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I was wondering how the Bohm interpretation of quantum mechanics would affect the Schrodinger's cat experiment. Would it imply that the cat would be in one state and never in superposition, so the cat would always be dead or alive whether or not the observer looks or not?

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In Bohmian mechanics, the state of the system is not just given by the wave function $\psi$: it is given by the pair $(\psi, Q)$ where $Q$ is the position in the configuration space. Here $Q$ collectively denotes the positions of all the particles the cat is made of, their spin, … you name it. So $\psi$ is still the superposition

$$\psi=\psi_\text{dead}+\psi_\text{alive}$$

where the two wave functions on the right-hand side are well separated in the configuration space, but then $Q$ is either in the domain where $\psi_\text{dead}$ is non-zero, or in the domain where $\psi_\text{alive}$ is non-zero, domains which have no overlap to hammer the point home. Thus the cat is definitively either dead or alive, where "or" is exclusive. But we don't know the initial position of $Q$, so we can't know until we make a measurement (by opening the box in this case).

To go a little bit deeper, since Toby commented on this, a useful analogy is indeed that of the surfer: the configuration $Q$ of the cat is like a surfer and the wave function like a wave, the former riding on the latter. In the cat example, the wave function has two crests separated by a humongous trove. So the surfer riding one crest has no chance to end up on the other one (note that this implies what I wrote above, but add a dynamic aspect to it, by talking about the evolution of the configuration of the cat as time goes): if the cat is dead, it can't become alive again and vice versa.

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  • $\begingroup$ Not quite sure if I understand, I have very little knowledge of quantum mechanics. Also, how would the hidden variables play into this? $\endgroup$ – blueblast Sep 5 '17 at 14:25
  • $\begingroup$ The hidden variables are $Q$. Could you be a bit more specific about what you don't understand? I could then improve on my answer. $\endgroup$ – user154997 Sep 5 '17 at 14:40
  • $\begingroup$ Let me see if I got it: When the box is closed, we do not know the initial state of the cat(Q). But although we do not know it, it has to be either dead or alive(not both, like Copenhagen). Once we open the box, we can see which state it is in. $\endgroup$ – blueblast Sep 5 '17 at 21:48
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    $\begingroup$ Think of particles riding about on top of a wave - they are the 'hidden' variable. The wave is multidimensional on configuration space (3 dimensions for every particle in the system), so if the particles happened to be riding this wave in another part of the configuration space its movement would be different. If we could know the state of a system at any one time we would know exactly how it evolves forever. We can't possibly know this though as in BM there is only one wave function and so everything that measures the state of a system is a part of that system, thus affecting the measurement. $\endgroup$ – Toby Hawkins Sep 5 '17 at 22:42
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    $\begingroup$ This is it indeed. The other important piece is that even so we do not know the configuration of the cat for sure, we can assign a probability to it, which is computed from the wave function. In the classic Schrödinger cat, the probabilities are 50/50. Note that eventually, this is just metaphysics: there is no experiment we can use to distinguish Bohmian mechanics from the Copenhagen interpretation. $\endgroup$ – user154997 Sep 5 '17 at 22:45
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Since this came up again, here are my two cents:

I was wondering how the Bohm interpretation of quantum mechanics would affect the Schrodinger's cat experiment.

Bohmian mechanics give the same predictions as the usual non relativistic quantum mechanics, using a different model. AFAIK special relativity has not been successfully incorporated, so Bohmian predictions for the plethora of high energy experiments are moot.

Schodinger's cat is a proposed non relativistic experiment that could be carried out in a lab , except it is a sadistic one. In effect the cat is used as a geiger counter: if a count comes the cat is dead ,if it does not it is alive. In effect it is an instant in measuring the decay distribution of a radioactive substance. Bohmian mechanics deals with the probability of decay distributions as successfully as non relativistic quantum mechanics so there would be no change in the interpretation.

Would it imply that the cat would be in one state and never in superposition, so the cat would always be dead or alive whether or not the observer looks or not?

IMO this superposed cat gedanken experiment is just a gross exageration of a simple measurement. We do not angst whether the geiger counter exists, or whether a tick means a decay has happened, and not putting on the geiger counter it is both ticking and not ticking. It is all metaphysics and philosophy , again IMO

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  • $\begingroup$ Actually, I think special relativity is not the problem. Bohmian mechanics can be extended to deal with the fields of Quantum Fields Theory [1]: these fields become the hidden variables replacing particle positions. But then it is apparently unclear whether this extension can correctly deal with fermionic fields. Playing with fire here if our Motl sees this comment 😀 [1] Ward Struyve. Pilot-wave theory and quantum fields. Reports on Progress in Physics, 73(10):106001, 2010. $\endgroup$ – user154997 Sep 26 '17 at 20:20
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My superficial understanding is it exists in one state only since it does away with duality and the measurement problem.

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