# Can any vector in a given vector space be an eigenvector of a Hermitian operator in that space?

I have been reading David Albert's book "Quantum Mechanics and Experience", and came across the following statement in chapter 2 page 41: "(5) Any vector whatever in a given space will invariably be an eigenvector of some complete Hermitian operator on that space. That....will entail that any quantum state whatever of a given physical system will invariably be associated with some definite value of some measurable property of that system."

I am having some difficulty understanding this statement intuitively. I am particularly puzzled by the use of the words "invariably" and "whatever". It seems to me that it is saying that given an arbitrary vector in a vector space, I can always find some operator (corresponding to a measurable/observable) for which it is an eigenvector. What is an example of such an observable? Can there be (infinitely) many such observables? Does it depend on the chosen vector space?

I would like to understand this intuitively, if possible, with some examples. Of course, a formal proof is also helpful. Please note that I do understand vector spaces and other formalism, but I have not come across anything like this particular statement before or when searching the literature.

Thanks in advance for any clarification.

• You can be more specific than that. Any vector is an eigenvector of the identity operator. – user247327 Jun 13 '19 at 20:28

In this abstract model of quantum mechanics the physics happens in some vector space. By definition, states of the system are vectors (actually, one dimensional subspaces) and and observables (things you can measure) are Hermitian operators.

To solve any particular physical problem - for example, to calculate the spectrum of the hydrogen atom - you have to specify a particular vector space and particular observables that make physical sense.

I think Albert's use of "invariably" and "whatever" in this discussion is his attempt to remind you of the abstraction.

Here's an example in the abstract setting. Given any nonzero vector $$v$$ you can find many operators for which $$v$$ is an eigenvector. The simplest such operator is the projection $$P$$ onto the one dimensional subspace consisting of multiples of $$v$$. Then $$v$$ is an eigenvector of $$P$$ with eigenvalue $$1$$, the orthogonal complement consists of all the eigenvectors with eigenvalue $$0$$ and the observable $$P$$ is "observing" the experiment that asks if the system is in state $$v$$.

Edit in response to comment.

A (unit) vector $$v$$ isn't the value of an observable, it's a state. To find the value of an observable $$A$$ in state $$v$$ you calculate the inner product $$\langle Av, v \rangle$$. If $$v$$ happens to be an eigenvector of $$A$$ then that is just the corresponding eigenvalue.

For any particular state, there are many observables for which that state happens to be an eigenvector - that is, observables for which that state has a particular value (the eigenvalue) with probability $$1$$. Think about a hydrogen atom in its ground state $$v$$. The projection onto the subspace spanned by $$v$$ is the observable that measures whether the atom is in its ground state. With probability $$1$$ the answer is "yes". $$v$$ is also an eigenvector for the Hamiltonian - the eigenvalue is the energy in that state. It's also an eigenvector for the projections on the vectors determined by the other pure states, with eigenvalue $$0$$, since with probability $$1$$ the ground state isn't any other state.

To construct "many" operators for which $$v$$ is an eigenvector you can specify an operator $$A$$ (i.e. an observable) by setting $$Av = \lambda v$$ for any value of $$\lambda$$ you wish, then arbitrarily on the orthogonal complement of the one dimensional space spanned by $$v$$. Most of those observables will have no physical interest, but they're still there.

• Thanks, Ethan. Please could you explain (or point me to a proof) of the statement you made (my emphasis added) "Given any nonzero vector v you can find many operators for which v is an eigenvector." I am having difficulty understanding how an arbitrary vector can represent specific measurable values of multiple observables. Some examples perhaps? Thanks in advance for your patience in explaining this. – Dyons Jun 16 '19 at 13:33