direct sum of anyons?

Question

In the topological phase of a fractional quantum Hall fluid, the excitations of the ground state (quasiparticles) are anyons, at least conjecturally. There is then supposed to be a braided fusion category whose irreducible objects are in 1-1 correspondence with the various types of elementary quasiparticles.

The tensor product of objects has an obvious physical meaning: it's the operation of colliding (fusing) quasiparticles...

... but what about direct sum?

• The tensor product of two irreducible objects might be a direct sum of irreducible ones: what does this means physically in terms of the outcome of a collision of quasiparticles?

• Let $X$ be an irreducible object of the fusion category. Is there any physical difference between (the physical states corresponding to) $X$ and to $X\oplus X$?

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Answer 1

This was originally a comment on Joe's excellent answer, but it got too long. I'm trying to address the question of what φ ⊕ φ means.

Suppose you look at the equation

φ ⊗ φ ⊗ φ = φ ⊕ φ ⊕ I.

What this says is that when you fuse three φ particles, there are two different ways of producing φ, and one way of producing I. The two ways are (a) and (b) below; the one way to produce I is (c):

(a) fuse φ ⊗ φ to get I, and then fuse φ ⊗ I to get φ;

(b) fuse φ ⊗ φ to get φ, and then fuse φ ⊗ φ to get φ.

(c) fuse φ ⊗ φ to get φ, and then fuse φ ⊗ φ to get I;

These three states are orthogonal, and you can take them to be basis states of the Hilbert space φ ⊗ φ ⊗ φ. When counting these different ways, you have to keep the order in which you fuse the particles fixed. If you want to change this order, you have to apply what physicists call the F matrix (possibly repeatedly) to make this basis change.

One way of thinking about this is that the tensor product corresponds to the joint state of two systems, and the direct product to the superposition of states. When you fuse particles, you do a measurement. The above equation implies that if you have three Fibonacci anyons, their Hilbert space breaks up into two sectors. In one of these (the two-dimensional one), when you fuse all three anyons, you'll get a Fibonacci anyon. In the other (the one-dimensional one), when you fuse all three anyons, you'll get the vacuum state. What you get when you fuse all three anyons does not depend on the order in which you do it (unless you braid these anyons with other anyons before fusing them, which is how you do quantum computation with anyons).

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Comments

+1: A far clearer answer than mine.

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This is again a very nice answer indeed and it's getting closer to what I'm looking for. But I still have the feeling that you are more explaining what φ ⊗ φ ⊗ φ means, and not so much what φ ⊕ φ means. I'm looking for something of the form: "If you have a physical system whose Hilbert space is φ ⊕ φ, then you can do the following measurement and you'll get result A, whereas if you have a physical system whose Hilbert space is φ and you do that same measurment you'll get result B".

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If you have a physical system whose Hilbert space is A ⊕ A, you can think of this as the direct sum of two Hilbert spaces, each of which is a copy of A. That is, the system is either in a state in one of the Hilbert spaces, in a state in the other, or in a superposition of two states from each of them. As Joe Fitzsimons said, it's essentially a two-fold degeneracy of the space.

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@Joe: Is that something physical, or is it unphysical? Here's the analogy that I have in mind (which is maybe the reason why I'm so confused): in quantum mechanics, a vector $v$ in my Hilbert space $H$ represents a state of the system. I can take the sum $v+v=2v$, and that's still a vector in $H$, but those two vectors $v$ and $2v$, even though they're different in $H$, represent the same physical state. Does something similar happen with $\phi$ versus $\phi+\phi$?

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Think of an electron in some Hilbert space $H$. Either it has spin up or spin down. If you take the Hilbert space $H$ to be the set of possible states of the electron, given that it has spin down, then the state of the electron with arbitrary spin is the Hilbert space $H \oplus H$; two copies of the Hilbert space $H$. For $\phi \oplus \phi$, you are essentially adding some extra (possibly hidden) qubit to the system. For $\phi \oplus \phi \oplus\phi$, you are adding some extra qutrit.

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@Peter. Yes, I understand this. But in the case of the electron, there exist experiments that can distinguish the state "spin down" from the state "either spin down or spin up" (and, ok, one needs to measure probabilities, so one should actually repeat the same experiment many times).

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@André: For anyons, there are indeed experiments which would distinguish between the two copies in φ ⊕ φ for any case where φ ⊕ φ arises naturally. Let's look at the case φ ⊗ φ ⊗ φ = φ ⊕ φ ⊕ I above. As I described it in my answer, the two copies of φ are the results of fusing anyons. If you haven't fused the anyons yet, there are experiments distinguishing these states. After you've fused the anyons, the results of the fusions (if you've recorded them) would distinguish the states. If you haven't recorded them, then they're indistinguishable.

