Particle-Wave Function: Questions & Answers

[blue2script]

Hello,

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!

Blue2script

 

[eljose79]

Dear Blue2script,

Thank you for your questions regarding the wave function of particles. I will do my best to address each of your questions in turn.

1) The position-wavefunction is always symmetric because it describes the probability of finding a particle at a given position. In quantum mechanics, particles are described as waves and the probability of finding the particle at a certain position is determined by the amplitude of the wave at that position. Since the amplitude of a wave is always symmetric, the position-wavefunction must also be symmetric.

2) The reason that no 1/2+ \left|uuu\right\rangle is allowed is due to the Pauli exclusion principle. This principle states that no two fermions can occupy the same quantum state simultaneously. In the case of the 1/2+ \left|uuu\right\rangle, the flavor-wavefunction and the spin-wavefunction are both symmetric, which violates the Pauli exclusion principle. This is why the spin-wavefunction for 1/2 is mixed symmetric, as it allows for the overall wavefunction to be antisymmetric and satisfy the Pauli exclusion principle.

3) The wavefunction for the Rho+-Meson is indeed antisymmetric. The G-parity, which is a symmetry operation in particle physics, is used to ensure that the overall wavefunction is antisymmetric. This is necessary for the Rho+-Meson to be a boson, as bosons have symmetric wavefunctions. The minus sign that you mention when changing the places of the two quarks is due to the flavor-wavefunction, which is antisymmetric.

I hope that helps to answer your questions. If you have any further questions or need clarification, please don't hesitate to ask. Keep up the good work with your studies!

 

[denian]

,

1) The position-wavefunction is always symmetric because in quantum mechanics, the position of a particle is a continuous variable and it can take on any value within a given range. This means that the probability of finding the particle at any position is equal, leading to a symmetric wavefunction.

2) In quantum mechanics, the overall wavefunction of a system must be antisymmetric for fermions due to the Pauli exclusion principle. This means that the total wavefunction must change sign when the positions of any two particles are exchanged. In the case of 1/2+ baryons, the flavor-wavefunction and the position-wavefunction are both symmetric, so the spin-wavefunction must be antisymmetric to satisfy the overall antisymmetry requirement. Symmetrizing the spin-wavefunction would violate this requirement and lead to an incorrect wavefunction for a fermion system.

3) The wavefunction given for the Rho+-Meson is indeed antisymmetric. This is because the overall wavefunction must be antisymmetric for a boson system as well, unlike in the case of fermions. The G-parity is a symmetry operation that exchanges particles with their antiparticles, and this is taken into account when constructing the wavefunction. The minus sign in the flavor-wavefunction comes from the exchange of particles with their antiparticles, leading to an overall antisymmetric wavefunction.

I hope this helps clarify your questions. Keep exploring and asking questions!