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...

  • 2
    $\begingroup$ 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. $\endgroup$
    – Meng Cheng
    May 11, 2022 at 12:55

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ May 11, 2022 at 12:39
  • 1
    $\begingroup$ @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. $\endgroup$
    – ACuriousMind
    May 11, 2022 at 13:11

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