When you come from classical hamiltonian mechanics (which is based on the phase space), observables are introduced as functions $f$ on the phase space $(q, p)$. There can't be a classical observable that isn't a function of $q$ and $p$ by definition. In quantum mechanics, however, $\hat{X}$ and $\hat{P}$ are operators, acting on an infinite dimensional hilbert-space, so it seems at least imaginable to me that there are operators that can't be expressed as a "function" of $\hat{X}$ and $\hat{P}$.

I put the "function" into quotation marks because I don't know how to rigorously define such a function on the space of operators, put asside a taylor expansion using $\hat{X}$ and $\hat{P}$ as factors.

Of course I can just make the hilbert-space "bigger", for example by introducing a new spatial dimension $\hat{Y}$ with associated momentum $\hat{P}_y$, but these new observables would automatically commute with the former ones ($\hat{X}$ and $\hat{P}_x$) because they act on another subspace of the hilbert-space. I have a hard time imagining an operator which doesn't commute with ($\hat{X}$ and $\hat{P}_x$), while at the same time still not depending on each of them. Can somebody provide either an example for such an operator, or give a proof why such an operator can't exist?

EDIT: To clear 2 misconceptions that did arise:

  • When talking about operators, I only ask about cases where the operators are linear
  • By "function" I mean any operations that you can perform on operators (multiply them, add them, exponentiate them, have infinite sums or for the sake of the argument as well integrals). The question is about an operator that can't be expressed as "function" of $\hat{X}$ and $\hat{P}$ alone (that means I CAN'T FIND function in the above sense to relate these operators), but still does not commute with at least one of them.

EDIT 2: Since there is more than one answer now with (at first sight) contradicting content, it seems to me that the answers to the question do depend on further assumptions:

  • Do we consider $X$ (or $P$) to be a complete set of operators, or equivalently, can I express any state as a linearcombination of $|x\rangle$, where the coefficients do depend solely on $x$.
  • Do I require them to be self-adjoint, symmetric, and bijective?
  • Am I restricting to "local" Operators?
  • $\begingroup$ Maybe this is related $\endgroup$
    – slaaidenn
    Oct 4, 2019 at 5:16
  • 1
    $\begingroup$ A simple example would be the orthogonal projection on any vector $\psi\in L^2$: $A=\lvert \psi\rangle\langle \psi\rvert$. $\endgroup$
    – yuggib
    Oct 8, 2019 at 7:37
  • $\begingroup$ I hadn't explicitly stated it, but I already ruled this one out because it's not bijective (I made another EDIT to take this into account). $\endgroup$ Oct 8, 2019 at 8:11
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    $\begingroup$ If you don't require the operator to be local, then spatial inversion is one such operator $\hat{\mathcal{I}}: \psi(x) \to \psi(-x)$ you are looking for. It anticommutes with both $\hat x$ and $\hat p$, but cannot be expressed by the operator function $f(\hat x,\hat p)$. If you require the operator to be local, then it is a more interesting question. I think the answer would be no, but cannot prove it now. I feel that one needs a precise definition of locality to begin with. $\endgroup$ Dec 2, 2020 at 0:20

5 Answers 5


In classical mechanics, any "observable" has to be a function of $x,p$ because we define $x,p$ to be the degrees of freedom of the system. If we experimentally find an observable that does not depend on $x,p$, it means there were extra degrees of freedom that we forgot to include. If so, we just enlarge our phase space, i.e., we include these extra degrees of freedom among the $x,p$. So in the end, any observable can always be written as a function of the phase-space variables.

In quantum mechanics, the philosophy is exactly the same. Here, "funcional dependence" is replaced by the notion of completeness: a set of operators $\{\mathcal O_i\}$ is said to be complete if $[A,\mathcal O_i]=0$ implies $A\propto 1$, the identity operator.

When specifying a classical system, one must declare what the coordinates are. When specifying a quantum system, one must declare what a complete set of operators is. A typical example is $X,P$, which is often assumed to be complete. Some problems require extra degrees of freedom, such as the intrinsic spin $S$, in which case a complete set of operators would be $X,P,S$. One may imagine systems that require more degrees of freedom, but also ones that require less (say, finite-dimensional systems).

Being "complete" is the formalisation of the notion of expressing an operator as a function of other operators. In particular, if we declare that $\{\mathcal O_i\}$ is complete, then any other operator $T$ can be expressed as a function of the $\mathcal O_i$, by definition.

