Today I came across a paper, "Experiments testing macroscopic quantum superpositions must be slow," by Mari et al., which proposes and analyzes a thought experiment involving a first mass mA placed in a position superposition in Alice’s lab, the mass mA producing a gravitational field that potentially affects a test mass mB in Bob’s lab (separated from Alice’s lab by a distance R), depending on whether or not Bob turns on a detector. The article concludes that special relativity puts lower limits on the amount of time necessary to determine whether an object is in a superposition of two macroscopically distinct locations (versus a mixed state).

But the problem is that, as far as I understand, there is no way to determine whether an object is in a superposition at all (versus a mixed state)!

A superposition is determined by doing an interference experiment on a bunch of “identically prepared” objects (or particles or masses or whatever). The idea is that if we see an interference pattern emerge (e.g., the existence of light and dark fringes), then we can infer that the individual objects were in coherent superpositions. However, detection of a single object never produces a pattern, so we can’t infer whether or not it was in a superposition. Further, the outcome of every interference experiment on a superposition state, if analyzed one detection at a time, will be consistent with that object not having been in superposition. A single trial can confirm that an object was not in a superposition (such as if we detect a blip in a dark fringe area), but no single trial can confirm that the object was in a superposition. Moreover, even if a pattern does slowly emerge after many trials, every pattern produced by a finite number of trials – and remember that infinity does not exist in the physical world – is always a possible random outcome of measuring objects that are not in a superposition. We can never confirm the existence of a superposition, but lots and lots of trials can certainly increase our confidence.

In other words, if I’m right, then every measurement that Alice makes (in the Mari paper) will be consistent with Bob's having turned the detector on (and decohered the field) -- thus, no information is sent! No violation of special relativity! No problem!

Am I wrong? Is there a way to test whether a particular object is in a coherent superposition? If so, how? If not, then why do so few discussions of quantum superpositions mention this?

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    $\begingroup$ Of course you can determine if something is in a superposition. For example, a particle with definite momentum is in a superposition of different positions. If you know the momentum, you know that its position is in superposition. Or, if you start with a particle with spin up, and then tilt the spin, then it ends up in a superposition of up and down. $\endgroup$
    – knzhou
    May 19, 2020 at 21:36
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    $\begingroup$ That's at least what the standard formalism says. You could also say "bullshit, there's no such thing as real superpositions, everything has a definite value and we just happen to not know it". That's an alternative that has been investigated in thousands of papers, called a "hidden variable theory". The problem with this approach is that it ends up incompatible with special relativity, and also all known ways of setting this up are pretty mathematically clunky. $\endgroup$
    – knzhou
    May 19, 2020 at 21:39
  • $\begingroup$ If you think you've suddenly found a way to turn quantum mechanics into straightforward classical mechanics, you haven't -- you just haven't understood the reasons that it's hard! Physicists have made thousands of attempts. Nobody wants it to be hard; there is no conspiracy going on, nature is just how it is. $\endgroup$
    – knzhou
    May 19, 2020 at 21:41
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    $\begingroup$ This is a very condescending comment. I am not interested in mathematical discussions of noncommuting observables. Nor am I pretending to understand QM. I am simply asking a (pretty clear) question about whether there is a way to test whether an object is in a position superposition, versus a mixed state. $\endgroup$ May 19, 2020 at 22:45
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    $\begingroup$ @AndrewKnight you could probably make your question clearer (and shorter) and get better answers. E.g., your question doesn't mention mixed states. $\endgroup$
    – innisfree
    May 19, 2020 at 22:50

3 Answers 3


Straightforward Answer

Every state is obviously in a superposition of eigenstates of some observable. If you prepare your state as an eigenstate of an observable and claim that now it's not in a superposition, it's simply wrong because an eigenstate of one observable is not an eigenstate of many many other observables who do not commute with your observable, and thus, it still is in a superposition of eigenstates of some observable. As @knzhou mentions in the comment, if you prepare a definite momentum state, it would be in a superposition of infinitely many position eigenstates (and vice-versa). You cannot get around superposition in quantum mechanics because non-commuting observables exist in quantum mechanics (otherwise, it'd be classical mechanics).

OK, but what if I don't believe non-commuting observables exist?

Well, you'd be hard-pressed to believe so if you want to maintain sanity. Borrow a single spin $\frac{1}{2}$ particle from Stern (Gerlach wouldn't respond because he's mad about not getting a Nobel ;)) and do some experiments with it.

  • Measure its spin state in $z$ direction. Let's say you find it to be up. Measure it again (as many times as you like), it would still be up. OK, so it is definitely in the spin-up state in $z$ direction.

