# Understanding spin 1/2 in a projection valued measure perspective

Situation: I am looking at a 2-level or spin 1/2 system from the perspective of a projection valued measure. I thus get a set $$\Omega = \{ \uparrow, \downarrow \}$$, a pretty simple $$\sigma$$ algebra $${\cal A} = 2^\Omega$$ and for every element $$S \in {\cal A}$$ I get an orthogonal projection $$\mu (S)$$ in a suitable Hilbert space $$H$$ such that for vectors $$\vec{x}, \vec{y}$$ the function $$\mu_{\vec{x}, \vec{y}}(\cdot) = \langle \vec{x} | \mu (\cdot) | \vec{y} \rangle$$ is a complex-number valued $$\sigma$$-additive function. Fine. I can describe my Stern-Gerlach experiments.

Now I change the spatial orientation of my Stern-Gerlach device. Formally this means that I make a different experiment. Obviously $$\Omega$$ no longer is $$\{ \uparrow, \downarrow \}$$ but rather $$\{ \uparrow_y, \downarrow_y \}$$ with some connections between the $$\uparrow, \downarrow$$ basis (for the old axis) and the $$\uparrow_y, \downarrow_y$$ basis (for the new axis).

Question: I am looking for a formal setting in which I can describe the relationship between the two $$\Omega$$ sets $$\{ \uparrow, \downarrow \}$$ and $$\{ \uparrow_y, \downarrow_y \}$$ in the language of projection valued measures.

Background: Of course, I could go back to the classical setting of observables where I start with a state space in form of a projective Hilbert space and hermitian observables. Then I would obtain measurement values in the spectrum of the observables (which then, obviously, are real numbers, and thus fairly restricted in their outcome). I intentionally switched to the PVM setting to avoid real-valued measurement results; I want a setting where measurement results come from a $$\sigma$$ algebra setting. Then, I want to reconstruct structures on the measurement results. For example, in case of spin I should get some $$SU(2)$$ symmetry on all the different $$\Omega$$ sets for the manifold of possible experiments. I am looking for theoretical frameworks for this, prior art and references.

Clarification: Somewhere in the back of my mind I am thinking of spin 1/2 measurements in terms of a bundle over $${\mathbb S}^2$$: For every direction ($$\vec{x} \in {\mathbb S}^2$$) I get one of two measurement results (UP, DOWN). The trivial bundle $${\mathbb S}^2 \times \{ \uparrow, \downarrow \}$$ will not work, since it has the wrong topology (two components, not path connected). In order not to be tricked into thinking along the lines of a trivial bundle I do this "set dressing" - it ensures that the fibres all are different. So I get $$\cup_{\vec{r}\in {\mathbb S}^2} \{\uparrow_{\vec{r}}, \downarrow_{\vec{r}}\}$$ as set for this bundle. Now I am looking how I can impose some kind of bundle-structure on this set in a more technical sense. A notion of base or rather local section seems clear, but what about topology, measure, differentiability and more? It looks like a double cover of the sphere. While moving around the sphere somewhere we will jump from a lower leaf to an upper leaf and we probably will not get a consistent UP or DOWN.

Ok. Conceptually this is very similar to the Hopf bundle - but there the fibre is an $${\mathbb S}^1$$ instead of a two-element set. This is nice to understand spin description QM theory but it does not reflect the two-valued-ness in spin measurements which I want to grasp.

Ok. We might dream of a double covering of the $${\mathbb S}^2$$. But the $${\mathbb S}^2$$ is simply connected and so we will not find a double covering.

Searching on...

• Regarding the bundle issue, if you consider the rank-1 bundle defined as the state with $\mathbf{n}\cdot\mathbf{S}=1/2$ over the sphere (so $\mathbf{n}$ is a unit vector), this is indeed a Hopf bundle. However, if we consider the other state as well, the one with $\mathbf{n}\cdot\mathbf{S}=-1/2$, then it is also a Hopf bundle but with opposite Chern class. Together, we have a direct sum of two bundles with opposite first Chern class over sphere, and it is equivalent to a trivial bundle. May 11, 2022 at 12:55

You're looking at the wrong part of the measure to change: $$\Omega$$ doesn't change, it's just a two-element set, and writing one of its elements as $$1$$ or $$\uparrow$$ or $$\uparrow_y$$ is just set dressing.
When you change the axis, you change the map $$\mu$$, and if $$R\in\mathrm{SO}(3)$$ is the rotation between your two axes, then $$\mu' = U(R)\mu U(R)^\dagger$$ is the relation between the two measures, where $$U(R)$$ is the (projective) representation of the rotation on the Hilbert space $$H$$.
So if we start from one $$\mu$$, we get a family of projection-valued measured $$\mu_R = U(R)\mu U(R)^\dagger$$, and if $$\mu$$ represented the spin measurement relative to an axis $$x\in S^2\subset \mathbb{R}^3$$, then $$\mu_R$$ represents the spin measurement relative to the axis $$Rx$$. There's your Hopf fibration - for any initial $$x\in S^2$$, you get the action of $$S^3 \cong \mathrm{SU}(2)$$ on it (as rotations), and the stabilizer is the $$S^1$$ of rotations with $$x$$ as their axis - this is one of the classic ways to construct the Hopf bundle.
For any given $$\mu_R$$, you can look at the two states left invariant by $$\mu_R(\uparrow)$$ and $$\mu_R(\downarrow)$$ - the eigenstates of the corresponding spin operator. If we label these as $$\lvert \uparrow_R \rangle, \lvert \downarrow_R\rangle$$ (and here the explicit $$R$$ label is justified since these are genuinely different states in $$H$$, depending on $$R$$ through $$\mu_R$$), we get an action of $$\mathrm{SU}(2)$$ on the Hilbert space, and the spin-1/2 projective Hilbert space is the Bloch sphere, so this is a Hopf fibration, too, where the $$S^3\to S^2$$ map is the map $$R\mapsto \lvert \uparrow _R\rangle$$, and you get the same for $$\lvert \downarrow_R\rangle$$.
But note that the map that sends $$\lvert \uparrow_1 \rangle$$ to $$\lvert \uparrow_R\rangle$$ is just $$\lvert \uparrow_R\rangle = U(R)\lvert \uparrow_1\rangle$$, so passing through the language of projective-valued measures didn't gain us anything here - all the structure is essentially due to the special properties of $$\mathrm{SU}(2)$$'s two-dimensional spin-1/2 representation.
• I see your point, which is helpful on $\mu$. With regard to the set, I added a clarification why I am doing it the way I am. It also may indicate the direction I am chasing this and my motivation. May 11, 2022 at 12:39
• @Nobody-Knows-I-am-a-Dog I amended the answer to explain how my $\mu_R$ is consistent both with constructing the Hopf bundle of rotations, and how you can construct two Hopf bundles from the two eigenstates. Meng Cheng correctly remarks in a comment that the combination of the two eigenstate bundles is just a trivial bundle. May 11, 2022 at 13:11