Are the photons traveling between two linear polarizing filters in the "cat state"?

Question

If I have two linear polarizing filters that are at 0º and 45º, then 50% of the photons from a source of non-polarized light will pass through the first first, and 50% of these will pass through to the second filter:

It is correct to say that the photons between the 0º filter and the 45º filter are in the "cat state" ?

Wikipedia defines the so-called cat state as "a quantum state that is composed of two diametrically opposed conditions at the same time, such as the possibilities that a cat be alive and dead at the same time." Part of my confusion is that it would seem that Schrödinger's wave equation allows every particle to be described as a linear combination of any number of other wave equations, so it would seem that every particle is always in a cat state, and that the term in meaningless. I gather that there isn't anything fundamentally different between the photons about to hit the filter that's at 0º and the filter that's at 45º—in both cases only 50% of the photons will proceed. But that's this diagram. If the second filter was also at 0º, then all of the photons passing through the first would pass through the second. So would they be in the cat state then?

Answer 1

By "cat state", I suppose you mean a superposition of quantum states. After the unpolarized light passes through the first polariser, its quantum state collapses to the eigenstate corresponding to the first filter. Initially, the unpolarized light was in a superposition of eigenstates corresponding to the set of eigenkets of polarisation direction of the first filter, say $\frac{1}{\sqrt{2}}\left(\ \left| 0º,+\right> + \left| 0º,-\right> \right)$. Let us say the polariser picks the $\left| 0º,+\right>$ ket, rejecting the $\left| 0º,-\right>$ states. The light that now passes to the second polarizer is in the state $\left| 0º,+\right>$ which can be broken up into a superposition of eigenstates corresponding to the polarization direction of the second filter: $\left| 0º,+\right> =\frac{1}{\sqrt{2}}\left(\ \left| 45º,+\right> + \left| 45º,-\right> \right)$.

So yes, it is in a "cat state" or superposition of 2 states between the first and second filter if you chose the $45º$ polarization direction as your basis kets. Whereas it is in the single state $\left| 0º,+\right>$, if you choose the $0º$ polarization direction as your basis states.

Answer 2

A "Schrodinger's cat state" or "cat state" refers to a state of a macroscopic system in which the macroscopic observables are highly delocalized. For an example of how this is used, see Zurek, "Decoherence, einselection, and the quantum origins of the classical," http://arxiv.org/abs/quant-ph/0105127 . So your example is not a cat state, because it's a single photon.

Part of my confusion is that it would seem that Schrödinger's wave equation allows every particle to be described as a linear combination of any number of other wave equations, so it would seem that every particle is always in a cat state, and that the term in meaningless.

This related to the preferred basis problem in decoherence. Decoherence addresses this because the process of decoherence occurs through an interaction with the environment. The characteristics of the interaction determine a preferred basis. In the simplest case, if the interaction is an interaction at a point, then the x basis becomes the preferred basis, and it is in this basis that the density matrix will become diagonal. This prevents us from observing interference between states that are cat states in terms of the x basis.