Suppose we can make an arbitrarily precise preparation of a Schrödinger's cat (and isolate it arbitrarily well so that decoherence is not a problem). If we prepare lots of cats in this state, what measurement can tell us whether these cats were in a superposition of dead/alive states or just in a mixture of them? Of course, I mean experiment not on the poison-triggering decaying particle, but on the cat itself.

I guess we need some sort of double-slit or other experiment which would let us see interference between dead/alive states, but I can't seem to come up with an experiment suitable for such a large "particle" as a cat.

To clarify: let's model our "cat" as an object with two possible states, like a supercooled fluid and a crystal. So the supercooled fluid would correspond to the cat which is alive, and crystal would be a dead cat. I.e. in this model the decaying particle would create a nucleation site.

The question then is how to experimentally distinguish the state $\alpha\left|\mathrm{liquid}\right\rangle+\beta\left|\mathrm{crystal}\right\rangle$ from the state $\alpha\left|\mathrm{liquid}\right\rangle\left\langle\mathrm{liquid}\right|+\beta\left|\mathrm{crystal}\right\rangle\left\langle\mathrm{crystal}\right|,$ given that we can prepare arbitrary number of copies of the system in that state.

  • 1
    $\begingroup$ <rant> Dead is not a state vector, and neither is Alive. The common statement that the cat is in a superposition of dead and alive states is misleading.</rant> It might in principle be possible to detect a superposition of macroscopic objects, but I agree that it's hard to see how this could actually be done. $\endgroup$ Feb 28, 2015 at 8:18
  • $\begingroup$ Why the downvote? Aside from my rant about dead/alive superposition this seems a perfectly reasonable question. $\endgroup$ Feb 28, 2015 at 11:18
  • $\begingroup$ Your rant is probably exactly why it got a downvote (although it wasn't from me). Schroedingers Cat is so misrepresented in popular culture and media, we have a bunch of people walking around thinking that it is, in practice, possible to prepare dead and alive cat $\endgroup$
    – Sean
    Feb 28, 2015 at 12:47
  • $\begingroup$ A problem with the question is how do you define dead/alive? Isn't this quite difficult even in classical context? $\endgroup$
    – jinawee
    Feb 28, 2015 at 12:52
  • 1
    $\begingroup$ OP, please clarify if you are asking: "Are there measurement techniques to detect whether a single state is a superposition?" or "Are there measurement techniques to detect whether a macroscopic object is in a superposition?" or "Are there techniques to detect whether Schrödingers cat is in a superposition of dead and alive?". As John Rennie says, the latter question is nonsense, but the first two would be equally fine. $\endgroup$
    – ACuriousMind
    Feb 28, 2015 at 14:05

1 Answer 1


The problem is that

suppose we can make an arbitrarily precise preparation of a Schrödinger's cat

is not quite what you need to do. You need to extend your capabilities so that you can make an arbitrarily precise implementation of the unitary which takes $\newcommand{\k}[1]{\left|\,\mathrm{#1}\right\rangle}\k{alive}$ to $U\k{alive}=\tfrac1{\sqrt{2}}(\k{alive}+\k{dead})$. If you want to be able to detect this, you need to stay reasonably confined to a small dimensionality, which then implies by orthogonality that $$U\k{dead}=-\tfrac1{\sqrt{2}}(\k{alive}-\k{dead})$$

The idea is then to run your cat twice through the black box, which will invariably give $$U^2\k{alive}=\k{dead}.$$ This is impossible to achieve with purely incoherent population driving: if your box kills live cats 50% of the time and revives dead cats 50% of the time, then a 50-50 dead/alive mixture can only result in a 50-50 dead/alive mixture. In essence, you're running Rabi oscillations instead of incoherent decay and driving.

This method is going to be pretty hard to implement with an actual cat, insofar as our current technology prevents us from raising the dead. With a more reversible transition, like your liquid-crystal one, there's more hope - you just need to be careful and coherent in your control of both directions of the transition.

  • 1
    $\begingroup$ Or, even better, define $U$ so that $U\k{dead}=+\frac1{\sqrt2}\left(\k{alive}-\k{dead}\right)$, then you'd get $U^2\k{alive}=\k{alive}$, i.e. 100% revival :) . $\endgroup$
    – Ruslan
    Jul 2, 2015 at 9:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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