Thus, if we declare that $X,P$ is complete, then any other operator must be expressible as a function thereof. In order to have an operator that cannot be expressed as such, one must assume that $X,P$ is not complete. So complete the set: $X,P,Q$, for some $Q$. Now we are free to define the $Q$'s to behave as we want. Say, the $Q$'s could be the position and momenta of extra dimensions, in which case there is a somewhat canonical (pun indented) notion of commutator, namely $[Q,X]=[Q,P]=0$. (But note that this is not forced upon us; it is perfectly consistent to assume that the extra dimensions do not commute with the old variables. For example, we could assume a curved configuration space, and so $[X_\mu,X_\nu]=\omega_{\mu\nu}$ for some form $\omega$).

But we could also take some $Q$'s that do not, by hypothesis, commute with $X,P$, in which case we would have $[Q,X]\neq 0\neq [Q,P]$. And we assumed that $X,P,Q$ is complete, but $X,P$ alone is not, which means that $Q\neq Q(X,P)$. So the answer to the question in the OP is: yes, you can definitely have an operator that does not commute with the canonical variables, yet is not expressible as a function thereof.

(At least, at a matter of principle it can exist: there is nothing inconsistent about the existence of such an operator. But experience tells us that the canonical variables are always enough to encode all the relevant degrees of freedom, and so they are always in practice complete. If we find a system where it is not, it means we did not realise there were some extra degrees of freedom, and we have to include those together with the canonical variables, i.e., they become canonical variables themselves.)

  • $\begingroup$ Do I understand right, that in the moment I declare $X$ and $P$ to be not complete, (the step where I introduce $Q$ to not be a function of $P$ and $X$), there will be more than one eigenvector for a given value $x$, and the basis of the subspace spanned by the eigenvectors with the eigenvalue $x$ consists of another (new) operator that commutes with $X$? $\endgroup$ Oct 7, 2019 at 5:03
  • $\begingroup$ @Quantumwhisp Yeah, that's the idea. Take for example a system with spin; a basis is given by $|x,s\rangle=|x\rangle\otimes|s\rangle$, where $x\in\mathbb R$ and $s\in[-S,-S+1,\dots,S-1,+S]$. Note that the eigenvalue $x$ has degeneracy $2S+1$. $\endgroup$ Oct 8, 2019 at 1:49
  • $\begingroup$ So in that case there could be an operator which can't be expressed as a function of $X$ and $P$ alone, because it might be some weird combination of $X$, $P$ and $S$, which (as three Observables) form a complete set of operators. $\endgroup$ Oct 8, 2019 at 7:10
  • $\begingroup$ @AccidentalFourierTransform I agree on the general philosophy of you answer but I disagree on the fact that your notion of completeness of operators can be considered equivalent to the fact that every operator is a "function" of the complete set of operators. At least thinking of functions of operators as Taylor expansions and similar bad-defined tools. $\endgroup$ Oct 8, 2019 at 8:51
  • $\begingroup$ The notion of completeness you used (which actually should be stated in terms of spectral measures) is equivalent to saying that every selfadjoint operator is the strong limit of operators algebraically (sum and product) constructed out the spectral measures of $X$ and $P$ (in $H= L^2(\mathbb{R},dx)$). In general, however, there is no way to make explicit this dependence in the form $f(X,P)$ using some, e.g., Taylor expansion intepretation. $\endgroup$ Oct 8, 2019 at 8:56

Let's say we're working in a Hilbert space of square-integrable functions $f(x)$. Indeed it's natural to talk about the operators $X$ and $P$ which act in a simple way on $f(x)$. In principle, you can talk about any linear operator that acts on $f$. For instance, you could try to solve the Schrödinger equation $$(P^2 + X^2 + T)\psi(x) = E \psi(x)$$ where $$T\psi(x) = \int dy\, Q(x,y) \psi(y)$$ and $Q(x,y)$ is your favorite function. If $Q(x,y)$ is proportional to a delta function, say $$Q(x,y) = \delta(x-y) V(x)$$ then this is nothing but the standard Schrödinger equation with a potential $V(x)$. But for a generic function $Q(x,y)$, you can't express the operator $T$ as a finite sum of $X$es and $P$s.

Often in physics, we rule out such Hamiltonians. The reason is that the interaction is not completely local: at a fixed time $t$, operators at points that are very far away (spacelike separated) interact. In a fundamental theory, say in particle physics, this type of interaction would violate causality. But in effective theories (say in condensed matter), such Hamiltonians can and do arise.

