# Schrodinger's cat and consistent histories

I was reading Wikipedia's article on Schrodinger's cat:

https://en.wikipedia.org/wiki/Schr%C3%B6dinger%27s_cat#Many-worlds_interpretation_and_consistent_histories

Quote: "When opening the box, the observer becomes entangled with the cat, so "observer states" corresponding to the cat's being alive and dead are formed; each observer state is entangled or linked with the cat so that the "observation of the cat's state" and the "cat's state" correspond with each other. Quantum decoherence ensures that the different outcomes have no interaction with each other."

My question is... what concretely it means for different outcomes to interact with each other.

So suppose decoherence was not happening. Then could we have:

1) The cat is observed dead, but the cat is alive (so the observer from one outcome is interacting with the cat from the other outcome).

2) At one point in time "the cat is dead and observed dead", and at a later point in time "the cat is alive, and observed alive" (ie: in reality there's a wave equation which is a superposition of both possibilities... at different points in time we may observe different outcomes but the world is internally consistent at any point in time and the wave equation continues to evolve as it does regardless).

Is decoherence needed to prevent 1), 2) or both?

Thanks.

• Schroedinger's cat died in 1935, when Schroedinger came up with this nonsense. A cat is a living being and living beings can't exist in a perfectly closed box. They die a few minutes after you put them in there for thermodynamic reasons. If Schroedinger had spent even a few minutes thinking about this before he wrote it down, he would have noticed, that the irreversible radioactive process happens in the interaction between the nucleus and the electromagnetic quantum field, everything after that is classical. – CuriousOne Mar 2 '16 at 23:29
• 1) If the cat is in state $\alpha$(alive)+$\beta$(dead) then the observer is in state $\alpha$(sees cat alive)+$\beta$(sees cat dead). You are positing that $\alpha=1$ and $\alpha=0$ simultaneously. Obviously this can't happen. 2) Presumably the time evolution of the state (dead) involves at most a phase shift. Dead cats do not come alive. – WillO Mar 2 '16 at 23:30
• @WillO, so decoherence has no effect here? – Ameet Sharma Mar 2 '16 at 23:33
• Of course it does... on the first $10^{-18}-10^{-15}m$ where the actual quantum process happens. – CuriousOne Mar 2 '16 at 23:48
• @CuriousOne, so what if decoherence did not happen... what would be the state of things? – Ameet Sharma Mar 3 '16 at 0:27

## 1 Answer

Think about it like this: Say Schrodinger's Cat is in his box in a windowless, soundproof room. An observer enters the room, closes the door, checks on the cat, closes the box back up. If the cat's alive, he stands there. If the cat's dead, he shoots himself. Now, before another observer opened the door, the first observer would be, like the cat, in quantum superposition and alive and dead at the same time until observed.

• Take the cat and everything else out of the equation. Enter a completely isolated room just barely large enough to contain you. How long are you going to be alive? – CuriousOne Apr 20 '16 at 16:22
• Can we get past the cat? It's an intentional absurdity. It can be Schrodinger's LED for all it matters. – Franklin Kuzenski Apr 22 '16 at 12:23
• That's what I just said. Take the cat out and analyze the problem correctly. When does decoherence happen? Schroedinger had it all wrong and so does everybody who believes in the cat being a useful teaching tool for quantum mechanics. – CuriousOne Apr 22 '16 at 18:21