# General quantum operator

Is it true that any operator can be expressed as (e.g. in one dimension) $$\hat{A}=\sum_{n=0, \, m=0}^{\infty}c_{n,m}\hat{x}^n\hat{p}^m \, ?$$ It seems true because any classical observable is a function of coordinates and momentum, by the correspondence between classical and quantum argument, any quantum operator is a function of its position operator and momentum operator, up to commutation. How to prove it in the sense any Hermitian operator can be expressed in this form?

Let's only focus on the space spanned by $$\lvert x \rangle$$.

• Well, that sounds like a horribly general statement without further qualifiers. What about spin, for example? Dec 1, 2020 at 20:24
• my apology. let's only focus on the space spanned by |x> Dec 1, 2020 at 20:26
• Dec 2, 2020 at 1:58

It is literally (mathematically) false. E.g., the observable defined by the operator $$|X|$$ cannot be written that way. This observable has also a direct operational definition: I measure the position of the particle on the real line and next I take the absolute value if the outcome.

However, it is true from the point of view of Von Neumann algebra theory. Physically speaking, the fact can be considered true. See Is there a Hermitian operator on $L^2(\mathbb{C})$ which is outside the C*-algebra generated by $\hat{x}$ and $\hat{p}$?.

• Strictly speaking, yes, of course--who would argue; however, the OP assumed such a representation existed classically. And $|X|$ may be approached by the large k limit of $X\tanh kX$. Dec 2, 2020 at 20:13
• Well, I think that maybe this way one could argue almost everything. I admit that I am too mathematically minded, but I prefer to think that the statement is false... Dec 2, 2020 at 20:38
• @Cosmas Zachos I found an approach that permits me to agree with you. I changed my answer correspondingly. Dec 3, 2020 at 19:23
• Relieved; but as indicated, I believe the OP is asking about a generic function of $\hat x$ and $\hat p$, and not the power series representation he adduces as a placeholder requiring analyticity... I'm sure he is not asking about analyticity, but whether he is missing something egregious, as discussed in linked questions. Dec 3, 2020 at 19:30

Your statement is basically true for the conventional Hilbert space without spin or other bells and whistles. I don't wish to get into proofs, as flakey proofs are an excuse to miss the point in such contexts, more often than not, and reduce the issue to their assumptions, better sheltered from your inquisitiveness.

You may appreciate this is true and natural in QM in finite-dimensional spaces, as Weyl nicely illustrates in his book, using Sylvester's clock and shift matrices, which provide a complete, orthogonal basis of $$GL(N,\mathbb{C})$$. The careful $$N\to \infty$$ limit of this construction amounts to your statement.

In actual practice, the Wigner-Weyl transform maps phase-space functions to Hilbert space operators invertibly, and any other ordering prescription (like your "normal" ordering, with momenta on the right), is formally equivalent to Weyl ordering, and so can be systematically converted to it.

Crudely, all phase-space functions are expandible in such polynomials, and their Weyl transforms are perfectly symmetrized polynomials in $$\hat x$$ and $$\hat p$$, as detailed in phase-space quantum mechanics books, like this one. you may normal order those to your reference expression. If you had a fetish for rigor, you might try Wong's book, which is clearly overkill. Learning how to use the formalism correctly for routine problems, instead, should assuage your anxieties.

People have objected to your expansion which excludes functions non analytic at the origin, but I'll give you a well-meaning extension of it, such as "a sensible function of $$\hat x$$ and $$\hat p$$" : I am assuming you are not fussing about analyticity and such, which you may always fix by smoothing of sorts in practice.

• Geeky Aside on specific question that arose in a linked question.

The question arose there how an integral operator such as $$A_x \psi(x) = \int\!\!dy~ K(x,y) \psi(y)$$ is represented as a function of $$\hat x$$ and $$\hat p$$.

Observe $$\int\!\!dy~ K(x,y) \psi(y)= \int\!\!dy~ K(x,y+x) \psi(y+x)\\ = \int\!\!dy~ K(x,y+x)~e^{y\partial_x} \psi(x),$$ hence $$\hat A =\int\!\! dy~ K(\hat x, y+\hat x)~ e^{iy\hat p/\hbar},$$
with self-evident matrix elements.