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A many-body wavefunction of identical fermions must be antisymmetrized because of fermionic statistics. We don’t antisymmetrize the meson wavefunction because it contains a quark and an antiquark, and they are not identical fermions. But in particle physics, we do antisymmetrize the baryon wavefunction even though it may consist of 3 non-identical quarks of different flavors.

What is the fundamental reason for this? Why is the baryon wavefunction anti-symmetrized? The different colors of quarks have exact SU(3) color symmetry, so it is probably okay to regard them as identical fermions [And we write it as a color singlet, fully antisymmetric, see below.] But the different flavors of quarks have no exact SU($N_f$) flavor symmetry, so why are different flavors of quarks regarded as identical?

There is some discussion in Griffiths Introduction to Elementary Particles p.184,

The wave function of a baryon consists of several pieces; there is the spatial part, describing the locations of the three quarks; there is the spin part, representing their spins; there is a flavor component, indicating what combination of u, d, and s is involved; and there is a color term, specifying the colors of the quarks: (space) (spin) (flavor) (color). It is the whole works that must be antisymmetric under the interchange of any two quarks.

Notice that a subtle extension of the notion of “identical particle” has implicitly been made here, for we are treating all quarks, regardless of color or even flavor, as different states of a single particle.

Unfortunately, Griffiths does not give any further reasoning.


Notes on the question:

  1. I am not asking why the baryon color wavefunction is a singlet. I already knew it had to be $(| rgb \rangle -| rbg \rangle + | gbr \rangle - | grb \rangle + | brg \rangle -| bgr \rangle )/\sqrt{6}$.

  2. Explicitly, consider the baryon octet $\Delta^+$ (from the $10$ in $3 \otimes 3 \otimes 3= 10 \oplus 8 \oplus 8 \oplus 1$) with a total spin $3/2$ and spin $z$ as $-1/2$, with wavefunction: $$\begin{aligned} &(| rgb \rangle -| rbg \rangle + | gbr \rangle - | grb \rangle + | brg \rangle -| bgr \rangle )/(\sqrt{6})\\ \otimes&(uud+udu+duu)/(\sqrt{3})\\ \otimes&(\downarrow \downarrow \uparrow +\downarrow \uparrow \downarrow +\uparrow \downarrow \downarrow )/(\sqrt{3}) \end{aligned}$$ The answer should argue why this wavefunction needs to be antisymmetrized, though $u$ and $d$ are non-identical.

  3. My question is related to this nice Phys.SE post (which I happily vote it up +1), but the accepted answer does not explain anything new, only re-iterate the known fact.

  4. Related older Reference: Phys. Rev., Vol. 50, 846 1936, On Nuclear Forces by B. CASSEN AND E. U. CONDON

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  • $\begingroup$ Have you seen: physics.stackexchange.com/q/219980 ? $\endgroup$ – Darkseid Sep 30 '17 at 17:42
  • $\begingroup$ @ Fedor Indutny, please make sure that I am asking the why the flavor symmetry is treated exact. The color sector is not a problem. Read my p.s.0. $\endgroup$ – wonderich Sep 30 '17 at 17:55
  • $\begingroup$ @ Fedor Indutny, But thanks for your post and concern -- I apprecaite it $\endgroup$ – wonderich Sep 30 '17 at 18:51
  • $\begingroup$ You are probably asking why the anticommutator of any two fermion fields vanishes, as opposed to just the anticommutator of a given fermion field with itself? This is a statement on the Lorentz-transformation properties of such fields. $\endgroup$ – Cosmas Zachos Oct 1 '17 at 14:21
  • $\begingroup$ For the generalized Pauli principle, Hogaasen & Sorba Hogasen 2007 might be useful. $\endgroup$ – Cosmas Zachos Oct 1 '17 at 16:54
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The extension that Griffiths is proposing isn't as drastic as you think. When you first learn about identical particles, you antisymmetrize the spatial wavefunction alone. Later, you add the spin and antisymmetrize the combined spin and spatial wavefunctions. But this is a major conceptual leap, because a spin up and spin down electron are not identical particles. The Pauli exclusion principle doesn't apply to them; you can put a spin up and spin down electron in the same state.

Now the jump to including the 'flavor wavefunction' is the same. An up quark and a down quark are not identical particles, and they can indeed be put in the same state. You might argue that including spin is different because you can flip it with a rotation, but the up and down quark states can be flipped by an isospin rotation; the situation is really exactly analogous.

So spin and flavor are on the same grounds; neither of them are correctly justified with the standard 'quantum mechanics' argument of interchanging the positions of two particles, since that only applies to the spatial wavefunction. Instead, we have to go to quantum field theory. The reason the spatial wavefunction is antisymmetric is because the creation operators anticommute, $$a_x^\dagger a_y^\dagger |0 \rangle = - a_y^\dagger a_x^\dagger |0 \rangle.$$ But the spatial indices $x$ and $y$ are just like any other kind of index; in reality we should have indices for flavor, color, and spin as well. Then by the same logic, the overall antisymmetry of the flavor/color/spin/space wavefunction follows from the fact that any fermionic creation operators anticommute.


So, why we don't antisymmetrize the meson wavefunctions? Consider a box of 10 electrons and an 11th electron a light year away. The wavefunction for all 11 electrons must be antisymmetrized by the general argument above, but nothing changes if we only antisymmetrize the first 10. Concretely, that's because the only observable thing the antisymmetrization does is add an exchange force, and the 11th electron will never be close enough to feel it. Conceptually, we can say the 11th electron is distinguishable by virtue of its position -- it's 'the one that's far away'.

Similarly, if we have an atom with 10 spin up electrons and 1 spin down electron, we can treat the spin down electron as separate from the antisymmetrized wavefunction of the other 10, distinguishable by its spin. The true wavefunction of all 11 is still fully antisymmetric by the argument given above, but nothing goes wrong here if we forget that; we won't run afoul of the Pauli exclusion principle by accident.

Now we turn to mesons and baryons.

  • In the case of mesons, the quarks are always effectively distinguishable because only one of them is an antiparticle, or equivalently, only one has anticolor.
  • For baryons with all three quarks distinct, the same reasoning holds; the particles are effectively distinguishable by their flavors. For example, consider $uds$ baryons. If we treat the particles as distinguishable, there are $2^3 = 8$ spin assignments, and accordingly there are $8$ low-energy $uds$ baryons. (They're the four spin states of the ${\Sigma^*}^0$ and the two spin states of the $\Sigma^0$ and $\Lambda$.)
  • For baryons with some quarks the same, none of the quarks are effectively distinguishable; we must account for the full antisymmetry. For example, there are only $4$ baryons with quark content $uuu$, not $8$. (They're the four spin states of the $\Delta^{++}$.) Similarly there are only $6$ with quark content $uud$. (They're the four spin states of the $\Delta^+$ and the two of the proton.)

So the particles in a meson are always effectively distinguishable, but the particles in a baryon are only sometimes. It's more economical to treat all the baryons the same way, so that's where the rule to 'antisymmetrize baryons but not mesons' comes from.

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