Can spin be related to a shift in angle?

Question

If $\hat{T}(\Delta x) = e^{-\frac{i}{\hbar}\hat{p}\Delta x}$ is the spatial translation operator, then there exists a function $f$ from $\mathbb{R}$ to the ket space $V$ such that $\hat{T}(\Delta x) f(x) = f(x+\Delta x)$. Namely, the function that sends $x$ to the position eigenstate $|x\rangle$.

Similarly if $\hat{U}(\Delta t) = e^{-\frac{i}{\hbar}\hat{H}\Delta t}$ is the time evolution operator (for time-independent Hamiltonians), then there exists a function $f$ from $\mathbb{R}$ to $V$ such that $\hat{U}(\Delta t) f(t) = f(t+\Delta t)$. Namely, the function that sends $t$ to $|ψ(t)\rangle$, the state of a particle at time $t$.

And similarly if $\hat{R}_z(\Delta\theta) = e^{-\frac{i}{\hbar}{\hat{L}_z\Delta \theta}}$ is the orbital rotation operator, then there exists a function $f$ from $\mathbb{R}$ to $V$ such that $\hat{R}_z(\Delta \theta)f(\theta) = f(\theta+\Delta\theta)$. Namely the function which sends $\theta$ to $|r,\theta,\phi\rangle$, an eigenstate of the position operator $\hat{\theta}$ in spherical coordinates.

But my question is, if $\hat{R}_z(\Delta\theta) = e^{-\frac{i}{\hbar}\hat{J}_z\Delta \theta}$ is the intrinsic rotation operator, then does there exist a function $f$ from $\mathbb{R}$ to $V$ such that $\hat{R}_z(\Delta \theta)f(\theta) = f(\theta+\Delta\theta)$? By intrinsic rotation operator I mean the rotation operation related to spin angular momentum.

I suspect the answer is no, because there is no operator corresponding to $\theta$, the parameter of the intrinsic rotation operator, since in quantum mechanics it doesn't really make sense to think of spin as a particle rotating about its own axis. But then again, there is no time operator in non-relativistic quantum mechanics, and yet the time evolution operator satisfies the property.

In any case, assuming that the answer to my question is no, I'd like a formal proof that there cannot exist such an $f$.

EDIT: I posted a follow-up question here.

Answer 1

There is no such function for spin. Here's why:

Orbital angular momentum is not fixed. The space of states $L^2(S^1,\mathrm{d}\theta)$ decomposes into the infinite sum of spherical harmonics $\bigoplus_{\ell\in\mathbb{N}} H_\ell$ where $H_\ell$ is the finite-dimensional irreducible representation of $\mathrm{SO}(3)$ associated to the Casimir value $L^2 = \ell(\ell +1)$. Since $L^2(S^1)$ is infinite-dimensional, we can have the operator $\theta$ on which which has continuous spectrum $[0,2\pi]$ and to which the "eigenstates" $\lvert \theta_0\rangle$ belong. Note that these states do not actually lie in the Hilbert space of states - they are the delta distributions $\delta(\theta-\theta_0)$, which are neither functions nor square-integrable and therefore don't lie in $L^2$. The same goes for the position eigenstates: The kets you are acting on do not actually lie within the Hilbert space of states, but in the larger space that's part of the notion of a rigged Hilbert space, see also this answer by user1504. Crucial to the existence of the continuously labeled kets $\lvert x\rangle$ and $\lvert \theta\rangle$ is that we have an unbounded linear operator whose eigenvectors they are supposed to be.

Spin angular momentum is fixed. Each particle only transforms in one of the finite-dimensional representations $H_s$ of $\mathrm{SU}(2)$, where $s$ can now also be half-integer. There is no rigged Hilbert space, no unbounded linear operator here to which we could associate continuously labeled kets. This is because all linear operators on finite-dimensional spaces are automatically continuous and bounded.

Note that the time evolution operator works differently: We don't have kets $\lvert t\rangle$, we have that in the Schrödinger picture we define the time evolution of a state to be $\lvert \psi(t)\rangle := U(t)\lvert \psi\rangle$. This is an equation of motion, not an intrinsic property of two kets called $\lvert \psi(t)\rangle$ and $\lvert \psi\rangle$. The object $\lvert \psi(t)\rangle$ does not exist prior to this definition.

