# 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?

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}

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.

• Here's the exact quote: "the values of the four observables lying along the horizontal line [i.e. the set mentioned in my post above] always multiply to -1, no matter what the three-photon polarization state." – objectivesea Oct 30 '19 at 16:36
• And there's a footnote on this, saying "The first argument showed this only for the state $|GHZ\rangle$ [a superposition state with all photons linearly polarized +1 or all -1]. I have omitted the proof that the product of values of the observables on the horizontal line is always -1 since it involves knowledge of the explicit form of the operators $\bigcirc$, $\diagup$. For experts, these mirror those of the Pauli spin matrices and the result follows from their anticommutation relations (see Mermin [1993])." – objectivesea Oct 30 '19 at 16:36
• Now that I pay closer attention to the footnote, I think you're right that Healey must not quite mean that the product: $\diagup\diagup\diagup \times \diagup\bigcirc\bigcirc \times \bigcirc\bigcirc\diagup \times \bigcirc\diagup\bigcirc$ is -1. As I understand the argument referred to in the footnote, it proves that for the $|GHZ\rangle$ state, each of these observables would have value -1, and their product would be +1. – objectivesea Oct 30 '19 at 16:43
• That helps, but it raises further issues. I can't see why the conclusions Healey reaches form a contradiction: Roughly, I think he's saying that there's no way to assign +1 or -1 to each of the 10 observables so that for any triple of compatible $\diagup$ or $\bigcirc$ operators, we have that: 1) their product is -1, and 2) The product of the three observables with the value of their tensor product is +1. Why isn't this a counter-example to that claim? $\bigcirc_2 =\diagup_2 = -1$ and $\bigcirc_1 = \bigcirc_3 = \diagup_1 = \diagup_3 = +1$, – objectivesea Oct 30 '19 at 16:59
• To clarify, Healey says the following triples are the only compatible sets of $\diagup$ and $\bigcirc$ operators: $\{\diagup_1,\bigcirc_2,\bigcirc_3\}, \{\bigcirc_1,\bigcirc_2,\diagup_3\},\{\bigcirc_1,\diagup_2,\bigcirc_3\}\{\diagup_1,\diagup_2,\diagup_3\}$ – objectivesea Oct 30 '19 at 17:02