1
$\begingroup$

I learned that for the lowest lying states,

  • $uuu$, $sss$, $ddd$ can only have $J=\frac{3}{2}$
  • $uus$, $ddu$, $ddu$, $ssu$ etc can only have $J=\frac{1}{2},\frac{3}{2}$
  • $uds$ can have $J=\frac{1}{2}, \frac{1}{2},\frac{3}{2}$

where $J$ is the spin.

I also learned that the above results are deduced from requiring the color-flavor-spin-obirtal wavefunction of the state to be antisymmetric (by Pauli's exclusion principle). However, I still find myself confused about how to deduce the allowed $J$ values for each of the three cases.

What is the systematic way to deduce the spin of a given lowest lying baryon state?

$\endgroup$
2
$\begingroup$

You should be able to answer your question directly by directly inspecting the baryon wavefunctions in your textbook. They are the unique solution to the constraints you mention, so, since they are color singlets, they must be color antisymmetric: consequently they are spin-flavor-space symmetric. Lowest mass baryons are in an S state, parity +, so, symmetric: your problem reduces to mapping out all flavor-spin symmetric states.

You are studying flavor SU(3), and (by inspection of the evident Young tableaux) three triplets (quarks) of SU(3) combine to a Symmetric decuplet 10; an Antisymmetric singlet 1; and two octets 8 of mixed symmetry. The symmetric combination of three spin 1/2 doublets is the spin 3/2 quartet, while the doublets are always of mixed symmetry.

Review them here.

Within a spin multiplet or flavor multiplet, raising and lowering spin, Isospin, U-spin, and V-spin operators interchanging quark spins and flavors will keep you inside the multiplet and so will not change its symmetry. So, as always, you start with the extreme multiplet states, and you ladder yourself around the combined multiplet.

uuu is flavor symmetric, so it must be in the flavor decouplet, and so it must also be spin Symmetric (for total symmetry), and so must be in the spin quartet: spin 3/2.

It is the $\Delta^{++}$ in the diagram, and the flavor ladder operators can move it to its spin 3/2 decuplet confreres $\Delta^-$ and $\Omega^-$. Note the flavor ladder operators will also move it to $\Delta^+,\Sigma^{*+},\Sigma^{*-}$, in your second rung, also spin 3/2; and, further, $\Sigma^{*0}$ in your 3rd rung, also spin 3/2. All spin 3/2 states are now accounted for, and everything else remaining must be orthogonal to them, and so must be spin 1/2. For example, lowering the $S^z$ of the $\Delta^+$ state, we evidently get $$ |\Delta^+_\uparrow\rangle= \frac{1}{3} [ | u_\uparrow d_\downarrow u_\uparrow \rangle + | u_\uparrow u_\uparrow d_\downarrow \rangle +| d_\downarrow u_\uparrow u_\uparrow \rangle \\ + | u_\uparrow u_\downarrow d_\uparrow\rangle +| u_\uparrow d_\uparrow u_\downarrow\rangle +| u_\downarrow d_\uparrow u_\uparrow\rangle +| d_\uparrow u_\downarrow u_\uparrow\rangle +| d_\uparrow u_\uparrow u_\downarrow\rangle +| u_\downarrow u_\uparrow d_\uparrow\rangle ]. $$

We are left with mixed-symmetry octets and the flavor antisymmetric singlet, all necessarily spin 1/2, since they must be orthogonal to the spin 3/2 states. I frankly don't have a slick way to separate the octets, but it is straightforward to verify the orthogonality of the proton wavefunction, below, to the $\Delta^+_\uparrow$ of the previous paragraph, $$ |p_\uparrow\rangle= \frac{1}{\sqrt {18}} [ 2| u_\uparrow d_\downarrow u_\uparrow \rangle + 2| u_\uparrow u_\uparrow d_\downarrow \rangle +2| d_\downarrow u_\uparrow u_\uparrow \rangle \\ - | u_\uparrow u_\downarrow d_\uparrow\rangle -| u_\uparrow d_\uparrow u_\downarrow\rangle -| u_\downarrow d_\uparrow u_\uparrow\rangle -| d_\uparrow u_\downarrow u_\uparrow\rangle -| d_\uparrow u_\uparrow u_\downarrow\rangle -| u_\downarrow u_\uparrow d_\uparrow\rangle ]. $$ This one is fully symmetric, even though its flavor and spin parts are of mixed symmetry, individually! You now get to its confreres in the spin 1/2 flavor octet through flavor and spin ladder operators. In particular, you may also further descend to the octet spin 1/2 uds, called $\Sigma^0$, on your 3rd rung.

Finally, a state flavor-orthogonal to it in the same flavor octet is the iso-singlet spin-1/2 $\Lambda$ baryon, $$ |\Lambda_\uparrow\rangle= \frac{1}{\sqrt{12}}[u_\uparrow d_\downarrow s_\uparrow -u_\downarrow d_\uparrow s_\uparrow -d_\uparrow u_\downarrow s_\uparrow +d_\downarrow u_\uparrow s_\uparrow + \hbox{perms}]. $$ Marvel at its full symmetry, despite its flavor and spin mixed-symmetries.

This has now populated your three rungs. In summary, the states discussed above are a spin quartet of flavor decuplet, and a spin doublet of flavor octet, 56 states in all, (fitting into the 56 symmetric representation of SU(6), but that is another story, yet...)

You may now further inspect the rarer different parity P-wave (L=1) states of higher energy, but I doubt you are intrigued by them at this stage.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.