# What is significance of $\vec{S}^2$ of spin in quantum mechanic?

The main question is as the title. I have read about the spin in quantum mechanics. I know that the $$[\vec{S}^2,\vec{S}_i]=0$$, they commute, they can share the same Eigenvector, and so what? I just wonder that just $$\vec{S}_z$$ is not enough? The other thing is what is the meaning of this $$\vec{S}^2$$, what is it measures?

• $\uparrow$ which book? which page? Jan 19, 2020 at 17:44

You're quite right that the desire to use $$\hat{S}^{2}$$ is because the label $$s_{z}$$ (eigenvalue of operator $$\hat{S}_{z}$$) is not enough.

Mathematical preliminaries (skip if you want to get straight to the point)

To see why we must consider the idea of different representations of the spin algebra $$\mathfrak{su}_{2}$$. By a representation (I will restrict myself to finite dimensional representations that in mathematical language are irreducible) I mean a set of three matrices $$\mathrm{S}_{x}$$, $$\mathrm{S}_{y}$$ and $$\mathrm{S}_{z}$$ that "represent" the operators $$\hat{S}_{x}$$, $$\hat{S}_{y}$$ and $$\hat{S}_{z}$$ [$$\dagger$$]. "Represent" means that the algebra is respected by matrix multiplication: in this case if $$[\hat{S}_{i}, \hat{S}_{j}] = \hbar i\epsilon_{ijk}\hat{S}_{k}$$ then the matrices of the representation stisfy $$[\mathrm{S}_{i}, \mathrm{S}_{j}] = i\hbar \epsilon_{ijk}\mathrm{S}_{k}$$ where the commutator is calculated using matrix multiplication.

For the spin 1/2 system we get a $$2$$-dimensional representation where $$\mathrm{S}_{i} = \frac{\hbar}{2} \sigma^{i}$$ with $$\sigma^{i}$$ being the Pauli matrices. For example, the matrix elements of $$\hat{S}_{z}$$ in the basis $$|\pm \frac{1}{2}\rangle$$ are $$\langle \pm \frac{1}{2} | \hat{S}_{z} | \pm \frac{1}{2} \rangle = \pm \frac{\hbar}{2}; \qquad \textrm{and} \qquad \langle \mp \frac{1}{2} | \hat{S}_{z} | \pm \frac{1}{2} \rangle = 0$$ so that $$\mathrm{S}_{z} = \frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$ It's two dimensional because there are two states in the basis which implies the matrices are two dimensional. In the same basis we get matrix elements of $$\hat{S}^{2}$$ to be $$\langle \pm \frac{1}{2} | \hat{S}^{2} | \pm \frac{1}{2} \rangle = \hbar^{2} \frac{1}{2} (\frac{1}{2} + 1); \qquad \textrm{and} \qquad \langle \mp \frac{1}{2} | \hat{S}_{z} | \pm \frac{1}{2} \rangle = 0$$ so that $$\mathrm{S}^{2} = \frac{3\hbar^{2}}{4}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$ Note that this matrix is diagonal and proportional to the identity. In fact Schur's Lemma gives us the result for free: if we have a representation (irreducible) of a Lie group and there is a matrix that commutes with all elements of the representation then that matrix must be proportional to the identity. Here since $$[\hat{S}^{2}, \hat{S}_{i}] = 0$$ for all $$i$$ we get that $$\hat{S}^{2}$$ must be represented by a matrix $$\mathrm{S}^{2}$$ that is proportional to the $$2\times 2$$ identity.

Systems with spin greater than $$1/2$$

We know that spin = $$\frac{1}{2}$$ is not the only option -- in other words there exist systems with values of $$s_{z}$$ greater than $$\pm \frac{1}{2}$$. The easiest way to get to this is to consider the addition of angular momentum. You've probably seen that the "sum" of angular momenta of two particles (or the combination of spin and orbital angular momentum) can be determined as illustrated in the following way (see my answer here for more info): $$\textrm{Spin 1/2 with Spin 1/2 gives Spin 1 and Spin 0}$$ It's easy to think geometrically: when the spins are aligned $$(s_{z} = \pm\frac{1}{2}) + (s_{z} = \pm\frac{1}{2})$$ gives $$s_{z} = \pm1$$. When the spins are anti-aligned we get $$(s_{z} = \pm\frac{1}{2}) + (s_{z} = \mp\frac{1}{2}) \Longrightarrow s_{z} = 0$$. So in the combination of two spin 1/2 particles (in fact the tenor product space of the Hilbert spaces of the individual particles) we have some states with $$s_{z} = \{ 1, 0, -1\}$$.

