# Identicalness and Indistinguishability in quantum mechanics

I've been reading chapter 10.3 'Identical Particles' in Shankar's book on quantum mechanics and also looked through some of other books on this subject and one rather subtle objection started bothering me.

They all argue that unlike classical physics, in which you can follow the trajectories of the particles without disturbing the states, there exists no physical basis for distinguishing between identical particles in quantum mechanics where any measurement leads to collapse of the state. And they straight go to saying that this in turn implies that there should be restriction imposed on the system of identical particles that two configurations related by the exchange of identical particles must be treated as one (=invariant under exchange operator) and be indistinguishable.

But as far as I understood, to be Identical only means that all the Internal Tags such as spin, mass, charge etc., which allows the observer to distinguish between particles without referring to their positions or momentums, are all the same. And here's what's been bugging me. I thought that whereas it is necessary to be 'Identical' in order for the system to be 'Indistinguishable' under the exchange of the particles, particles being 'Identical' is not sufficient for the particles to be 'Indistinguishable' .

Say, there existed a state corresponding to two identical particles and that is some nontrivial superposition of symmetric and antisymmetric states, i.e.,

$$|\psi\rangle$$= $$\alpha|\omega_1\omega_2,S\rangle+\beta|\omega_1\omega_2,A\rangle$$

then it is clear upon acting the exchange operator that this state is a 'Distinguishable' yet being one possible state of two identical particle system.

So it is clear from this that for the system to be indistinguishable, it requires them to follow either "Fermi-Dirac statistics" or "Bose-Einstein statistics" (either completely symmetric or completely antisymmetric) in addition to them being identical in their intrinsic properties such as spin, mass, charge, etc., and that Indistinguishability and Identicalness better be kept separate. I do agree that it can be postulated that all systems of identical particles must choose to be totally symmetric or antisymmetric (thereby becoming indistinguishable) and verify this postulate by experiments after experiments, but this indistinguishability should not be taken as something naturally derived from identicalness of the system.

Please verify if I'm right here or convince me with the legitimate way to address this issue if I'm wrong.

+)

I edit my question cause I found myself with a video from MIT opencourseware that supports the idea that it should be made an extra postulate.

https://youtu.be/G-5KHKrNPMs

It starts on 1:10

Check this out!

• Ty, edited the title!
– 류민석
Feb 15 at 14:48
• The main argument with the symmetric and antisymmetric combinations is unclear. If particles are indistinguishable, their wave function must be either symmetric or antisymmetric... since otherwise they would be distinguishable. Ale, is it philosophical or physical question? (i.e., is it really on topic?) Feb 15 at 14:53
• Yeah, for particles to be indistinguishable, their wavefunctions must be either symmetric or antisymmetirc. I'll give you that. But what i'm saying here is, for particles to be just identical but not indistinguishable, combination of symmetric and antisymmetric will do.
– 류민석
Feb 15 at 15:11

Somewhat along the lines that you suggest, "indistinguishable" means the $$n$$-body state transforms by a 1-dimension representation of the symmetric group $$S_n$$. There's a discussion of this - including a discussion of the distinctions between the classical and quantum concepts - in

Bach, Alexander. "The concept of indistinguishable particles in classical and quantum physics." Foundations of physics 18, no. 6 (1988): 639-649

a more mathematically-oriented discussion in

Kaplan, Inna G. "The exclusion principle and indistinguishability of identical particles in quantum mechanics." Soviet Physics Uspekhi 18, no. 12 (1975): 988

and a really hard-core-math discussion of this in

Hudson, Robin L., and Graham R. Moody. "Locally normal symmetric states and an analogue of de Finetti's theorem." Zeitschrift für Wahrscheinlichkeitstheorie und verwandte Gebiete 33, no. 4 (1976): 343-351.

