Making sense of tensor product of observables I'm trying to understand an argument against naive realism in Richard Healey's book, The Quantum Revolution (The second argument in Appendix B). The book is pitched at philosophers, but discusses some of the math. And I'm very new to the math myself, so it's a bit of a challenge. In any case, here's what's confusing me: 
The argument works by assuming that an arbitrary system of three photons is such that each photon has both a definite circular and linear polarization, corresponding to the dynamical variables $\bigcirc_1, \bigcirc_2, \bigcirc_3$ and $\diagup_1,\diagup_2,\diagup_3$ (whose possible values are 1 or -1) and operators $\hat{\bigcirc}_1, \hat{\bigcirc}_2, \hat{\bigcirc}_3$ and $\hat{\diagup}_1,\hat{\diagup}_2,\hat{\diagup}_3$. 
The idea behind the proof, I think, is that by determining which collections of operators commute, you determine all the possible joint measurements that could be made on the system, but there's no way to assign definite values to the six dynamical variables that doesn't result in a contradiction with what has (apparently) been found in experiments. 
This is what confuses me most: Healey claims that the following set of observables commute: 
$$\{\widehat{\diagup\diagup\diagup},\widehat{\bigcirc\bigcirc\diagup},\widehat{\diagup\bigcirc\bigcirc},\widehat{\bigcirc\diagup\bigcirc}\}$$
Where these observables are identical to the tensor product of the smaller observables mentioned above, so that for instance: 
$$\widehat{\diagup\diagup\diagup} = \hat{\diagup}_1 \otimes \hat{\diagup}_2 \otimes \hat{\diagup}_3$$
And Healey claims that when such measurements are observed, the product of the measurements is always -1, indicating that an odd number of the observables in the set have value -1. 
But if that's so, I have no idea how to make sense of what observables like $\widehat{\diagup\diagup\diagup}$ represent. My original thinking was that the dynamical variable $\diagup\diagup\diagup$ should have as its value the product of the photons' linear polarizations. But then it would be impossible for the product $\diagup\diagup\diagup \times \bigcirc\bigcirc\diagup \times \diagup\bigcirc\bigcirc \times \bigcirc\diagup\bigcirc$ to be -1, since each $\pm1$ circular/linear polarization observable appears in this product twice; the product would have to be 1. 
In certain cases, though, I thought I had good reasons for thinking that $\widehat{\diagup\diagup\diagup}$ would have as its value the product of the photons' linear polarization. Suppose the three photons each have definite linear polarization, in which case we would find that the individual photon states are eigenvectors for the $\diagup$ operators. Wouldn't this then imply the following?: 
$$\widehat{\diagup\diagup\diagup}|\psi_\text{all}\rangle = (\hat{\diagup}_1 \otimes \hat{\diagup}_2 \otimes \hat{\diagup}_3)(|\psi_1\rangle|\psi_2\rangle|\psi_3\rangle = \lambda_1\lambda_2\lambda_3|\psi_1\rangle|\psi_2\rangle|\psi_3\rangle = \lambda_1\lambda_2\lambda_3|\psi_\text{all}\rangle$$
I tried to think through the case where the photons are in a superposition of opposite linear polarizations, but I'm not sure how the math works in that case. Would that change the above result? 
 A: I'm answering the specific question posed in the final paragraph of the post. If this doesn't clear up your overall confusion, please feel free to comment or edit your question.
I will denote eigenvectors of $\widehat{\diagup}$ with eigenvalues of $+1$ and $-1$ as $\left|+\right>$ and $\left|-\right>$ respectively. I'll also use the notation $\left|+++\right> = \left|+\right>_1 \otimes \left|+\right>_2 \otimes \left|+\right>_3$, etc.
Consider the superposition state
\begin{align}\left|\psi\right> = \frac{\left|+++\right> + \left|---\right>}{\sqrt{2}}.\end{align}
Applying $\widehat{\diagup\diagup\diagup} = \widehat\diagup_1 \otimes \widehat\diagup_2 \otimes \widehat\diagup_3$ to $\left|\psi\right>$,
\begin{align}
\widehat{\diagup\diagup\diagup} \left|\psi\right> &= \frac{1}{\sqrt{2}}\left[\widehat{\diagup\diagup\diagup}\left|+++\right> +  \widehat{\diagup\diagup\diagup}\left|---\right>\right]\\
&= \frac{1}{\sqrt{2}}\left[(+1)(+1)(+1)\left|+++\right> + (-1)(-1)(-1)\left|---\right>\right]\\
&= \frac{1}{\sqrt{2}}\left[\left|+++\right> - \left|---\right>\right].
\end{align}
Evidently, $\left|\psi\right>$ is not an eigenvector of $\widehat{\diagup\diagup\diagup}$.
Re-reading your question, I think you may have instead intended the following product state:
\begin{align}
\left|\chi\right> &= \left(\frac{\left|+\right>_1 + \left|-\right>_1}{\sqrt{2}}\right)\left(\frac{\left|+\right>_2 + \left|-\right>_2}{\sqrt{2}}\right)\left(\frac{\left|+\right>_3 + \left|-\right>_3}{\sqrt{2}}\right)\\
&=\frac{1}{2^{3/2}}\left[\left|+++\right> + \left|++-\right> + \left|+-+\right> + \left|+--\right>\right.\\ &\left.\hspace{3.5em}+ \left|-++\right> + \left|-+-\right> + \left|--+\right> + \left|---\right>\right],
\end{align}
or something similar.
This yields
\begin{align}
\widehat{\diagup\diagup\diagup} \left|\chi\right> &= \frac{1}{2^{3/2}}\left[\left|+++\right> - \left|++-\right> - \left|+-+\right> + \left|+--\right>\right.\\ &\left.\hspace{3.5em}- \left|-++\right> + \left|-+-\right> + \left|--+\right> - \left|---\right>\right].
\end{align}
A: Indeed, the eigenvalues of a tensor product of operators are just the product of the individual operators' eigenvalues, for the reason that you state.
Do you have Healey's exact quote? I think when he says "the product of the measurements" he means each of the four products of the three single-particle measurements, not the product of the four three-particle joint measurements.
