For this quantum mechanical system, is there an operator that corresponds to "position" the same way that there are operators corresponding to angular momentum and energy?

My first guess for what the operator would be is $$ \hat \Phi \big[ \psi(\phi) \big] = \phi \, \psi(\phi) $$ which would be analogous to the position operator for the "particle in a box" system, but I am getting confused on the following issue:

If the particle has a 50% chance of being in one place on the ring and a 50% chance of being in the place that is exactly opposite the first place, what should the "expectation value" of the position be? With this operator it seems like the expectation value would depend on where $\phi = 0$ is chosen to be, which makes me doubt the validity of this operator since its expectation value changes depending on which coordinate system you use.

Also if the particle has some probability to be at $\phi = \epsilon$ and the rest of the probability to be at $\phi = 2 \pi - \epsilon$ then it seems like the "expectation value" should be around $\langle \hat \Phi \rangle = 0$ but this operator would put it near $\pi$.

  • 3
    $\begingroup$ You might be interested in this question. physics.stackexchange.com/q/338044 $\endgroup$ May 4, 2021 at 14:14
  • $\begingroup$ The simple answer is yes, the two issues you raise (lack of translational invariance & expectation not being physically possible) arise with the normal position operator $\hat{x}$ as well, and are not fundamental to "being an operator". $\endgroup$ May 5, 2021 at 0:28

3 Answers 3


This question is a great setup for explaining a better way of describing particles in quantum theory, one that bridges the traditional gap between single-particle quantum mechanics and quantum field theory (QFT).

I'll start with a little QFT, but don't let that scare you. It's easy, both conceptually and mathematically. In fact, it's easier than the traditional formulation of single-particle quantum-mechanics, both conceptually and mathematically! And it makes the question easy to answer, both conceptually and mathematically.

To make things easier, here's a little QFT

In QFT, observables are tied to space, not to particles. That's the most important thing to understand about QFT. Instead of assigning a "position observable" to each particle (which would be impossible in many models because particles can be created and destroyed), we assign detection observables to regions of space. Let $D(R)$ denote an detection observable associated with region $R$. In nonrelativistic QFT, the eigenvalues of $D(R)$ are natural numbers ($0$, $1$, $2$, ...) representing the number of particles found in $R$ when $D(R)$ is measured. If the model has more than one species of particle, then we have different detection observables for each species.

By the way, the traditional bit about "indistinguishable particles" is really just an obtuse way of saying what I said above: observables are tied to space, not to particles. Very simple.

Now, the difference between nonrelativistic QFT and nonrelativistic single-particle quantum mechanics is almost trivial: in the latter, the eigenvalues of $D(R)$ are restricted to $0$ and $1$.

As an example, consider ordinary nonrelativistic single-particle quantum mechanics, in one-dimensional space for simplicity. The traditional formulation uses a position operator $X$, whose measurement magically returns the particle's spatial coordinate. That formulation is convenient for some purposes, but it also causes trouble:

  • First, it causes conceptual trouble, because this is not how real-world measurements work: there is no measuring device that magically tells us the particle's coordinates no matter where it is in the universe. Real-world measurements are localized — they correspond to observables that are tied to regions of space, not tied to particles.

  • Second, it causes mathematical trouble, because $X$ has a continuous spectrum — it doesn't have any (normalizable) eigenstates, so it can't be perfectly measured even on paper.

Fortunately, we don't really need $X$. We can do better, both conceptually and mathematically, by using a collection of projection operators $D(R)$ instead. For any given spatial region $R$, the definition of $D(R)$ is simple:

  • If the particle's wavefunction is concentrated entirely within $R$, then it's an eigenstate of $D(R)$ with eigenvalue $1$.

  • If the particle's wavefunction is concentrated entirely outside of $R$, then it's an eigenstate of $D(R)$ with eigenvalue $0$.

In other words, $D(R)$ counts the number of particles in $R$, and since the model only has one particle, the answer can only be $0$ or $1$. If the particle's wavefunction is partly inside $R$ and partly outside $R$, then the result of the measurement will be $1$ with probability $$ p = \frac{\langle\psi|D(R)|\psi\rangle}{\langle\psi|\psi\rangle}, \tag{1} $$ and the result will be $0$ with the opposite probability $1-p$.

That's nicer conceptually, because it's closer to how we do things in the real world. It's also nicer mathematically, because $D(R)$ is a bounded operator with a discrete spectrum.

Note that the observables $D(R)$ for different regions $R$ all commute with each other (this should be obvious from the definition), so we can measure them all simultaneously if we want to — just like we can use an array of detectors in the real world.

The traditional position operator

How are the operators $D(R)$ related to the usual position operator $X$? Simple: $$ X \approx \sum_n x_n D(R_n) \tag{2} $$ where the sum is over a set of regions $R_n$ that partition all of space into non-overlapping cells, and $x_n$ is the coordinate of some point in the $n$-th cell. In the limit as the size of the cells approaches zero, this approaches the usual position operator $X$. Conceptually, we're just filling space with an array of little detectors. We know where each one them is located, so if the $n$-th one detects the particle, then we know where the particle is.

