7
$\begingroup$

I am told that the most general form of a spin rotation invariant Hamiltonian for two systems 1 and 2 both with spin $S$, i.e., the spin operators

\begin{align} (\hat{S}_1^x)^2 +(\hat{S}_1^y)^2 + (\hat{S}_1^z)^2 = (\hat{S}_2^x)^2 +(\hat{S}_2^y)^2 + (\hat{S}_2^z)^2 = S(S+1)\hbar^2 \end{align} is given by

\begin{equation} \mathcal{H} = \sum_{j=0}^{2S} a_j \bigg(\frac{\mathbf{\hat{S}_1}\cdot\mathbf{\hat{S}_2}}{\hbar}\bigg)^j \end{equation}

I understand that it should be a function of $\mathbf{\hat{S}_1}\cdot\mathbf{\hat{S}_2}$ but I do not understand why should the sum terminate at $j=2S$. Can someone explain it to me. Thank you.

$\endgroup$

1 Answer 1

12
$\begingroup$

$\newcommand{\bm}[1]{\mathbf{#1}}$ You need to look at this in terms of spin quantum numbers (i.e., eigenvalues).

$(\bm S_1+\bm S_2)$ can take values $S_\mathrm{tot} = 0,\dots,2S$. Now if we restrict to the subspace with total spin $S_\mathrm{tot}$, we have that \begin{equation} \begin{aligned} 2S_\mathrm{tot}(2S_\mathrm{tot}+1) = (\bm S_1+\bm S_2)^2 &= \bm S_1\cdot \bm S_1 + \bm S_2\cdot \bm S_2 + 2\, \bm S_1\cdot \bm S_2 \\ &= S(S+1) + S(S+1) + 2\,\bm S_1\cdot \bm S_2\ , \end{aligned} \end{equation} and thus $$\bm S_1\cdot \bm S_2 = S_\mathrm{tot}(2S_\mathrm{tot}+1)-S(S+1) \tag{1} $$ can take $2S+1$ possible values. (Note that this means that $\bm S_1\cdot \bm S_2$ and $\bm S_1+\bm S_2$ are diagonal in the same basis, that is, we can reason about them as if they were just numbers which can take the corresponding set of values.)

A $\mathrm{SU}(2)$ invariant Hamiltonian of the two spins will take a different value of each subspace of total spin $S_\mathrm{tot}$, i.e., it is of the form (with $\Pi_{S_\mathrm{tot}}$ the projector onto the subspace with total spin $S_\mathrm{tot}$) $$ \mathcal H = \sum_{S_\mathrm{tot}=0}^{2S} E_{S_\mathrm{tot}} \Pi_{S_\mathrm{tot}}\ . \tag{2} $$ Since there is a one-to-one relation between the total spin and the value of $\bm S_1\cdot \bm S_2$ -- Eq (1) --, each projector $\Pi_{S_\mathrm{tot}}$ can be expressed as a function of $\bm S_1\cdot \bm S_2$: $$\Pi_{S_\mathrm{tot}}=f_{S_\mathrm{tot}}(\bm S_1\cdot \bm S_2)\ . \tag{3} $$ This function must be $f_{S_\mathrm{tot}}(\bm S_1\cdot \bm S_2)=1$ for the desired $S_\mathrm{tot}$ (using (1)), and $f(\bm S_1\cdot \bm S_2)=0$ for all other values the total spin can take (again using (1)). This means that we only need to fix the value of $f$ at $2S+1$ points, and thus, it can be choosen to be a polynomial of degree $2S$: $$ f_{S_\mathrm{tot}}(\bm S_1\cdot \bm S_2) = \sum_{j=0}^{2S} a_{j,S_\mathrm{tot}} (\bm S_1\cdot \bm S_2)^j\ . $$ Substituting this into (3) and then into (2) gives that $$ \mathcal H = \sum_{j=0}^{2S} a_{j} (\bm S_1\cdot \bm S_2)^j\ . $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.