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Schrödinger's cat: a cat, a flask of poison, and a radioactive source are placed in a sealed box. If an internal monitor detects radioactivity (i.e., a single atom decaying), the flask is shattered, releasing the poison that kills the cat. - Wikipedia

Schrödinger's cat experiment is claiming that putting the whole this "cat killer radioactive decay system" to a closed box(blocking photons) will disconnect all communication methods between me(observer) and radioactive source inside the box.

But, isn't that possible to communicate with the radioactive source through gravitational waves? I mean, suppose it's an Uranium-238 atom and it decayed to Uranium-234 and radiated an alpha particle (Helium atom), so isn't the decrease of mass for Uranium atom and position change for Helium atom will create gravitational waves? And I think I will be able to capture this information just by detecting the gravitational waves.

So where is the uncertainty on this experiment?

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    $\begingroup$ Well, I'm not well knowledged in these things, but I always thought the cat experiment was an analogy, so don't take it too literally. To what? I have no idea. $\endgroup$ – The Great Duck Oct 27 '16 at 5:17
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    $\begingroup$ @TheGreatDuck It was not an analogy - it was a ridicule. It's a bit sad how everyone picked it up entirely wrong and loads of sci-fi shows just keep repeating it as an example of how quantum physics works. Mind you, just because something is a ridicule (the name of the Big Bang theory was also an attempt at ridicule) doesn't make the premise wrong. $\endgroup$ – Luaan Oct 27 '16 at 7:17
  • $\begingroup$ @Luaan That's my point though, it may have been ridiculing quantum physics, but it's also an analogy. Just as the cat cannot be neither alive nor dead so can the particles not be in some "in between state". $\endgroup$ – The Great Duck Oct 27 '16 at 14:50
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All the paraphernalia of cat, box, etc are classical, i.e. obey classical mechanics and electrodynamics. They are a big fancy "detector" set up in order to illustrate the quantum mechanical probabilistic nature of a decaying nucleus.

The observation is made by the "internal monitor" which interacts with a decay product of the nucleus, and triggers the release of the poison. Not the outside observers of the whole setup.

If one had the whole system on a very accurate scale the change in mass due to the decay would register in the weight , no need for gravitational waves. Also the fall of temperature of the whole system as the dead cat cooled, from the shape of the black body radiation of the setup, one would know when the cat died.

The radioactive decay could be replaced with a random number generator, throwing of dice, for the decision of poison, a classical probability distribution The poor cat is just a detector of one instance in the probability distribution and black body radiation monitoring would show when the cat died.

It is a muddled, not a very good experiment imo, and confuses people.

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    $\begingroup$ It was not intended to be a good experiment. It was intended to use the absurdity of macroscopic superposition to show the absurdity of quantum superposition and thereby refute the Copenhagen interpretation. 81 years later, we see this as a quixotic exercise. $\endgroup$ – Eric Towers Oct 27 '16 at 5:34
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    $\begingroup$ Well, if we look at real examples of superposition, I would expect that if such an experiment was possible (and it very likely isn't - again, it was an attempt at ridicule, not a serious experiment), the mass of the system in a superposition would be observably different from the mass of the system "before decay" and "after decay" classically, just like the mass of the ammonia molecule is higher than it "should" be due to a superposition of two possible configurations of the molecule. $\endgroup$ – Luaan Oct 27 '16 at 7:21
  • $\begingroup$ Also it is not a very ethical experiment, but it was a good allegory at the time to get things started and people to grasp the basic concepts. $\endgroup$ – Mindwin Oct 27 '16 at 11:09
  • $\begingroup$ @annav My ultimate resolution is quantum level uncertainty cannot apply to the macroscopic scale. Because when a quantum decision made, it immediately creates gravitational waves and through gravity you can detect this quantum decision. Because nothing can block the effect of gravity right? $\endgroup$ – mertyildiran Oct 27 '16 at 19:51
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    $\begingroup$ Only in a thought experiment, since gravitational waves are so very very weak. But you do not need them. The shcrodinger cat experiment could be set up with another origin of the probabilistic nature, a random number generator, or a machine throwing dice for the poison to be released. The same dead or alive conundrum arises, the condition of the cat just one measurement from the probability distribution, which can be either quantum or classical and can be detected classically as I said . I will edit above. $\endgroup$ – anna v Oct 28 '16 at 4:24
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I think you are confusing the observer effect and the uncertainty principle. Schrödinger's cat is about the former, not the latter. Also, a point could be made about the fact that, according to general relativity, energy does bend space-time.... although one could then make a counter point about the fact that the center of gravity could have changed?

