1
$\begingroup$

Assume that the walls of the box prevent any interaction between the inside and the outside. Are the different quantum states within the box (living cat, dead cat, and all permutations thereof) coherent before the box is opened? If so, then does this mean that the states can interfere? If not, then what made the quantum states decohere? Does this mean that quantum states can decohere without interacting with another system?

$\endgroup$
1
  • $\begingroup$ The "other systems" are within the box, not outside. As the answer reminds you, any QM interference is lost and gone by the time the poison is released following decay observation, or not; and a live cat will never interfere with a dead cat: this is the root of the absurdity of the specious cat paradigm. $\endgroup$ Aug 9, 2020 at 20:55

1 Answer 1

4
$\begingroup$

Schrodinger's cat is a bad example for coherent and incoherent quantum states. If the experiment is carried out the only quantum dependence is on the quantum probability for the radioactive decay which will activate the poison. All the rest are the detector effects, which are incoherent as far as quantum mechanics goes, including the cat which is used as a geiger counter for the decay of the isotope.

Are the different quantum states within the box (living cat, dead cat, and all permutations thereof) coherent before the box is opened?

So the answer is no, the cat states are not coherent with the basic quantum mechanical interaction of decay, except as a detector. In the same way the tracks in a bubble chamber photograph have nothing to do with the main quantum interaction, whether one looks at the film or not. The main quantum mechanical probabilistic interaction is at a point where the tracks come out.

bubble chamber

Bubble chamber picture showing dark proton tracks

This picture was taken in the CERN 2-metre hydrogen bubble chamber expose to a beam of negative kaons with an energy of 4.2 GeV, entering from the bottom These beam particles produce parallel trails of bubbles.

If so, then does this mean that the states can interfere?

The outgoing tracks (analogue cat) do not interfere with each other once they have left the quantum mechanical point interaction of a K- hitting a proton. By interacting with the hydrogen in the chamber and leaving small bubbles of their path, all quantum coherence is lost.

By interacting with the poison mechanism and releasing the gas all quantum coherence is lost. The death of the cat whether seen by an observer or not is irrelevant to the basic quantum mechanical probabilities, except as a detector.

If not, then what made the quantum states decohere?

Quantum states of many particles( there are $10^{23}$ per mole of matter are decohered except for special cases as superfluidity, superconductivity ...

Many particle states are described by the density matrix formalism.

Does this mean that quantum states can decohere without interacting with another system?

Interactions are necessary for decoherence of quantum mechanical states. As far as the cat experiment, the interaction is between the decay particle and the poison mechanism that introduces decoherence. The cat was never in a quantum mechanical state.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.