To be fair, the traditional position operator (2) has some advantages over the detection observables $D(R)$. One advantage is that it allows us to use summary-statistics like averages (think of Ehrenfest's theorem) and standard deviations (think of the traditional uncertainty principles). This is just like the usual situation in statistics: the information is in the distribution (the results of measuring lots of $D(R)$s), but choosing a convenient labeling scheme lets us define things like averages and standard deviations to convey some incomplete but concise information about the distribution.

The particle on a circle

Now, the answer to the question should be obvious. Does a particle living on a circle have a position operator? Well, it certainly does have the operators $D(R)$, where now $R$ is any portion of the circle. Those operators are defined just like before. We can also define something analogous to $X$ if we really want to, like this: $$ \Theta \approx \sum_n \theta_n D(R_n) \tag{3} $$ or like this: $$ U \approx \sum_n e^{i\theta_n} D(R_n), \tag{4} $$ where the sum is over a partition of the circle into non-overlapping intervals $R_n$, and $\theta_n$ is an angular coordinate within the $n$-th interval. We can take the limit as the size of the intervals goes to zero, if we want to.

The operator (3) is most closely analogous to (2), and it's perfectly well-defined mathematically, but it's unnatural because the angular coordinate must have a discontinuity ($2\pi$ jump) somewhere on the circle. That makes the operator (3) less useful for talking about things like averages and standard deviations, but that's not quantum theory's fault. It's just mundane statistics: averages and standard deviations are most useful when using a monotonically increasing coordinate, which we can't do everywhere on a circle if we want the coordinate to be single-valued.

The operator (4) is more natural mathematically, because $e^{i\theta}$ is continuous everywhere around the circle. Beginners might complain that (4) is not hermitian, but observables do not need to be hermitian (see this question), contrary to what some introductions say. Unitary operators like (4) can also be used as observables, because quantum theory doesn't care about the coefficients of the projection operators, it only cares about the projection operators. The coefficients are useful for talking about averages and standard deviations, but we don't really need to talk about those things. They're convenient, but they're not necessary. Again, this is just like in ordinary statistics. Labels can be convenient, but they're not necessary.

So... does a particle on a circle have a position operator? It depends. What properties do you want the position operator to have? It does have the observables that really matter, which are the detection-observables $D(R)$. And we can use those to construct operators like (3) and (4) that each resemble the position operator (2) in different ways, but of course we can't have an operator that is completely like (2) because a circle is not completely like the real line. So, whether the answer is "yes" or "no" depends on exactly what you want, but at least the conceptual obstacles should be gone now.

  • 31
    $\begingroup$ "To make things easier, here's a little QFT" I would've never thought to hear that sentence in my life. $\endgroup$ May 4, 2021 at 11:01
  • $\begingroup$ "angular coordinate must have a discontinuity (2π jump) somewhere on the circle." Why not just use R mod 2 pi as your space? No discontinuity. I guess that doesn't fix the average problem, but the problem is no longer the discontinuity... but rather that your space doesn't have much of a notion of "average" anymore. Ok, I guess that fixes nothing. $\endgroup$
    – Yakk
    May 5, 2021 at 14:20
  • $\begingroup$ This seems like it corresponds to constructing the Newton-Wigner position operator for this system. $\endgroup$ May 6, 2021 at 4:06
  • $\begingroup$ @RobinEkman The Newton-Wigner position operator is an attempt to construct a position operator in relativistic QFT. I wrote another answer that elaborates on the situation in relativistic QFT. Here, I immediately specialized to nonrelativistic QFT (which is consistent with the question), where constructing a position operator for a single particle is not a problem. Maybe there is some sense in which the Newton-Wigner paradigm reduces to this one in the nonrelativistic limit, but I haven't checked that. $\endgroup$ May 6, 2021 at 12:44
  • $\begingroup$ @ChiralAnomaly yes that's exactly what I'm wondering, if this is the appropriate limit/generalisation of the Newton-Wigner operator for this case. The characterisation of N-W has having localised eigenstates and a formulation in terms of detection probabilities $D(R)$ feel similar enough that this should be the case, morally speaking. $\endgroup$ May 9, 2021 at 7:16

For concreteness, define $S^1 = \mathbb R \bmod{2\pi}$ and let the Hilbert space under consideration be $L^2(S^1)$. We can certainly define the operator $\hat \Phi$ such that $\big(\hat \Phi \psi\big)(\phi) = \phi\cdot \psi(\phi)$. $\hat \Phi$ is clearly bounded - $\Vert \hat \Phi \Vert= 2\pi$ - which means that we can define its action on the entire Hilbert space. Furthermore, it is straightforwardly self-adjoint, so it's a perfectly fine observable.

With this operator it seems like the expectation value would depend on where ϕ=0 is chosen to be, which makes me doubt the validity of this operator since its expectation value changes depending on which coordinate system you use.