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Comment continued: This operation of fusing anyons would in some sense correspond to measuring the spin of the electron in the Hilbert space $H \otimes H$, and then putting it in the spin down state. If you don't record the result of your measurement, it's lost.

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@André: That's not what the notation means. $\phi + \phi$ is not adding two vectors, but rather creating a 2D vector space from 2 1D vector spaces.

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@André: the resulting space $\phi\oplus \phi$ can be thought of as an internal, but non-local part of the Hilbert space. Certain operations will induce internal rotations in this extra space. That means we can have state vectors in this internal space which are orthogonal. Since probabilities are related to squared amplitudes, this multidimensionality of the Hilbert space will be important when you look at interference-like experiments (much like the double-slit experiment).

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The difference with ordinary spin states is that these topological Hilbert spaces are not a localized property of one particle. The internal space is like a topological part of the Hilbert space, for lack of a better phrasing. The state of the internal space cannot be determined with a local measurement, but requires some sort of non-local process, which is usually braiding.

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In this comment thread people seem to be referring to $\phi$ as though it were a Hilbert space. In all interesting examples it is *not even a vector space*. The Hilbert spaces are $\mathrm{End}(\phi)$ and $\mathrm{End}(\phi \oplus \phi)$ (which are 1 and 4 dimensional respectively).

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@Noah: If End(φ ⊕ φ) has dimension 4, what does it correspond to physically? And what is the two-dimensional vector space corresponding to φ that you get from the equation φ ⊗ φ ⊗ φ = φ ⊕ φ ⊕ I?

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I think that equation says that the Hilbert space for a 3-particle system is 5 dimensional (which is roughly what it should be since each particle has golden ratio internal degrees of freedom). One of these states is the state that you'd get if annihilated then created them all from the vacuum. The other four correspond to ways of fusing them down to a single particle and then unfusing them back up (there are 2 ways of doing each, so 4 total).

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For anyons, the two ways of fusing them down to a single particle is a 2-dimensional vector space which is important in the theory. Does this thing have a name I can use to talk to mathematicians?

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Look at the first two sentences of chapter 3 of Pachos's lecture notes and the last sentence on page 10.

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Ways of fusing down to a single particle is $\mathrm{Hom}(\phi \otimes \phi, \phi)$ which is indeed a vector space (unlike $\phi$ itself). This is the point I was trying to make: it's the Hom spaces that are important physically, more than the objects themselves.

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@NoahSnyder: That is incorrect. For Fibonacci anyons $\phi\otimes\phi\otimes\phi = \mathbb{I} \oplus \phi \oplus \phi$ which is exactly 3 levels. Each $\phi$ in the direct product corresponds to a single dimension. (see for example p17 of http://rockpile.phys.virginia.edu/trieste08.pdf ).

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So by $\phi \otimes \phi \otimes ... \otimes \phi$, it seems that we mean the what Pachos calls the fusion Hilbert space ${\cal M}_n$, which splits into sectors corresponding to $I$ and $\phi$. Pachos never mentions Hom in defining this, as far as I can tell.

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@JoeFitzsimons Good point. I somehow got thrown off by 5 being a Fibonacci number as well. So the Hilbert space for an n-particle system is $\mathrm{Hom}(\phi^{\otimes n}, \phi) \oplus \mathrm{Hom}(\phi^{\otimes n}, 1)$?

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@Noah: It depends on your exact definition of "Hilbert space for an $n$-particle system" whether it's $\mathrm{Hom}(\phi^{\otimes n}, \phi) \oplus \mathrm{Hom}(\phi^{\otimes n}, 1)$ or just $\mathrm{Hom}(\phi^{\otimes n}, 1)$. Pachos's "fusion Hilbert space" ${\cal M}_n$ is the first.

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@Peter Looking at simple concrete models, like topological p-wave superconductors, it seems that the n-particle Hilbert space should be a direct sum over all sectors. But since Braiding can't mix states in different sectors, one sector seems to be sufficient for quantum computation. Maybe that's the reasoning behing Pachos' definition. But I'm not sure.

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I'm sure I've seen it defined both ways. Which one is more useful probably depends on what you want to use it for, but I imagine that defining it as the sum over all sectors is less likely to lead to confusion among people learning the material.

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Answer 2

There is a very nice set of lecture notes on the subject by Jiannis Pachos here. (see specifically section 1.3 on fusion and braiding properties).

As regards the first question, the tensor product and direct product are basically different ways of divvying up the Hilbert space (see John Baez's illuminating discussion here). When you have a relation like $\phi \otimes \phi = \mathbb{I} \oplus \phi$ (as for Fibonacci anyons), what this is saying is that when two anyons fuse they create either the vacuum or a single anyon. Physically the direct sum is basically enumerating possibilities, whereas the tensor product is basically describing the single possibility for a system composed of several subsystems. So an equation like this is saying that fusing two anyons produces either a single anyon or the vacuum state.

