Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
    
Similar to real numbers. –  C.R. Feb 11 '12 at 8:12
5  
I think one point is that you need the completeness to get a reasonable spectral (resp. functional) calculus... In physics vector spaces with inner products are sometimes called "pre-hilbert spaces" to emphasize that they are like hilbert spaces but without the completness of a hilbert space, perhaps you could google for pre-hilbert spaces and quantum mechanis and probably find some references how far you can go using only pre-hilbert spaces... –  student Feb 11 '12 at 13:40
add comment

4 Answers 4

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 defined1 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.


1As V Moretti remarks, this definition of the exponential isn't really well-defined for unbounded operators. Nevertheless it's typically used by physicists... and indeed ok if simply taken as a shorthand for the expression applied to a suitable state.

share|improve this answer
    
What exactly you mean by finite energy, that the energy has lower and upper bound or simply that it is conserved? –  physicsGuy Mar 27 at 8:32
    
@Thomas: lower and upper bound, as in, there exists no sequence of states $\psi_i$ in the Hilbert Space for which $\hat{H} | \psi_i \rangle$ approaches infinity. Actually, that is not granted at all: momentum operators are unbounded – you can obtain arbitrarily high results from normalised, but fast-oscillating states. But it's the kind of thing where we can confidently argue on physical grounds that really we're only dealing with a finite-energy subspace of the mathematical Hilbert space. –  leftaroundabout Mar 27 at 11:26
1  
@leftaroundabout The definition of exponential of an operator you gave does not work, generally, if the operator (assumed to be at least normal) is not bounded. Instead, the right definition arises from the spectral theorem. Nevertheless it is true that if $\psi$, spectrally decomposed with respect to a self-adjoint operator $A$, defines a bounded set in the spectrum of $A$ (as it seems you are saying), then $$e^{iA}\psi = \sum_{n=0}^{+\infty} \frac{i^n}{n!}A^n\psi\:.$$ Everything in a Hilbert space. –  V. Moretti Mar 29 at 19:18
    
@V.Moretti: good point. Though I suspect the majority of physicists wouldn't care, and just pretend all operators are bounded... –  leftaroundabout Mar 29 at 19:24
1  
Yes I think so. However, in QM almost none operator is bounded. Boundedness is equivalent to boundedness of the spectrum, that it turn, is equivalent to boundedness of the set of values taken by an observable. Almost all natural observables are therefore unbounded, in spite of the believes of physicists and just for physical reasons! –  V. Moretti Mar 29 at 19:34
add comment

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

There are many physically fundamental properties that would not hold if the completeness requirement were dropped. First of all the spectral theorem for self-adjoint operators would not hold. So, an observable (let us assume to deal with observables with pure point spectrum) would not have a complete set of eigenstates. There is a fundamental idea in quantum physics that, given an observable and a pure state, the state can always be realized as a coherent superposition of states where the observable is defined. Without completeness this basic idea of quantum theory would not be valid any more.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.