On the Hilbert spaces for $\hat L$ and $\hat S$

Question

Are there convincing arguments that suggest taking $\mathbb C^{2s+1}$ as Hilbert spaces for the spin angular momentum operator $\hat S$?

For what concerns $\hat L$, the choice ${\scr H} \equiv L^2(\mathbb R^3)$ seems reasonable:

1. $\psi$ has to be a function (a wave function precisely);
2. according to the Copenhagen interpretation, $|\psi|^2 d^3{x}$ is a probability density, so it has to be normalized: $\displaystyle \int_{\mathbb R^3} |\psi|^2 d^3{ x} = 1$. A necessary condition is therefore $\psi \in L^2(\mathbb R^3) $ .

But what about ${\scr H} \equiv \mathbb C^{2s+1}$ for $\hat S$? Why a (column) vector instead of a "spin-function"? Since nothing is really "spinning" in a classical sense, it is pointless to see this vector as some sort of rotation axis. So what does it really mean?

Answer 1

The way one does this is to start from the symmetry group $\mathrm{SO}(3)$ of rotations in 3-dimensional space, consisting of the $3\times 3$ real, orthogonal matrices with determinant +1. The infinitesimal generators of this group constitute the Lie algebra $\mathfrak{so}(3)$, which is the set of $3\times 3$ antisymmetric matrices. The standard basis for this algebra is

$$\mathbb J_1=\pmatrix{0 & 0 & 0 \\ 0 & 0 &-1\\0&1&0} \quad \mathbb J_2=\pmatrix{0 & 0 & 1 \\ 0& 0 &0\\-1&0&0} \quad \mathbb J_3=\pmatrix{0 & -1 & 0 \\ 1 & 0 &0\\0&0&0}$$ with commutation relations $[\mathbb J_i,\mathbb J_j]=\epsilon_{ijk}\mathbb J_k$.

If you are unfamiliar with any of this, it suffices to say that if we have some Hilbert space $\mathscr H$ and we can find a set of three self-adjoint operators $\{\hat J_1,\hat J_2, \hat J_3\}$ with the same commutation relations (that is, $[\hat J_i,\hat J_j]=\epsilon_{ijk}\hat J_k$), then those operators can be understood as the angular momentum operators on $\mathscr H$.

The map $\rho: \mathbb J_i \mapsto \hat J_i$ which takes an abstract element of $\mathfrak{so}(3)$ to a concrete operator on some Hilbert space is called a representation; a classification of all such representations would tell us how to implement angular momentum operators on whatever Hilbert space we want.

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The representation theory of the rotation group $\mathrm{SO}(3)$ and its Lie algebra $\mathfrak{so}(3)$ is a topic which could occupy a large section of a textbook in its own right, so a full and honest explanation is far beyond the scope of a single PSE answer. However, we can quote the salient information.

As a preliminary note, all finite-dimensional complex Hilbert spaces are isomorphic to $\mathbb C^n$ for some $n$, so those are the finite-dimensional Hilbert spaces we'll speak of. Secondly, you can always chain together two representations to get a larger one - for example, you could take the operators $J^{(2)}_i$ which act on $\mathbb C^2$ and $J^{(3)}_i$ which act on $\mathbb C^3$ and put them together in a block-diagonal way to produce operators which act on $\mathbb C^5$ like this:

$$ \pmatrix{J_i^{(2)} & \matrix{0&0&0\\0&0&0}\\\matrix{0&0\\0&0\\0&0} & J^{(3)}_i}$$ However, that doesn't really give us anything new, so we will ignore all such so-called reducible representations and focus on the irreducible ones.

With those preliminaries out of the way, the important results are as follows:

1. For each $n>0$, there is (up to change-of-basis) exactly one way to represent $\mathfrak{so}(3)$ as self-adjoint operators on $\mathbb C^n$.
2. For any finite-dimensional irreducible representation $\{\hat J_1,\hat J_2,\hat J_3\}$, the operator $\hat J^2 := \hat J_1^2 +\hat J_2^2 + \hat J_3^2$ commutes with each $\hat J_i$ and is in fact proportional to the identity operator, $\hat J^2 = c \mathbb I$. The constant $c$ depends on the representation, and can be shown to be equal to $j(j+1)$ for some non-negative integer or half-integer $j$; furthermore, the eigenvalues of each $j_i$ can be shown to range from $-j$ to $j$ in integer steps.
3. The aforementioned $j$ turns out to be related to the dimensionality $n$ of the representation space via $n = 2j+1$. We use $j$ to label the particles in question, calling them "spin-$j$."
4. The infinite dimensional case is a bit more subtle, but we can also represent $\mathfrak{so}(3)$ on infinite dimensional Hilbert spaces like $L^2(\mathbb R^3)$. Such representations can be obtained e.g. by implementing rotations as families of unitary operators and then finding the corresponding self-adjoint generators via Stone's theorem. However, note that $J^2$ is no longer proportional to the identity operator in the infinite-dimensional case.

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Are there convincing arguments that suggest taking $\mathbb C^{2s+1}$ as Hilbert spaces for the spin angular momentum operator $\hat S$?