None of this appeals to the idea of overlapping wavefunctions or states. In the specific case of the $$\vert\psi\rangle$$ that you give, where the state has amplitudes in both the symmetric and antisymmetric parts, the state is then partially symmetric. The obvious example example is the product state $$\vert \omega_1\rangle \vert\omega_2\rangle = \frac{1}{\sqrt{2}}\left( \vert \omega_1\rangle \omega_2\rangle+ \vert \omega_2\rangle \vert\omega_1\rangle\right) +\frac{1}{\sqrt{2}}\left( \vert \omega_1\rangle \omega_2\rangle- \vert \omega_2\rangle \vert\omega_1\rangle\right)\, .$$ In fact, when the probability of finding the state for each irrep (there may be more than one irrep for $$S_n$$ with $$n\ge 3$$) is the same (counting multiple copies of an irrep separately), then the state is fully distinguishable, as your $$\vert\psi\rangle$$ above. See

Tillmann, M., Tan, S.H., Stoeckl, S.E., Sanders, B.C., De Guise, H., Heilmann, R., Nolte, S., Szameit, A. and Walther, P., 2015. Generalized multiphoton quantum interference. Physical Review X, 5(4), p.041015

for a discussion of the $$S_3$$ case.

• Thanks a lot for the answer. I'll try to work on your references but it DOES seem to require quite a lot involved understanding in the group theory first to comprehend these concepts well enough in the general n-body identical case....
– 류민석
Feb 16 at 1:58
• The 1-dimensional irreps are easy (symmetric or antisymmetric). Kaplan gives an example where a 2-dimensional one (of $S_3$) fails a common sense test. Feb 16 at 4:11

yet being one possible state of two identical particle system.

Possible mathematically, but rejected in quantum theory based on the principle that identical particles in a single atomic system are indistinguishable.

Just like helium atom having 1st particle in 1s state and 2nd particle in 2p state is possible mathematically, but is rejected in calculations (in favor or symmetric/antisymmetric superpositions). This rejection is sometimes argued to be a matter of principle, but really the reason is calculations of psi functions and other quantum chemistry results work better when we limit the psi functions (including spin) to those that are antisymmetric.

You are correct that particles being identical does not imply necessarily that they are also indistinguishable. Electron in microwave oven and electron on the Moon are distinguishable. But for atomic systems there is no way to distinguish them and we assume, with good results, they are not.

• Thank you for answering! So are you saying that it's just a matter of convenience to treat identical particles that are close together as totally symmetric or antisymmetric?
– 류민석
Feb 16 at 1:50
• Only partially, it does make calculations more efficient, because we or the computer does not need to consider big class of functions. The main reason is that without the restriction, many results would be hard or impossible to obtain, such as energy levels and spectra of many-electron atoms and molecules. Feb 16 at 9:53

In non-relativistic quantum mechanics, you are correct. It is logically possible to have (in your language) identical but not indistinguishable particles. Indistinguishability is an extra assumption.

The way I think about it is that indistinguishability is a logical possibility that is allowed in quantum mechanics, but not in classical mechanics, and Nature has chosen to take advantage of this possibility.

Of course this assumption has many observable consequences which are very striking and have been observed, so there's no doubt it is in fact a good assumption. The Pauli Exclusion principle and the idea of bosons "bunching" together, or more generally the notion of an "exchange force," come to mind.

In relativistic quantum mechanics (quantum field theory), there is a deeper structure from which both identicalness and indistinguishability follows. It's not possible to have a relativistic quantum theory with a fixed number of particles, and so one quantizes fields which can describe an indefinite number of particles. The indistinguishability (ie, the fact that the state has to be totally symmetric or antisymmetric under exchange) and identicalness (equivalence of "internal tags") of the particle states constructed from the quantum field operators fall out naturally of the formalism (if you want to dig into this, google "Fock Space"). Furthermore the spin-statistics theorem, which states that integer spin particles must be bosons and half integer spin particles must be fermions, can only be proven in quantum field theory, and not non-relativistic quantum mechanics.

• Yet a fundamental assumption of classical statistic mechanic is that identical particles are distinguishable from one another (as in Gibbs’ paradox of mixing). Feb 16 at 4:22
• Well I might have to wait till I learn QFT then but this was the answer I was looking for!! Appreciate it!
– 류민석
Feb 16 at 5:59
• @ZeroTheHero Indeed, the fact that indistinguishable particles in quantum mechanics resolve the Gibbs paradox is a strong theoretical reason for studying them. Feb 16 at 12:30