I hope someone (may be you) can realize it. ;)

]]>Thank you very much! Glad that this word-press not just helping my poor memory and also other people!! Have a good day.

]]>I ended up switching my major from physics after changing schools, but I’ve been meaning to try learning physics on my own because I miss it. I’m glad I found this reading list.

Cheers

]]>I am sorry, I did not know. But bear in mind that isospin is an abstract idea and it may be not “real”. Isospin help us think, but Isospin can never be measured. And there are many isospin breaking examples.

]]>Hi Erik,

For K^\pi = 0+ states, I think the J can be even or odd. not really related. because J formed by orbital and spin angular momentum. And only the orbital angular momentum tells you the parity.

For the parity of the rotation wave function, you can check by applying the Wigner-D parity transform rule ( stated in the post ).

For the |N n_z m_l K> state, if we look at the solution on the page https://nukephysik101.wordpress.com/2018/01/23/axial-harmonic-oscillator-nilsson-orbit/

The solution is a function of (z, \rho, \phi) , parity transform rule is

(z, \rho, \phi) –> ( -z , \rho, \phi+\pi)

so, the term exp(i m_l \phi) is like the spherical harmonic that, it introduce (-1)^(m_l) for aparity transform.

For the Hermite polynomial, when we replace z->(-z), it introduce (-1)^n_z for parity transform. ( or you can check wiki page and see that the Hermite polynomial has parity depends on n_z.

Thus, in total, the |N n_z m_l K> will have parity of (-1) ^( n_z +m_l ).

However, the N, n_z, m_l, and K are not independent, they obey some rules. https://nukephysik101.wordpress.com/2019/07/26/quantum-number-of-nilsson-orbital/

For N = even (odd) , n_z+m_l = even (odd), so, overall, the rotation Lab frame wave function preserved its own parity.

In fact, when I expand the Nilsson orbital into spherical Harmonic. For N=odd (even), only odd (even) parity spherical Harmonic matters.

The Key point is that, in the term ( D_K +- D_{-k} ) keep the parity. And the rest depends on the nature of Nilsson orbital. Since the wave function must obey parity, thus, |J M K > only mixed with even or odd Nilsson orbital.

I hope this is the answer you seek. But feel free to ask if I misunderstand your question.

Ryan

]]>i have a question about the parity of the Lab-frame wavefunction. The parity of this wavefunction is determined by the + or – between the two Wigner-D-matrixelements. Why, for example, do K(pi)=0(+) states must have a even J quantumnumber. If i transform the Lab-frame wave function with a parity transformation by transforming the Wigner-D’s, only the + or – between the two Wigner-D’s determine the parity of this wavefunction. The J and K quantumnumbers do not influence the parity transformation, the two Wigner-D terms just swap their positions.

What am i missing? Do the |N n_z m_l K> states influence the parity?

Kind regards

Erik

Yes, by definition, but how about in the reality?

]]>[Lz, Sz] = (Lz⊗1)(1⊗Sz) – (1⊗Sz)(Lz⊗1) = 0⊗0

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