  • 2
    $\begingroup$ What about infinite sums? Do you think there is a proof that there are functions which cannot be represented like this? $\endgroup$
    – lalala
    Oct 2, 2019 at 15:50
  • $\begingroup$ The sketch of such a proof is as follows. In a basis of eigenstates $\psi_n$ of the harmonic oscillator, the operator $P$ acting on $\psi_n$ gives a linear combination of $\psi_{n-1}$ and $\psi_{n+1}$, so $P^k$ at most gives you states in between $\psi_{n-k}$ and $\psi_{n+k}$. The same holds for $X$. If you find just a single $T$ that mixes states that are arbitrarily far apart (e.g. it maps $\psi_0$ to a linear combination of $\psi_n$ with arbitrarily high $n$) then you're done. $\endgroup$ Oct 3, 2019 at 7:17
  • $\begingroup$ I wrote a separate answer and afterwards realized it is related to your answer. But I reached an opposite conclusion. I explicitly provided an expression for your $T$ in terms of $x$ and $p$. $\endgroup$ Dec 2, 2020 at 1:51
  • $\begingroup$ Integral operators T for a generic function 𝑄(𝑥,𝑦) can, in general, be expressed as functions of X and P; the "finite sum" requirement is an unnatural red herring. I'd be glad to answer a separate, dedicated question on the subject. $\endgroup$ Dec 3, 2020 at 12:56

An operator can be defined by the comutators it has with respect to x and p. Suppose we have an operator $y$ that has a non-trivial commutation relation with $x$, i.e.

$$ [x,y]=f(x,p,y),\qquad [p,y]=g(x,p,y) $$

Note that ordering is determined by $f$ and $g$. We can always solve this equation by the ansartz $y(x,p)$, up to some integrability condition.

  • $\begingroup$ How can an operator defined by the commutators it has with x and p? For this definition to behave well, the action the operator has on an abitrary state has to be completely fixed by giving the 2 commutators. $\endgroup$ Oct 2, 2019 at 16:26
  • $\begingroup$ Solving the equation above by $y(x,p)$, we know how they act on states by choosing a basis, say the one that diagonalize x. For this basis $p=i\frac{d}{dx}$, and $y(x,i\frac{d}{dx})$ will act on wave functions $\psi(x)$. The main obstruction here is that it might be the case that these equation does not have a solution $y(x,p)$, so you need to satisfy some integrability condition. $\endgroup$
    – Nogueira
    Oct 2, 2019 at 23:06
  • $\begingroup$ Do I guess right that you assume $X$ to be a complete Set of operators / that the eigenstates of $X$ don't show any degeneracy? I'm trying to make sense of your and of Hans Molemans answers, because they seem to contradict themselfes. $\endgroup$ Oct 8, 2019 at 7:30
  • $\begingroup$ They don't. The integrability condition to this equation might not hold. In that case $y$ cannot be $y(x,p)$. $\endgroup$
    – Nogueira
    Oct 9, 2019 at 12:44

The total angular momentum, involved in the spin-orbit interaction: \begin{equation} {\bf J} = {\bf L} \oplus {\bf S} \end{equation}

Cannot be expressed as a function of $X$ and $P$ (because of the spin part ${\bf S}$), and it doesn't commute with them either (because ${\bf L}$ doesn't).

  • $\begingroup$ But $\mathbf L$ is a function of those operators, so overall it's still a function of them. A simpler example, $0$ isn't a function that depends on $x$, but $0 + x$ is a function of $x$ $\endgroup$ Oct 4, 2019 at 4:44
  • $\begingroup$ @AaronStevens Right, ${\bf L}$ is a function of $X$ and $P$, but ${\bf J}$ is not (How could you write the spin part with $X$ and $P$? It's not possible) $\endgroup$
    – slaaidenn
    Oct 4, 2019 at 4:50
  • $\begingroup$ @AaronStevens But if you have an operator with no dependence on $X$ and $P$ at all, it trivially commutes with them, and it can still make sense, e.g. spin $\endgroup$
    – slaaidenn
    Oct 4, 2019 at 5:03
  • $\begingroup$ ${\bf J(X,P)} = {\bf L(X,P)} \oplus {\bf S}$. Just like how we could say $f(x)=x+5$. $\endgroup$ Oct 4, 2019 at 11:29

I think there does exist such a general expression: $$ \hat O = \int dy \, O(y,\hat x) \exp\left[i\hat p(y-\hat x)\right], $$ where $y$ is a c-number and $O(y,x)$ is a function satisfying $O(y,x) = \langle y|\hat O|x \rangle$.

Edit — it’s pretty easy to prove actually, and I leave that as a homework problem. Use the fact that $\{|x \rangle\}$ is a complete set of basis, and show the two sides gives the exact matrix elements.


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