This is not a duplicate, the other answers do not specifically solve the contradiction, nor do they give an exact answer.

I have read this question:

Are we so sure about superposition?

How do we know superposition exists?

Will there be two cats in the box (superposition) if I use virtual particles (magnetic field) to open the box in Schrodinger's experiment?

where Stéphane Rollandin in a comment says:

We never observe a system in superposition. What would "two cats" even look like? Superposition happens only when there is no observation.

If we can prove that superposition exists, then why can't we measure it?

where Cort Ammon says:

Superposition does not "exist," as much as it is a mathematical tool that we use to describe the world that does exist. We can't measure it because its not something to be measured. Superposition is a concept in mathematics.

and where knzhou says:

We do measure superpositions all the time. We can measure if light is diagonally polarized with a polarizing filter, but diagonal polarization is an equal superposition of horizontal and vertical polarization. Northeast is a superposition of north and east. But it's also just an ordinary vector on its own. Similarly, north is a definite direction, but it's a superposition of northeast and northwest. Totally analogously, quantum measurements will give you definite results. But the wavefunction after a measurement is still a superposition.

So this is a contradiction. One of them says we never observe superposition, the other one says we do, because even if the measurement gives us a definite result, it is still a superposition of many states.


  1. which one is right, can we (do we or did we ever) observe (measure) a quantum system in superposition?
  • 3
    $\begingroup$ The reason you can have apparently contradictory answers to this question is that the question is ill-defined. Different people have different ways of formulating and interpreting quantum mechanics. In a typical presentation using the Copenhagen interpretation, "measurement" is undefined, but there is an axiom saying that measurements always result in an eigenvalue of the thing being measured. This is essentially a quick and dirty way of approximating the behavior of decoherence. $\endgroup$
    – user4552
    Jun 24, 2019 at 3:14
  • 2
    $\begingroup$ What is "observation of superposition"? If you think "we see the same atom twice", we don't. If you think "ammonia has anomalous mass due to superposition of its two possible configurations", we do observe that. $\endgroup$
    – Luaan
    Jun 24, 2019 at 7:17

3 Answers 3


To answer the question, you have to specify what observable’s eigenstates the “superpositions” you’re talking about are superpositions of.

If you measure the energy, you always get one single energy eigenstate, not a superposition of energy eigenstates. The same applies for any other observable quantity.

But that one single energy eigenstate might be a superposition of eigenstates for a different observable, such as the $x$-component of angular momentum, at least when the other observable doesn’t commute with the one you measured.

The point is that “a superposition” is a meaningless phrase. Asking “Is this quantum state a superposition?” is like asking “Does this vector have multiple components?” It depends on the coordinate system you’re using! (This was knzhou’s point.) Different observables’ eigenstates provide different bases (coordinate systems, basically) for Hilbert space.

By contrast, “a superposition of eigenstates of observable $O$” is a meaningful, well-defined phrase.

  • 1
    $\begingroup$ So we measure eigenstates? I thought we measured eigenvalues... $\endgroup$ Jun 24, 2019 at 9:58
  • $\begingroup$ @thermomagneticcondensedboson you measure a state, which gives you a value. $\endgroup$
    – OrangeDog
    Jun 24, 2019 at 14:29
  • $\begingroup$ @OrangeDog in that case G. Smith got it backward, as he claims that one can measure the energy and get an energy eigenstate out of it. Do I misunderstand him? To me the energy is a value, and an eigenstate is a state. $\endgroup$ Jun 24, 2019 at 14:48
  • $\begingroup$ @thermomagneticcondensedboson you can also measure a value and infer a state. The word "measure" has two usages: you can measure yourself to get your weight, or you can measure your weight to infer how heavy you are. $\endgroup$
    – OrangeDog
    Jun 24, 2019 at 14:50
  • 1
    $\begingroup$ @thermomagneticcondensedboson - Eigenvalues are associated with eigenspaces, which consists of every eigenvector associated with the value. When you measure the eigenvalue, it is the same as measuring the entire eigenspace. I disagree with G. Smith saying this is not physically meaningful, because other measurements could then be performed to reduce the space to a single state (otherwise they would not be separate states). However, the distinction is meaningless for this particular measurement. $\endgroup$ Jun 24, 2019 at 16:25

As emphasized in G. Smith's answer, every state-vector is a superposition of other state-vectors. A Hilbert space is, among other things, a vector space. Any vector can be written as a linear combination (superposition) of other vectors, in uncountably many different ways.