  • Now, measure its spin state in $x$ direction. It doesn't matter what you get. Now, measure it in $z$ direction again. OK, let's say you are lucky and you get it to be up again. But we want to be sure of this phenomenon that if you measure the spin state to be up in $z$ direction, then measure its spin state in $x$ direction, and then again in $z$ direction then you'd again get the spin state to be up in $z$ direction. So, we repeat the experiment many many times. And you'd quickly notice that this doesn't always happen. In fact, it happens exactly $50\%$ of the times. Other $50\%$ of the times, you start with a spin-up state in $z$ direction, measure its spin state in $x$ direction, and then when you measure it again in the $z$ direction, it would come out flipped.

  • This conclusively shows that the spin state in $z$ and $x$ directions cannot be observed simultaneously. In other words, the two observables don't commute. This immediately implies that observing the one necessarily prepares the particle in a state of superposition of eigenstates of the other.

So, respecting your wish, I didn't use any identically prepared multiple copies of the same state. I just took one state, did a bunch of experiments with it, and concluded that observables don't commute, and thus, each state is necessarily a superposition of some observable.

Finally, I'd mindread here. What you seem to be confused about is that you cannot determine what a given unknown quantum state is if you only have a single state. For example, if you want to know what the state is, say, in the momentum basis, you won't be able to know which all momentum eigenstates are participating in the superposition with what coefficients to produce the given state. And when you measure the momentum, the superposition over the eigenstates of momentum would be lost and you'd only get a specific momentum eigenstate. If this is what you're saying, I have a few words to add.

If you're given an unknown quantum state, it's obviously true that you cannot determine what the state is unless you have multiple identical copies of the state. But that is the point of quantum mechanics that you can't just observe a state. There is a distinction between the state and the observables (the distinction arises out of non-commuting observables and thus, is absent in classical physics). You can simply observe the observables and the post-observation state would be the projection of the initial state onto the eigensubspace of the observable corresponding to the observed value. And the projection won't tell you anything about the initial state. You will need multiple identical copies of the state to determine what an unknown given state was, this is not a "gotcha" on quantum mechanics, it's what necessarily arises out of non-commuting nature observables.

Additional Comment

Of course, the post-measurement state wouldn't be a superposition of eigenstates (with different eigenvalues) of the measured observable. But this doesn't mean that the post-measurement state wouldn't be a superposition (for all the reasons described above). The fact that upon measurement, you don't get a superposition of eigenstates (with different eigenvalues) of the measured observable isn't something weird, rather, it's an obvious requirement of having a consistent definition of what a measurement is. It'd be completely meaningless to say that I measured the state in a $z$ direction and find it to be in a superposition of spin-up and spin-down states in the $z$ direction.

  • $\begingroup$ Sorry, but I'm annoyed by your answer and @knzhou. OBVIOUSLY a momentum eigenstate implies a superposition of positions. That is CLEARLY not the point I was making, or the question I was asking, in the above. It seems obvious that you both only read the title question and ignored the rest. What I meant, which I thought was obvious, was whether there was a way to test whether a given object is in a superposition VERSUS a mixed state. $\endgroup$ May 19, 2020 at 22:44
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    $\begingroup$ @AndrewKnight It was very unclear that you were asking about the distinction between a quantum state and a mixed state as evident by the fact that you mentioned "mixed state" exactly zero times in your question. $\endgroup$
    – user87745
    May 19, 2020 at 22:51
  • $\begingroup$ And, by the way, unlike the reply by @knzhou, your reply did NOT seem condescending -- it seemed like you were genuinely trying to answer my question, and I do appreciate that. $\endgroup$ May 19, 2020 at 22:51
  • $\begingroup$ @AndrewKnight I suppose I ruined my modesty repo with my last comment 🤦🏽‍♂️😛 In any case, if you can explicitly address how your question relates to the discussion of pure vs. mixed states by editing your question, that'd be super helpful for people to address your question more clearly. $\endgroup$
    – user87745
    May 19, 2020 at 23:01
  • $\begingroup$ Done. thanks... $\endgroup$ May 20, 2020 at 0:42

Is it ever possible to determine whether an object is in a quantum superposition?

I think you need to make the question more precise. Otherwise it's susceptible to more than one interpretation, and the answer could be either yes or no.

First of all, you need to distinguish between coherent and incoherent superpositions, as in the Mari paper that you reference.

Second, you need to think about when and relative to what state this is being determined.

Let's discuss several interpretations.

(A) If I put a silver atom through a Stern-Gerlach spectrometer, and it deflects in a certain way, then I have measured its spin. Say I've measured its spin to be $s_z=+1/2$, now. Note also that the SG apparatus does not change the spin of a particle, so there are no concerns about whether the object is so delicate that the measurement process has disturbed it. The fact that it's in the state $s_z=+1/2$, now, means that it's automatically in a superposition of $s_x=\pm 1/2$, now, and this superposition is coherent. So in this sense, I can definitely determine that it's in a coherent superposition of these two $s_x$ states, now.

