Completeness of Norm in Hilbert Space

Question

I am not sure what it really means for the norm to be complete in a Hilbert Space. Can you provide me a proper definition? I am aware of the formula $||\Psi|| = <\Psi|\Psi>^{1/2}$.

What are its implications?

Answer 1

Consider a normed vector space $H$, equipped with some norm $\Vert \bullet\Vert$.

A sequence $\{\psi_i\}$ of elements $\psi_i\in H$ is called Cauchy if, given any $\epsilon>0$, there exists some natural number $N$ such that for all $n,m>N$, $\Vert \psi_n - \psi_m \Vert <\epsilon$. Loosely speaking, this means that the terms in the sequence eventually get (and remain) arbitrarily close together.

A sequence $\{\psi_i\}$ of elements $\psi_i\in H$ is said to converge to some element $\Psi\in H$ if, given any $\epsilon>0$, there exists some natural number $N$ such that for all $n>N$, $\Vert \psi_n - \Psi \Vert < \epsilon$. Loosely speaking, this means that the terms in the sequence eventually get (and remain) arbitrarily close to $\Psi$.

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Question: These definitions look similar. Is a sequence being Cauchy the same as being convergent? It's not difficult to show that if a sequence converges, then it is Cauchy. However, the reverse is not generally true - one can have Cauchy sequences which do not converge to any element of the space.

As an example, consider the rational numbers $\mathbb Q$ equipped with the absolute value as a norm. It is not difficult to come up with a Cauchy sequence of rational numbers which nevertheless does not converge to a rational number; consider the sequence of partial sums for the alternating harmonic series: $$ \begin{aligned} a_1 &= 1\\ a_2 &= 1 - \frac{1}{2}\\ a_3 &= 1 - \frac{1}{2} + \frac{1}{3}\\ \end{aligned} $$ so on and so forth. This sequence can be shown to be Cauchy (exercise for the reader), but does not converge to a rational number (it converges to $\ln(2)$).

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If $H$ has the property that all Cauchy sequences of elements in $H$ are also convergent, then $H$ is called complete - more specifically, complete with respect to the metric induced by the norm. A Hilbert space is, by definition, a complete inner product space.

As for the implications of this, they are mostly technical$^\dagger$ (at least in the sense that they rarely show up explicitly in physics applications). I would say that a good general idea is that the completeness is intimately tied to convergence, which is important if you need to write some element of the Hilbert space as an infinite series of basis elements.

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$^\dagger$ As per Valter Moretti's comment, I don't mean to downplay the importance of this completeness requirement to the framework of quantum mechanics. The spectral theorem for self-adjoint operators, which is fundamentally the basis for our association between self-adjoint operators and physically measurable quantities, requires completeness of the underlying space. Similarly, Stone's theorem which provides, among other things, a way to associate self-adjoint operators with continuous symmetries, requires completeness as well.

Answer 2

Completeness with respect to the norm means that every Cauchy sequence converges to a point in the space.

A typical example of an incomplete inner product space is the polynomials on the interval $(0,1)$ with the inner product $(f,g)=\int_0^1 dx \; f(x) g(x) $.

Any linear combination of polynomials is a polynomial. But an infinite (Cauchy) sequence of polynomials (i.e. a Taylor series) can converge to a non-polynomial function. Thus polynomials form a vector space with inner product, but not a Hilbert space.

On the other hand, a Cauchy sequence of square-integrable functions on (0,1) remains square-integrable. So the square-integrable functions do form a Hilbert space in this inner product.