Anyway, I think Griffiths answers your question when he writes, in "Introduction to Quantum Mechanics":

The cat is neither alive nor dead, but rather a linear combination of the two, until a measurement occurs—until, say, you peek in the window to check. At that moment your observation forces the cat to "take a stand": dead or alive. And if you find him to be dead, then it's really you who killed him, by looking in the window. Schrodinger regarded this as patent nonsense, and I think most physicists would agree with him. There is something absurd about the very idea of a macroscopic object being in a linear combination of two palpably different states. An electron can be in a linear combination of spin up and spin down, but a cat simply cannot be in a linear combination of alive and dead. How are we to reconcile this with the orthodox interpretation of quantum mechanics? The most widely accepted answer is that the triggering of the Geiger counter constitutes the "measurement," in the sense of the statistical interpretation, not the intervention of a human observer. It is the essence of a measurement that some macroscopic system is affected (the Geiger counter, in this instance). The measurement occurs at the moment when the microscopic system (described by the laws of quantum mechanics) interacts with the macroscopic system (described by the laws of classical mechanics) in such a way as to leave a permanent record. The macroscopic system itself is not permitted to occupy a linear combination of distinct states. I would not pretend that this is an entirely satisfactory resolution, but at least it avoids the stultifying solipsism of Wigner and others, who persuaded themselves that it is the involvement of human consciousness that constitutes a measurement in quantum mechanics. Part of the problem is the word "measurement" itself, which certainly carries a suggestion of human participation. Heisenberg proposed the word "event," which might be preferable. But I'm afraid "measurement" is so ingrained by now that we're stuck with it. And, in the end, no manipulation of the terminology can completely exorcise this mysterious ghost.

The takeaway from the above paragraph is that the experiment is about the fact that there doesn't need to be a sentient observer for the "measurement effect" to occur.

I'd like to add to what Anna wrote: it doesn't matter if you're certain or uncertain about whether the cat is dead: whether or not you weigh the box, the cat is not in a superposition of states, because the actual "measurement" was made by the Geiger counter, not by the human observer.

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    $\begingroup$ Of course, there could be a third state: "Greebo had spent an irritating two minutes in that box. Technically, a cat locked in a box may be alive or it may be dead. You never know until you look. In fact, the mere act of opening the box will determine the state of the cat, although in this case there were three determinate states the cat could be in: these being Alive, Dead, and Bloody Furious." - Terry Pratchett, Lords and Ladies $\endgroup$ – Tim Oct 27 '16 at 3:11
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Gravitational waves emitted by the contents of the box would not cause significant decoherence. If you start with an initial superposition of the contents of the box of the form:

$$\left|\psi\right>=\frac{1}{\sqrt{2}}\left[\left|\psi_1\right> + \left|\psi_2\right>\right]$$

where the $\left|\psi_j\right>$ will evolve into different macroscopic states of the box, then adding a coupling of the contents of the box to the outside World, will lead to a state at time $t$ of the form:

$$\left|\psi_{\text{universe}}(t)\right>=\frac{1}{\sqrt{2}}\left[\left|\psi_1(t)\right>\left|\phi_1(t)\right> + \left|\psi_2(t)\right>\left|\phi_2(t)\right>\right]$$

where the $\left|\phi_j(t)\right>$ describes the physical state outside the box if the inside of the box is in state $\left|\psi_j(t)\right>$. If the overlap between the states $\left|\phi_1(t)\right>$ and $\left|\phi_2(t)\right>$ is zero (or very close to zero), then that amounts to the rest of the universe effectively having measured the contents of the box. So, we're then in one of the two MWI branches. Now, if the box only allows gravitational waves to leak out then the overlap between these two states will be almost equal to 1, the amplitude for gravitational wave emission is almost zero and the most dominant emissions will involve the cat moving in a different way in one branch compared to the other, also even the components of the states containing gravitons will have some overlap.

One can ask a similar but more down to Earth question. Why doesn't the emission of soft Bremsstrahlung photons cause interference patters to be erased in double slit experiments involving electrons? Whenever a charge accelerates, a large number of very low energy photons is emitted, in fact this number is infinite; the more low energy photons you take into account the more of these photons there are. The answer is that the states containing all these photons corresponding to the electron going through either one of the slits have a significant overlap.

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protected by Qmechanic Oct 26 '16 at 21:13

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