(B) Or a different way of interpreting your question is that maybe before the atom went into the SG apparatus, it could have been in some state like $s_x=+1/2$, which is a coherent superposition of $s_z=\pm 1/2$. So then in the language of the Copenhagen interpretation, we've caused the wavefunction to collapse into $s_z=+1/2$, and we can never recover the information about what the state was before the measurement (because measurement is a nonunitary process). Or in MWI, what has happened is that we're now entangled with the spin, and because of decoherence, we will never be able to tell that our own wavefunction also contains a part that observed $s_z=-1/2$. So in this way of stating the question, the answer is that you can find out about the state after measurement, but you will never know about the state before measurement.

(C) Determining that the particle used to be in an incoherent superposition of states does seem impossible to me, for the reasons described in your question: this can only be determined for a statistical ensemble, not for a single particle that someone gives us. In MWI, this would be the kind of thing that would happen if someone has already measured the state of the particle, decoherence has already happened, and you want to know what the state of the particle used to be before the first measurement, which caused the split between two worlds. That's impossible in standard quantum mechanics, although maybe if we had a viable nonlinear version of quantum mechanics, we could do this, because the nonlinearities could allow the two worlds to interact rather than just superposing.

  • $\begingroup$ Notice that non-linear quantum mechanics is fiction at this point, and it is very likely will continue to be so. The framework of quantum mechanics is an isolated island. It's proven to be nearly impossible to tweak any of its features even slightly and obtain something even remotely resembling quantum mechanics. This is a very important piece of information. To contrast, for example, you can tweak GR and still get something that reasonably looks like GR. $\endgroup$
    – user87745
    May 19, 2020 at 23:15

I’ll start my answer with the quote of a comment:

If you think you've suddenly found a way to turn quantum mechanics into straightforward classical mechanics, you haven't -- you just haven't understood the reasons that it's hard! Physicists have made thousands of attempts.

The double slit experiment is the experiment showing the wave behavior of light. The interference pattern shows the superposition of light, from the amplification of rectified wave crests to the cancellation of light wave crests and troughs. Does it?

No, it doesn't. Light consists of photons and they do not interact. To photons intersect each over and do not cancel out or double their energy. It is discussed on PSE many times, that no cancelation in the black area of the fringes takes place.
But please take in mind that the calculation with sinusoidal functions works well. This are easy to use equations to conclude from the fringes about the frequency of the used light.

Furthermore, the fringes also appear behind a single slit. No interaction of light from two slits. The explanation of the interference was transferred to both sides of the single slit. Now how about: no slit at all, but a single edge? Are you surprised that even behind a single edge the light deflection and by this of fringes can be seen?

The attempt 1

If you have no problem with the last sentence, there is a simple explanation for the “interference” pattern. Light is deflected by edges, and in the interaction between light and edges this deflection could be uneven, sometimes stronger and sometimes weaker. This is because photons actually have a wave behaviour; their electric and magnetic field components are sinusoidal functions.

The photon field components change periodically over time. When interacting with the surface electrons of an edge, the photon is deflected in a way that depends on its (varying) field strength at the time of interaction. By the way, this is the reason why photons are deflected behind edges, while electrons are only deflected away from edges.

I am not interested in mathematical discussions of noncommuting observables. Nor am I pretending to understand QM. I am simply asking a (pretty clear) question about whether there is a way to test whether an object is in a superposition ...

Please read about How is the orientation of quantum spin testing of entangled particles made the same between two distant tests?.

The attempt 2

The pairwise generation of entangled photons (by their spin) has a part of uncertainty. The point is that the parallel-antiparallel orientation of the two particles could point in any direction (until now we have no better conditions for the pair production). The measuring instrument (e.g. a grid), in turn, must also be oriented in any direction, no matter where by 360°. After a series of measurements we obtained a correlation between the two entangled particles. After many experiments the entanglement is assumed to be a fact. It is derived empirically and always has this statistical component. Only in some cases we measure the entanglement, in the other cases the result is unknown.

The answer to your question is that statistically we know whether two particles are superimposed. Since there are no better instruments, the uncertainty of our not knowing collapses after a series of measurements. That is what Zeilinger does. He stabilizes the result, shields it from the environment and filters out the coincidences. That's how they tried to regulate quantum computing.

Is it ever possible to test whether an object is in a quantum superposition (versus a mixed state)?

Take it the other way around. Numerous experiments are carried out to create superpositions, in particular with the help of the polarization of photons or the spin coupling of electrons and atoms. The results of these experiments show that we are able to create superpositions.


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