The question is in the title,

Let's say I have one boson (Mike) somewhere on earth and another one (Fatima) somewhere in proxima centauri. They are identical.

My friend argues that whatever the configuration I should write a symmetric wavefunction to describe this system. I argue that it is not the case, and that I should include symmetry only when the overlap between the spatial part of their wavefunctions is non negligeable and continuously go from a non symmetrized wavefunction to a symmetrized one, when the boson from Proxima centauri approaches my boson on earth.

What's your take ?

Personally I think that symmetrizing the wavefunctions entails that I don't know which boson is where, but for exmaple, I know for a fact that Fatima couldn't have travelled the distance between earth and proxima centauri simply because it cannot travel faster than light, so there is no uncertainty in the case where bosons are isolated...

  • 5
    $\begingroup$ You claim the bosons are identical. How do you know the one on Earth is "Mike" and not "Fatima" or vice versa? In your argument, you seem to be suggesting that by performing a measurement you can somehow tell that the boson you are measuring is the one on Earth or is the one that came from PC. If you can tell them apart, they are not identical and you should not symmetrize. $\endgroup$
    – Prahar
    Commented Feb 16, 2022 at 11:04
  • $\begingroup$ I don't need to make a measurement of any kind to know they are identical I believe. When you claim that two bosonic isotopes of any atom are identical, you don't measure their mass, spin, or whatever, you just say all these isotopes are identical. otherwise I could claim also that electrons are not identical unless Imeasure all of their properties, which obviously is not what is done in everyday physics. And surely there are identical bosonic atoms in PC and earth.. $\endgroup$
    – DarkBulle
    Commented Feb 16, 2022 at 12:20
  • $\begingroup$ For the sake of clarity, assume that the two bosons I speak are elementary aprticles like gluons. $\endgroup$
    – DarkBulle
    Commented Feb 16, 2022 at 12:23
  • 2
    $\begingroup$ You can't just claim two things are identical until you prove that there is absolutely no experiment that one can possibly do to distinguish between those two things. Now, in the context of your question since you are proposing a thought experiment, I can assume the particles are identical for argument's sake. Now if that is the case, my question to you is how do you know that the boson on Earth is Mike and not Fatima? $\endgroup$
    – Prahar
    Commented Feb 16, 2022 at 12:23
  • $\begingroup$ Well, say I prepare the state mike in earth and fatima in PC, because I am god. If I make a measurement just after my preparation, I know that I will find the same state, that's at the base of the quantum Zenon effect for example. Moreover finding the state fatima in earth and Mike in PC would imply that they have changed their position faster than allowed by relativity theory (?). So after my preparation, I know for a fact (?) that if I look at the boson on earth, it is necessarily Mike and not fatima $\endgroup$
    – DarkBulle
    Commented Feb 16, 2022 at 12:27

3 Answers 3


In quantum mechanics, there simply are no unsymmetrized states, in the Hilbert space for identical bosons. By definition, all states are completely symmetric between the exchange of identical bosons. There is no arbitrary rule that says that some states are symmetrized, and others not.

Therefore, your friend is correct: for identical bosons, the state is always symmetrized, even if the state involves two boson excitations that are localized and spacelike separated from each other.

However, the error you would make in ignoring the fact that the state is symmetrized will be vanishingly small if the overlap between the two localized wavefunctions is small. So, in practice, we often ignore bosons in distant galaxies when talking about bosons on Earth, and it doesn't cause any problems.

  • $\begingroup$ "So, in practice, we often ignore bosons in distant galaxies when talking about bosons on Earth, and it doesn't cause any problems." When would you ever not ignore them? $\endgroup$
    – D. Halsey
    Commented Feb 16, 2022 at 14:02
  • 1
    $\begingroup$ @D.Halsey Two examples that come to mind are if you wanted to prove their effect didn't matter, or if you were doing some modeling of the whole Universe, like in early Universe cosmology. I'll admit though, "often" here means "in essentially every imaginable application" :-) $\endgroup$
    – Andrew
    Commented Feb 16, 2022 at 14:08

I agree the answer from Andrew: the Hilbert space for identical bosons only has symmetric states. Another point worth making is that if two particles are identical then any Hamiltonian describing their dynamics will be symmetric with respect to exchange of those particles and it follows that the joint state will never evolve between symmetric and antisymmetric forms. So if it is ever symmetric then symmetric it will stay.

To get another perspective on this, it may be useful to invoke the concept of a quantum field. When we treat particles one at a time, and have conservation of particle number, we are treating the quantum fields in a simplified way. This can be useful, but ultimately field theory is the more precise and general model. And in field theory we do not need to label individual particles. Rather, we label modes and we can have degrees of excitation of those modes. Roughly speaking, what we call a particle is the presence of one degree of excitation of a mode.

In the example under consideration, we have a bosonic field in a state which can be written $$ |\psi \rangle = a_i^\dagger a_j^\dagger | 0 \rangle = |0,0,\cdots 0, 1, 0,0, \cdots 0, 1, 0, 0, \cdots\rangle $$ where $i$ and $j$ label the modes, for example one localized on Earth and one localized on Proxima Centuri, and the 1's in the list on the right hand side are in the $i$'th and $j$'th positions. In this expression it is indicated that the mode localized on Earth has an excitation (there is one 'thing' there) and the mode localized on Proxima Centuri has an excitation (there is one 'thing' there) but there is no further physical distinguishing-mark attached to each of the two things individually. The notation does not allow it, and the notation reflects the physical nature of the things under discussion. If one tries to say something like 'the one on Earth, called Mike, went on a trip to Promixa Centuri' then one is adopting words which do not succeed in mapping to the mathematical assertions, so such ways of speaking are not conveying good physical intuition. It would be a bit like saying 'I see you have some joules of energy there, and by the way, the first joule is called Mike and the second joule is called Fatima'. The analogy with energy helps us to get an intuition about why it is a mistake to attach labels to the particles. Indeed in this respect these identical co-called 'particles' are quite unlike the 'particles' which appear in Newtonian physics.

In this formulation, the difference between bosons and fermions comes into the commutaton relations among the raising and lowering operators, which result in $(a^\dagger)^2 = 0$ for fermions but not for bosons.


I should include symmetry only when the overlap between the spatial part of their wavefunctions is non negligeable and continuously go from a non symmetrized wavefunction to a symmetrized one, when the boson from Proxima centauri approaches my boson on earth.

That can't work in general because with two or more spatial dimensions, it's possible to observe bosonic/fermionic statistics even if the particles never get anywhere near each other.

Say Mike and Fatima start 4 light years apart, and then based on a quantum coin flip you either leave them where they are or start them moving on opposite half-circular paths at the same speed $v$ so that they exchange places in $2πc/v$ years. After that time has passed, you will observe constructive, destructive, or no interference depending on whether they are identical bosons, identical fermions or nonidentical. The particles in this experiment are always 4 light years apart. That is, the state at every time is a superposition of position-basis states in all of which they're 4 light years apart.

It would be incredibly difficult to do an experiment of that sort, even scaled down, but it's allowed by the rules of quantum mechanics.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.