Moretti
asked Dec 17, 2013 in Theoretical Physics by Valter Moretti (2,025 points) [ no revision ]

4 Answers

+ 4 like - 0 dislike

The answer can probably be summarized in two points:

1) As discussed in a beautiful paper by Leinaas and Myrheim, the configuration space of a system of $N$ identical particles in $n$ dimensions is not $\mathbb{R}^{nN}$, but $\mathbb{R}^{nN}/S_N$ where we mod out the action of the permutation group $S_N$ (and also remove the singularities that happen when two particles occupy the same point).

2) Quantum mechanics is not about functions from the configuration space to the complex numbers, $\psi : \mathbb{R}^{nN}/S_N \to \mathbb{C}$, but, in modern terms, about sections in vector bundles on the configuration space. In classic terms, one would argue that the phase of the wave function is unobservable, and hence can be multi-valued, as discussed in Dirac's paper on magnetic monopoles.

It turns out that in $n=2$ dimensions, the configuration space of identical particles supports not just two, but many different interesting vector bundles, which corresponds to anyons.

answered Apr 16, 2016 by Greg Graviton (665 points) [ no revision ]
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In quantum mechanics on nonsimply connected spaces, we can use wave functions living on the universal covering space. In order to see that, we must remember that the symplectic potential of identical particles must include an extra piece to the free particle symplectic potential given by a flat connection represented by a magnetic vector potential. This also happens for example in the Aharonov-Bohm case. In the two dimensional identical particle case where anyons exist, this vector potential can only have the following form:

$$\mathbf{A} =\frac{\theta}{2 \pi} \frac{(-y, x )}{\left | x-y \right |^2}$$

where $ \mathbb{r_{12}} = (x,y)$ is the relative particle position.

As a form this vector potential is closed but not exact

Please see, for example, Wu's article:  http://content.lib.utah.edu/utils/getfile/collection/uspace/id/4293/filename/3674.pdf

The solution of the Schrödinger equation with this type of magnetic potential is equivalent to taking following solution without the vector potential:

$$\Psi(z_1, z_2) = (z_1-z_2)^{\frac{\theta}{2\pi}}\psi(z_1, z_2)$$

where $z_{1,2}$ are complex coordinates on the plane and $\psi(z_1, z_2)$ single valued. (Please see Wu for the details).

Actually, the 2D case is actually conceptually simpler than in higher dimensions, where only bosons and fermions exist, because in higher dimensions the flat connection is torsion and has no vector potential representative in the de-Rham cohomology, but only in the Čech cohomology.

answered Apr 17, 2016 by David Bar Moshe (3,845 points) [ no revision ]
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The best way to answer the question "How are anyons possible" is to use the "dynamical" path integral formalism, rather than the "static" wave function formalism. The permutation group action on the wave function is "static" in the sense that only initial and final states are specified. It will be ambiguous if there are more than one non-equivalent ways to perform the exchange process, which is the key for the "possibility" of anyons.

Consider the amplitude from the initial state $|i\rangle$ to final state $|f\rangle$ in the path integral formalism $$\langle f|i\rangle = \int_\gamma \mathcal{D}x(t) e^{iS[x(t)]},$$ where $\gamma$ is a path from the initial configuration to the final configuration (they are set to the same). The confituration manifold will be discussed later. When two paths $\gamma_1$ and $\gamma_2$ are not equivalent to each other homotopically, we can assign a phase factor $e^{i\theta([\gamma])}$ to the path integral amplitude for each homotopy class $[\gamma]$: $$\langle f|i\rangle = \sum_{[\gamma]\in \pi_1(M)} e^{i\theta([\gamma])}\int_\gamma \mathcal{D}x(t) e^{iS[x(t)]},$$ where $\pi_1(M)$ denotes the fundamental group of the configuration space $M$. The phase factors $\{e^{i\theta([\gamma])}\}$ form an one dimensional representation of the group $\pi_1(M)$ because of the multiplication property of the propagator: $\langle f|i\rangle=\sum_m \langle f|m\rangle\langle m|i\rangle$. If we absorb the phase $\theta$ to the action $S$, it will be called a topological term as it depends only on the homotopy class.

The next task is to calculate the one dimensional representation of the fundamental group of the configuration space. For $N$ identical particles in $d$ space dimension, the configration space is $M=(\mathbb R^{Nd}\backslash D)/S_N$, where $D=\{(r_1,...,r_N)|\ \exists i\neq j,\ s.t. r_i=r_j\}$ is the space where two particles occupy the same point, and "$/S_N$" means the order of the particles is neglected.

(1) $d=1$. No exchange process can happen, and the notion of statistics is meaningless.

(2) $d=2$. $\pi_1(M)=B_N$ is the braiding group. The one dimension representation of $B_N$ is characterized by an angle $\theta$ which corresponds to the statistical angle of the Abelian anyon.

(3) $d\geq 3$. $\pi_1(M)=S_N$ is the permutation group. It means that, we only need to specify the order of particles in the initial and final states, to determine which homotopy class the path $\gamma$ belongs to. Therefore, only in this case, the wave function formalism can be used without ambiguity.

To describe the non-Abelian anyons, one only need to replace the phase factor $e^{i\theta}$ by an unitary matrix. The result is that non-Abelian anyons are determined by the higher dimension representations of the fundamental group of the configuration space.

This post imported from StackExchange Physics at 2014-04-11 15:20 (UCT), posted by SE-user Tengen
answered Dec 19, 2013 by Tengen (105 points) [ no revision ]

May I inquire you how to prove the claim "$d=2\ \pi_1(M)=B_N$" and "$d=3\ \pi_1(M)=S_N$"

+ 1 like - 0 dislike

The point $(11)$ is not correct, by doing $2$ successive "exchanges", you may have a global phase factor, such as $\psi'(x,y) = e^{i\alpha}\psi(x,y)$. The two wave functions describe the same physical state. The correct considerations are topological, inside of considering a discrete operation, consider a continuous operation, so that it is equivalent to keep one particle fixed, and to make a rotation of the other particle of $2\pi$, the solution is in fact to look at the fundamental group ($1$st homotopy group) of $SO(d)$, where $d$ is the number of spatial dimensions (we suppose here only one time dimension). The structure and the dimension of the fundamental group (the number of different classes of paths) is correlated to the number of possible statistics. Of course, the fundamental group for $SO(d)$ with $d\geq 3$ is $\mathbb Z_2$, while the fundamental group of $SO(2)$ is $\mathbb Z$. This explains why we found different statistics (anyons) in $2$ spatial dimensions.

This post imported from StackExchange Physics at 2014-04-11 15:20 (UCT), posted by SE-user Trimok
answered Dec 17, 2013 by Trimok (950 points) [ no revision ]

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