2019 Award

## Main Question or Discussion Point

Commonly there is a lot of imprecision in talking about ”indistinguishable” (or ”identical”) particles, even in serious work. This Insight article clarifies the issues involved in a conceptually precise way.
Classical mechanics. Historically, indistinguishable particles were introduced in order to explain the failure of the thermodynamics of a Newtonian ##N##-particle system to describe the absence of an entropy increase when mixing two volumes of the same substance. This assumption, which has no logical basis in classical mechanics but appears as an ad hoc device to save the theory, drastically changes the state space of the multiparticle system, the phase space of functions of positions and momenta, to a much smaller space of symmetric functions of positions and momenta.
Nonrelativistic quantum mechanics. The first quantum manifestation of indistinguishable particles was the Pauli exclusion principle, which states that wave functions of a nonrelativistic...

troglodyte, Klystron, vanhees71 and 5 others

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PeterDonis
Mentor
2019 Award
Excellent article!

Regarding the discussion of entanglement and when it makes sense to use that term: consider an atom with electrons in multiple energy levels in its ground state (for example, beryllium, with two 1s electrons and two 2s electrons). Would it make sense to say that the electrons in a particular energy level (e.g., 1s) are indistinguishable and therefore it's not useful to talk about them being entangled, whereas the electrons in different energy levels (e.g., 2s vs. 1s) are distinguishable and therefore it is useful to talk about them being entangled (so the overall ground state of the Be atom could be described as an entangled state of two 1s electrons and two 2s electrons)?

2019 Award
Would it make sense to say that the electrons in a particular energy level (e.g., 1s) are indistinguishable and therefore it's not useful to talk about them being entangled, whereas the electrons in different energy levels (e.g., 2s vs. 1s) are distinguishable and therefore it is useful to talk about them being entangled (so the overall ground state of the Be atom could be described as an entangled state of two 1s electrons and two 2s electrons)?
The notion of an elecron being in a particular energy level is already a notion of a particular approximation scheme that describes the multielectronic wave function in the full Hilbert space in terms of a reduced, simplified description based on a tensor product of Hilbert spaces for electrons with fixed energy levels - a degenerate extension of the Hartree-Fock approach.

In this reduced description, the two electrons in a particular degenerate energy level are indistinguishable, and the electron pairs in different energy level are distinguishable and entangled. However, it is not clear to me how useful the terminology is in this case.

For example, multiplication by xkxk, the standard position observable for particle kk in the distinguishable case, does not have this property. And if one correctly (anti)symmetrizes the result, the resulting operator only describes the center of mass of the system, not a property of individual particles.
By this, do you mean that if we took an operator acting on two antisymmetric particles like

$$X_2 = X\otimes I + I\otimes X,$$ and calculated its exepctation value on an arbitrary antisymmetric state ##|A\rangle = |\psi\rangle|\phi\rangle - |\phi\rangle|\psi\rangle,## we would get $$\langle X_2\rangle = 2 \langle\psi|X|\psi\rangle + 2 \langle\phi|X|\phi\rangle$$

And that it's impossible to conjure up any legal operator which could be used to probe just one of the factor states ##|\psi,\phi \rangle##?

2019 Award
By this, do you mean
I mean that the physical Hilbert space ##\cal H## is not a dense subspace of the tensor product, and the position operator of the first particle, as an operator on the tensor product, does not map a dense subspace of ##\cal H## into ##\cal H##. But the latter is required for operators with a physical meaning.

Can the (anti)symmetrization of the total wavefunction of fermions/bosons be proved somehow?

2019 Award
It is a direct consequence of quantum field theory.

vanhees71
Gold Member
2019 Award
Well, but you put the (anti-)symmetrization postulate in by hand when quantizing the fields with (equal-time) (anti-)commutation relations.

There is a topological argument by Laidlaw and deWitt why in ##\geq 3## spatial dimensions one can have only boson or fermion statistics of indistinguishable particles, while in 2 spatial dimensions you can have "anyon" statistics. Indeed anyonic quasiparticles are observed in condensed-matter physics, e.g., related to the fractional quantum Hall effect.

M. G. G. Laidlaw, C. M. DeWitt, Feynman Functional
Integrals for Systems of Indistinguishable Particles, Phys.
Rev. D 3 (1970) 1375.

The relativistic QFT description of anyons is developed in

Fröhlich, J., Marchetti, P. Quantum field theory of anyons. Lett Math Phys 16, 347–358 (1988).
https://doi.org/10.1007/BF00402043

DrDu
Very nice article!
I am not sure, whether particle statistics really forbids to introduce position operators.
For me, the (anti-) symmetrization is somehow secondary.
E.g., in one dimension, indistinguishability means that it suffices to consider ordered coordinates x1<x2<x3 ...<xn. In higher dimensions this generalizes to sectors. The (Fermi, Dirac or para-) statistics is hidden in the boundary conditions on the sector borders (for 2d, there exists also the possibility of anyons).
A position operator for particle i can then be formulated as ##\int dx x\delta(x_i-x)##.
Working in a larger but redundant space with any order of the xi may offer other advantages, but it is not driven by mathematical necessity, at least not in non-relativistic QM.