However, it turns out there are 2 states with $$s_{z} = 0$$! In conventional notation we can write the four states as $$\textrm{Spin 1:} \begin{cases} |\uparrow \uparrow\rangle & s_{z} = +1 \\ \frac{1}{\sqrt{2}} \left[ |\uparrow \downarrow \rangle + | \downarrow \uparrow \rangle\right] & s_{z} = 0 \\ |\downarrow \downarrow \rangle & s_{z} = -1\end{cases} \qquad \textrm{and} \qquad \textrm{Spin 0: } \frac{1}{\sqrt{2}} \left[ |\uparrow \downarrow \rangle - |\downarrow \uparrow \rangle \right] \qquad s_{z} = 0$$ where the $$\uparrow$$ and $$\downarrow$$ in positions $$1$$ and $$2$$ represent the spin states $$1/2$$ and $$-1/2$$ of the individual particles. Note the spin $$s_{z} = 0$$ states are symmetric (anti-symmetric) in the Spin 1 (Spin 0) sectors respectively.

If we continue we can form even larger values of spin: part of the combination of 3 spin 1/2 particles will contain states with $$s_{z} = \{ 3/2, 1/2, -1/2, -3/2\}$$. Furthermore there are two further copies of states with $$s_{z} = \{1/2, -1/2\}$$! Now here are two questions that answer your query:

How do we distinguish between the $$s_{z} = 0$$ states of the Spin 1 and Spin 0 sectors? How do we distinguish between the $$s_{z} = 1/2$$ or $$s_{z} = 1/2$$ states of the single Spin 1/2 particle and the states with $$s_{z} = \pm 1/2$$ of the addition of three angular momenta?

Resolution: Casimir operator

It won't surprise you to hear that we distinguish the states using their eigenvalues of the operator $$\hat{S}^{2}$$. This operator, commuting with the other elements of the algebra, is known as the Casimir. Its explicit form can be computed from the tensor product of operators from each subsystem but I won't go into that now. Suffice it to say that the three states of Spin 1 enumerated above are eigenstates of $$\hat{S}^{2}$$ with eigenvalue $$\hbar^{2} \times 1 \times (1+1) = 2\hbar^{2}$$. Likewise, the single state of Spin 1 is an eigenstate of the operator $$\hat{S}^{2}$$, but this time with eigenvalue $$\hbar^{2} \times 0 \times (0 + 1) = 0$$. Thus we choose to label those states by the eigenvalues of the commuting operators $$\hat{S}_{z}$$ and $$\hat{S}^{2}$$. In particular the Spin 1/2 states are denoted $$|\frac{1}{2}, \pm \frac{1}{2}\rangle$$.

Going to the Spin 1 and Spin 0 states we write them as $$\begin{cases} |\uparrow \uparrow\rangle & |1, 1\rangle \\ \frac{1}{\sqrt{2}} \left[ |\uparrow \downarrow \rangle + | \downarrow \uparrow \rangle\right] & |1, 0\rangle \\ |\downarrow \downarrow \rangle & |1, -1\rangle\end{cases} \qquad \textrm{and} \qquad \frac{1}{\sqrt{2}} \left[ |\uparrow \downarrow \rangle - |\downarrow \uparrow \rangle \right] \qquad |0, 0\rangle$$ In the same language we can enumerate the states of the combination of three spins: $$\textrm{Spin 3/2: } \begin{cases} |3/2, 3/2\rangle \\ |3/2, 1/2\rangle \\ |3/2, -1/2\rangle \\ |3/2, -3/2\rangle \end{cases} \qquad \textrm{Spin 1/2: } \begin{cases}|1/2, 1/2\rangle \\ |1/2, -1/2\rangle\end{cases}$$ (advanced note: there are, as mentioned above, in fact two copies of the states with Spin 1/2).

