Let's say we have a wavefunction $\psi$ and a measurement operator $\hat A$.

I understand how eigenvalues and eigenvectors of $\hat A$ describe the possible outcomes of the measurement.

I also understand that the average measurement can be computed as $\langle \psi|\hat A|\psi\rangle$.

It's still not clear to me what the direct meaning of $\hat A |\psi\rangle$ is. It is a wavefunction; how does its corresponding quantum state relate (in physical terms) to the original state $\psi$ and the measurement $\hat A$?

  • 2
    $\begingroup$ It's a projection of the original wavefunction to the subspace of possible outcomes of that measurement. Is there a physical meaning to this mathematical operation? No. We can do the math independently of whether we do the measurement, or not. The math simply tells us the possible outcomes of the measurement which, in some sense, is a description of possible futures in case we decide to do the corresponding measurement. Only when we do does nature tell us which of these possible futures has become the new present, and no amount of math can predict that "choice". $\endgroup$
    – CuriousOne
    Sep 9, 2015 at 9:39
  • $\begingroup$ No need for a direct physical meaning. All formalisms for quantum mechanics have an equivalent for $\langle \psi \vert \hat{A} \vert \psi \rangle$, but not all of them have an equivalent for the wavefunction, Hilbert spaces or operators. It is just part of the mathematical formalism. $\endgroup$
    – Slereah
    Sep 9, 2015 at 9:46
  • 1
    $\begingroup$ @CuriousOne Technically $A\psi$ is not a projection, for $A$ is not necessarily a projector. $\endgroup$
    – yuggib
    Sep 9, 2015 at 9:46
  • $\begingroup$ @yuggib: True. I'll retract my statement. $\endgroup$
    – CuriousOne
    Sep 9, 2015 at 9:54
  • $\begingroup$ @CuriousOne The statement is not wrong in its essence, it was just a small technical remark ;-) $\endgroup$
    – yuggib
    Sep 9, 2015 at 9:55

3 Answers 3


I do not think that the action $A\psi$ has a direct physical meaning, when $A$ is a generic observable.

This is because the interpretation of a quantum system as a mathematical model yields the wavefunction and its corresponding Hilbert space as a sort of byproduct. In fact, the state may not always be a wavefunction: without entering too much into details, let's say it is just a mathematical object suitable to evaluate observables.

The mathematical objects with direct physical relevance are observables and states; and the action of an observable on the state (or vice-versa) is assumed to be the evaluation (averaging) process.

Nevertheless, since this (abstract) mathematical system that has QM as a model corresponds exactly to the Hilbert structure of wavefunctions and self-adjoint operators, it may be useful and important to study the behavior of $A\psi$, in order to improve the knowledge of the system, as well as to make physical predictions.

For example, the behavior of $H\psi$, where $H$ is the Hamiltonian operator (energy observable), is directly related with the time evolution of the system (by Schrödinger equation).

