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Particle-wave function

  1. Jul 23, 2008 #1

    again a question, this time concerning the wave function of particles. I am currently working through the Berger.

    1) Why is the position-wavefunction always symmetric?

    2) It is said that no 1/2+ [TEX]\left|uuu\right\rangle[/TEX] is allowed because this flavor-wavefunction is symmetric as is the position-wavefunction. The color-singlett is antisymmetric and since a baryon is a fermion, the spin-wavefunction would have to be symmetric. This is the case for 3/2 but not for 1/2. That is true. However, the spin-wavefunction for 1/2 is mixed symmetric, e.g.

    [TEX]\frac{1}{\sqrt 2}\left|\uparrow\right\rangle\left(\left|\uparrow\downarrow\right\rangle - \left|\downarrow\uparrow\right\rangle \right)[/TEX].

    What prevents me to just symmetrize the spinwave-function to get an overall antisymmetric wavefunction? In similar other cases this is done as well so why can't we do that here?

    3) Berger also gives an explicit expression for the Rho+-Meson:

    [TEX]\left|\rho^+\right\rangle = \frac{1}{\sqrt 3}\left(\left|R\bar R\right\rangle + \left|B\bar B\right\rangle + \left|G\bar G\right\rangle\right)\frac{1}{\sqrt 2}\left(\left|\u\bar d right\rangle - \left|\bar d u\right\rangle\right)\frac{1}{\sqrt 2}\left(\left|\uparrow\downarrow\right\rangle + \left|\downarrow\uparrow\right\rangle\right)\frac{\leftR\left(r\right)\right\rangle}{\sqrt{4\pi}}[/TEX].

    A meson is a boson, but isn't this wavefunction antisymmetric? He develops this formula by taking into account the G-parity - but for me this is antisymmetric. If I change the places of the two quarks, the flavor-wavefunction gives a minus, doesn't it?

    A big thanks beforehand! Every answer to any of the three questions above would help me a lot!

  2. jcsd
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