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Direct product of space and spin in hadrons

by geoduck
Tags: hadrons, product, space, spin
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geoduck
#1
Jul29-14, 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 r1, another at r2, and another at r3, 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 r1, r2, and r3 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)?
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Einj
#2
Jul30-14, 09:08 AM
P: 328
If you exclude color, the quantum numbers of each quark are spin up and u-flavor, which means that the wave function for the i-th 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/anti-symmetrize anything (if you don't admit the presence of color).
ChrisVer
#3
Jul30-14, 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...

geoduck
#4
Jul30-14, 07:28 PM
P: 231
Direct product of space and spin in hadrons

Quote Quote by Einj View Post
If you exclude color, the quantum numbers of each quark are spin up and u-flavor, which means that the wave function for the i-th 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/anti-symmetrize anything (if you don't admit the presence of color).
I was thinking a spatial wavefunction along the lines of:
$$
\psi(r_1,r_2,r_3)=r_1 r_2^2-r_2 r_1^2+r_2r^2_3-r_3r_2^2+r_3r_1^2-r_1r_3^2
$$
It is already antisymmetric in r1, r2, r3.

That is, the probability amplitude for finding a spin up, u-quark, at locations r1, r2, r3, 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.
Orodruin
#5
Jul30-14, 07:51 PM
P: 672
The requirement to be able to (completely) anti-symmetrize 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 life-time).
geoduck
#6
Jul30-14, 11:09 PM
P: 231
Quote Quote by Orodruin View Post
The requirement to be able to (completely) anti-symmetrize 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 life-time).
It's true that color allows all quarks to be in the lowest one-particle eigenstate.

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.
Orodruin
#7
Jul31-14, 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.
ChrisVer
#8
Jul31-14, 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...
Orodruin
#9
Jul31-14, 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.
geoduck
#10
Jul31-14, 10:26 PM
P: 231
Quote Quote by Orodruin View Post
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.
Do we really observe the energy state of each individual quark in uuu? I thought we can only see the composite particle and not the individual constituents, so a uuu spin 3/2 in an excited state looks like a different baryon. In that case there can be a spin 3/2 uuu baryon with the lowest energy that comprises antisymmetrizing 3 different energy eigenstates, and we would see it as the lowest energy baryon, but this would not imply that all u quarks are in the lowest 1-particle energy eigenstates (which is automatically symmetric and hence needs color).
geoduck
#11
Aug1-14, 01:01 AM
P: 231
Quote Quote by ChrisVer View Post
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...
From what I understand (for example these notes, slide 222: http://www.hep.phy.cam.ac.uk/~thomso...out_7_2011.pdf )

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.
ChrisVer
#12
Aug1-14, 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...
Orodruin
#13
Aug1-14, 04:59 AM
P: 672
Quote Quote by geoduck View Post
Do we really observe the energy state of each individual quark in uuu? I thought we can only see the composite particle and not the individual constituents, so a uuu spin 3/2 in an excited state looks like a different baryon. In that case there can be a spin 3/2 uuu baryon with the lowest energy that comprises antisymmetrizing 3 different energy eigenstates, and we would see it as the lowest energy baryon, but this would not imply that all u quarks are in the lowest 1-particle energy eigenstates (which is automatically symmetric and hence needs color).
We do not observe the individual quarks, no. However, my point was just that - in order to have the mass corresponding to all quarks in the lowest energy state, the lowest energy state must be degenerate in order to allow anti-symmetrization. Hence the need for a new quantum number to accomodate this.


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