Radioactivity and quantum superpositions

Question

In the Schrödinger's cat experiment 'there is a tiny bit of radioactive substance, so small that perhaps in the course of the hour, one of the atoms decays'. The rest of the experiment magnifies this into a macroscopic superposition, but I want to know more about the claim that the radioactive decay produces a superposition.

Firstly, has this been experimentally tested? Something along the lines of accelerating radioactive ions so there is a chance that they will decay while in flight (and so change trajectory), and then combining the decayed and undecayed parts to look for interference.

Secondly, the tiny bit of radioactive substance will still contain large numbers of atoms. Won't this cause problems? If the atoms were in a Bose-Einstein condensate, then I would expect that there could be a superposition of 'one (unspecified) atom decayed' and 'no atoms decayed', but they're not, so a specific atom will decay. Won't that mess things up?

Answer 1

The existence of superpositions of the type $$|\psi\rangle = \exp(-\Gamma t/2)|{\rm undecayed}\rangle+|\text{decayed pieces}\rangle$$ is a fact that follows from quantum mechanics and other tests demonstrating that it's right.

However, it's hard if not impossible to measure the interference - i.e. the relative phase etc. - between the undecayed and decayed pieces because all measuring apparatuses I may imagine (or we may imagine) that determine something about the decaying particle or its products do also measure the number of particles at the same moment.

What you would need is a measurement of the probability that $|\psi\rangle$ is found in the state of the type $$a|{\rm undecayed}\rangle + b|{\rm decayed}\rangle$$ i.e. you need a measurement sensitive on the inner product of $|\psi\rangle$ with the superposition state above. However, the superposition state above gets first "collapsed" either into the decayed piece or the undecayed piece at the beginning of the measurement. The measurement may also find something else but I don't believe there is a doable way to avoid the measurement of the "has the particle decayed" at the same moment.

I don't have a rigorous proof. But I may tell you why it's different from the normal interference experiment. In the double-slit experiment, you have a basis composed of states "slit A" and "slit B". The relative phase of the amplitudes in front of these two states may be measured on the photographic plate of the interference pattern because the electrons in the "slit A" and "slit B" may reinterfere behind the slits. Both states "slit A" and "slit B" may evolve into the electron at a given place on the screen, with a nonzero probability amplitude.

However, a decayed particle has no chance to "undecay" again so the final states you could measure, analogous to the photographic plate, simply have to deal with the decay products. At most, you could measure the interference between the decays that occur at different times. This is actually a possible positive answer to your question because similar things may be measured. Imagine Feynman diagrams for a decay. The interaction point must be integrated over some region. If you may guarantee that the decay only occurs at some regions of spacetime, these points will interfere with each other. One must first calculate the amplitude of the process by the integral and then square its absolute value to get the decay probabilities. This is effectively the same thing as you suggested because histories where the particle was decayed at some moment will interfere with histories where it wasn't decayed yet at the same moment. We can "ban the decay" in some condensed-matter-like, composite situations. By construction, it's obvious that the interference will work as quantum mechanics predicts because these experimental setups are made out of pieces that are known to behave according to quantum mechanics.