What is the physical meaning of a “complete” Hilbert space in QM?

Question

What does the word "complete" means from the physical point of view? I do not understand what it physically means to say that a Hilbert space is a complete vector space.

Answer 1

To give an example where convergence of Cauchy sequences is important: time-evolution is typically calculated as $$ |\psi(t)\rangle = e^{i\hbar^{-1} \hat{H}\cdot t}|\psi_0\rangle $$ now, the exponential of an operator is defined by $$ e^{\hat{A}} = \sum_{i=0}^\infty\frac{\hat{A}^i}{i!} $$ where the sum in turn is defined by $$ \Bigl(\sum_{i=0}^b\frac{\hat{A}^i}{i!}\Bigr)|\psi\rangle = \sum_i\frac{\hat{A}^i|\psi\rangle}{i!} =:S_b $$ which is simply a superposition of vectors, no problem there – for finite sums. But if the sum is infinite, it is defined as the limit of the sequence of partial sums. Is that a Cauchy sequence? It can be shown under quite reasonable assumptions (essentially, finite energy) that it is. So in a Hilbert space, we normally have a nice well-defined expression for time evolution, which is obviously quite handy if you want to verify any theoretical model with experimental results. In a non-Hilbert inner product space we can't be sure that the result is well-defined. Bad!

Another thing that I imagine to be a big problem: the Riesz representation theorem doesn't hold on general inner product space. Though this theorem is seldom explicitly mentioned in physics, it is the reason that you can do lots of things that are often taken for granted; in particular it's required for uniqueness of the Hermitian adjoint. That, I expect, might cause considerable havoc if you're working with stuff like ladder operators.

Answer 2

It is an idealization that allows one to do calculations that are otherwise impossible, and in this way accounts for the power of quantum mechanics.

It is the same sort of idealization that makes physicists work with real numbers in place of rational numbers, although all raw numbers measured are rational. However, the rational numbers lack most of the useful properties of the real numbers. Restricted to rationals you dont even have an exponential or trigonometric function. Thus rational numbers can express only very limited physics.

In the same way, working in a Hilbert space allows one to do many operations (such as talking about $e^{itH}$) that don't make sense in an incomplete vector space.

Hilbert spaces (or variants of it such as rigged Hilbert spaces) are therefore absolutely necessary for having conceptual precision in quantum mechanics.

Answer 3

I think your question is "why wasn't quantum mechanics formulated on normed vector spaces?" i.e "why was the completeness criterion required?"

I don't know a rigorous answer, but it seems reasonable for the following reason:

Completeness means that every Cauchy sequence of elements of H converges to an element of H. The QM postulate says that physical states are represented by vectors (strictly speaking rays) in H, so if I had an infinite sequence of physical states which were getting "physically" closer and closer together - in the sense that the characteristics of the physical quantities encoded in the states were converging, then it seems reasonable to require that the thing they're converging to is also a physical state. Mapping this over to H, then the Cauchy completeness criterion will take care of this.

The reason I'm worried that this is a bit of a weak answer, however, is that not all elements of H necessarily represent realizable physical states. For example the sums of vectors in different superselection sectors certainly doesn't. So maybe the Hilbert space criterion where every Cauchy sequence converges to an element of H is sufficient but not necessary.