- #1
RedX
- 970
- 3
I still don't understand antisymmetry and fermions.
Is the proton wavefunction equal to this:
[tex]|\psi_p>=\frac{1}{\sqrt{6}}\left(2|u\uparrow u\uparrow d \downarrow \rangle -
|u\uparrow u\downarrow d \uparrow \rangle -
|u\downarrow u\uparrow d \uparrow \rangle \right)[/tex]
or this:
[tex]\frac{1}{\sqrt{18}}\left(2|u\uparrow u\uparrow d \downarrow \rangle
+2|u\uparrow d \downarrow u\uparrow \rangle
+2|d \downarrow u\uparrow u\uparrow \rangle
-
|u\uparrow u\downarrow d \uparrow \rangle
-
|u\downarrow u\uparrow d \uparrow \rangle
-
|u\uparrow d \uparrow u\downarrow \rangle
-
|u\downarrow d \uparrow u\uparrow \rangle
-
|d \uparrow u\uparrow u\downarrow \rangle
-
|d \uparrow u\downarrow u\uparrow \rangle
\right)[/tex]
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: [tex]\frac{1}{\sqrt{2}}\left(| \uparrow \downarrow \rangle- \downarrow \uparrow \rangle \right)[/tex].
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.
Is the proton wavefunction equal to this:
[tex]|\psi_p>=\frac{1}{\sqrt{6}}\left(2|u\uparrow u\uparrow d \downarrow \rangle -
|u\uparrow u\downarrow d \uparrow \rangle -
|u\downarrow u\uparrow d \uparrow \rangle \right)[/tex]
or this:
[tex]\frac{1}{\sqrt{18}}\left(2|u\uparrow u\uparrow d \downarrow \rangle
+2|u\uparrow d \downarrow u\uparrow \rangle
+2|d \downarrow u\uparrow u\uparrow \rangle
-
|u\uparrow u\downarrow d \uparrow \rangle
-
|u\downarrow u\uparrow d \uparrow \rangle
-
|u\uparrow d \uparrow u\downarrow \rangle
-
|u\downarrow d \uparrow u\uparrow \rangle
-
|d \uparrow u\uparrow u\downarrow \rangle
-
|d \uparrow u\downarrow u\uparrow \rangle
\right)[/tex]
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: [tex]\frac{1}{\sqrt{2}}\left(| \uparrow \downarrow \rangle- \downarrow \uparrow \rangle \right)[/tex].
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.