direct sum of anyons? 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 $X\oplus X$? 
 A: 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


*

*Topological quantum computation (2010) By Zhenghan Wang
Last but not least, the canonical reference for non-abelian anyons is the review paper


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

A: 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).
A: 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 $X_i$ 
$$\bigoplus_{X_i} \mathrm{Hom}(X_{a_1} \otimes X_{a_2} \otimes \ldots \otimes X_{a_k}, X_i).$$
A: 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.
A: I feel that I finally understand the physical meaning of composite (ie non-simple) objects 
like $\phi\oplus\phi$. It is explained in the section II of my paper with Tian Lan arxiv.org/abs/1311.1784 .
We know that putting a few anyons (ie the objects in tensor category) on a Riemann surface may generate degenerate states (ie the fusion space of the objects in tensor category). 
Physically, "putting a few anyons on a Riemann surface" means that using a certain local Hamiltonian $\Delta H$ to localize the anyons at certain fixed locations. The above mentioned degenerate states are the
ground states of the total Hamiltonian $H_0+\Delta H$.
If the degeneracy is independent of any perturbation of $\Delta H$ near a particle, then that particle is of the simple type. If the degeneracy can be split by a perturbation of $\Delta H$ near a particle, then that particle is of the composite type. 
In other words, the particle of composite type correspond to accidental degeneracy in physics.
