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According to Wikipedia,

Hilbert space [...] extends the methods of vector algebra and calculus from the two-dimensional Euclidean plane and three-dimensional space to spaces with any finite or infinite number of dimensions

However, the article on Euclidean space states already refers to

the n-dimensional Euclidean space.

This would imply that Hilbert space and Euclidean space are synonymous, which seems silly.

What exactly is the difference between Hilbert space and Euclidean space? What would be an example of a non-Euclidean Hilbert space?

user56834
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    "Euclidean space" means simply "finite dimensional Hilbert space". Nothing more nothing less. So, Hilbert space is more general than Euclidean space, since a Hilbert space may not be finite dimensional. – Crostul Aug 25 '16 at 13:28
  • @Crostul That's quite clear now! – Yu Gu Sep 09 '18 at 12:06

4 Answers4

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A Hilbert space essentially is also a generalization of Euclidean spaces with infinite dimension.

Note: this answer is just to give an intuitive idea of this generalization, and to consider infinite-dimensional spaces with a scalar product that they are complete with respect to metric induced by the norm. Clearly, there are finite-dimensional Hilbert spaces, as $\mathbb{R}^n$, with the standard scalar product and Euclidean metric.

user288972
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Hilbert space: a vector space together with an inner product, which is a Banach space with respect to the norm induced by the inner product

Euclidean space: a subset of $\mathbb R^n$ for some whole number $n$

A non-euclidean Hilbert space: $\ell_2(\mathbb R)$, the space of square summable real sequences, with the inner product $((x_n),(y_n)) = \sum_{n=1}^{\infty}x_n y_n$

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    Not a 'subset' of $\mathbb{R}^n$. Nor necessarily is a Euclidean space $\mathbb{R}^n$ for any $n$: for instance, the space of linear functionals on $\mathbb{R}^n$ is not $\mathbb{R}^n$, although it is naturally isomorphic to it, as indeed any Euclidean space is isomorphic to $\mathbb{R}^n$, once a basis has been specified. But it is better to be basis-free as much as possible. – Calum Gilhooley Aug 25 '16 at 13:49
  • @CalumGilhooley May I ask you what is the definition of Euclidean space(I'm confused by what wiki said)? Does it necessarily have a inner-product structure(i.e. Euclidean space is necessarily a inner-product space)? Or it just something that satisfy some "geometric" axioms, such like Euclid's or Hilbert's? – Eric Jan 15 '17 at 04:38
  • @Eric I confess to also being confused by the term "Euclidean space", and the Wikipedia article does not resolve my confusion. I think the most common use of the term, and also the clearest, is that given by Crostul's comment on the OP, viz. a Euclidean space is any finite-dimensional real inner product space. (All such spaces are complete, i.e. are Hilbert spaces.) But the definition of the term in this article is more restrictive. So I think one has to be careful, and look to the context, and define the term if using it oneself. – Calum Gilhooley Jan 16 '17 at 16:24
  • @CalumGilhooley Thanks. So how the Hilbert's or Euclid's axioms come into play in modern geometry? Logically speaking, if we choose the definition of a euclidean space to be real inner product space, meaning that we don't follow/need their(Euclid's or Hilbert's) axiom anymore right? For example, there's a Hilbert's axiom "For every two points A, B there exists a line a that contains each of the points A, B." but in inner-product space terminology, it can be directly proved by trivial pre-calculus techique. – Eric Jan 16 '17 at 17:13
  • I also asked a new post here, maybe you can move there to help me. :) – Eric Jan 16 '17 at 17:15
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  1. A Hilbert space does not have to be infinite dimensional (it could be).
  2. The Euclidean space is an example of a finite dimensional (n- dimensional) Hilbert space where the scalar field is the set of real numbers, i.e., $\mathbb{R}^n$.
  3. It is best to leave out quantum mechanical discussions out of this since it will generally confuse the issue. But, I feel compelled, for the sake of students in QM, to correct Luther4. (i) In quantum mechanics the wavefunction (state vector) of a single particle system is an element of a Hilbert space. The scalar field used in QM is the set of complex numbers. In general the Hilbert space of a single particle system is infinite dimensional. In many example, a finite dimensional subset is sufficient to describe the system. (ii) There is no connection between the number of particles and the dimension of the Hilbert space used in QM. A m-particle system can be described by a cartesian product of m single particle Hilbert spaces. If quantum indistinguishability of the particles is imposed (Fermions Bosons etc,), a subset of the Hilbert space that satisfies certain symmetry with regards to particle exchange is used.
Pand
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An Euclidean space is a normed linear space, that is, it has a norm and its elements are linear functions.

An Euclidean space has an inner product (scalar product):

$$(x_\alpha,x_\beta)=0$$ for orthogonal elements

and

$$(x_\alpha,x_\alpha)=1$$

This scalar product must satisfy the following:

$$ \begin{array} f\bullet\ being\ not\ negative (x,x>0)\\ \bullet\ being\ symmetric (x,y)=(y,x)\\ \bullet\ being\ linear: \lambda x, y)=\lambda(x,y)\\ \bullet\ (x,y+z)=(x,y)+(x,z) \end{array} $$

We say therefore that any space equipped with an inner product satisfying these properties is called Euclidean space.

Euclidean space is therefore a general term for a normed linear space with an inner product, where the norm can be

$$d(x,y)=||x-y||=\sqrt{(x-y)^2}$$

The norm is a distance in the linear normed space, between two points.

Now, a Hilbert space is an example of an Euclidean space. It has an inner product, however it is of infinite many dimensions. For instance, considering a particle in quantum mechanics as a dimension, one can form a Hilbert space of 10 dimensions for a system of 10 particles, where the particle must satisfy the orthogonality principle of the inner product stated at the top. Therefore, all the wavefunctions for the particles given above give an inner product equal to zero, since they have the same angle between one another. Have a look at the atomic orbital chart and see how they have the same angle preserved between each wavefunction.

This wavefunction model is then said to satisfy the conditions for the inner product, and you can have infinitely many of them. If you used an Euclidean space to describe a system of particles, you were restricted to 3 particles, that is, the three dimensions Euclidean space is restricted to. Therefore, you will see that Hilbert space is commonly used in quantum mechanics, with all its properties of the Euclidean space, but with its infinitely many dimensions.

Luthier415Hz
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