As regards the second question, since the direct sum is constructing the Hilbert space by combining the Hilbert spaces of the arguments, $\phi+\phi$ is not the same as $\phi$, but rather is a larger Hilbert space of single anyons. You may want to look at page 17 of the Fibonacci link. You will notice that $\phi\otimes\phi\otimes\phi = \mathbb{I} \oplus \phi \oplus \phi$, which is a 3 dimensional Hilbert space, where as $\phi\otimes\phi=\mathbb{I} \oplus \phi$ which is a 2 dimensional Hilbert space.

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Comments

+1 A much more precise and cleaner answer than mine.

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For physical relevance, let me mention that the Fibonacci anyons are conjectured to show up in the $\nu = 12/5$ plateau in FQH systems. This state is much harder to control experimentally than the $\mu = 5/2$ state, because the gap above the ground state is small. But contrary to the anyons I mentioned, these can perfom universal quantum computation.

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@Heidar: Sorry, hadn't seen your answer when I posted this. I guess we were writing them at the same time.

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No need to apologize, I like your answer much more than mine!

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Nice answer, and thank you for the references... I am still very confused as to what the physical difference should be between $\phi+\phi$ and $\phi$... (also, I think that the dimension -- rather "statistical dimension" -- of the Hilbert space of $\phi$ should be the golden ratio as opposed to one... whatever that means). Could you explain what an experimental setup might be that distinguishes $\phi+\phi$ from $\phi$?

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@André: Check out Chapter 2 of the first link. They are specifically using qubits encoded in $\mathbb{I} + \mathbb{I}$. Specifically these correspond to the two different ways to fuse four anyons to obtain the vacuum. There are several ways this can be achieved. (see for example quant-ph/0703143 where anyons are encoded as topological defects in a lattice of qubits)

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Answer 3

The simple objects in the braided fusion category correspond to the possible particle types. In the simplest important example there are two particle types 1 and $\phi$. (Well, 1 is the vacuum so it's a slightly odd sort of particle type.)

The non-simple objects don't have any intrinsic physical meaning, $\phi \oplus \phi$ just means any system "that can be a single particle but in two different ways" but makes no claims about what those two different ways are.

Tensor product of simple objects does have an intrinsic meaning, it means looking at a system with several particles in it.

Since the underlying category only has finitely many objects, any time you have a multi-particle system you can break up the Hilbert space as a direct sum of states where you've fused them all together into a single particle (either 1 or $\phi$). For example, since $\phi \otimes \phi \otimes \phi \cong \phi \oplus \phi \oplus 1$ this means that the Hilbert space for the 3 particle system is 3-dimensions, and splits up into a two-dimensional space of things that behave like a single particle (this is the $\phi \oplus \phi$ part) and a one-dimensional space of things that behave like the vacuum (this is the 1 part). In this case $\phi \oplus \phi$ has a physical meaning imbued by virtue of its appearing as a summand of $\phi^{\otimes 3}$, but other appearances of $\phi \oplus \phi$ inside other tensor products have different physical meanings.

In general, the Hilbert space assigned to the system of k particles $X_{a_1} \otimes X_{a_2} \otimes \ldots \otimes X_{a_k}$ is the direct sum over all particle types $X_i$ $$\bigoplus_{X_i} \mathrm{Hom}(X_{a_1} \otimes X_{a_2} \otimes \ldots \otimes X_{a_k}, X_i).$$

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Comments

From our discussions in the comments on my answer, the Hilbert space of a system with $n$ identical $X$-type particles is not $\mathrm{End}(X^{\otimes n})$ but $\mathrm{Hom}(X^{\otimes n}, I)$.

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I totally rewrote the answer based on the above discussion. Hopefully it's less wrong now.

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@Peter Just a minor detail. Not all anyons are self-dual and thus $\text{Hom}(X^{\otimes n},I)$ can be trivial. For fusion rules $X\otimes X = \bigoplus_Y Y$, the full Hilbert space for $n$ anyons of type $X$, is given by $V_{n} = \bigoplus_Y\text{Hom}(X^{\otimes n},Y)$.

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Damn, I'm too slow.

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@Heidar You didn't take your nitpick far enough, as $X^{\otimes n}$ can also contains summands which don't appear in $X \otimes X$. I think the formula in my answer is right: you want to sum over *all* particle types.

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@Heidar, Noah: you're absolutely right. My mistake.

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After the edit, definitely +1.

