Spin of Particle: Charge & Neutrality Effects

  • Thread starter Thread starter shakeel
  • Start date Start date
  • Tags Tags
    Particle Spin
shakeel
Messages
23
Reaction score
0
how the spin is associated with a particle. is there a difference in spin for charge particle and a neutral particle, i mean difference in spin due to charge or spin is independent of charge. what is procedure to assign spin to a particle? p help.
 
Physics news on Phys.org
The procedure which introduces spin formally, requires you to know some group theory : irreducible representations (IR) of a symmetry group etc. Do you know this ? If not, answering your question is going to be difficult.

In short : spin is assigned because of symmetry. More specifically, in QM the following concepts are invariant under spatial rotations:

1) the normalization of the wavefunction
2) the expectation value of any observable
3) the Hamiltonian



For this symmetry to be respected, the wavefunctions need to "behave" in a certain way under such rotations. Mathematically one says : the wavefunctions need to transform under certain IR of the rotational symmetry group.

One set of wavefunctions that respects this demand are the spinors. They are defined based upon their behaviour under rotatios. If A is a spinor and you rotate it 360 degress you get -A.

To read the full story, go check https://www.physicsforums.com/blogs/marlon-13790/what-is-spin-152/

marlon
 
Last edited by a moderator:
Thread 'Why is there such a difference between the total cross-section data? (simulation vs. experiment)'
Well, I'm simulating a neutron-proton scattering phase shift. The equation that I solve numerically is the Phase function method and is $$ \frac{d}{dr}[\delta_{i+1}] = \frac{2\mu}{\hbar^2}\frac{V(r)}{k^2}\sin(kr + \delta_i)$$ ##\delta_i## is the phase shift for triplet and singlet state, ##\mu## is the reduced mass for neutron-proton, ##k=\sqrt{2\mu E_{cm}/\hbar^2}## is the wave number and ##V(r)## is the potential of interaction like Yukawa, Wood-Saxon, Square well potential, etc. I first...
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Back
Top