Do Eigenvectors of an unbounded operator form a complete basis?

Question

It is common to work with unbounded operators in quantum mechanics and quantum field theory (such as position and momentum operators in QM and field operators and their conjugate momenta in QFT) and it is common practice to use their eigenvectors as a complete orthonormal basis for the Hilbert space. However, since these are unbounded operators, they can only be defined on a dense subspace of the Hilbert space (even if we close them). Hence, It is not clear to me if it would be a valid assumption to consider the eigenvectors of these operators as a complete basis! I would appreciate your comments on this assumption and if there is a way to fix this.

Answer 1

I assume you mean to ask if eigenvectors of (essentially) self-adjoint operators form a complete (dense) basis. Without the self-adjointedness requirement the statement is not even true in finite dimension. Instead I suppose you are interested to see how the finite dimensional result generalizes.

The short answer to your question is no. A self-adjoint operator on an infinite dimensional Hilbert spaces can have no eigenvector. The operator itself need not be unbounded either. The multiplication operator by a bounded function is bounded but has no eigenvector.

If you ask for a generalization of a finite dimensional version of the spectral theorem for infinite dimensional Hilbert space there is one which makes sense (also mentioned in the comments). Namely, you know that a finite dimensional Hermitian matrix can be diagonalized by a unitary operator. This version essentially still holds in infinite dimensions. Namely, any self-ajoint operator is unitary equivalent to a multiplication operator in a suitable measure space.

If you think about it long enough you should realize that the operator multiplication by a function is the equivalent of multiplication by a diagonal matrix in finite dimension, except that "position basis" has a different meaning in the two contexts.

A simple example of how this works is given by the following operator on $\ell^2(\mathbb{Z})$:

$$ H |n\rangle = |n+1\rangle + |n-1\rangle $$

where $|n\rangle$ is the vector with one at position $n$ and zero everywhere else. This operator (defined on an infinite dimensional Hilbert space) is unitarily equivalent to multiplication by $2 \cos(k)$. You can see this answer for the explicit construction of the unitary map.