# What is the definition of particle(s) in QM?

1. Nov 19, 2011

### nonequilibrium

What is the definition of "particle(s)" in QM?

Hello,

It is said that "identical particles are indistinguishable", but I'm beginning to think such a weird sentence is only a result of butchering the word "particle". More concretely: what is actually meant with the word "particle" in the QM formalism?

"One particle", I suppose, can be defined as "a solution of the regular three-dimensional Schrödinger equation". Would someone agree? To go on, one can then define a certain experimental apparatus (defining in the sense of building it) and define position as what is measured by it and postulate $|\psi(\textbf r)|^2$ as the distribution of this "position".

Then for the definition of "particles": analogously one can define "a system of n identical particles" as a solution of
$$i \hbar \frac{\partial}{\partial t} \psi(\textbf r_1, \dots, \textbf r_n) = \left( - \sum_{i=1}^n \frac{\hbar^2}{2m} \nabla_i^2 + V(\textbf r_1, \dots, \textbf r_n) \right) \psi(\textbf r_1, \dots, \textbf r_n)$$
but from then on I'm not sure what to do. Should we again define position from scratch operationally? Or should it somehow relegate back to the one-particle case? Or is the measuring apparatus somehow taking into account all the n "particles" at the same time? Isn't a "n particle system position measuring apparatus" simply the measuring apparatus for the one particle case multiplied by n?

Basically, what I'm struggling with is knowing what meaning to give to a "n particle system", while sentences like "identical particles are indistinguishable" already presume everything is well-defined, as if the one particle case said everything you needed to know (which was the case in classical mechanics, anyway).

I hope there are some out there who understand what I am getting at. If not, then let me simply ask: what do you understand under "a system of n (identical) particles"?

2. Nov 20, 2011

### The_Duck

Re: What is the definition of "particle(s)" in QM?

Defining an n-particle system as a solution to the n-particle Schrodinger equation you wrote seems reasonable.

To add a small point to what you are saying above, when someone talks about a "system of n identical particles" they mean a solution of the n-particle Schrodinger equation, as you wrote above, *with the additional stipulation* that the function psi is either totally symmetric (for bosons) or totally antisymmetric (for fermions) in its arguments. This symmetry condition is the mathematical content of "indistiguishability of particles." We could also talk about a system of n distinguishable particles, in which case we would not impose this symmetry requirement.

If we have an n-particle wave function psi(r1, ..., r_n), the natural analog of the 1-particle position space probability distribution |psi(r)|^2 is the n-particle position space probability distribution |psi(r1, r2, ... r_n)|^2. This tells you the probability to observe a particle at r1, and simultaneously a particle at r2, ..., and simultaneously a particle at r_n. Here we are imagining an experimental apparatus that, as you say, basically consists of n copies of the 1-particle position measuring apparatus. The apparatus looks at an n-particle system and outputs n positions.

In the n-particle case, you can also use this joint probability distribution for the positions of all the particles to get the probability distribution for the i'th particle's position, by integrating |psi|^2 over the position variables of all the other particles. Of course if psi is totally (anti)symmetric, this probability distribution is the same for any i. This probability distribution will correspond to the output of a regular 1-particle position measurer set to look for the i'th particle (though if all the particles are indistinguishable it can't look only for the i'th particle; it will then just measure the position of *a* particle).

Ultimately, I think, "particle" refers to a discrete packet of mass, or mass/energy. Practically, if you are discussing an n-particle system, you mean that it contains n discrete components which you could in principle extract and count one by one. We find that such systems are correctly described by the n-particle Schrodinger equation. "Indistinguishability" of particles can be viewed as the empirical fact that to get the right answer when you calculate the behavior of an n-particle system using the n-particle Schrodinger equation, you need to assume that the wave function is either totally symmetric or totally antisymmetric, depending on the kind of particle.

Last edited: Nov 20, 2011
3. Nov 20, 2011

### kith

Re: What is the definition of "particle(s)" in QM?

I'm not sure, if I get your problem. One of QM's postulates is that the states of composite systems live in the direct product of the individual Hilbert spaces. In addition, there is the postulate that the state has to be either anti-symmetric (called fermionic) or symmetric (called bosonic) under particle exchange. So if you have a well-defined concept for a single particle, the many-particle case should follow naturally from these postulates.