I've been studying some Quantum Mechanics recently. I am a mathematics student but I've always been interested in physics so I am currently learning this material from a more mathematical treatment of the subject. I've come across two propositions:

Proposition: Let $H$ be a separable Hilbert space and let $A,B,C: H \to H$ be self-adjoint linear maps such that $[A,B] = iC$. Then, the following inequality holds: $$E_{\psi}(A^2) E_{\psi}(B^2) \geq \frac{1}{4} E_{\psi}(C)^2$$ for any unit vector $\psi$. Over here, we define: $$\forall \psi \in H \setminus \{0\}: E_{\psi}(A) := \frac{1}{\left|\left|\psi\right|\right|^2}\langle \psi, A\psi \rangle$$

This can be used to prove the following theorem.

Theorem: Let $P$ denote the momentum operator and $X$ denote the position operator. Then: $$\Delta_{\psi}(P) \Delta_{\psi}(X) \geq \frac{1}{2}\hslash$$ for any unit vector $\psi$. In this case, we define: $$\Delta_{\psi}(A) = \sqrt{E_{\psi}(A^2)-(E_{\psi}(A))^2}$$

This is the Heisenberg Uncertainty Principle and let's forget about the fact that both of these theorems are not stated in the most explicitly precise way they could have been stated. The author has intended on creating a book in Quantum Mechanics for mathematics students who have gone through a course or two in rigorous Linear Algebra, so there are obviously going to be parts which are lacking.

Now, what isn't lacking is the fact that the author refrains from making certain statements that are "over the top". In particular, my understanding of the position and momentum operators is that they are just that; they're operators and they don't correspond to any sort of classical notion of position/momentum. On the other hand, I've read certain sources which make the claim that the Uncertainty Principle places a bound on how accurately we can measure "momentum" and "position".

My problem with this statement is that it seems to suggest that there exist things like "position" and "momentum" that can be measured. I'm supposing that what's being referred to here is the "classical momentum" and "classical position"? But what do these operators have to do with their classical counterparts and how does the inequality above actually affect anything about these classical observables?


Some of the commenters have misunderstood the point of the question entirely. Let me give a quote from Eugene Merzbacher's book on Quantum Mechanics (this was the source of my original confusion):

The uncertainty relation limits the precision with which position and momentum can be simultaneously ascribed to the particle. Generally, both quantities are fuzzy and indeterminate.

This quote implies that momentum and position, as classical observables, still do exist. Moreover, in the statement above, he seems to be referencing the uncertainty principle in the following form: $$\Delta x \Delta p \geq \frac{1}{2} \hslash$$ where $\Delta x$ and $\Delta p$ are the uncertainties in the classical position & momenta. These are real numbers in their standard definitions in Classical Mechanics & there is a nice way to multiply them such that their multiplication commutes.

On the other hand, the position and momentum operators do not commute. That's a standard result. Moreover, the two propositions I've stated above talk about the expectations of operators, not real numbers or classical observables. In the definition of these operators, there's no indication that they have anything to do with the notion of any sort of classical observable. The theorems above deal with statistical statements about operators.

So, my questions are these:

  1. What is the connection, if any, between these operators and their classical counterparts?

  2. How does Heisenberg's inequality above, which is a statement about operators, affect what we do with classical observables?

The reason why I am asking this is because there are clearly big differences between the classical observables that we are all familiar with and their operator counterparts. In classical mechanics, you can obtain the position from the momentum. The claim being made over here is inconsistent with the idea that we can obtain position from the momentum.

I have also been referencing some books written by mathematicians on Quantum Mechanics, as well as a lecture series on Youtube by Frederic Schuller. Their general claim is that classical observables simply do not exist and that, in certain situations, we can approximate quantum observables as being classical (for instance, the Ehrenfest Theorem seems to suggest that this can be done in some situations). If so, that is a solution to the issues I raised above.

However, there seems to be an opinion among physicists that classical observables do exist and that they exist alongside quantum observables, with there being a correspondence between the two. If so, then the above is still a logical inconsistency.


