# Is a large system just a set of smaller systems?

All particles exhibit wave-particle duality. And I have a strange question.

Why does a larger system, liken an atom that is just a set of smaller systems, itself exhibit wave-particle duality?

In principle all large systems can be defined as a set of smaller systems. An atom is a set of nucleus (a smaller set of up and down quarks held together by the strong interaction) and some electron(s) bound to the set of quarks by the electromagnetic force.

How is this set of smaller systems being able exhibit wave-particle duality as a whole as if it is a particle itself?

Does this imply that electron and all other elementary particles are indeed just another set of smaller sets and any sets can exhibit wave-particle duality?

And here comes the question: how do you define a set? We naturally define composition such as atom, molecule or "elementary particle" like electron as a set. Can the composition of electrons and up quarks be defined as a set (a system with wavefunction) and the down quarks that present in the atom be defined as a separated set (another system)?

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Can you please clarify on which theory you are running? Why can't a composite system show same attributes as of its subsystem? –  SS-3.1415926535897932384626433 Jan 24 '13 at 19:11
@SachinShekhar Quantum theory. I have this doubt not because I'm thinking that a composite system can't show same attributes as of its subsystem. I'm thinking about the nature of a system itself. –  user8302 Jan 25 '13 at 3:19
Oh, the last paragraph? Fine.. nice question. –  SS-3.1415926535897932384626433 Jan 25 '13 at 6:04
@SachinShekhar Thanks! –  user8302 Jan 25 '13 at 6:06

According to Quantum mechanics, every particle has a wavefunction which completely describes it. The behavior of the particle, including its time evolution and the distribution of outcomes to any measurement performed on the particle, is determined by its wavefunction (Edit: Michael Brown correctly notes in the comments that if the particle is part of a bigger composite system, we often cannot use a wavefunction to describe it, and we need to use a density matrix instead).

Some properties of these quantum particles are similar to properties of classical particles. Some properties are similar to classical waves. The coexistence of properties of both kinds is what some people describe as "wave-particle duality" (I have to say that personally I dislike this term).

A composite system, like an atom which is composed of a nucleus and electrons, also has a wavefunction (or density matrix), and therefore also has both kinds of properties (particle-like and wave-like), and therefore can also be said to have "wave-particle duality".

The question of whether a certain elementary particle is indeed elementary or is composed of smaller particles is interesting, but unrelated. As of now mainstream physics considers electrons elementary particles, but of course we have no way of knowing whether or not this will change in the future.

Edit: To answer your last question - yes, in principle you can take any set of particles and consider them as one system. The only question is whether it is useful to do so. For example the nucleus and the electrons of an atom are often easy to treat as one system - an atom. You can in principle consider, like you suggest in your question, only part of the quarks in the nucleus together with the electrons as one system, but you are not likely to achieve any useful insight by doing that. The rest of the quarks in the atom interact with the quarks in your system very strongly, and any meaningful description will need to include them.

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Usually, yes. Sometimes there are subsystems which interact weakly with each other, and then some of them can be left out, and their influence on the main system is accounted for in a different way. –  Joe Jan 25 '13 at 9:54
My advisor fancies to picture it to undergraduate students in this briefly and possibly literary way: "The whole is more than the simple sum of its parts". –  user17581 Jan 25 '13 at 10:04
The first part of your answer: No no no... There is one wavefunction which is a function of the configuration variables of all the particles, e.g. $\Psi\left(\vec{r}_1,\vec{r}_2,\ldots,\vec{r}_N\right)$ in an $N$ particle system. Sometimes the wavefunction factorises $\Psi=\phi_1\left(\vec{r}_1\right)\phi_2\left(\vec{r}_2\right)\ldots\phi_N\left(‌​\vec{r}_N\right)$ and you can think of the $\phi_i$s as individual wavefunctions for the individual particles. In general this factorisation doesn't happen and you have "entanglement" - which is what @user17581's advisor is most likely refering to. –  Michael Brown Jan 25 '13 at 10:09
That exactly the case, Michael Brown. I think this issue of the separability of the wavefunction should be adressed in some way or another in any of the answers, but I don't see where this concept clashes with the answer that Joe has given contradicts. Perhaps I've misread his answer :-/ –  user17581 Jan 25 '13 at 10:39
@MichaelBrown - you're absolutely right, but I don't think this is the information the OP was asking for. To be more accurate I should of said that every system has a density matrix which describes it, but the rest of the answer would stay the same. –  Joe Jan 25 '13 at 11:06

If you know Classical Mechanics of two interacting particles, then it is easy to explain.

So, let us consider two interacting particles in a bound state, like the Earth and the Moon, whatever. Instead of writing their "individual" coordinates $r_1$ and $r_2$, we can introduce the center of mass $R$ and the relative motion $r=r_1-r_2$ coordinates. It's just a reversible variable change. But in terms of $R$ and $r$ we have another equation system - one describes the system as a whole and the other does its "internal" motion.

In QM it is the same: the common wave function $\Psi(r_1,r_2)$ can be represented as a product of $\psi(R)$ and $\varphi(r)$. So, the system as a whole is described as a wave too - with $\psi(R)$.

If the bound state is tight and needs a certain amount of energy to get excited, then there may be processes of interaction with one of constituent particles which are "too weak" to excite the bound state, so the system behaves as an "unbreakable" unity. For atoms $\psi_A(R)$ describes the wave packet of an atom as a whole and it may change, whereas $\varphi_A(r)$ describes the relative position of atomic constituents and it may stay unchanged. Then the dynamics of a "set" is a dynamics of its center of mass solely.

The nucleus has also its own constituents, but if the energy of an external interaction is not sufficient to excite the nucleus, the latter will be "seen" as something whole. As long as the nucleus $\varphi_N(r)$ does not change, you can use the nucleus $\psi_N(R)$ as if the nucleus was point-like, with no sets "inside".

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This answer is more clear than the other one. –  user10001 Jan 25 '13 at 8:28