Doubt 1:
If two particles are identical, you can not distinguish between them. Then, I think, permutation operation is meaningless.
Because you can not distinguish them, how can you tell if they are permuted?

Doubt 2:
Suppose we have two fermions located at x and y.
If their positions are exchanged, the wave function will get a minus sign. But in this case, we can distinguish these two fermions, because their location x and y are distinguishable.
Usually the above description is half-cooked, we need to (anti-)symmetrize them in the final step, to make the identical particles theory fully consistent.
My doubt is,
Sometimes it seems not necessary to (anti)symmetrize identical particles, and people just use the half-cooked picture for granted. why?
for examples:
1. solid crystal atoms, few people symmetrize the wave function of atoms.
2. for two far-distance electrons, no need to antisymmetrize them; but for two close electrons, it is necessary

My current understanding is:
Leggett's book or Jain's book (I forgot which) has a chapter says, when two particles are able to permute, we have to consider the fully-cooked picture, to symmetrize them! If they can't permute easily (like crystal atoms), then the half-cooked picture also works.
Easily is an adjective, I think, something need to be compared with $\hbar$ to determine whether we must use fully-symmetrized picture or not, like a classical-quantum choice problem.

  • $\begingroup$ Please read this question and the answers. It will definitely help you understand this issue. $\endgroup$
    – DanielSank
    Oct 20, 2015 at 18:35
  • $\begingroup$ Note that position is not always a property to distinguish particles due to Heisenberg's uncertainty principle. $\endgroup$
    – Zetaman
    Aug 5, 2016 at 18:09

4 Answers 4


The description of identical particles using a wavefunction with ordered particles is somewhat unnatural: As you can not distinguish them except by their properties, calling one particle the first one, another one the second one,... is usually meaningless. Therefore in many-body physics, most people prefer a fully second-quantized description: Instead of saying "Particle 1 has energy $E_1$, particle 2 has energy $E_2$,..." you simply say "There is one particle with energy $E_1$, one with energy $E_2$,..." which captures the indistinguishability except for the properties much more natural. The (anti-)symmetrisation is only needed because a not (anti-)symmetrized version of the ordered wavefunction does not properly treat the indistinguishability.

About the cases where one does not (anti-)symmetrize: For various cases, I have seen different justifications: One for sure is laziness, or as some call it, simplification. Another one is that one can sometimes simply assign labels to particles using a (perhaps classical) property: A common example would be ions on an ion chain that usually are labelled from one end to the other. In condensed matter systems, that usually doesn't work: The particles change their location and looking would imply a measurement including the wavefunction collapse - some would say in most condensed matter systems, there is no which-way-information. Perhaps that fits to your distinction between far away and close particles. Furthermore I remember a justification, that there are cases where a (anti)symmetrisation does not change expectation values. However, I do not remember the context.

  • 1
    $\begingroup$ Welcome on Physics SE and thank you for your answer :) Our editor has a wonderful TeX functionality to write things like $E_1$ :) $\endgroup$
    – Sanya
    Aug 5, 2016 at 17:27
  • $\begingroup$ Thanks very much for the welcome. The indices are now written with LaTeX. Couldn't find a LaTeX-button, but luckily two dollar-signs just as in LaTeX-documents work for inline math code. $\endgroup$
    – QuantumAI
    Aug 6, 2016 at 7:26
  • 1
    $\begingroup$ and two dollar signs before and after give you a mathline :) $\endgroup$
    – Sanya
    Aug 6, 2016 at 7:29

Doubt 1: you can't, that's the point. You have to make a wave function that, if you try to permute the particles, nothing happens. You end up with the same state you started with.

Doubt 2: I don't understand what you are saying. Fermions are no less distinguishable than bosons. What do you mean $x$ and $y$ are distinguishable? What do you mean by "half baked".

The wave function of electrons in crystals are certainly anti-symmetrized. The ion cores are usually taken as external potentials, so symmetrization doesn't apply. Furthermore, the effect of symmetrization is manifested when the individual wave functions overlap. For distantly separated electrons they don't overlap.

  • $\begingroup$ In Doubt 2, half-cooked means non-symmetrized wavefunctions(eg: a product state of N particles) ; fully-cooked means (anti)symmetrized wavefuntions. The half-cooked picture is not the right many-body wavefunction, but is widely used for simplicity. $\endgroup$
    – Jian
    Oct 20, 2015 at 19:52
  • $\begingroup$ @wwwjjj I wouldn't use the words "half-baked", but you have the reason already: simplicity. Everything physicists do is a simplification. Exact solutions are impossible for $10^{23}$ particles. In fact exact solutions are impossible for three particles. A symmetrized wave function will be needed when the un-symmetrized one gives results that are not accurate enough for the application or context. $\endgroup$
    – garyp
    Oct 21, 2015 at 2:25

Doubt 1: If two particles are identical, you can not distinguish between them.

