- #1

- 164

- 2

**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)In order to write down the Hamiltonian function of the nucleus the following variables are useful: Each particle in the nucleus is characterized by five quantities: the three position-coordinates (x,y,z) = r, the spin σ^{z}along the z-axis and a fifth number ρ^{ς}which can be +/- 1. ρ^{ς}= +1 means the particle is a neutron, ρ^{ς}= -1 means the particle is a proton. Because of the exchange there appear in the Hamiltonian transitional elements changing ρ^{ς}= +1 to ρ^{ς}= -1 and it is useful to introduce the following matrices

[tex]

\rho^{ξ} = \left( \begin{array}{cc} 0 & 1\\ 1 & 0 \end{array} \right)

\rho^{η} = \left( \begin{array}{cc} 0 & -i\\ i & 0 \end{array} \right)

\rho^{ς} = \left( \begin{array}{cc} 1 & 0\\ 0 & -1 \end{array} \right)

[/tex]

The space [itex](\xi, \eta, \zeta)[/itex] is, of course, not the ordinary space.

In terms of these variables the complete Hamiltonian function of the nuclei (M proton-mass, ##r_{kl} \rightarrow |r_k-r_l|##, ##p_k## momentum of particle k) is

[itex]

\begin{align}

H &= 1/2M \sum_k p_k^2 \\

&- 1/2 \sum J(r_{kl})(\rho^{\xi}_k\rho^{\xi}_l + \rho^{\eta}_k\rho^{\eta}_l) \\

&+ 1/4 \sum_{k>l}K(r_{kl})(1 + \rho^\zeta_k)(1 + \rho^\zeta_l) \\

&+ 1/4 \sum_{k>l}e^2/r_{kl}(1 - \rho^\zeta_k)(1 - \rho^\zeta_l) \\

&- 1/2 D \sum_k (1+\rho^\zeta_k)

\end{align}

[/itex]

^{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?