If you were a unentangled observer, would you see the correct state of Schrödinger's cat?

Question

If you were able to peer into the box without opening it and without interacting with the wave function, I would think that you can see the true state of the cat.

Therefore, I assume the cat is at a definite state and not either dead or alive. Does this not mean that the cat can never be in the dead/alive state.

Does Schrödinger's example pertain to the observers known information and interaction perspectives of a entangled observer through taking a sample?

I am relating this to the "many worlds" theory that spans off this example, isn't it incorrect to generate a theory off the observers limited data and assume it projects to physical phenomenon, ie. Just because an observer doesn't know that it generates a branch in reality.

Any clarrification and explinations will be appreciated, thanks in advance.

score 2 · Answer 1 · answered Mar 17 '20 at 09:57

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If you were able to peer into the box without opening it and without interacting with the wave function, I would think that you can see the true state of the cat.

You can not observe it without interacting. Even just "looking" at it requires photons to be bounced of the cat / elementary particle back to your eye, where the bouncing is the interaction.

answered Mar 17 '20 at 09:57

infinitezero

1,224

But if it was possible for a unentangled observation, would the cat be in a definite state? – leopardxpreload Mar 17 '20 at 09:59
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If I can do then . – simon at rcl Mar 17 '20 at 14:52

score 0 · Accepted Answer · answered Mar 17 '20 at 11:10

Schrodinger's cat is a misleading example of quantum mechanics, I will never be tired to explain. The quantum mechanical state is that of a nucleus which has a given decay probability. How do we measure the QM decay probability?

We take a sample of the nuclei and by weight we know how many nuclei there are. Then we have a set up with a Geiger counter recording the numbers of particles hitting it per second. The plot will let us measure the lifetime. Each nucleus has a probability to decay. Before it decays , it is whole, after it decays it has become decay fragments.

Let us suppose we are so good with nanotechnology that we can isolate a single nucleus . We then wait for the geiger counter to blip. We do not know if the nucleus is whole or not, because it is a quantum mechanical probability of decay; until we hear/record the blip. Before it blips the nucleus is whole, after it blips it is in pieces.

The inhumane cat experiment replaces the geiger counter with the cat. The blip instead of being recorded , unlocks poison which kills the cat. If we wait enough lifetimes the cat will be dead with high probability.

It is a bad experiment and a bad display of quantum mechanics uncertainties, confusing the basic quantum mechanical interaction with macroscopic measurement. There is no entanglement anyplace in the cat experiment.

For more details see my answer here.

Fantastic answer @anna v - this is exactly how I imagined the whole concept of "measuring" that affects the measurement. I think of it as using a thermometer to measure a temperature, the thermometer can affect the temperarure its measuring if there is a difference and also if it takes or adds energy to the system. The measurement of the thermometer will not exactly represent the system, because its uses the energy of the system to represent the thermal energy. I was watching a video that concluded that "because we don't know the state that it can be in any state and therefore is in both" — leopardxpreload, Mar 17 '20 at 11:44
I on the contrary like the thought experiment, although I would today pick another victim in place of the cat. It nicely eliminates interpretations and leaves only the ensemble or statistical interpretation as reasonable. — my2cts, Mar 17 '20 at 14:44

If you were a unentangled observer, would you see the correct state of Schrödinger's cat?

2 Answers2