  • $\begingroup$ The action of am observable is most definitely not the averaging process. The averaging process is done by a completely separate second device that records the cumulative frequency of the various outcomes for repeated interactions on identically prepared systems. $\endgroup$
    – Timaeus
    Sep 9, 2015 at 15:11
  • $\begingroup$ @Timaeus Mathematically, the states are characterized defining the evaluation of each observable on it. This is the modern (well, it dates back to the 1950s at least) mathematical way of defining QM systems, but may not be the only way (or the best one). Still, it is in good agreement with physical observations. I am not talking about measurement processes or devices, just on how states and observables are defined mathematically (and the fact that it does not strictly require the introduction of an Hilbert space). $\endgroup$
    – yuggib
    Sep 9, 2015 at 15:35
  • $\begingroup$ If the question was labelled mathematical-physics then your answer would be understandable. When a physicist asks for the direct meaning they are asking about reality not a out math, especially not math that ignores the fact that you get individual results. You are getting your agreement by ignore the parts you ignore (getting individual results) and frankly you make it sound like you can do more measurements than you have evidence you can do. $\endgroup$
    – Timaeus
    Sep 9, 2015 at 15:57
  • $\begingroup$ @Timaeus The question is asking about the (eventual) physical meaning of a mathematical construct. I am not ignoring the fact that you could get individual results; simply it is not directly a part of the question. Anyways, not every observable has a discrete spectrum, i.e. only a discrete set of admissible results. The probability of a value to be measured is nevertheless encoded in the spectral decomposition of the operator (by a mathematical point of view). $\endgroup$
    – yuggib
    Sep 9, 2015 at 16:09
  • $\begingroup$ You explicitly said the action of the observable on the state was the averaging process. That's a lie people tell when they want to pretend like their story is just as good as the correct story about how observations really happen. The real story is that things evolve according to the Schrödinger equation. So you need a Hamiltonian that allows the observable in question to couple to something that can separate. This is physics not story time. And stories can be great if labeled as such and if they help people remember, learn, visualize, compute, understand, etc. $\endgroup$
    – Timaeus
    Sep 9, 2015 at 16:35

To add to Yuggib's Answer, which I am in complete agreement with: I have never particularly liked the name "operator" for an "observable", because the former implies a mapping and, therefore, that the image $\hat{A}\,\psi$ has a direct physical meaning. As in Yuggib's Answer, there is in general no direct physical meaning. Rather, an "observable", as I like to think of things, is an operator together with a recipe for how to interpret its predictions when state $\psi$ prevails, namely, that:

  1. The probability distribution of the measurement modelled by the observable has $n^{th}$ moment $\langle \psi|\hat{A}^n|\psi\rangle$, whence, with all the moments calculated thus, we can derive the distribution itself.

  2. Immediately after the measurement, the quantum state $\psi$ is an eigenvector $\psi_{A,\,j}$ of $\hat{A}$, the measurement outcome is the corresponding eigenvalue and the "choice" of eigenvector is "random", with the probability of its being $\psi_{A,\,j}$ given by the squared magnitude $|\langle \psi | \psi_{A,\,j}\rangle|^2$ of the projection of the state $\psi$ before the measurement onto the eigenvector $\psi_{A,\,j}$ in question.

Given point 1. above, another useful quantity to calculate is $\mathscr{P}(k)=\langle \psi|\exp(i\,k\,\hat{A})|\psi\rangle$, which is the characteristic function of the probability distribution.


Observables correspond to particular things you can do in the lab (or observe in nature). So let's first talk about something you can do in the lab.

You can take a particle with spin and subject it to an inhomogeneous magnetic field. A particle with spin has a magnetic moment proportional to the spin and we know to Hamiltonian for a particle with a magnetic moment in an externalmagnetic field. So we get a term in the Hamiltonian proportional to $B_x\hat\sigma_x+B_y\hat\sigma_y+B_z\hat\sigma_z.$ At each point, this is an Hermitian operator. By setting up the magnetic field to point in the $\hat z$ direction and to be inhomogeneous in that direction we can send a beam in that is eigen to $\hat\sigma_z$ and it will be deflected up or down (depending on the eigenvalue and on how we made the field vary in the z direction) and this deflection literally happens because of the evolution determined by the Schrödinger equation when you have a term in the Hamiltonian proportional to $B_x\hat\sigma_x+B_y\hat\sigma_y+B_z\hat\sigma_z.$

