# Does any basis belong to a set of eigenfunctions of some observable?

In the Quantum Mechanics courses in ocw.mit.edu it is said that any wave function can be expanded as a superposition of the eigenfunctions of any observable, i.e the eigenfunctions form a eigenbasis for our space.

However, if we take an arbitrary basis for our space, is that space an eigenbasis corresponding to an observable? In other words, is any basis a set of eigenfunctions of some observable?

• – Qmechanic Nov 22 '18 at 18:00

is any basis a set of eigenfunctions of some observable?

This depends on what exactly you want to count as an 'observable' and what you want to rule out.

The first example you need to consider is simply the identity operator $$\mathbb I$$, for which any nonzero vector is an eigenvector with eigenvalue $$1$$. (Physically, this corresponds to the number $$1$$ $$-$$ so, in the same sense that observables in classical mechanics are functions $$f(q,p)$$ of position and momentum, the identity operator corresponds to the function $$f(q,p) \equiv 1$$.) So the answer here is yes, though that might require what some people consider to be a 'trivial' observable.

In response to that, the next tempting step is to alter the request to

is any basis a set of eigenfunctions of some non-constant observable?

which rules out the identity operator and its multiples.

Once you do that, the answer is no. The way you've phrased it, the basis set need not be orthogonal, and if the basis is non-orthogonal then it's not possible to build a hermitian operator with that basis as its eigenfunctions, since any eigenbasis of a hermitian operator must be orthogonal.

(That's not quite true, as it happens, because you can have non-orthogonal basis vectors inside any degenerate eigenspace, but that returns us to the case above - the observable just acts like the identity within that subspace. In terms of the question, it means that if the basis $$\beta$$ is non-orthogonal but it does contain two subsets $$\beta_1=\{u_i\}$$ and $$\beta_2=\{v_j\}$$ which are mutually orthogonal, then you can build a non-constant observable $$A$$ such that $$Au_i=u \: u_i$$ and $$A v_j = v\:v_j$$ for two fixed numbers $$u$$ and $$v$$. Again, though, that's basically a trivial cop-out, and it doesn't bring anything new that the identity operator didn't already.)

So, v2 of the statement needs to be fixed:

is any orthogonal basis a set of eigenfunctions of some non-constant observable?

Here the answer is yes: let $$\beta = \{ v_n : n=1,2,\ldots \}$$ be a countable orthonormal basis. (And yes, it needs to be countable if you want any shot at a workable solution.) Then there exists a unique linear self-adjoint operator $$A$$ which extends $$A v_n = \frac1n v_n$$ to the entire Hilbert space, and that operator has $$\beta$$ as its unique eigenbasis with no degeneracies anywhere.

However, the solution above will produce a self-adjoint operator which (although it is perfectly well-defined mathematically) can be completely inaccessible to any imaginable physical experiment on your system, so it can again be seen as not particularly satisfactory. The solution to that is to introduce the concept of a physically-accessible observable, which you construct by starting with some core set (say, $$\hat x$$ and $$\hat p$$) and then allowing any linear combinations, powers of existing operators, and commutators of existing operators. (This represents a shift in thinking, which is well exemplified in the shift between §2.3.1.b and §3.2.1.a of arXiv:1211.5627.)

So, with that, you may ask

is any orthogonal basis a set of eigenfunctions of some non-constant physically-accessible observable?

and here the answer is no, there's some perfectly-reasonable systems with perfectly-reasonable state spaces and perfectly-reasonable algebras of observables, which allow for perfectly-reasonable bases that are not eigenbases of any observable in the chosen algebra. (Unfortunately, though, I don't have any concrete examples at hand. But within that narrow definition, it makes for a good follow-up question if you want to post it.)

So, what's the answer? Yes. And no. Depending on what you mean by "basis" and by "observable".

• Very nice answer, Emilio. I'm curious about something though: in the second case you discuss, you present a basis set $\beta$, and you state that the cardinality of the set must be countable to have a workable solution. I know that in principle we can construct vector spaces with uncountable bases, mathematically there's no problem there, but is your point that such an uncountable basis is impenetrable to the tools of a theoretical physicist? – N. Steinle Nov 22 '18 at 22:16
• @N.Steinle No. But if you have an uncountable basis, then either (i) your Hilbert space is not separable, in which case you're way off the beaten track of what QM usually handles, and you're not actually doing anything that's physically applicable at all, or (ii) your continuous "basis" is something like the momentum states in the 'canonical' QM, and the "basis states" don't actually live in the Hilbert space (and you'd need a rigged Hilbert space, a.k.a. a Gelfand triplet, to describe them rigorously) and you're a ways off from what the mathematical sophistication shown by OP can handle. – Emilio Pisanty Nov 22 '18 at 22:44
• (while, at the same time, not actually changing the content of the answer at all - the results are exactly the same, it's just more hassle to prove them correctly.) – Emilio Pisanty Nov 22 '18 at 22:45
• @EmilioPisanty Consider the superselection rule of the electrical charge and a selfadjoint operator which does not commute with that charge. It is not an observable and its Hilbert basis of eigenvectors is a possible counterexample you were serching for. Another possibility is the presence of a non Abelian gauge gruoup: selfadjoint ooerators which do not commute with the rep of gauge group are not observables... – Valter Moretti Nov 23 '18 at 19:27
• @ValterMoretti I don't understand what you're trying to say. The main point I'm trying to make here is that there are mathematically well-defined versions of the question that do have a positive answer, but that sometimes the 'observables' required by the chosen basis might not be fully satisfactory, despite being perfectly-legitimate hermitian operators. If I understood your comment correctly, you're just providing more ways for hermitian operators to be unsatisfactory as observables? – Emilio Pisanty Nov 23 '18 at 20:59

In other words, is any basis a set of eigenfunctions of some observable?

IN general, no, since we require that physical observables have certain properties. Mathematically speaking, however, a function can be expressed in any orthogonal basis, but that doesnt mean that the basis is physically meaningful.

A wavefunction may be written as a linear combination of any well defined basis (that is, it is orthonormal), and of course in principle one can define an operator that corresponds to that basis. However, this operator does not necessarily correspond to a physical observable. For instance, in the example of the harmonic oscillator potential, we may express the wavefunction of the system as a superposition of coherent states, but the operator you'd define from the basis of coherent states (a raising or lowering operator) does not represent a physical observable.

So, formally speaking, we require that physical observables be Hermitian, so that their corresponding basis is orthonormal over some interval and their eigenvalues are real-valued (so they provide real-valued expectation values). This is an axiom of the theory of quantum mechanics.

Probably not, but it may depend on what you mean by an observable.

Any hermitian operator will have a complete basis set of eigenfunctions. Some hermitian operators have been found to be associated with observables. Operators representing observables are required to be hermitian because they must have real expectation values.

Any linear sum of hermitian operators would also be hermitian, and if made up of observables that quantity could also be observed. I am unsure how to prove that the set of possible complete bases states is larger than the set that you could construct from arbitrary linear sums of existing observables, but you probably can prove that.

• The OP's question is essentially does every basis correspond to an observable. If you say that there exist Hermetian operators with orthonormal eigenbases that don't correspond to observables, then isn't the answer to the question just "no"? – Aaron Stevens Nov 22 '18 at 17:06