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In thermodynamics and statistical mechanics, we learn that many subatomic particles, such as electrons, are indistinguishable. But are they really? I can understand from one perspective that if we swapped one electron for another, we wouldn't be able to tell the difference because of its mass, charge, size, etc.

However, I have an argument against this and I would like to obtain some clarity on the topic. My argument is simply the "identity of indiscernibles" which states that there cannot be separate entities that have all their properties in common. In other words, if multiple particles share all of their properties, then they are the exact same particle, no distinction whatsoever.

Suppose we have two electrons (one in our left palm and the other in our right palm), then they share all of their properties except one: spatial coordinates, where they are located. Thus, they are in fact two separate electrons, not one. I do agree if we swapped one for the other, there would be no distinction, but as long they occupy different positions, they are distinguishable.

So how can we say that particles, such as in a lattice, are truly indistinguishable since each one is located at a different position? Any clarity would be appreciated.

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I do agree if we swapped one for the other, there would be no distinction, but as long they occupy different positions, they are distinguishable.

You are simply using the word "distinguishable" in a different way than it is used in the sense of elementary particles. In its technical sense, the fact that electrons are indistinguishable means that the state of a quantum mechanical system is invariant under their interchange, or equivalently that the electron exchange operator is unitary (in fact, proportional to the identity operator).

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  • $\begingroup$ I completely agree here and understand the exchange operator. To clarify though, are you just saying I'm using a different definition of "distinguishable" than what is typically used in QM? It just seems like you cannot exclude the property of position when describing a particle. Am I wrong about that? Thanks! $\endgroup$
    – RJ Onyx Moonshadow
    Commented Apr 23, 2020 at 11:56
  • $\begingroup$ @RJOnyxMoonshadow Yes, that's what I'm saying. You cannot exclude the property of position when describing the state of a particle at some moment in time, but this is not what we are doing here. $\endgroup$
    – J. Murray
    Commented Apr 23, 2020 at 12:14
  • $\begingroup$ @j-murray Can you explain further? What are we doing? I thought we were describing the states of both particles. $\endgroup$
    – RJ Onyx Moonshadow
    Commented Apr 23, 2020 at 12:20
  • $\begingroup$ @RJOnyxMoonshadow No. I am once again using the word state in its technical sense. The state of a system in any particular moment (by which I mean, in which element of the underlying Hilbert space the system currently exists) does not encode, for example, the mass of the particles in question, but we would certainly take mass as an important property when we describe them. $\endgroup$
    – J. Murray
    Commented Apr 23, 2020 at 12:38
  • $\begingroup$ @j-murray I believe I get what you are saying. Suppose we have 2 X's and 1 O. There are 6 states total, but only 3 are distinguishable: {XXO, XOX, OXX}. However, if we actually labelled each particle with a position, there are essentially 6 states: {X1X2O, X2X1O, X1OX2, X2OX1, OX1X2, OX2X1}. Is this what it really boils down to? $\endgroup$
    – RJ Onyx Moonshadow
    Commented Apr 24, 2020 at 5:05
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This question is answered by QFT. Indeed, the position of a particle is not an observable of the theory, but rather "parameters". Therefore, it makes no sense to attach a position label to a particle, as this is something that cannot be observed.

Consider the purely relativistic effect of a particle decay and it's potential reappearance later on. Once a particle has decayed, what's its position? On the other hand, such a problem does not exist when you have enough fields that can act as "reservoirs" of energy and momentum. Particles are then just "ripples" of these fields, some of which we can actually observe.

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  • $\begingroup$ Thank you for answering, but I still don't understand. How can position not be a property when it's one of the important concepts studied in quantum mechanics, such as in Heisenberg's Uncertainty Principle? I understand particles are waves, but it's clear that the particle across the street is not the particle in my hand. They are excitations of fields at two separate locations. Can you explain further please? $\endgroup$
    – RJ Onyx Moonshadow
    Commented Apr 23, 2020 at 4:32
  • $\begingroup$ In nonrelativistic quantum mechanics, where position and momentum are operators, the Heisenberg principle is easily derived from [x, p]=i*hbar where the [,] is a commutator of operators. In fact, the uncertainty principle generalizes to any pair of operators and gives the lower bound on uncertainty in terms of their commutator. In fact, to even start discussing the quantum theory of some nonrelativistic system, one imposes the above "canonical commutation relation" on the operators representing conjugate position and momentum. $\endgroup$
    – BRSTCohomology
    Commented Apr 23, 2020 at 12:59
  • $\begingroup$ However in quantum field theory, where the Lagrangian is that of a field and not that of a particle, the commutation relation you impose must be between the field $\phi(x)$ and corresponding canonical momentum $\frac{\partial \phi}{\partial t}(x)$, in order to quantize the theory. So we actually have a canonical commutator for each point in space when we do quantum field theory, and we've only ever used $x$ as a label instead of an operator. I certainly haven't done this explanation justice so I recommend you read Peskin and Schroeder, chapters 2 and 3 as it certainly answers your questions. $\endgroup$
    – BRSTCohomology
    Commented Apr 23, 2020 at 13:01
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So long as the wave functions of the particles in your hands do not overlap, you can say that they did not swap places. But this does not describe the quantum state, which describes the particles themselves as identical. While the wave functions of the particles does not overlap, there will be no change to calculations.

For the particles in a lattice, you have a tensor product consisting of wave functions for the individual particles. Suppose you detect the position of one of the particles. All you can say is "I found a particle at that position". You cannot actually say which particle it was, beyond the fact that it was at that position. This is reflected in the (anti-)symmetrising of the state. It does not say "all the particles are at all the positions", just that the state does not distinguish which particle is at which position.

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