A macroscopic object (let's call it "stone") may incidentally be a boson, right?

But identical bosons are "allowed" to have the same quantum state.

From this I conclude that two identical stones may share the same place.

However this urgently contradicts to our experience: Two stones cannot be in one place.

What is my error?

(I am not a quantum mechanics expert at all.)

  • $\begingroup$ You've more or less answered your own question. Is a quantum state also a place? $\endgroup$
    – Gert
    Commented Oct 28, 2015 at 23:50
  • 2
    $\begingroup$ This is similar to a question I asked previously here. $\endgroup$ Commented Oct 29, 2015 at 0:16

2 Answers 2


I will give you two answers for why this is, one simple and the other rather involved.

The simple answer: quantum statistics are only relevant for truly indistinguishable objects. Fundamental particles and simple composites (like atoms or some molecules) can be indistinguishable in this way, but a macroscopic object is so complex that there is a vanishing chance of two being identical. So no identical objects means no Bose condensation (that is, no macroscopic occupation of the same state).

The longer answer: you've been lied to about what exactly a Bose condensation really is. The closest system we have to an ideal Bose condensate is in a trapped ultracold gas, of the type that won the 1997 Nobel prize. In these systems all the atoms have the same probability distribution for their locations, and one can see macroscopic coherence effects like interference between two clouds. But the atoms are not in the exact same place, in the sense that if you made a very precise measurement of all the atomic positions at the same time you would find that no two are overlapping.

What's going on here? There are two important length scales in a quantum atom: the distance between the electrons and nucleus (usually less than 1 nm), and the space over which the total wavefunction of the atom itself is distributed, which must be at least as large but has no upper limit. For example, in these ultracold gases, the wavefunctions might be about 100,000 times larger than the electronic radius. Both of these are, in some sense, the "size" of the atom.

Bose condensation of atoms is only a sensible notion when the size of the wavefunction is much larger than the electronic radius. What happens, really, is that on the scale of the wavefunction each of the atoms has approximately the same distribution of possible positions, but this approximation (and it is an approximation) would break down if you confined the atoms' wavefunctions to not much larger than the electronic radius. At that point, you can no longer treat the atoms as point-like effective bosons, and must instead study the system in terms of their internal structures.

Macroscopic objects, on the other hand, never have any significant delocalization of their wavefunctions, which is why we think of things as having a definite position. It would take some time to explain why this is, but it is due to a phenomenon known as decoherence that you can look around for more information about. As a result, it is impossible for two macroscopic objects to overlap in their wavefunctions while having their internal structures be well-separated. And, of course, as CuriousOne says the internal structure of matter features Pauli exclusion and electrostatic interactions when things get too close. So even if you had two truly identical bosonic stones, you would never be able to delocalize them enough for their wavefunctions to overlap without their internal structures being near each other.


Being bosonic or fermionic is a property of a quantum field, not of every object. In particular, being a boson or a fermion is defined by whether or not the Fourier modes of the free field, which are the creation and annihilation operators for the particle associated to the field, commute or anticommute. Only by the spin-statistic theorem is this then translated into spin values of the particles created by the field.

There is no quantum field that has stones as particles. Hence, we cannot speak of stones being bosons or fermions, and, in particular, the Pauli exclusion principle is neither true nor false naively - we cannot tell whether two "stones" are able to occupy the same state without looking at the constituents.

In fact, the Pauli exclusion principle together with electromagnetic repulsion explain why matter cannot occupy the same state/pass through other matter, see also Why doesn't matter pass through other matter if atoms are 99.999% empty space?.

  • $\begingroup$ en.wikipedia.org/wiki/Boson "Composite particles (such as hadrons, nuclei, and atoms) can be bosons or fermions depending on their constituents". It looks like you contradict to Wikipedia. Who to believe? $\endgroup$
    – porton
    Commented Oct 28, 2015 at 23:55
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    $\begingroup$ @porton: There are effective field theories where you can describe e.g. mesons as modes of a quantum field. You can then assign the status of boson or fermion based on the behaviour of that effective field. There is no effective field theory in this sense for macroscopic objects. $\endgroup$
    – ACuriousMind
    Commented Oct 28, 2015 at 23:58
  • $\begingroup$ @ACuriousMind that seems like a statement that requires justification. $\endgroup$
    – Rococo
    Commented Oct 29, 2015 at 0:07
  • $\begingroup$ @Rococo the justification comes from the dimensional regime where quantum mechanics holds, this is commensurate to h_bar , a very small number. For macroscopic objects like stones h_bar is effectively zero and quantum mechanical rules morph into classical physics rules. The road is through the density matrix formalism en.wikipedia.org/wiki/Density_matrix $\endgroup$
    – anna v
    Commented Oct 29, 2015 at 5:31

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