3 Answers 3


The operators X and P in quantum mechanics are postulated to lead to probabilities for real-life measurements of position and momentum. In real experiments, those quantities are measured in a pretty consistent way with classical mechanics or classical electromagnetism. For example, a "microchannel plate" which measures position has a bunch of holes, wherein an electron hitting the wall within any hole will cause a visible voltage. Based on which hole triggered a voltage drop, you can infer roughly the position of the incoming electron as it entered the plate.

Momentum measurements you might find even more "suspiciously classical". One of the methods used at the LHC is to gather many measurements on the position of a particle in a constant magnetic field. In this field, the particle (classically) would move along a circle with radius $p/qB$, and $p, q, B$ being its momentum, charge, and the magnetic field's strength. The points are fit to a circle to find the radius, and the radius is used to infer a value for the momentum.

If I have understood your question in any way incorrectly, feel free to comment any additional questions.

Response to edit:

What is the connection, if any, between these operators and their classical counterparts?

Even in the context of QM, it's not like we "measure the momentum operator" as opposed to measuring a number for the momentum. There is nothing measured, upon one run of any momentum experiment, besides the real number which gives the momentum itself. QM addresses such a situation by positing that the probability of measuring a momentum within some range $p_1 < p < p_2$ is given by

$$\int_{p_1}^{p_2}|\langle p,\psi\rangle|^2 dp$$

where the momentum eigenstate $|p\rangle$ corresponding to some value of momentum $p$ can be found by solving the eigenvalue equation for the momentum operator.

It is true that we can measure real values for quantities like x and p, but this does not mean that they behave classically. The difference in the orthodox QM framework is that there is no hidden value of position which we measure, rather that this value is generated probabilistically upon some event which we call measurement.

How does Heisenberg's inequality above, which is a statement about operators, affect what we do with classical observables?

The measurement examples I gave in my first answer are not classical observables, they are just the results of experiments on quantum mechanical particles. Nothing to change about them to make them any more "quantum". The version of heisenberg's uncertainty principle which you wrote here only makes a relatively weak statement in this case, namely that if we were to estimate a probability density for both position and momentum by performing this measurement many times, each time on a new particle which nonetheless starts in exactly the same state, we would see the standard deviations of their distributions tend to relate according to $\sigma_x \sigma_p \geq \hbar/2$.

  • $\begingroup$ Right, so this is a pretty nice response. I've edited my question significantly so perhaps you might want to have a look at that because it partially addresses what you've written here. So, from what I can see, you seem to be asserting that there are certain situations in which we can attempt to approximate quantum observables so that they behave classically. So, the momentum $p$ of the particle in the LHC is actually its momentum operator BUT it can be treated as a real number in this instance and can be used within classical formulas because the conditions allow that approximation. $\endgroup$
    – Mousedorff
    Commented Jul 31, 2022 at 1:15
  • $\begingroup$ Would that be a correct interpretation of what you're saying? If so, then would it be correct to say that there isn't a "classical observable" called the "momentum" of a particle? Instead, there is a quantum observable called the momentum operator that describes the evolution of a particle's wavefunction and this operator can be approximated as classical in certain settings but a priori, there is no classical momentum? $\endgroup$
    – Mousedorff
    Commented Jul 31, 2022 at 1:18
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    $\begingroup$ The experiments I gave are not some kind of approximation to quantum observables, they are real examples of what we mean by "measurement" in quantum mechanics. Check out chapter 4 of Shankar's principles of quantum mechanics, I think that would help out a bunch. There he lists the postulates of quantum mechanics from which most things are derived. The momentum of the LHC particle is not its momentum operator, it just simply is a real number, however the real numbers which you get from measurement are found with probabilities which come from the operators as shankar describes. $\endgroup$ Commented Jul 31, 2022 at 1:28
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    $\begingroup$ @MordeusMorgenstern we do not attempt to approximate quantum observables by their classical cousins. The starting point is nearly always the classical quantity. Finding an operator (acting on states in a Hilbert space etc) that will reproduce the correct experimental result is the problem of quantizing this observable, for which there is no definite solution. $\endgroup$ Commented Jul 31, 2022 at 2:15
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    $\begingroup$ I say nearly always because some quantum operator like spin do not have a classical analogue. $\endgroup$ Commented Jul 31, 2022 at 2:17

My problem with this statement is that it seems to suggest that there exist things like "position" and "momentum" that can be measured. I'm supposing that what's being referred to here is the "classical momentum" and "classical position"? But what do these operators have to do with their classical counterparts and how does the inequality above actually affect anything about these classical observables?

