# direct sum of anyons?

+ 10 like - 0 dislike
1045 views

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 $X\oplus X$?

This post has been migrated from (A51.SE)
asked Oct 15, 2011

+ 11 like - 0 dislike

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).

This post has been migrated from (A51.SE)
answered Oct 16, 2011 by (790 points)
+1: A far clearer answer than mine.

This post has been migrated from (A51.SE)
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.

This post has been migrated from (A51.SE)
@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.

This post has been migrated from (A51.SE)
@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).

This post has been migrated from (A51.SE)
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.

This post has been migrated from (A51.SE)
@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.

This post has been migrated from (A51.SE)
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.

This post has been migrated from (A51.SE)
+ 9 like - 0 dislike

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.

This post has been migrated from (A51.SE)
answered Oct 16, 2011 by (3,575 points)
+1 A much more precise and cleaner answer than mine.

This post has been migrated from (A51.SE)
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.

This post has been migrated from (A51.SE)
@Heidar: Sorry, hadn't seen your answer when I posted this. I guess we were writing them at the same time.

This post has been migrated from (A51.SE)
No need to apologize, I like your answer much more than mine!

This post has been migrated from (A51.SE)
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 $\phi$?

This post has been migrated from (A51.SE)
@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)

This post has been migrated from (A51.SE)
+ 6 like - 0 dislike

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 $X_i$ $$\bigoplus_{X_i} \mathrm{Hom}(X_{a_1} \otimes X_{a_2} \otimes \ldots \otimes X_{a_k}, X_i).$$

This post has been migrated from (A51.SE)
answered Oct 16, 2011 by (60 points)
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 $\mathrm{Hom}(X^{\otimes n}, I)$.

This post has been migrated from (A51.SE)
I totally rewrote the answer based on the above discussion. Hopefully it's less wrong now.

This post has been migrated from (A51.SE)
@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 $V_{n} = \bigoplus_Y\text{Hom}(X^{\otimes n},Y)$.

This post has been migrated from (A51.SE)
Damn, I'm too slow.

This post has been migrated from (A51.SE)
@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.

This post has been migrated from (A51.SE)
@Heidar, Noah: you're absolutely right. My mistake.

This post has been migrated from (A51.SE)
After the edit, definitely +1.

This post has been migrated from (A51.SE)
+ 4 like - 0 dislike

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.

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

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

This post has been migrated from (A51.SE)
answered Oct 16, 2011 by (855 points)
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?

This post has been migrated from (A51.SE)
@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)

This post has been migrated from (A51.SE)
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 $X\oplus X$.

This post has been migrated from (A51.SE)

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ys$\varnothing$csOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.