Experiments show us that certain particles are not accurately modeled by wavefunctions which inhabit the Hilbert space $L^2(\mathbb R^3)$. For example, an electron inhabiting the $\mathrm{1s}$ orbital of the hydrogen atom possesses non-zero angular momentum despite the fact that the angular momentum associated to its spatial wavefunction is equal to zero. Furthermore, there are only two distinct eigenvalues for this intrinsic angular momentum possessed by an electron, which leads us to surmise that the correct thing to do is to take the full Hilbert space to be $L^2(\mathbb R^3)\otimes \mathbb C^2$.

From there, we understand angular momentum as having an extrinsic part $\hat L_i$ associated to the wavefunction $\in L^2(\mathbb R^3)$ and an intrinsic part $\hat S_i$ associated to the internal spin state $\in \mathbb C^2$.

Answer 2

A few of the answers mention the key word "irreducible". I want to focus on this concept a bit more because I think this concept is really at the heart of the answer to your question.

The Hilbert space for $\bf L$, which consists of the complex-valued square-integrable functions on the sphere, is not an irreducible representation of the rotation group $SO(3)$. It can be decomposed into subspaces that are acted on independently by all the components of $\bf L$. Explicitly, the decomposition is into the eigenspaces of $L^2$. As you know, the eigenspaces of $L^2$ may be labelled with the quantum number $\ell = 0, 1, 2, \ldots$ corresponding to the eigenvalues $\hbar^2 \ell(\ell + 1)$. The eigenspace with a particular $\ell$ value has dimension $2\ell + 1$, and can be thought of as the "spin $\ell$ subspace". These subspaces are irreducible representations of $SO(3)$ (assuming the ground field is $\Bbb C$): it is not possible to leave such a subspace through application of the $\bf L$ operators.

In the case of spin, we focus on the irreducible representations. If we use an irreducible representation, then it follows that the different possible spin states (which are really just normalized vectors that belong to the Hilbert space in question) are all "reachable" from each other through the action of $SO(3)$. For example, an electron whose spin is up can be converted into an electron whose spin is down, simply by rotating the electron 180 degrees around an axis perpendicular to the z-axis.

If we have a reducible representation, then we have mutually "unreachable" states of the same particle. If there was no way to get a spin-up electron to become a spin-down electron, by any combination of rotations and linear combinations (which represent the particle interfering with itself), then we would say there are two different types of electron. We wouldn't say that there's one type of electron whose spin lives in a reducible representation. If the "electron"'s spin state were decomposable into an infinite number of irreducible representations (the way the Hilbert space for $\bf L$ is), then there would be infinitely many types of electron. This is not what we observe in nature.

So basically, as a consequence of how we define the concept of a particle, it follows that the spin state of a particle lives in an irreducible representation, not a reducible one. Then, in order to determine what these spin Hilbert spaces look like, all we have to do is find all the irreducible representations.

There is a slight complication, which is that two vectors that differ only by a phase factor represent the same physical state. Therefore, spin states may not necessarily live in irreducible representations of $SO(3)$; they are allowed to belong to irreducible "projective representations" where a 360-degree rotation may change the phase (but not the direction) of the vector.

So in the end our objective is to find all ordinary and projective representations of $SO(3)$ that are irreducible. When we do that, we discover (without having started with any assumptions about spaces having dimension $2s+1$) discover that there is one irreducible representation for each possible dimension, and that the odd-dimension representations are isomorphic to the previously described "integer spin" spaces (i.e., ones that have the eigenvalue $\hbar^2 s(s+1)$ for integer $s$). The even-dimension representations are the half-spin representations, which are projective.

Answer 3

Your last question: One does not need a classical spinning for example in a spin 1/2. The "rotation" is usually over $\mathbb{H}=\mathbb{R}^4$, the space of quaternions. This is because there exists a function between $\mathbb{H}$ and $SU(2)$:

$$ q = q_0+q_1\hat{i}+q_2\hat{j}+q_3\hat{k} \leftrightarrow\begin{pmatrix} q_0-iq_3& -q_2-iq_1\\ q_2-iq_1 & q_0+iq_3\end{pmatrix}, $$

where the $q_i$'s are real and $\hat{i},\hat{j},\hat{k}$ obey the "quaternionic multiplication rules". An element of $SU(2)$ (rotation of spin!) is usually given by $\exp(-i\frac{\theta}{2}\vec{w}\cdot \vec{\sigma})$, with $\sigma_j$ the Pauli matrices. Under the above identification, you have your axis of rotation $\vec{w}$.

Your first question: If you want to know that for a fixed $s$, you have an irreducible subspace of dimension $2s+1$ you need to first realize that you are dealing with a representation of $SU(2)$, $\pi(g), g\in SU(2)$. [As J. Murray points out, this usually occupies one or two chapters in the QM books.]

Last but not least: That $SU(2)$ can be related to $SO(3)$ is more like an accident...

Overall: So, if you have a spin, it is better to treat its state as a member of $\mathbb{C}^2$. Measuring the spin up or down amounts, in an experiment, to measure number of times you measured spin-up $n_+$, similarly for spin-down $n_-$ divided each measurement by the total number of measurements you do $n_+ + n_-$.

If you have an atom (hydrogen), then you can find the probability density of being around a certain orientation using something like $\langle \theta, \phi| l, m\rangle dV = Y_l^m(\theta,\phi)dV$.

Answer 4

It means the components of $\hat{\mathbf{S}}$ have exactly $2s+1$ eigenstates, differently from $\hat{\mathbf{L}}$, which has an infinite amount of them.