I Problems with Paper on QM Foundations

[Morbert]

Call the first, middle, and last filters E G and F respectively.

A conventional description of the photon as is passes through the three filters would present these 4 possibilities*:

i) A photon is absorbed by filter E
ii) A photon passes through filter E and is forced into a vertical polarisation, and is absorbed by filter F
iii) A photon passes through the filter E and is forced into a vertical polarisation, passes through the filter F and is forced into a 45 degree polarisation, and is absorbed by the filter G
iv) A photon passes through the filter E and is forced into a vertical polarisation, passes through the filter F and is forced into a 45 degree polarisation, and passes through the filter G and is forced into a horizontal polarisation

We can see that, in these descriptions, the interaction with a filter is what creates the corresponding polarisation property.

QM let's us write down each of these possibilities as a string of time-ordered projectors. If the time of interaction for E G and F are t1, t2, and t3 then the possibilities can be written down as

i) $\left[E',F_0,G_0\right]_{t_1+\delta t}\odot I_{t_2 + \delta t}\odot I_{t_3+\delta t}$
ii) $\left[\uparrow,E,F_0,G_0\right]_{t_1+\delta t}\odot\left[E,F',G_0\right]_{t_2 + \delta t}\odot I_{t_3 + \delta t}$
iii) $\left[\uparrow,E,F_0,G_0\right]_{t_1+\delta t}\odot\left[\nearrow,E,F,G_0\right]_{t_2 + \delta t}\odot\left[E,F,G'\right]_{t_3+\delta t}$
iv)$\left[\uparrow,E,F_0,G_0\right]_{t_1+\delta t}\odot\left[\nearrow,E,F,G_0\right]_{t_2 + \delta t}\odot\left[\rightarrow,E,F,G\right]_{t_3+\delta t}$

QM will return probabilities for these possibilities no problem. But a consistent historian** would say you can move the polarisation projectors to before the filter interactions like so:
i) $\left[E',F_0,G_0\right]_{t_1+\delta t}\odot I_{t_2 + \delta t}\odot I_{t_3+\delta t}$
ii) $\left[\uparrow\right]_{t_1-\delta t}\odot\left[E,F_0,G_0\right]_{t_1+\delta t}\odot\left[E,F',G_0\right]_{t_2 + \delta t}\odot I_{t_3 + \delta t}$
iii) $\left[\uparrow\right]_{t_1-\delta t}\odot\left[E,F_0,G_0\right]_{t_1+\delta t}\odot\left[\nearrow\right]_{t_2-\delta t}\odot\left[E,F,G_0\right]_{t_2 + \delta t}\odot\left[E,F,G'\right]_{t_3+\delta t}$
iv)$\left[\uparrow\right]_{t_1-\delta t}\odot\left[E,F_0,G_0\right]_{t_1+\delta t}\odot\left[\nearrow\right]_{t_2-\delta t}\odot\left[E,F,G_0\right]_{t_2 + \delta t}\odot\left[\rightarrow\right]_{t_3-\delta t}\odot\left[ E,F,G\right]_{t_3+\delta t}$

These, according to CH, would correspond to the possibilities:

i) A photon is absorbed by filter E
ii) A photon passes through filter E already having a vertical polarisation and is absorbed by F.
iii) A photon passes through the filter E already having a vertical polarisation, passes through F already having a 45 degree polarisation, and is absorbed by the filter G
iv) A photon passes through the filter E already having a vertical polarisation, passes through F already having a 45 degree polarisation, and passes through G already having a horizontal polarisation.

and QM will still return consistent probabilities for these possibilities.

So what happens when you remove the filter F? You lose the ability to infer anything about 45 degree polarisation between filters E and G because the possibilities will no longer decohere, and QM will refuse to return consistent probabilities, analogous to the way that, in the double-slit experiment, paths that specify one slit or the other will not decohere when a detector is not present. It's not that removing the filter retroactively changes the polarisation of photons. It's that you lose the ability to address polarisations at certain times without certain filters in place.

* For expediency I've omitted possibilities that are consistent but return probability 0

** I should make it clear that Griffiths's CH account of measurements revealing pre-existing properties is not standard among all CH proponents.