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From the Pauli's Exclusion Principle no two electrons in a bound system have all same quantum numbers. This means that an electron can be uniquely specified by the four quantum numbers and hence can be distinguished from others.

I understand we still will not be be able to tell 'which electron' has that set of quantum numbers. So for instance, we cannot tell apart ; electron $1$ having $\{n,m,l,+s\}$ and electron $2$ having $\{n,m,l,-s\}$ from electron $1$ having $\{n,m,l,-s\}$ and electron $2$ having $\{n,m,l,+s\}$.

But why does that matter? We don't need 'electron-$1$' and 'electron-$2$' labels . We can just tag the electron as 'electron -$\{n,m,l,+s\}$' and so on.

To J Murray's comment:

This is from the Tony Genault book on statistical mechanics and this is what I meant in my comment.

In this chapter we shall treat the other type of assembly, in which the particles are distinguishable. The physical example is that of a solid rather than that of a gas. Consider a simple solid which is made up of N identical atoms. It remains true that the atoms themselves are indistinguishable. However, a good description of our assembly is to think about the solid as a set of N lattice sites, in which each lattice site contains an atom. A ‘particle’ of the assembly then becomes ‘the atom at lattice site 4357 (or whatever)’. (Which of the atoms is at this site is not specified.) The particle is distinguished not by the identity of the atom, but by the distinct location of each lattice site. A solid is an assembly of localized particles, and it is this locality which makes the particles distinguishable.

I am interested specifically in the line:

The particle is distinguished not by the identity of the atom, but by the distinct location of each

I can simply rephrase it as:

"The particle is distinguished not by the identity of the atom, but by the distinct quantum numbers of each."

Since even in the initial case of solid, we don't care whether its the same atom at a particular lattice when we look at it the second time but that it is the 'atom at lattice 1435 etc'

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  • $\begingroup$ What is the question? $\endgroup$ Apr 1, 2021 at 23:50

2 Answers 2

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So, from the Pauli's Exclusion Principle no two electrons in a bound system have all same quantum numbers. This means that an electron can be uniquely specified by the four quantum numbers and hence can be distinguished from others. I understand we still will not be be able to tell 'which electron' has that set of quantum numbers.

Your last sentence is what we mean when we say that two particles are indistinguishable. More precisely, two particles are called indistinguishable if swapping their quantum numbers doesn't affect the state of the composite system, or equivalently if swapping their quantum numbers only causes an overall phase shift of the composite wavefunction, $\Psi \mapsto e^{i\theta}\Psi$ for some $\theta$.

The spin-statistics theorem says that under broad conditions, there are only two types of particle - those for which $\theta =0$, which we call bosons, and those for which $\theta = \pi$, which we call fermions. Under certain conditions (e.g. in 2D) this does not hold, but we'll put that aside for now.

The point is that indistinguishability doesn't mean that every particle in a system has the same properties, but rather that if any two particles are interchanged, the state of the composite system is unaffected. In the specific case of fermions, this means that $\Psi \mapsto-\Psi$ when two fermions are interchanged; if they were to reside in the same state, we would also have that $\Psi\mapsto \Psi$ (since nothing has changed), which implies that $\Psi=0$. As a result, it follows that a system of indistinguishable fermions cannot have any two particles in the same state - i.e. the Pauli exclusion principle.

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  • $\begingroup$ What about the case of localized atoms in a crystal lattice? We consider them distinguishable by their 'locality' but exchanging any two atoms would not make any difference (assuming lattice of same kind of atoms) to the system whatsoever. Whats going on here then? $\endgroup$
    – Lost
    Mar 31, 2021 at 16:12
  • $\begingroup$ @Lost I'm not sure I understand what you mean. Are you talking about the nuclei in a lattice? If so, then they are not distinguishable, but since their equilibrium separation is far larger than the extent of their individual wavefunctions (to the extent that they have individual wavefunctions), their indistinguishability does not actually have a noticeable influence on the system. $\endgroup$
    – J. Murray
    Mar 31, 2021 at 17:19
  • $\begingroup$ I have made an edit to my question. Hope my query in the comment is clearer now. Please give it a read $\endgroup$
    – Lost
    Apr 1, 2021 at 12:20
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    $\begingroup$ @Lost The wording isn't ideal, but the author doesn't consider them distinguishable: "It remains true that the atoms themselves are indistinguishable." The states in which the atoms live can obviously be distinguished from one another, but you are confusing yourself by mixing up the indistinguishability of particles - which has a technical meaning related to exchange symmetry as I wrote in my first paragraph - and the distinguishability of the states in which those particles live. $\endgroup$
    – J. Murray
    Apr 1, 2021 at 18:53
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    $\begingroup$ @Lost Indistinguishability is a purely quantum mechanical effect, insofar as classically it doesn't even make sense to talk about the state of a system being symmetric or antisymmetric under particle interchange. I don't think it's necessarily helpful to think of this as being a result of the uncertainty principle, since the latter certainly doesn't imply the existence of exchange symmetry. $\endgroup$
    – J. Murray
    Apr 1, 2021 at 19:03
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The distinguishability requires not only associating partciles with a different sets of quantum numbers, but also tracing these particles. In classical physics we can follow the trajectories of the two particles and knwo which came from which initial position. However, the two electrons in the OP could switch their places and we would never know this.

However, note that there is a more serious flaw with the reasoning in the OP: it assumes the validity of the Pauli principle, which follows from the idnistinguishability of the electrons, to question this very indistinguishability. In other words, this is a circular reasoning.

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  • $\begingroup$ If I am not wrong, the Pauli Exclusion Principle historically predates the indistinguability derivation of it i.e. Pauli postulated spin to explain certain experimental results which is to imply that the principle can be taken as an independent postulate (probably) . Also, bosons are indistinguishable but they don't follow Pauli exclusion principle. What's going on here then? $\endgroup$
    – Lost
    Mar 31, 2021 at 16:18
  • $\begingroup$ @Lost Historically this is probably the case, but logically Pauli principle follows from particles being identical, but not necessarily the other way round. If this is the aspect that interests you, perhaps you coudl reformulate the question to this end. $\endgroup$ Mar 31, 2021 at 16:22
  • $\begingroup$ Okay, but your statement "Pauli principle follows from particles being identical" doesn't apply on Bosons. They are identical but just because they are identical doesn't make them follow Pauli. $\endgroup$
    – Lost
    Apr 1, 2021 at 11:56
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    $\begingroup$ Oh okay. I probably understand now. So basically in other words what you are saying is that the indistinguishability is not always a sufficient condition for Pauli to hold but it still is one of the necessary ones. So, a particle following Pauli has to be indistinguishable since thats a necessary constraint. $\endgroup$
    – Lost
    Apr 1, 2021 at 12:13
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    $\begingroup$ @Lost yes, this is a good way to put it. $\endgroup$ Apr 1, 2021 at 12:17

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