I think the main problem you are facing is your supposing that Quantum Mechanics (QM) is saying something about the classical counterparts of the operators.

QM has connections with classical Physics at different levels:

  1. QM results must agree with classical theories in the appropriate limit;
  2. At some level, measurements of QM quantities imply measurement equipments described by classical physics;
  3. Classical physics provides a guide for the choice of the QM operators that should be associated with a specific physical quantity.

A part of these connections, once one has a self-adjoined operator on a Hilbert space representing a quantum observable, its spectrum is directly connected to the possible results of a measurement of that quantity in QM, with no reference to any classical counterpart. Quantum states are defined to provide a convenient way to extract the probabilities of the possible (quantum) values. Therefore, neither in the operator spectrum nor in the quantum state classical counterparts play any role.

This is all required to understand the meaning of Uncertainty Relations (UR).

In passing, I would note that the first part of Merzbacher's statement you cited ("The uncertainty relation limits the precision with which position and momentum can be simultaneously ascribed to the particle.") goes a little too far from what can be stated on the basis of the mathematical result. But this is a different story. You may find more on that in my answer to another question.

  • $\begingroup$ Right okay, nice, that seems to line up well with what Peter Shor said and with what I've read in other resources. I was just a little confused because some of these sources, like Merzbacher's book, seem to be heavily implying that QM is saying something about the classical counterparts of these operators and I fail to see any justification for that. $\endgroup$
    – Mousedorff
    Commented Jul 31, 2022 at 7:45
  • $\begingroup$ I've also read your answer on the uncertainty principle and that, once again, seems to line up well with the way I phrased my question, where I stated two theorems that were entirely mathematical statements about operators. So, it just seemed like a massive jump to conclude that these statements were saying anything about the real world or, at the very least, about the observables that we like to think of as the "momentum" or "position" of the particle. They seem to be entirely statistical statements. $\endgroup$
    – Mousedorff
    Commented Jul 31, 2022 at 7:50
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    $\begingroup$ @MordeusMorgenstern still those statements are about the statistical behavior of single measurements of quantum quantities whose possible values are described by the spectrum of the relevant operators. In this sense, they are about momentum and positions of particles. It is the meaning and the possible values which are affected by the quantum character. $\endgroup$ Commented Jul 31, 2022 at 7:58
  • $\begingroup$ Right, so once there is an affect on the possible values of these classical observables, then it changes their meaning entirely, yes? Essentially, we expect that classical observables have a continuous spectrum (a continuum of momenta, energy and so on) but the quantum observables may not necessarily have a continuous spectrum so you can't interpret them to be the same as the classical observables that you used as inspiration to actually build those quantum observables. $\endgroup$
    – Mousedorff
    Commented Jul 31, 2022 at 8:03
  • $\begingroup$ @MordeusMorgenstern I almost agree. Almost because I would not consider the classical variables as just "inspiration". In the proper limit quantum observable must go to their classical counterpart (if it exists). $\endgroup$ Commented Jul 31, 2022 at 13:08

The operators are just that- operators. Their relationship with observable properties (according to the interpretation of QM I studied in the 1980s) is that the properties are eigenvalues of the operators, so that the wave function associated with a particle that has, say, a given momentum, will be the eigenfunction of the momentum operator that corresponds to that momentum eigenvalue. The uncertainty principle follows directly from that interpretation, since if a particle is 'in' a momentum eigenstate it does not have a well-defined position, and if it is 'in' a position eigenstate it does not have a well-defined momentum.


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