Sure, but the same is true for the particle on a line, or even a particle in a box. If the particle's wavefunction is symmetric with respect to the center of the box, the expectation value $\langle x \rangle$ depends on whether the box extends from $-L/2$ to $L/2$ or from $0$ to $L$.

Also if the particle has some probability to be at ϕ=ϵ and the rest of the probability to be at ϕ=2π−ϵ then it seems like the "expectation value" should be around ⟨Φ^⟩=0 but this operator would put it near π.

Remember that the expectation value is the weighted arithemetic average of the possible measurement outcomes. It is not where one would actually expect to find the particle; indeed, for observables with discrete spectra, the expectation value generally isn't even a possible measurement outcome. If the expectation value of the $z$-component of an electron's spin is zero, that doesn't mean that we expect to measure it to be zero - it means that after many identical measurements, we'd expect the average of our results to be zero.

Similarly here, if the expectation value of the position is $\pi$, that doesn't mean we expect to find the particle near $\pi$ - it means that after many identical measurements, we expected the average of our measurement results to be near $\pi$.

  • $\begingroup$ Isn't there some issue with this operator being multivalued, making it better to work with $e^{i \hat{\Phi}}$ instead? $\endgroup$
    – d_b
    May 4, 2021 at 1:56
  • 5
    $\begingroup$ @d_b Well, by considering $\mathbb R\bmod 2\pi$ we ensure that $\phi \in [0,2\pi)$ so nothing is multivalued. That being said, unless $\psi(0)=0$, the action of $\hat \Phi$ does introduce a discontinuity in the wavefunction. This isn't necessarily a problem, but it does make the position/momentum uncertainty relation more subtle than it is on the line because the domain of the operator $\hat P_\Phi \circ \hat \Phi$ is not the same as the domain of the operator $\hat \Phi \circ \hat P_\Phi$, c.f. this nice answer by ACuriousMind. $\endgroup$
    – J. Murray
    May 4, 2021 at 2:13
  • $\begingroup$ Hi J. Murray, but if your Hilbert space is $L^2(S^1)$, then you operate on equivalence classes of periodic functions -so the operator must map such an equivalence class to another one. But this is not possible in general, due to the periodicity requirement. As far as I see, this has nothing to do with the interval you choose. Do I miss something obvious here? $\endgroup$ Aug 5, 2023 at 22:06
  • $\begingroup$ Hello @TobiasFünke - are you perhaps unconsciously imposing continuity on these periodic functions? For example, the function $\psi(x) = x$ on $[0,2\pi)$ can be extended to a periodic function by copying and pasting it over the whole real line. It won't be continuous (hence the subtlety mentioned in my comment above) but that's not a general requirement of the Hilbert space. $\endgroup$
    – J. Murray
    Aug 6, 2023 at 4:52
  • $\begingroup$ Hi, thanks for the reply. You are right, I agree here: If you use the periodic extension, it makes sense, because the product of two periodic functions (with the same period) is periodic again. Now everything is well-defined. This is also used in one of the answers you've linked. But then I find the notation a bit strange, which however was my misinterpretation. $\endgroup$ Aug 6, 2023 at 8:38

I don't have enough reputation to comment on Chiral Anomaly's answer so I am putting this here.

I think it is important to emphasise that nothing in this answer is specific to quantum field theory, and that this collection of operators $D(R)$ is a special case of the approach a quantum information theorist would use to answer this question.

A more general approach to quantum measurement, which is very widely used in the quantum information literature is the idea of a positive operator valued measure, which assigns positive operators to the measurable sets of a measure space and obeys some axioms which look a lot like the axioms of a probability measure (POVMs in particular satisfy an axiom of countable additivity so the operator assigned to the union of two disjoint sets is just the sum of the operator assigned to each of the sets individually).

This paper https://arxiv.org/abs/1604.00566, for example employs exactly the position observable mentioned in Chiral Anomaly's answer.

I mention this mainly because although in this case the natural position observable happens to be a projection valued POVM, in general the values of our POVM are free to be any positive operators which sum to the identity.

  • $\begingroup$ Church of the larger Hilbert space means you only really need to think about projection-valued though. As usual there's lot of different bi-interpretable ways of viewing things foundationally, and which is more useful can depend on the question. $\endgroup$
    – wnoise
    May 4, 2021 at 23:06
  • $\begingroup$ @wnoise I think this viewpoint can obscure a lot of interesting stuff though, my favourite examples are things like first measuring one observable then measuring a second (incompatible) observable and looking at what the effective observable is you measured the second time. Another thing you can do is take a pair of incompatible observables and ask what observables you can come up with which approximately measure the two of them. Of course a more practical thing is that there is classical noise in the real world, so real measurements we do in the lab are not projection valued observables. $\endgroup$
    – ors
    May 5, 2021 at 9:45
  • $\begingroup$ sure. My point is not that this is the right way to view things, but that there are multiple interoperable perspectives, and none of them are the right way. It's not fundamentally more general, just dividing things differently -- which is useful for many problems, of course. $\endgroup$
    – wnoise
    May 8, 2021 at 20:16

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.