Does the Pauli exclusion principle apply to mesons?

Question

According to the Pauli exclusion principle, two identical fermions cannot occupy the same quantum state simultaneously, but two bosons can. Mesons are bosons, but composed of two quarks, and quarks in turn are fermions. If two identical mesons were simultaneously at the same quantum state, there would thus be two pairs of identical quarks in the same quantum state, right? But on the other hands, mesons are bosons... this looks like a contradiction to me.

So, what is the truth about this matter? Can two identical mesons be in the same quantum state simultaneously or not?

Answer 1

So I guess the answer here is both yes and no. The best example of this is probably the existence of bose einstein condensates in laboratories. These are (almost) always made up of atoms that exhibit boson statistics as composite particles made up of fermionic particles.

My guess would be that this viewpoint is valid (at least to extremely high precision) as long as forces/densities are not high enough to resolve the internal structure of these particles. E.g. in the case of BECs if the densities are low enough such that each atom-atom interaction is represented not by the interactions of the constituents but rather by a single interaction of the composite particles.

The same should hold for two mesons:

At the most fundamental level they can never be in the same state, however at this level it is also not enough to treat the state of the meson as the valid degree of freedom. Rather the state should describe the constituent particles and their interactions.

Edit: As pointed out in the comments the first sentence of this paragraph is poorly worded, as it very much depends on the context. (See second answer for an example where they are indeed in 'the same state' to some degree)

Now in many applications, we can describe the state of the system by the dof of the meson rather than the quarks they are made of. This state will exhibit bose einstein statistics and thus viewed in these dof two meson can indeed be in the same state.

Answer 2

The subtlety in putting two identical mesons in the "same state" is that not only are the mesons indistinguishable from each other, but they aren't well-defined composites at all: We can't say which quarks are paired together.

Given two up quarks and two anti-down quarks, we can make two $\pi^+$ mesons, but they can equally well be $u_1 \bar{d_1}$ and $u_2 \bar{d_2}$, or $u_1 \bar{d_2}$ and $u_2 \bar{d_1}$. The two-meson state must contain both possibilities (with opposite signs since they differ by one fermion exchange).

So the closest thing to putting both mesons in the "same" one-meson state $f$ would be $F = f(u_1 \bar{d_1}) f(u_2 \bar{d_2}) - f(u_1 \bar{d_2}) f(u_2 \bar{d_1})$. This is symmetric under meson exchange (simultaneous exchange $u_1 \leftrightarrow u_2$ and $\bar{d_1} \leftrightarrow \bar{d_2}$).

If two identical mesons were simultaneously at the same quantum state, there would thus be two pairs of identical quarks in the same quantum state, right?

No, there is no inconsistency because the quarks are generally entangled and not in well-defined individual states; we do not obtain $F = 0$ except in special cases like $f(u \bar{d}) = g(u) h(\bar{d})$.

A simple example in operator notation, neglecting degrees of freedom other than spin: The one-pion state is $(u_\uparrow^\dagger \bar{d_\downarrow}^{\!\dagger} - u_\downarrow^\dagger \bar{d_\uparrow}^{\!\dagger})|0\rangle$. So with two pions we have $$(u_\uparrow^\dagger \bar{d_\downarrow}^{\!\dagger} - u_\downarrow^\dagger \bar{d_\uparrow}^{\!\dagger})(u_\uparrow^\dagger \bar{d_\downarrow}^{\!\dagger} - u_\downarrow^\dagger \bar{d_\uparrow}^{\!\dagger})|0\rangle = -2 u_\uparrow^\dagger u_\downarrow^\dagger \bar{d_\uparrow}^{\!\dagger} \bar{d_\downarrow}^{\!\dagger}|0\rangle \ne 0.$$