Consider the Hamiltonian:
$$H=D\bigg(S_z-\frac{1}{3}S(S+1)\bigg)$$
Where $S_z$ is the spin-$z$ operator (one half the Pauli matrix for a doublet state) and the matrix representation of $S$ is the unit-matrix times the spin of the system considered. Now my question is, is $S$ classified as an operator, like $S_z$ is? If that is so, then I guess the '$+1$' (the unit matrix) is classified as an operator as well?
The reason why I ask is that the typical spin states $|S\ M_S\rangle$ are eigenvectors of $S^2$, but not of $S$, so the one above cannot represent actual angular momentum.
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The identity is in fact an operator, but it's a trivial one. Every state is an eigenstate of the identity operator with eigenvalue one.
The $S$ in your hamiltonian is not an operator, that is just a number, but the combination $$S(S+1)\mathbb{1}$$ is an operator and, again, is a trivial operator since it's just the identity rescaled by some quantity $S(S+1)$. The eigenvalue of this new rescaled identity operator is just $S(S+1)$.
Since the eigenvalue of the operator $S^2$ are in fact $S(S+1)$ you'd be better rewriting your hamiltonian in the following manner $$\hat H=D\bigg(\hat{S}_z-\frac{1}{3}\hat{S}^2\bigg)$$ and now there's no doubt about the operator nature of the hamiltonian.
answered Jun 28, 2020 at 15:19
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$\begingroup$ You say that $S$ is just a number, but it is a number multiplied by the unit matrix, i.e. for a triplet state the matrix representation is: $S=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$. $\endgroup$– JacobCommented Jun 28, 2020 at 15:23
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$\begingroup$ That's not correct. You have $$S(S+1)\mathbb{1} = \left(\begin{matrix}S(S+1)&0&0\\0&S(S+1)&0\\0&0&S(S+1)\end{matrix}\right)$$ you should not regard $S$ as an operator since it clearly is not, what is an operator is the product of the number $S(S+1)$ with the identity operator. $\endgroup$ Commented Jun 28, 2020 at 15:27
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$\begingroup$ Oh. I have always treated it as a matrix with the spin number along the diagonal, and the '$+1$' as the unit matrix (and '$+2$' as the unit matrix times 2) and this always gives me the correct matrices. $\endgroup$– JacobCommented Jun 28, 2020 at 15:29
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$\begingroup$ Can you tell me how I am supposed to interpret the $S^2$ in this Hamiltonian term: $\left(\frac{a}{6}\right)\left[S_{x}^{4}+S_{y}^{4}+S_{z}^{4}-\frac{1}{5} S(S+1)\left(3 S^{2}+3 S-1\right)\right]$? $\endgroup$– JacobCommented Jun 28, 2020 at 15:53
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$\begingroup$ You should ask this in another post as per the rules of the thread. $\endgroup$ Commented Jun 28, 2020 at 16:11