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Direct product of space and spin in hadrons 
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#1
Jul2914, 05:47 PM

P: 231

The uuu hadron doesn't violate Pauli's exclusion principle presumably because there is color.
But even without color, can't the uuu exist if spatial wavefunctions are different? Suppose one u quark is located at r_{1}, another at r_{2}, and another at r_{3}, and say that all three u quarks have spin up. Then you can construct the antisymmetric wavefunction: $$(r_1 r_2 r_3 \rangle  r_2 r_1 r_3 \rangle+...)\otimes \uparrow \uparrow \uparrow \rangle \otimes  u u u \rangle $$ where r_{1}, r_{2}, and r_{3} are permuted with negative sign if the permutation is odd. If you swap any two particles, you get a negative sign. Also in general, must we assume that the wavefunction is a direct product? This seems to say that spin is uncorrelated with flavor. If a baryon is made of a u, d, and s quark, why must each of these different quarks have the same amplitude of being found spin up or spin down (as implied by a direct product)? 


#2
Jul3014, 09:08 AM

P: 328

If you exclude color, the quantum numbers of each quark are spin up and uflavor, which means that the wave function for the ith quark is [itex]\psi_{\uparrow, u}(r_i)[/itex]. This means that, no matter of the position of the quarks the wave function is always going to be:
$$ \psi_{\uparrow, u}(r_1)\psi_{\uparrow, u}(r_2)\psi_{\uparrow, u}(r_3). $$ You can't symmetrize/antisymmetrize anything (if you don't admit the presence of color). 


#3
Jul3014, 11:21 AM

P: 1,023

Also in general [...] :
Because that's the most natural thing to say. Why would being an u quark imply which spin polarization it'll have? The same then for d and s... 


#4
Jul3014, 07:28 PM

P: 231

Direct product of space and spin in hadrons
$$ \psi(r_1,r_2,r_3)=r_1 r_2^2r_2 r_1^2+r_2r^2_3r_3r_2^2+r_3r_1^2r_1r_3^2 $$ It is already antisymmetric in r_{1}, r_{2}, r_{3}. That is, the probability amplitude for finding a spin up, uquark, at locations r_{1}, r_{2}, r_{3}, is given by the above probability amplitude, which is antisymmetric when you swap two of the quarks. Such wavefunctions might seem odd, but adding color is a drastic step. 


#5
Jul3014, 07:51 PM

P: 672

The requirement to be able to (completely) antisymmetrize the spatial part of the wave function is simply that all of the quarks are in different eigenstates of the system. This means more energy (they cannot all be in the lowest energy state) and thus larger mass (and typically shorter lifetime).



#6
Jul3014, 11:09 PM

P: 231

But if you take the analogy of atomic physics, wouldn't it be like saying that there must be an additional quantum number in addition to the spin, so that lithium can have all three electrons in a 1s state? You have Fermi particles, so they can't all occupy lowest single particle states, but you introduce color to do so. This seems ad hoc. 


#7
Jul3114, 04:27 AM

P: 672

But Lithium does not occur with all three electrons in the 1s state for precisely this reason. The logic is that if you observe the uuu baryon with spin 3/2 and all valence quarks in the lowest spatial energy state, then there must be an additional quantum number  not the other way around.



#8
Jul3114, 04:36 AM

P: 1,023

Also the existence of uuu hadron you are talking about, wasn't the experimental proof of the existence of color. It was just an evidence...



#9
Jul3114, 04:42 AM

P: 672

I am not sure that is a good formulation. You never prove anything in empirical sciences ... You collect evidence and reject theories that do not make correct predictions.



#10
Jul3114, 10:26 PM

P: 231




#11
Aug114, 01:01 AM

P: 231

the proton wavefunction seems to say that flavor and spin are a bit entangled. That is, assuming the color part is antisymmetric, and the spatial wavefunction is symmetric, the proton has a wavefunction: $$\left[\phi(M_s)\chi(M_s)+\phi(M_a)\chi(M_a)\right] \otimes \psi(r_1)\psi(r_2)\psi(r_3) \otimes RGB_a$$ where [itex]\psi[/itex] is the spatial wavefunction of the ground state (and this is symmetric), and RGB are the colors antisymmetrized. [itex]\phi[/itex] is the flavor/isospin and [itex]\chi[/itex] is the spin, and [itex]M_s[/itex] stands for mixed symmetry in the first two particle indices while [itex]M_a[/itex] stands for mixed antisymmetry in the first two particle indices. So it seems if you measure the spin of your 3 quarks and determine it's [itex]\chi(M_s)[/itex], then your flavor wavefunction is [itex]\phi(M_s)[/itex], so in some sense flavor does determine spin and vice versa. 


#12
Aug114, 04:09 AM

P: 1,023

No there is no entanglement due to that.
There you still don't know what spin goes to which flavor. For example if you tell me that you found x to be up,up,down spin then what do you know about phi? suppose you have uds quarks... you can assign the spins to each quark... 


#13
Aug114, 04:59 AM

P: 672




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