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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?

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    $\begingroup$ I asked ChatGPT this question too, and it initially said… Please don’t use computer-generated text for questions or answers $\endgroup$
    – Ghoster
    Commented May 23, 2023 at 19:31
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    $\begingroup$ I recommend editing your question to remove any reference to what chatbots have to say about physics, since this part of your post violates the site’s guidelines. $\endgroup$
    – Ghoster
    Commented May 23, 2023 at 19:35
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    $\begingroup$ The point of the avuncular advice above is that your ChatGPT banter has nothing to do with the essence of your question, except as a proof that you "tried" to get to the bottom of it. In any case, the wavefunction of two mesons is symmetric, so they don't exclude each other in an obvious way. You do know the reason a ρ cannot decay into two neutral pions, right? $\endgroup$ Commented May 23, 2023 at 19:42
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    $\begingroup$ This idea is behind superconductivity. Electrons are fermions, but a Cooper pair is a boson. $\endgroup$
    – mmesser314
    Commented May 24, 2023 at 0:06
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    $\begingroup$ related/possible dup: physics.stackexchange.com/q/705959/84967 $\endgroup$ Commented May 24, 2023 at 12:30

2 Answers 2

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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.

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    $\begingroup$ Is it really the correct answer? It seem to me that according to @nanoman's answer, at the most fundamental level they CAN be in the same state. $\endgroup$
    – agemO
    Commented May 25, 2023 at 7:01
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    $\begingroup$ @agemO I don't think that the answers contradict each other due to the reason that nanoman still treats the two mesons separately, which in close proximity is not valid as then $F = f(u_1 u_2 \bar{d_1} \bar{d_2})$ if I am not mistaken. $\endgroup$ Commented May 31, 2023 at 8:42
  • $\begingroup$ Indeed on second thought both answers say the same thing, but IMHO the "At the most fundamental level they can never be in the same state" sentence is misleading (and "they CAN be in the same state" is also misleading). $\endgroup$
    – agemO
    Commented May 31, 2023 at 12:31
  • $\begingroup$ Yeah I agree that is indeed worded poorly $\endgroup$ Commented May 31, 2023 at 13:21
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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 $$(u_\uparrow^\dagger \bar{d_\downarrow}^{\!\dagger} - u_\downarrow^\dagger \bar{d_\uparrow}^{\!\dagger})(u_\uparrow^\dagger \bar{d_\downarrow}^{\!\dagger} - u_\downarrow^\dagger \bar{d_\uparrow}^{\!\dagger})|0\rangle = -2 u_\uparrow^\dagger u_\downarrow^\dagger \bar{d_\uparrow}^{\!\dagger} \bar{d_\downarrow}^{\!\dagger}|0\rangle \ne 0.$$

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  • $\begingroup$ I think the state $F$ is equal to zero. If I write it in second quantized notation, it would be something like $(u_f^\dagger)^2(\overline{d}_f^\dagger)^2|0\rangle$, where $u_f^\dagger$ creates a $u$ fermion in the state of meson $f$ (and likewise for $\overline{d}$). This is zero from Pauli exclusion. If you disagree, I would find it helpful to see how you think this state should be written with creation and annihilation operators. $\endgroup$
    – Rococo
    Commented May 29, 2023 at 3:48
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    $\begingroup$ @Rococo See edit. $\endgroup$
    – nanoman
    Commented May 29, 2023 at 15:51
  • $\begingroup$ Thanks, with this clarification I think this is a nice answer. Entanglement and particle statistics work in such a way that you can make two copies of the same state, at least in a certain sense when represented in a certain way, but at the same time you will always see it so that if you measure the interior spin states of one of the mesons, the other one will be in a different state. $\endgroup$
    – Rococo
    Commented May 30, 2023 at 1:29
  • $\begingroup$ Why isn't this logic applicable to composite fermions? $\endgroup$
    – mavzolej
    Commented Jun 24 at 16:56

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