Identical does not mean "we cannot distinguish them". There is a reason why identical particles are sometimes mentioned as distinguishable and sometimes as indistinguishable.

In order to explain the difference between identical and distinguishable, let us consider simple example. Two coins from the same batch can be called identical; it means that all their intrinsic properties (weight, shape, material, minted pattern) are the same, so we cannot distinguish them based on these properties alone. However, we can still distinguish them based on their different position and orientation with respect to other bodies in the observed space.

The same goes for electrons; because all electrons have the same mass, charge and other intrinsic characteristics, they are all identical. The electron that oscillates in cellphone network tower is identical to the electron that oscillates in a mobile phone. But these two electrons are distinguishable, because they have different location.

If we discuss two electrons in an atom, we can no longer track and distinguish two electrons in such a way. That's why they are treated as indistinguishable. Even if we knew density function for presence of an electron in 3D space around the atom $\rho(\mathbf x)$ (which is a very detailed information that we do not have), we could not use it to distinguish between the two electrons.

Then, I think, permutation operation is meaningless. Because you can not distinguish them, how can you tell if they are permuted?

In an atom the electrons are indistinguishable, so indeed it is not possible to "take them and interchange them" in the usual sense (for example, interchange their state of position and momentum). However, in the expositions of the theory of atom and molecules, the permutation operation discussed is not done to the electrons, but to arguments of the $\psi$ function.

Let the two arguments be $\mathbf x$, $\mathbf y$. These are distinguishable because they have different name, so we can do the permutation and introduce two different functions:

$$ f(\mathbf x,\mathbf y) = \psi(\mathbf x,\mathbf y), $$

$$ f'(\mathbf x,\mathbf y) = \psi(\mathbf y,\mathbf x). $$

Sometimes it seems not necessary to (anti)symmetrize identical particles, and people just use the half-cooked picture for granted. why?

The requirement of antisymmetrization and symmetrization is useful in determining the solutions of the Schroedinger equations for small systems such as atoms and molecules. It certainly is not useful for description of large macroscopic systems such as electrons in cyclotron. This shows that applicability of the Schroedinger equation and the usual ideas around it is limited. This is probably related to strength of interaction; interaction of two electrons separated by large distance is much weaker than electron-electron interaction in a single atom.


For doubt 1: Interchanging identical particles is not a physical thing, there is no physical observable from which we can tell if an interchange happens or not. In this sense, you are right, we can't interchange identical particles.

For doubt 2: I think we can't use the location of two electrons to distinguish them. Let's name them A and B. Since their wavefunction is anti-symmetrized, when you find an electron at one location, it could either be A or B with equal probability. In fact, we can't even name them in the first place because they are identical. But we can name them once we have measured their locations, say, we call the electron at x "A", and that at y "B". If we do scattering of these two and catch them in two places, you can't tell who's A and who's B. You can give them names but you cannot keep track of them.

For atoms in the solid: I am not a condensed matter person, but I think atoms are sources of the EM fields to affect the states of electrons we are mainly interested in, in that regard it doesn't matter if we symmetrize atoms or not.

In general non-interacting multiparticle states transform as tensor products of each particle, even for those containing identical particles (e.g. angular momentum addition). So we start with a tensor product, and the tensor product comes with an order even for identical particles, and then we have to quotient out interchange of identical particles. For Bosons, that quotient can be done in any manner, for example we can order identical particles in the tensor product by z component of the momentum, but symmetrizing the wavefunction is perhaps the most elegant way. For Fermions, nature demands no two identical particles in the same state, so anti-symmetrizing the wavefunction is a natural way to do the quotient. After this mathematical setup (much of it for convention and convenience rather than physics), you can now interchange states in the tensor product and claim a plus or minus sign for symmetrized or anti-symmetrized ones. But this mathematical operation does not correspond to a physical operation.

  • $\begingroup$ I think I am pushing it too far. For a given state there is no interchange of identical particles that can be observed physically. But interchange of identical particles and the phase associated with it in a path integral is physically meaningful, see anyons. $\endgroup$ Apr 28, 2021 at 10:33

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