Matrix element of two nucleon interaction

Question

I have a couple of questions regarding the following problem:

Let the interaction between two nucleons given by: $\hat{V}_{1,2}=\boldsymbol{\sigma}^{(1)}\boldsymbol{\sigma}^{(2)}$.

What I need help with is the following: Evaluate the following two-body matrix element in the coupled base: $$<l_1s_1l_2s_2;LS;JM_J|\hat{V}_{1,2}|l_1s_1l_2s_2;LS;JM_J>$$ being the angular momentum couplings: $\mathbf{J=L+S}$ with $\mathbf{L=l_1+l_2}$ and $\mathbf{S=s_1+s_2}$.

Now, what I have so far is: $$\hat{V}_{1,2}=\hat{\sigma}_x^{(1)}\hat{\sigma}_x^{(2)}+\hat{\sigma}_y^{(1)}\hat{\sigma}_y^{(2)}+\hat{\sigma}_z^{(1)}\hat{\sigma}_z^{(2)}$$ which I think only affects the coordinates $m_{s_1}$ and $m_{s_2}$, since the $\sigma$'s are the Pauli matrices and the states are $<s_1=1/2,m_{s_1}=1/2>=\begin{pmatrix} 1\\ 0 \end{pmatrix}$ and $<s_1=1/2,m_{s_1}=-1/2>=\begin{pmatrix} 0\\ 1 \end{pmatrix}$, but these coordinates ($m_{s_i}$) don't appear explicitly on the matrix element, so I don't know how the operator should act.

On the other hand, I know the only bound state is the proton-neutron pair, which has $J^{\pi}=1^+$, so the only possibilities are $S=1,L=0(\implies J=1)$ and $S=1,L=2(\implies J=1,2,3)$. This I guess implies the state $|l_1s_1l_2s_2;LS;JM_J>$ should be decomposed into two possible states, but I don't know how to do that, or how the operator $\hat{V}_{1,2}$ will act on the state in the end.

My first guess is that $\hat{V}_{1,2}$ doesn't affect the coordinates $l_1,s_1,l_2,s_2,L,S,J,M_J$ so that would mean the matrix element is 1, but I suppose that is not correct.

Could anyone help me on getting the concepts clear?

It parses out the method. You need to clarify the question; it appears, but fails to be stated, that it concerns the deuteron. — Cosmas Zachos, Oct 02 '23 at 21:16
I edited the question. What I need to do is evaluate the matrix element. The question only talks about two nucleons, but I inferred it had to be the deuteron since it’s the only nucleon-nucleon bound state, right? — cut, Oct 02 '23 at 21:20

Cosmas Zachos · Answer 1 · 2023-10-03T14:28:30.853

I am not sure I understand your possibly malformed question, but, ignoring the gaps, I rush to review basic facts, $$\hat{V}_{1,2}=\boldsymbol{\sigma}^{(1)}\cdot \boldsymbol{\sigma}^{(2)}= \tfrac{1}{2}\left ((\boldsymbol{\sigma}^{(1)}+\boldsymbol{\sigma}^{(2)})^2-\boldsymbol{(\sigma}^{(1)})^2-(\boldsymbol{\sigma}^{(2)})^2 \right).$$ Note this has the wrong sign w.r.t. the standard interaction in the link provided to the question, so your statement on its significance is discounted.

Assuming you are talking about the deuteron, the r.h.s. evaluates to $\tfrac{1}{2}(4\cdot 2 -3-3)=1$ for S=1, L=0, J=1, i.e., the $^3S_1$ parallel-spin ground state, isosinglet;
or else, $\tfrac{1}{2}(-3-3)=-3$ for S=0, L=1, J=1, of lower energy in this unconventional flipped sign potential, for $^1P_1$.

In more detail, $\mathbf{S=s_1+s_2}$, where $\mathbf{s_1}=\boldsymbol{\sigma}^{(1)}/2,$ etc, and $\mathbf{S}^2=S(S+1)$, yielding the above standard resolution.

So, you see your matrix element only depends on S, as L and J are completely blind to this "potential", as evaluated above.

(It looks like a homework problem to me, but perhaps there are unacknowledged dimensions here...)

Matrix element of two nucleon interaction

1 Answers1