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.

  • $\begingroup$ It really depends on what you mean by "spin," so it might help to clarify that. There's no algebraic difference the spin operator algebra and the angular momentum operator algebra, so from one point of view whatever function you came up with for $L$ should also work for $J$. $\endgroup$ – Luke Pritchett Oct 31 '16 at 0:13
  • $\begingroup$ So any difference between angular momentum and spin that you're looking for must be a difference between the two associated ket spaces. Probably the difference is this: the ket space of angular momentum is the space of square-integrable functions on the sphere. The ket space of spin is usually a finite-dimensional irreducible representation of $SU(2)$ (with definite $j$). Does that sound right? $\endgroup$ – Luke Pritchett Oct 31 '16 at 0:33
  • $\begingroup$ @LukePritchett But the thing is that $e^{-\frac{i}{\hbar}{\hat{L}_z\Delta \theta}} |r,\theta,\phi\rangle = |r,\theta+\Delta\theta,\phi\rangle$, whereas $e^{-\frac{i}{\hbar}{\hat{J}_z\Delta\theta}} |r,\theta,\phi\rangle $ does not equal $|r,\theta+\Delta\theta,\phi\rangle$. The spin angular momentum operator has no effect on position. $\endgroup$ – Keshav Srinivasan Oct 31 '16 at 0:34
  • $\begingroup$ @LukePritchett Well, you can also think of the ket space of orbital angular momentum as a finite-dimensional irreducible representation of $SO(3)$ (with definite $l$). $\endgroup$ – Keshav Srinivasan Oct 31 '16 at 0:35
  • $\begingroup$ not if you want to think of position operators like $\hat{\theta}$ and $\hat{\phi}$. Then the ket space needs to be $L^2(S^2)$, or $L^2(R^3)$, a space of square-integrable functions. This space is infinite-dimensional and reducible. It decomposes into $0\oplus 1 \oplus 2\oplus \dots$ via decomposition into spherical harmonics: $\langle \theta,\phi|\ell,m\rangle \sim Y_\ell^m(\theta,\phi)$ An arbitrary function does not have definite $\ell$. $\endgroup$ – Luke Pritchett Oct 31 '16 at 0:40

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.

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    $\begingroup$ I'm not sure the boundedness argument really works - you'd get the same if you restricted the state space to a definite L. To make the playing field even you'd have to take the direct sum of all half-integer spins. $\endgroup$ – Emilio Pisanty Nov 3 '16 at 17:38

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