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.


1 Answer 1


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.

  • $\begingroup$ 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: $S=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]$. $\endgroup$
    – Jacob
    Jun 28, 2020 at 15:23
  • $\begingroup$ 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. $\endgroup$ Jun 28, 2020 at 15:27
  • $\begingroup$ 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. $\endgroup$
    – Jacob
    Jun 28, 2020 at 15:29
  • $\begingroup$ Can you tell me how I am supposed to interpret the $S^2$ in this Hamiltonian term: $\left(\frac{a}{6}\right)\left[S_{x}^{4}+S_{y}^{4}+S_{z}^{4}-\frac{1}{5} S(S+1)\left(3 S^{2}+3 S-1\right)\right]$? $\endgroup$
    – Jacob
    Jun 28, 2020 at 15:53
  • $\begingroup$ You should ask this in another post as per the rules of the thread. $\endgroup$ Jun 28, 2020 at 16:11

Your Answer

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

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