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Answer 4

If I remember correctly, isomorphism classes of simple objects correspond to different types of particles (which is assumed to be finite), furthermore more is structure is usually needed than a fusion category, for example braiding (which is the reason why anyons are so interesting). Let me be very concrete. A physically (and experimentally) relevant category has three isomorphism classes of simple objects $(\mathbf 1, \psi, \sigma)$ with the non-trivial fusion rules

$$ \psi\otimes\psi = \mathbf 1, \quad \psi\otimes\sigma =\sigma\quad \text{and}\quad\sigma\otimes\sigma = \mathbf 1 \oplus \psi,$$ where $\sigma$ is the so-called Ising anyon (and $\mathbf 1$ is the unit object). These quasi-particles are conjectured to show up in the $\nu = 5/2$ plateau in fractional quantum Hall systems and in $p+ip$ wave superconductors.

These fusion rules can be used to construct the ground state Hilbert space, which are given through the space of morphisms between simple objects. Defining $V_{ab}^c = \text{Hom}(a\otimes b,c)$, the Hilbert space for two Ising anyons is $V_2 = V_{\sigma\sigma}^{\mathbf 1}\oplus V_{\sigma\sigma}^{\psi}$ which is two-dimensional. For $2n$ anyons, the ground state is $\text{dim}V_{2n}= \text{dim}V_{2n\sigma}^{\mathbf 1} + \text{dim}V_{2n\sigma}^{\psi} = 2^{n-1}+2^{n-1} = 2^n$ dimensional (this is nicely seen using a graphical notation for morphisms, se references below). Using the fusion rule $\sigma\otimes\sigma = \mathbf 1 \oplus \psi$ one can solve the pentagon and hexagon equations for the $F$ and $R$ symbols, which when combined gives rise to a representation of the Braid group $B_{2n}$ (more preciely, the mapping class group of the n-punctured sphere=braids group + Dehn twists) on the ground state Hilbert space$V_{2n}$.

Thus one physical consequence of these direct sums is that the ground state is degenerate and the anyons have highly non-trivial statistics, the ground state wave function transforms under (higher dimensional) representation of the braid group when the particles are adiabatically moved around each other. This property of (non-abelian) anyons has given rise to the idea of using them for quantum computation (another property is their non-local nature, which partially saves them from decoherence).

To get a more physical idea of what fusion (or collision as you call it) of particles mean, one can look at the concrete $p+ip$ wave superconductors. In such superconductors zero (majorana) modes can be bound to the core of Abrikosov vortices, where for 2n vortices there will be $2^n = 2^{n-1} + 2^{n-1}$ fermionic states. This means that it takes two majorana fermions to get one conventional fermion. When the vortices are spatially separated, the state in the core of the vortex cannot be measured by local measurements. In the above notation; $\sigma$ is a vortex, $\psi$ an electron, and $\mathbf 1$ a cooper pair ("the trivial particle"). With this identification, the fusion rules say that fusing two electrons ($\psi\otimes\psi = \mathbf 1$) gives a cooper pair which vanishes in the condensates, while fusing two vortices ($\sigma\otimes\sigma = \mathbf 1\oplus\psi$) give either nothing or an electron.

Thus the physical meaning of these direct sums of simple objects has something to do with the possible outcomes when we measure the state after fusing two particles. In this way (non-abelian) anyons can be used to construct qubits, by braiding them one can do a computation, in the end one can fuse them and measure the resulting state.

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References: You can read appendix B and then chapter four in this thesis, to get a more precise description of how braided ribbon categories and anyons are connected. These lecture notes by John Preskill gives a more physical insight, in section "9.12 Anyon models generalized" category theory is used to formulate the physics (although category theory language is not used, and might be annoying if you are a mathematician). For a mathematician a better reference is

• Topological quantum computation (2010) By Zhenghan Wang

Last but not least, the canonical reference for non-abelian anyons is the review paper

• Non-Abelian Anyons and Topological Quantum Computation - Nayak, Simon, Stern, Freedman and Das Sarma (arxiv).

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Comments

First of all, thank you for your small correction: I modified fusion category --> braided fusion category in my post. Now concerning your sentence "Thus the physical meaning of these direct sums of simple objects has something to do with the possible outcomes when we measure the state after fusing two particles", this seems to imply that there is no physical difference between $X\oplus X$ and $X$ as $X$ **or** $X$ is really the same thing as $X$. Do you agree with that last statement?

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@Andre I think you need to think about it together with fusion. Assume you have the particles $(\mathbf 1, X, Y, Z)$ with fusion rules $Y\otimes Y = \mathbf 1\oplus X$ and $Z\otimes Z = \mathbf 1\oplus X\oplus X$. Then i guess one can rephrase your question to: is fusion of two $Y$ particles physically equivalent to fusing two $Z$ particles? Well, no. It is true that both gives the possibilities $\mathbf 1$ or $X$, but there are more ways to obtain $X$ when fusion $Z\otimes Z$, than $Y\otimes Y$. (cont)

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In other words, the Hilbert space associated to $n$ $Y$ particles is different from the one associated with $n$ $Z$ particles. Therefore there is a physical difference between $X$ and $X\oplus X$.

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