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Help with Heisenberg's "isospin" Hamiltonian

  1. Nov 6, 2014 #1
    I'm having some trouble grasping the meaning of the exchange term in the Hamiltonian Heisenberg gives in his classic 1932 paper (the one typically given as the first to describe nucleons via a spin-like degree of freedom; NOTE: I realize this isn't the same as what is today called isospin, but for lack of a better term that's what I'm going to call it). Since it's not widely available in English, here's the relevant bit (from D. M. Brink's Nuclear Forces):

    Now, up to a point, I perfectly understand it; he gives the number of protons as [itex]1/2 \sum_i(1 - \rho^{\zeta}_i) [/itex], the number of neutrons as [itex]1/2 \sum_i(1 + \rho^{\zeta}_i) [/itex], and a nucleus with equal numbers of protons and neutrons has [itex]\sum_i(\rho^{\zeta}_i) = 0[/itex] (assuming the "1" is actually the identity matrix, or that ##\rho^{zeta}## here refers to the eigenvalue rather than the matrix). He doesn't state it, but the ##\rho##s act on a two-component (iso-)spin vector, where the normalized eigenvectors of [itex]\rho^{zeta}[/itex] are [(1, 0)T and (0, 1)T] with eigenvalues [1, -1] respectively, so (1, 0)T would be a state that's definitely a neutron and (0, 1)T would be one that's definitely a proton, and linear combinations of them giving mixed states.

    However, where things start to get murky for me are when the other "axes" come in; I can see that the eigenvectors of the other ##\rho##'s both have components whose absolute value is ##\sqrt(2)/2## (e.g. equal odds of being a proton or a neutron), which also makes sense -- if they have a definite "isospin" along the [itex]\xi[/itex] "axis", then the value along the [itex]\zeta[/itex] axis will be completely unknown. But, did he come up with these? He says that it's helpful to introduce them because of the exchange, but I'm failing to see how that follows. It seems like the operator that would swap a vector ([a, b, c, d]) to ([c, d, a, b]) would be [itex]\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0&0&0&1 \\ 1 &0& 0& 0 \\ 0 & 1& 0& 0 \end{pmatrix}[/itex], and I don't see how that can come from the "exchange operator" in the above Hamiltonian (##1/2 \sum J(r_{kl})(\rho^{\xi}_k\rho^{\xi}_l + \rho^{\eta}_k\rho^{\eta}_l)##) (I tried kronecker products of the 2x2 matrices and expanding them into 4x4 block matrices and using normal matrix multiplication, neither of which gave anything looking like an exchange operator)...

    Further, I'm not even sure what the eigenvalues of the non-##\zeta## ##\rho##operators would even mean; the ##\rho^{zeta}_k##'s tells you whether the ##k^{th}## is a proton or a neutron, but what does the ##\rho^{xi}_k## tell you?
     
  2. jcsd
  3. Nov 12, 2014 #2
    Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Nov 29, 2014 #3
    Thanks for the encouragement; I think the problem stems from my incomplete understanding of spin, particularly with regards to multiple particles. I've since found a few references where they show that the two-particle spin operators are I⊗σ and σ⊗I, as opposed to the naive versions I tried (e.g. diag(σ, I) and diag(I, σ)); I still don't quite follow *why* mime were incorrect, but at least now I know where to look for the answers.
     
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