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The overall parity of a proton is 1 because the parity of a quark is 1. How does this go together with the proton's wave function being anti-symmetric? Is the reason for the proton's wave function's anti-symmetry the fact that in $SU(3)_C$ you consider the $u,d,s$ quark flavors to be identical for the strong interaction?

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  • $\begingroup$ You would need at least two identical protons to talk about a wavefunction being symmetric (or antisymmetric) since the permutation group acts on identical particles. $\endgroup$ Mar 28, 2017 at 22:11
  • $\begingroup$ It does not seem to be so easy. You can have a look at slide 16 of these lecture notes: link $\endgroup$
    – MmeTautou
    Mar 29, 2017 at 6:29
  • $\begingroup$ There it says that because quarks are fermions the total wave function has to be symmetric. Can you now simply neglect the charge and isospin which is different for up and down quark for example? And why can protons then have a parity of 1? $\endgroup$
    – MmeTautou
    Mar 29, 2017 at 6:30
  • $\begingroup$ A very clear example: the $\Delta^{++}$ resonance. There you have three up-Quarks with parallel spin. That's what motivated the color quantum number. There you try to make sure you have an antisymmetric wave function and yet the resonance has a parity of 1? $\endgroup$
    – MmeTautou
    Mar 29, 2017 at 7:15
  • $\begingroup$ The proton is a baryonic state comprising three identical valence quarks $qqq$. It is in this sense the description of a proton is amenable to Fermi dirac statistics which on the level of the quantum state means an antisymmetric wavefunction. The spin, colour, flavour degrees of freedom pin down the state of each quark and are such that no two quarks have the same set of these quantum numbers. $\endgroup$
    – CAF
    Mar 29, 2017 at 10:40

2 Answers 2

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The spatial parity of the proton (and the quark) is +1 by convention. Fermion and antifermion have opposite spatial parity. So the antiproton has spatial parity -1, by the same convention. But it could equally well be the other way about, provided you're consistent.

The proton wave function is antisymmetric under permutation of labels, as @CosmasZachos says. So is the antiproton wave function. That's not a convention, that's the spin statistics theorem.

There is no direct link (or conflict) between spatial parity and permutation symmetry here.

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The spin-flavor part of the proton wavefunction, $$ |p_\uparrow\rangle= \frac{1}{\sqrt {18}} [ 2| u_\uparrow d_\downarrow u_\uparrow \rangle + 2| u_\uparrow u_\uparrow d_\downarrow \rangle +2| d_\downarrow u_\uparrow u_\uparrow \rangle \\ - | u_\uparrow u_\downarrow d_\uparrow\rangle -| u_\uparrow d_\uparrow u_\downarrow\rangle -| u_\downarrow d_\uparrow u_\uparrow\rangle -| d_\uparrow u_\downarrow u_\uparrow\rangle -| d_\uparrow u_\uparrow u_\downarrow\rangle -| u_\downarrow u_\uparrow d_\uparrow\rangle ] $$ is completely symmetric under interchange of any two quarks with each other; even though, individually, flavor and spin interchanges are of mixed symmetry: this is the cornerstone of the model, really.

The position (orbital angular momentum) wavefunction is then an S-wave, symmetric, so, given the total antisymmetrization by the color, normally omitted, the three fermion constituent quarks are in a fully antisymmetric state, as they should be.

Consequently, since S contributes + to the overall parity, this parity then turns out to be ++++ $\leadsto$ + for the overall parity of the proton.

The idea linking generalized Pauli antisymmetrization of fermions to parity amounts to appreciating that there is no P wave spatial component, so the parity of the nucleon is the product of the parities of the three constituents.

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