Proton Wavefunction: Antisymmetry & Fermions Explained

[RedX]

I still don't understand antisymmetry and fermions.

Is the proton wavefunction equal to this:

|\psi_p&gt;=\frac{1}{\sqrt{6}}\left(2|u\uparrow u\uparrow d \downarrow \rangle -<br /> |u\uparrow u\downarrow d \uparrow \rangle - <br /> |u\downarrow u\uparrow d \uparrow \rangle \right)

or this:

\frac{1}{\sqrt{18}}\left(2|u\uparrow u\uparrow d \downarrow \rangle <br /> +2|u\uparrow d \downarrow u\uparrow \rangle<br /> +2|d \downarrow u\uparrow u\uparrow \rangle<br /> -<br /> |u\uparrow u\downarrow d \uparrow \rangle <br /> - <br /> |u\downarrow u\uparrow d \uparrow \rangle<br /> -<br /> |u\uparrow d \uparrow u\downarrow \rangle<br /> -<br /> |u\downarrow d \uparrow u\uparrow \rangle<br /> -<br /> |d \uparrow u\uparrow u\downarrow \rangle<br /> -<br /> |d \uparrow u\downarrow u\uparrow \rangle<br /> <br /> \right)

The difference between the two is that in the former only the u quarks are symmetrized (i.e., you can swap the first two positions 1 and 2) and in the latter every position is symmetrized (i.e., you can swap 1 and 2, 2 and 3, and 1 and 3).

If u and d quarks are separate particles, then the first wavefunction should be correct, since we only demand symmetry on swapping positions 1 and 2, since the d quark occupies position 3 and is distinguishable from the u-quarks that occupy the first two positions.

If u and d are isospin quantum numbers, then the latter is the correct wavefunction, since all 3 particles in the proton would be the same (they just have different spin and isospin quantum numbers), and you would have the symmetrize the exchange of each position.

If you have an electron that is spin up, and another electron that is spin down, then they have different quantum numbers, but they are the still the same type of particle, so the spin state would have to be antisymmetrized: \frac{1}{\sqrt{2}}\left(| \uparrow \downarrow \rangle- \downarrow \uparrow \rangle \right).

Similarly if quark flavors/isospin are just different quantum numbers and there is only one type of quark that can take on different flavors/isospin, then you have to symmetrize all positions.

 

[Meir Achuz]

Either one is correct in the appropriate context, but the first one is simpler and more general. If you consider the u and d as different particles, then the first is correct. If you want to introduce the Ispin formalism, then the second can be used. But the Ispin formalism is not necessary here, because any properties of the proton follow from just using spin, and keeping the u and d distinct.

"Quark flavors/isospin are just different quantum numbers and there is only one type of quark" is an unnecessary abstraction that is often used, but particles with different masses are different particles.

 

[zincshow]

Similar but related question, what is the wavefunction of the delta baryon?

 

[Meir Achuz]

The Delta^++ (3/2/(3/2) is uuu[\uparrow\uparrow\uparrow. The other spin states are found with lowering operators.
The D^+(3/2,3/0) is uud[\uparrow\uparrow\uparrow].
The uud does not have to be further symmetrized.

 

[Meir Achuz]

The D^+(3/2,3/2) is uud[\uparrow\uparrow\uparrow].
The uud does not have to be further symmetrized.

 

[RedX]

The D^+(3/2,3/2) is uud[\uparrow\uparrow\uparrow].
The uud does not have to be further symmetrized.

Can you symmetrize the uud if you wanted to? Can it be:

|\Delta^+ \rangle =(|uud \rangle+|udu \rangle+|duu \rangle)\otimes |\uparrow \uparrow \uparrow \rangle instead?

This wavefunction has the nice property that it transforms in the SU(3) decuplet. With just uud, then it no longer transforms in the decuplet.

But isn't there really only one wavefunction? Is there no possible experimental way to distinguish the two different proton wavefunctions:

<br /> |\psi_p&gt;=\frac{1}{\sqrt{6}}\left(2|u\uparrow u\uparrow d \downarrow \rangle -<br /> |u\uparrow u\downarrow d \uparrow \rangle - <br /> |u\downarrow u\uparrow d \uparrow \rangle \right)<br /> and <br /> \frac{1}{\sqrt{18}}\left(2|u\uparrow u\uparrow d \downarrow \rangle <br /> +2|u\uparrow d \downarrow u\uparrow \rangle<br /> +2|d \downarrow u\uparrow u\uparrow \rangle<br /> -<br /> |u\uparrow u\downarrow d \uparrow \rangle <br /> - <br /> |u\downarrow u\uparrow d \uparrow \rangle<br /> -<br /> |u\uparrow d \uparrow u\downarrow \rangle<br /> -<br /> |u\downarrow d \uparrow u\uparrow \rangle<br /> -<br /> |d \uparrow u\uparrow u\downarrow \rangle<br /> -<br /> |d \uparrow u\downarrow u\uparrow \rangle<br /> <br /> \right)<br /> ?

I understand that both proton wavefunctions show that the probability that the d-quark is found with its spin down is twice as likely as finding it with its spin up. Is that all you can measure in an experiment? So there is no way of finding the real ground state proton wavefunction?

The concept of isospin is really confusing. You say that particles with different masses are different particles. But is mass really an intrinsic property, since it is the result of interaction with the Higgs? Are the properties of particles determined entirely by their free-field properties, so that electric charge doesn't determine the free-field properties since it is the result of an interaction with another field, the photon field? The only fundamental property that I can think of would be spin.

 

[Meir Achuz]

Can you symmetrize the uud if you wanted to? Can it be:

|\Delta^+ \rangle =(|uud \rangle+|udu \rangle+|duu \rangle)\otimes |\uparrow \uparrow \uparrow \rangle instead?

This wavefunction has the nice property that it transforms in the SU(3) decuplet. With just uud, then it no longer transforms in the decuplet.

You can use Ispin if you want and then you can use group theory, but it is not necessaary, and it adds complication. Without Ispin or SU(3),There are only ten possible light three quark states with spin 3/2, and only eight with spin 1/2.

But isn't there really only one wavefunction? Is there no possible experimental way to distinguish the two different proton wavefunctions:

They give identical experimental predictions, unless additional SU(3) symmetry is ssumed. If too much SU(3) symmetry is assumed, the
predictions don't all agree with experiment.[/QUOTE]

 

[RedX]

You can use Ispin if you want and then you can use group theory, but it is not necessaary, and it adds complication. Without Ispin or SU(3),There are only ten possible light three quark states with spin 3/2, and only eight with spin 1/2.

That's true, so what's the point of introducing SU(3) flavor/isospin symmetry?

Is it true that the strong Hamiltonian is to a good approximation just a function of the SU(3) flavor Casmir, H=H(\Sigma_{i=1}^{8} T_{i}^2)? Or can one only say it is to a good approximation just a function of the SU(2) flavor Casmir, H=H(\Sigma_{i=1}^{3} T_{i}^2)?

Anyways, the reason is because then it might actually be useful to categorize particles into SU(2) and SU(3) flavor/isospin multiplets, as then the strong interaction won't take particles out of their multiplets (\langle \mbox{i-multiplet}|H|\mbox{j-multiplet}\rangle=0 \mbox{ for } i\neq j). So for example a proton and neutron can turn into a sigma⁺ particle, as the sigma⁺ has the same SU(2) Casmir (total isospin), but do they have the same SU(3) Casmir? Maybe it's possible for a neutron and a proton to add to be in the same multiplet as just a single proton and neutron (the 8 representation), i.e., p+n belongs to 8, as does p and n, and sigma⁺?