The question implicitly has another piece, though. In this answer, I'll address that other piece, in case it helps resolve some of the apparent contradictions: Regardless of the superposition issue, can we really observe a state-vector (or density matrix)?

If we're being picky about the wording...

If we really want to be fastidious, then the answer is no. Technically, the concept of observing a state-vector (or density matrix) isn't meaningful. The state-vector/density matrix is not something we observe; it is something we choose in order to account for what we have already observed.

What criteria do we use to choose it? We choose it so that it gives the correct expectation values for observables that have already been measured (or that will be enforced in the design of the experiment). The model's pattern of time-dependent observables allows us to use that state to relate expectation values of different observables to each other, in particular to relate things we've already observed (or enforced) to things we haven't observed yet. That's how we make predictions.

What about "quantum state tomography", or other experiments that allegedly use measurements to determine what the state-vector or density matrix is? When we look past the loose language and consider what is actually being done in those experiments, it is consistent with what I wrote above. This is discussed in some detail in Unknown Quantum States: The Quantum de Finetti Representation (https://arxiv.org/abs/quant-ph/0104088).

If we're not being so picky about the wording...

Most physicists (including me) aren't usually so picky about the way things are worded, because wording things perfectly all the time is expensive, maybe even impossible. Language like "measuring the state" is slang for something that really is legitimate, as illustrated in Hari's answer.

To illustrate the point about language in a classical context, consider a coin-tossing experiment. If we assign a probability 0.5 to the event "coin lands heads-up", then consistency demands that we assign a probability very close to 1.0 to the event "almost exactly 50% of the coins land heads-up when we toss a jillion of them." This derived probability assignment is what we call a prediction. "Prediction" refers to an event to which we are obliged to assign a probability close to 100% based on consistency with other inputs.

Have we "measured" or "observed" the original 0.5 probability assignment? No, strictly speaking that doesn't even make sense. But we can still use it to make a testable "prediction" as defined above, and in that roundabout sense, we can validate the original probability assignment through observation.

(Again, this is only to illustrate the point about how loose language can have a legitimate foundation. I'm not trying to suggest a hidden-variables model for quantum theory.)

Summary (picky-wording version)

So, to address the question in a strict sense, the state-vector or density matrix is technically something we choose, not something we observe, regardless of how it may be expressed as a superposition. Maybe this perspective can help resolve some of the apparent contradictions between the various comments.


Suppose there is a quantum system that is in a superposition of two states $\psi_{A}$ and $\psi_{B}$ with eigenvalues $a$ and $b$. Consider producing many copies of this state such that they are exactly identical, and making a measurement on it that the wavefunction collapses to either $\psi_{A}$ or $\psi_{B}$, giving $a$ or $b$ respectively as the observable.

For example, one can restrict the spin to point in a particular direction and then measure its projection along an orthogonal direction. To make it more mathematical, consider that you prepare 'up spins', with respect to the z-direction, i.e it gives an eigenvalue of $\frac{ℏ}{2}$ when its z component of spin is measured. Now if a measurement is made for its projection along the x-axis, it will give an eigenvalue of either $\frac{ℏ}{2}$ or $-\frac{ℏ}{2}$. If a sufficiently large number of measurements are made on the identical copies prepared, the probability for each observation would be 0.5.

Here, during the measurements, we observe the quantum state in a particular eigenstate of the operator, there is no superposition with respect to the eigenstates of this operator. But since we prepared the system to be in spin up state with respect to the z-axis, and due to our observation that it can collapse to either spin up or down with respect to the x-axis after a measurement, we can conclude that our initial state was a superposition of these two states. Hence, when one makes a measurement, we always observe an eigenstate of the operator for that measurement, but it could be a superposition of eigenstates of another operator. Nevertheless, from the resulting observations, one can still conclude that our initial state was in a superposition of some eigenstates, as in the example above.


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.