# Justification for “which path” inference in Hardy's paradox?

I'm going over Hardy's paradox again recently and whenever I try to think about it I always get confused with the classical part of the analysis. Just as a brief summary, the setup of the paradox is summarized in the following diagram: The state the $$c$$ detectors are tuned to measure can be written as:

$$i\left|c^{\pm}\right\rangle = \frac{\left|v^{\pm}\right\rangle + i\left|w^{\pm}\right\rangle}{\sqrt{2}}$$

A key observation in this single particle case is that if the $$w$$ path is disrupted, then the state arriving at the detector is just $$\left|v^\pm\right\rangle$$. This allows the following inference: A detection at $$d$$ implies an obstacle in either path. Specifically, on average, equal counts occuring at $$d$$ and $$c$$ implies that either $$v$$ or $$w$$ are completely blocked.

Lifting analogously from the one particle analysis, the following map is made in order to infer the "which path" information of the two particle system:

$$\begin{equation} \begin{split} \left|c\right\rangle & \rightarrow \left|c^+\right\rangle\left|c^-\right\rangle \\ \left|d\right\rangle & \rightarrow \left|d^+\right\rangle\left|d^-\right\rangle \end{split} \end{equation}$$

One can calculate that $$\left|\psi\right\rangle$$ has nonzero inner product with $$\left|d^+\right\rangle\left|d^-\right\rangle$$, specifically a probability of $$\frac{1}{16}$$. From here the deduction is made that because the inner product of these two states are nonzero, this implies that both particles must have been in the $$w$$ path.

This inference loses me completely. For one, $$\left|\psi\right\rangle$$ is clearly orthogonal to $$\left|w^+\right\rangle\left|w^-\right\rangle$$. Secondly, the "inferential" map between the $$c$$ and $$d$$ states of the one particle and two particle systems seems entirely unjustified and likely incorrect.

In the one particle system, it's more correct to say that $$d$$ infers a change in the relative populations of the $$v$$ and $$w$$ paths, as opposed to an obstruction. In the one particle case, this conflation is harmless as an obstruction and a change in the relative population are semantically equivalent. This is however not the case in the two particle system. An obstruction of one path is correlated with a change in the relative populations of the other possible paths. The exclusion of the $$\left|w^+\right\rangle\left|w^-\right\rangle$$ state results in a relative increase of the population of the mixed path states - even though there is no obstruction for just one of the particles - which $$\left|d^+\right\rangle\left|d^-\right\rangle$$ can be used to infer. We can write $$\left|d^+\right\rangle\left|d^-\right\rangle$$ explicitly in the path basis:

$$\left|d^+\right\rangle\left|d^-\right\rangle = \frac{1}{2}\big(\left|v^+\right\rangle\left|v^-\right\rangle - i\left|v^+\right\rangle\left|w^-\right\rangle - i\left|w^+\right\rangle\left|v^-\right\rangle - \left|w^+\right\rangle\left|w^-\right\rangle\big)$$

And if we take the inner product of this with $$\left|\psi\right\rangle$$, the remaining terms are:

$$\left \langle d^+, d^-|\psi\right\rangle= \frac{1}{4} \big( \left\langle v^+, v^+|v^+, v^+ \right\rangle - \left\langle v^+, w^- |v^+, w^-\right\rangle - \left\langle w^+, v^- |w^+, v^-\right\rangle \big)$$

Clearly $$\left|d^+\right\rangle\left|d^-\right\rangle$$ obtains a nonzero inner product because of the double $$v$$ and mixed paths, it has nothing to do with the double $$w$$ path being occupied.

So then I'm left wondering, what is the formal, valid, classical argument that justifies the correspondence of $$\left|d\right\rangle \rightarrow \left|d^+\right\rangle\left|d^-\right\rangle$$ as an inference specifically for $$\left|w^+\right\rangle\left|w^-\right\rangle$$? As far as I can tell, the classical inference presented by Hardy and on Wikipedia is logically invalid.

• I haven't closely studied this one, but in my experience the mistake in all quantum mechanical "paradoxes" is falling back on implicit classical thinking -- that even if we didn't measure something, there must have been a unique position or trajectory all along. In reality, measurement results simply don't exist unless we measure them. Assuming they do is tantamount to going back to classical mechanics. Assuming they do only sometimes quickly yields logical contradictions. – knzhou Sep 25 '18 at 21:57
• So while I'm not 100% about this case, it really looks like the same story again. The setup looks annoyingly complicated, probably because it helps obscure the fact that the same old simple mistake is being made. Keep track of the quantum state the whole time honestly, and there is no problem. – knzhou Sep 25 '18 at 21:58
• That is typically the case in these paradoxes but what's strange to me is that even with all of the "quantum weirdness" the path in question is clearly not occupied. My issue is with the way they use the detectors to infer this, coincident counts in the d detectors do not correspond with this path. You need a specific arrangement of coincident detections between all 4 detectors in order to infer that the w path is occupied. However, the prepared state is clearly orthogonal to this w-path detection state. – Daniel Kerr Sep 25 '18 at 22:08
• I might dig in later if I have time, but if you get no good answers ping me and I’ll drop a bounty. – knzhou Sep 25 '18 at 22:26