2019 Award
I am not sure, whether particle statistics really forbids to introduce position operators.
For me, the (anti-) symmetrization is somehow secondary.
E.g., in one dimension, indistinguishability means that it suffices to consider ordered coordinates x1<x2<x3 ...<xn.
Dimension 1 is special since particles can there be distinguished by their relative location, but only when coordinates are distinct. If two coordinates agree one cannot tell which particle moves left and whch moves right. Moreover, if you measure one particle position and don't know where the others are, you cannot tell which particle coordinate you measured. Thus the problem remains.
In higher dimensions this generalizes to sectors.
In 2D, there are no natural sectors, unless you fix a center. And even then there remains a rotational indistinguishability requirement. In 3D, sectors cannot even be naturally ordered.

In any case, for applications in chemistry there is no way to distinguish the hydrogen atoms in CH_4 or NH_3, say.

vanhees71
Gold Member
2019 Award
Very nice article!
I am not sure, whether particle statistics really forbids to introduce position operators.
For me, the (anti-) symmetrization is somehow secondary.
E.g., in one dimension, indistinguishability means that it suffices to consider ordered coordinates x1<x2<x3 ...<xn. In higher dimensions this generalizes to sectors. The (Fermi, Dirac or para-) statistics is hidden in the boundary conditions on the sector borders (for 2d, there exists also the possibility of anyons).
A position operator for particle i can then be formulated as ##\int dx x\delta(x_i-x)##.
Working in a larger but redundant space with any order of the xi may offer other advantages, but it is not driven by mathematical necessity, at least not in non-relativistic QM.
Well, as @A. Neumaier very convincingly argues in his very nice article, the most convenient way to deal with many-body systems of indistinguishable particles (in the relativistic case it's a necessity because of the possibility of particle creation and destruction already in two-body collisions) in terms of a quantum field theory.

The position operator can be defined for all massive particles in the relativistic case through the representation theory of the Poincare group. For massless particles it's only possible to define a position operator in the strict sense for spin ##\leq 1/2##. For details see @A. Neumaier 's theoretical-physics FAQ

https://www.mat.univie.ac.at/~neum/physfaq/topics/position.html

In the non-relativistic case it's simply the center of mass operator (here written in the Heiseberg picture
$$\hat{\vec{R}}=\int_{\mathbb{R}^3} \mathrm{d}^3 r \vec{r} \hat{\psi}^{\dagger}(t,\vec{r}) \hat{\psi}(t,\vec{r}).$$

DrDu
I don't deny that only considering sectors is probably inconvenient most of the time, nevertheless I find it illuminating. Btw. the first ab-initio calculation of a many electron atom, namely of Helium with spectroscopic accuracy was performed by Hylleraas using a sector wavefunction in radial coordinates so that only r1 <= r2 was considered. Evidently, there is no problem with rotational symmetry, here.

vanhees71
Gold Member
2019 Award
What is a "sector wave function"? I've never heard this. Of course for a two-electron system, it's not so inconvenient to use the 1st-quantization formalism with antisymmetrized wave functions, and I guess that's what Hylleraas used in his calculation (btw: do you have the reference? I'm always interested in historical papers).

DrDu
Here is the reference:
In this case I should have better spoken maybe of shell wavefunctions than of sector wave functions.
What I mean is that you consider the radial distance of electron 1 to be always lower than that of electron 2.

vanhees71
Gold Member
2019 Award
I see. Hylleraas calculates the spatial part of the Slater determinant which either symmetric or antisymmetric under exchange of the electrons, depending on the spin part being antisymmetric or symmetric (which leads to the distinction between para- and ortho-helium).

I still don't understand what "shell wavefunctions" or "sector wave functions" are...

DrDu
I see. Hylleraas calculates the spatial part of the Slater determinant which either symmetric or antisymmetric under exchange of the electrons, depending on the spin part being antisymmetric or symmetric (which leads to the distinction between para- and ortho-helium).

I still don't understand what "shell wavefunctions" or "sector wave functions" are...
Hi Hendrik,
he doesn't start with a Slater determinant but with a function where ##r_1 \le r_2##. Symmetry or antisymmetry of the spatial part is explicitly guaranteed by boundary conditions at ##r_1 = r_2##.

2019 Award
I don't deny that only considering sectors is probably inconvenient most of the time, nevertheless I find it illuminating. Btw. the first ab-initio calculation of a many electron atom, namely of Helium with spectroscopic accuracy was performed by Hylleraas using a sector wavefunction in radial coordinates so that only r1 <= r2 was considered. Evidently, there is no problem with rotational symmetry, here.
This is essentially a 1D calculation since only the radial coordinate is used. It does not work for neutral hydrogen or any other composite molecule.

vanhees71
vanhees71