Answer 2

Spin, as an intrinsic property, has nothing to do with position space. However, it can be parameterized on a state space (Hilbert space) in terms of a finite discrete basis. For example, in the case of the spin-half representation of SU(2), the state space is the Bloch sphere. The two-dimensional basis in this case can be represented by $|0\rangle$ and $|1\rangle$. Any arbitrary state on the Bloch sphere can be represented in terms of two angles $\phi$ and $\theta$, which are the two angle coordinates for the sphere. In terms of these angle, an arbitrary state on the Bloch sphere can be given by $$ |\psi\rangle = |0\rangle\exp(-i\tfrac{1}{2}\phi)\cos(\tfrac{1}{2}\theta) + |1\rangle\exp(i\tfrac{1}{2}\phi)\sin(\tfrac{1}{2}\theta) . $$

The spin operator$^*$ $\exp(-i \hat{\mathbf{J}}\cdot\alpha)$ is now a unitary operator and the generators are the Pauli matrices. Therefore, it represents spin transformations as rotation of the states on the Bloch sphere around the three principle axes. One of these transformation represents rotations around the $z$-axis. We can represent it as $$ \hat{U}(\Delta\phi) = \exp(-i \hat{\sigma}_z\cdot\Delta\phi) . $$ The result of such a transformation on an arbitrary state would then be $$ \hat{U}(\Delta\phi)|\psi\rangle = |0\rangle\exp(-i\tfrac{1}{2}(\phi+\Delta\phi))\cos(\tfrac{1}{2}\theta) + |1\rangle\exp(i\tfrac{1}{2}(\phi+\Delta\phi))\sin(\tfrac{1}{2}\theta) . $$ (Note sure about the signs.)

Now, if you want to define a function with this relation, then you can project the state onto a reference state $f(\theta,\phi)=\langle\psi_{ref}|\psi\rangle$, for example $$ |\psi_{ref}\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle+ |1\rangle) . $$ The result would be a function of the two angles that can be shifted by the operation of the transformation $$ f(\theta,\phi+\Delta\phi)=\langle\psi_{ref}|\hat{U}(\Delta\phi)|\psi\rangle . $$

This example can be generalized to arbitrary spin in higher dimensions in an obvious way.

Very important: the two angles have noting to do with angles in position space. They are parameters of the state space.

$*$ Note that I removed the Planck constant through a simple redefinition of the spin generator.

Answer 3

I would have to begin by saying that you have a minor mistake in your conception of those functions. They are not $\mathbb R$ to $V$ but rather $\mathbb R \times \mathcal H \mapsto \mathcal H$ because these unitary operators should act on kets living on Hilbert space to give rise to another ket in the Hilbert space.

After we are done with that nitpick, I do not see why we cannot have $$ \hat U _z ( \Delta \varphi ) \begin{pmatrix} e^{i \varphi/2} \cos \frac\vartheta2 \\ e^{-i\varphi/2} \sin \frac\vartheta2 \end {pmatrix} = \begin{pmatrix} e^{i(\varphi+\Delta\varphi)/2} \cos \frac\vartheta2 \\ e^{-i(\varphi+\Delta\varphi)/2} \sin \frac\vartheta2 \end {pmatrix} $$ and equivalent things for the spin down version of that. We even know that the Pauli matrices are the correct operators for the arbitrary rotations about arbitrary axes via the standard $\hat{\vec n} \cdot \vec{\hat \sigma}$ prescription.

The issue is that while there is a simple interpretation for the angle of rotation in orbital parts, the spin angle is kinda difficult to give a physical interpretation.

The above are all standard physics. Here comes some speculation. If you look at physics from the point of view of geometric algebra (kinda the better version of quaternions), you will see that half spin is naturally coming from the "square root" of rotation, and is coming purely from a classical context, no need for quantum theory. It is quite fashionable in standard physics to treat whatever behaves in the correct way as the definition of things, so we may also consider spin half objects as something causing rotations. If you do that, then Dirac and Pauli kind of spinors, where we always sandwich these spinors to get stuff out, namely we always do $\left < \psi \mid \vec \gamma \mid \psi \right >$ for some vector-like quantity in the middle, then it is clear that we can consider the wavefunction itself as a thing that causes rotation, (in GA lingo this is called a rotor), and then there is no mystery as to what spin half quantities actually should be. Then you can say, yes, there is a rotation angle being expressed.

However, again, this is beyond the realm of standard physics. Study this in your own frustration.