In short the answer is as follows. Recalling that $$s_{z}$$ is the projection of angular momentum in the $$z$$ direction,

The eigenvalues of the operator $$\hat{S}^{2}$$ allow us to distinguish states with the same values of $$s_{z}$$ but different values of total angular momentum.

In other words whilst $$s_{z}$$ measures the $$z$$-projection of Spin, the eigenvalue of $$\hat{S}^{2}$$ measures the total angular momentum of the state (the length of the angular momentum vector).

Returning to representations

If you followed the first section of this answer, it's worth returning to complete the discussion of representations. We saw the Spin 1/2 $$|\frac{1}{2}, \pm \frac{1}{2} \rangle$$ states provide a $$2 = (2 \times \frac{1}{2} + 1)$$ dimensional representation of the $$\mathfrak{su}(2)$$ algebra.

The Spin 1 states $$|1, \pm 1\rangle$$ and $$|1, 0\rangle$$ are 3 in number and we expect them to give a $$3 = (2*1 + 1)$$ dimensional representation of the algebra. I leave it as an exercise to find the matrix elements of the operators $$\hat{S}_{z}$$ and $$\hat{S}^{2}$$ in the three dimensional basis we constructed to verify that acting on these states we have $$3\times 3$$ matrices $$\mathrm{S}_{z} = \hbar \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1\end{pmatrix} \qquad \textrm{ and } \qquad \mathrm{S}^{2} = 2\hbar^{2} I_{3}.$$ Here $$I_{3}$$ is the $$3\times 3$$ identity matrix. Exercise is to construct the matrices for $$\hat{S}_{x}$$ and $$\hat{S}_{y}$$ and to verify that these $$3\times 3$$ matrices satisfy the algebra.

In general the subspace with Spin $$S$$ will have $$2S + 1$$ states and will provide us with a $$2S + 1$$ dimensional representation of the algebra!

Similarly the Spin 3/2 states $$|\frac{3}{2}, \pm \frac{3}{2}\rangle$$, $$|\frac{3}{2}, \pm \frac{1}{2}\rangle$$ are four in number: they give a $$4$$ dimensional representation of the algebra. Exercise to show we have $$4 \times 4$$ matrices $$\mathrm{S}_{z} = \frac{\hbar}{2} \begin{pmatrix} 3 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -3\end{pmatrix} \qquad \textrm{ and } \qquad \mathrm{S}^{2} = 4\hbar^{2} I_{4}.$$

[$$\dagger$$] In this way we have a map from the algebra to the set of invertible matrices, $$\mathfrak{R}_{n} : \mathfrak{su}(2) \rightarrow GL(n, \mathbb{C})$$ where $$n$$ is the dimension of the representation.

It measures the (square of the) magnitude of the spin angular momentum, ignoring the spin’s direction.

If we have a rotating body, it's classical rotational energy is $$E_{rot} = \frac{L^2}{2\theta}$$ Therefore, if we include a rotational particle in quantum mechanics it seams "natural" to use the same term and replace $$L^2$$ by it's operator -- the mathematical argument involves some group theory and so called generators [see e.g. the book of Sakurai for a simple derivation]. Since we consider to spin as a rotation, the same is true for a spinning particle.

Furthermore, there are different coupling themes like $$J = L + S$$ where the spin and the orbital angular momentum form the total angular momentum $$J$$. In this case the Hamiltonian concludes a $$J^2$$ term and a $$LS$$ term as well. Hence, the operators $$L$$, and $$S$$ gain importance as you progress with your book.

As an addendum to the other answers: When we say a particle is "spin 0" (e.g. Higgs boson), "spin 1/2" (e.g. electron), "spin 1" (e.g. photon), etc., we are talking about their different values for this operator. In general a "spin $$s$$" particle is described by eigenstates of $$\hat{S}^2$$ with eigenvalue $$\hbar^2 s (s + 1)$$.