And so now when you have a general state and again evolve it according to the Schrödinger equation when you have a term in the Hamiltonian proportional to $B_x\hat\sigma_x+B_y\hat\sigma_y+B_z\hat\sigma_z$ then the incoming spatial state splits its beam into two beams, one going up and one going down and the size of the two beams is such that the total current in each beam has a ratio equal to the ratio of the projection of the spin state onto the eigenstates of $\hat\sigma_z.$ Again, that fact is determined by the Schrödinger equation evolution for he actual state of the subject (the thing with spin) and the Stern-Gerlach device (the thing with the inhomogeneous magnetic field). And the spin state evolves so that the branch that is spatially deflected up becomes spin up and the branch that is deflected down becomes spin down. And again this cones out from the evolution determined by the Schrödinger equation when you have a term in the Hamiltonian proportional to $B_x\hat\sigma_x+B_y\hat\sigma_y+B_z\hat\sigma_z$ then the incoming spatial state splits its

So we know what causes measurements. Devices interacting with subjects according to the laws of physics. The Hamiltonian always has a term proportional to $B_x\hat\sigma_x+B_y\hat\sigma_y+B_z\hat\sigma_z$ when there is a particle with spin and an external magnetic field, there is no choice about whether it is there.

And we know the effects of measurements, they split states into a sun of states. Each each term in the sum has an eigenstate of the operator and that eigenstate is entangled with some other state. In this example the spin state become entangled with the position state of the particle. Deflected up entangled with spin up, and deflected down entangled with spin down.

So the only mystery is why we call it measurement. And that is because of that polarization, now if you put it on a similar device again it won't be split it will just be deflected. The real measurement effect happens later when these different states have a chance to affect the states of many other things so the two branches no longer can interact just because it is too hard to get them to ever overlap again.

So now to the question of operator. It isn't random operators, it is operators that have real eigenvalues and that have orthogonal eigenvectors. That is what is important. Why?

The real eigenvalues allow the continuous splitting of a state into the multiple eigenstates, which is what allows the Schrödinger equation to be able to split the state to entangle the projections with something else in a continuous manner (which is what the Schrödinger equation requires and we have to evolve according to the Schrödinger equation for the actual experimental setup there is no choice about that and no other options). To actually do it well, you have to find something else to couple with the eigenvectors to get the entanglement. And for that you can't actually just do anything, you are restricted to real terms in real Hamiltonians and nature only provides so many to choose from.

So why do we need the eigenstates to be orthogonal. That is key to getting that do it twice get the same answer. Otherwise it is just an interaction, not a measurement.

So to call something a measurement it basically has to be a real linear combination of orthogonal projections onto mutually orthogonal states. And so they do need to be operators, very particular operators. And you can only actually observer them if you can find something else to couple differently to those different states.

It's still not clear to me what the direct meaning of $\hat A |\psi\rangle$ is.

The point is that the different eigenvalues you get from $\hat A |\psi\rangle$ tell the other thing that is becoming entangled with your object to continuous change at different rates depending on the eigenvalue.

For instance if you have a general spin state then for a fixed magnetic field the spin zero doesn't couple at all so isn't deflected at all and the lowest non zero spins deflect but deflect less than larger spins so deflect in opposite directions but at a power rate of motion in the deflected direction. So spin zero goes straight, spin $\pm 1/2$ get sent a little bit up and down and spin $\pm 1$ get sent more up and down. And I mean the z component t being 0, 1/2, or 1 which I why I included the minus sign to make that clear. So the eigenvalue tell you how separated the other thing (the thing used the measure the state) becomes. This is literally why you project onto an eigenspace. When two things have the same eigenvalue the thing you couple with doesn't move differently for the things with the same eigenvalue so they don't get separated at all so the whole projection onto the eigenspaces gets entangled with the exact same state of the thing used to measure it.

It is a wavefunction

The eigenfunctions are the ones that don't change in repeated measurements son get deflected in different ways as a parameterized by the real umber eigenvalue.

how does its corresponding quantum state relate (in physical terms) to the original state $\psi$ and the measurement $\hat A$?

The original state can be decomposed as eigenstates and they evolve into entangled states as they separate. So when separated you have the different eigenstates entangled with different states of the thing used to measure it. Spin is easiest since it entangles with its own position so we don't really have to bring in anything else other than the inhomogeneous magnetic field.


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