What separates Hilbert space from other spaces?

[SemM]

Hi, I have the impression that the special thing about Hilbert space for Quantum Mechanics is that it is simply an infinite space, which allows for infinitively integration and derivation of its elements, f(x), g(x), their linear combination, or any other complex function, given that the main operators in QM, which are differentiation, multiplication and addition, are all linear self-adjoint operator which thus act orthogonally on orthogonal elements, and thus operate in infinitively many dimensions. This is not possible in $\mathbb{C}$ and $\mathbb{R}$, where both include coordinates as their elements.

What about Banach spaces however? Is my explanation given above on the Hilbert space sufficient to explain to anyone why that makes it ideal for QM?

Thanks!

 

[fresh_42]

Hi, I have the impression that the special thing about Hilbert space for Quantum Mechanics is that it is simply an infinite space, which allows for infinitively integration and derivation of its elements, f(x), g(x), their linear combination, or any other complex function, given that the main operators in QM, which are differentiation, multiplication and addition, are all linear self-adjoint operator which thus act orthogonally on orthogonal elements, and thus operate in infinitively many dimensions. This is not possible in $\mathbb{C}$ and $\mathbb{R}$, where both include coordinates as their elements.

What about Banach spaces however? Is my explanation given above on the Hilbert space sufficient to explain to anyone why that makes it ideal for QM?

Thanks!

There is no restriction on dimension. Hilbert as well as Banach spaces can be of any dimension, finite or infinite. Hilbert spaces have a dot product which induces a norm and Banach spaces just a norm, no inner product required.
https://www.physicsforums.com/insights/tell-operations-operators-functionals-representations-apart/

 

[SemM]

Thanks Fresh, I will print this out.

About Banach space, if no dot product is required, how are the observables being evaluated in a defined space? I have never seen using a norm to define the momentum operator in Quantum mechanics. Would that be relevant with the Heisenberg model of Quantum Mechanics, where he uses matrices rather than wavefunctions - for example?

 

[fresh_42]

I haven't enough knowledge about QM, but what I have seen on Wiki about the Heisenberg model, it uses the product of neighboring spin vectors as weights, which is a dot product. In Banach spaces you still have a norm, which means a length and a distance. What's lost are angles. This could e.g. be compensated by the use of other bilinear forms which provide a dot product, e.g. semisimple Lie algebras provide such a product by their Killing form and also allow a partially geometric handling, because one can define reflections. This might induce a different norm, so some care is needed. On the other hand, as far as I know, QM deals with Hilbert spaces from the start.

 

[SemM]

Yes, QM deals with Hilbert spaces from the start. This is based on that a particle is constrained to one dimension, $\mathbb{R}$, however, it must be transferred to a Hilbert space, should one derive the expectation values of its properties, momentum and position. This is where Hilbert comes in. However, it also becomes confusing: I suppose one can say that a Hilbert space in QM should not be considered as a geometrical space, but more as an infinite space made so for that the operators of QM to give the mean values by the inner product to the state $\psi$ selected, as it must be square integrable to give real expectation values with a physical meaning.

I was however wondering, can this also be done in a Banach space? If so, why is not Banach space selected in QM over Hilbert space?

Thanks!

 

[fresh_42]

I was however wondering, can this also be done in a Banach space? If so, why is not Banach space selected in QM over Hilbert space?

Because with $\int f^*(x)g(x)dx$ you automatically have a Hilbert space which is also a Banach space.

 

[mathman]

http://people.math.gatech.edu/~heil/handouts/banach.pdf

Above is a discussion of the concepts of Hilbert space and Banach space. Their definitions are completely mathematical, which happen fit in with Quantum mechanics.

 

[SemM]

Because with $\int f^*(x)g(x)dx$ you automatically have a Hilbert space which is also a Banach space.

So Banach space = Hilbert space? I guess this is not the case, but based on this, it either means that Banach Space = Hilbert space, or Banach space $\subset$ Hilbert space, which probably also is not true.

That formula you give is the inner product, if it can be used in both Banach space and Hilbert space, why is QM strictly applied in Hilbert space?

 

[fresh_42]

All (Hilbert spaces, $\langle .,.\rangle \Rightarrow ||.||$) are (Banach spaces, $||.||$).

That formula you give is the inner product, if it can be used in both Banach space and Hilbert space, why is QM strictly applied in Hilbert space?

If this product exists, we have a Hilbert space. In a Banach space, there is in general no inner product, for otherwise we would call it a Hilbert space. It is not forbidden for a Banach space to have an inner product, it is simply not required.

The rational numbers $\mathbb{Q}$ are

• an Abelian additive group
• a $\mathbb{Q}-$vector space
• a communtative ring with $1$
• an integral domain
• an associative $\mathbb{Q}-$algebra
• a $\mathbb{Q}-$Lie algebra
• a field
• a prime field
• a quotient field
• an Archimedean ordered field

Now make your choice. Your question sounds like: Why do we consider $\mathbb{Q}$ as a field and not as a vector space?

 

[SemM]

All (Hilbert spaces, $\langle .,.\rangle \Rightarrow ||.||$) are (Banach spaces, $||.||$).

If this product exists, we have a Hilbert space. In a Banach space, there is in general no inner product, for otherwise we would call it a Hilbert space. It is not forbidden for a Banach space to have an inner product, it is simply not required.

The rational numbers $\mathbb{Q}$ are

• an Abelian additive group
• a $\mathbb{Q}-$vector space
• a communtative ring with $1$
• an integral domain
• an associative $\mathbb{Q}-$algebra
• a $\mathbb{Q}-$Lie algebra
• a field
• a prime field
• a quotient field
• an Archimedean ordered field

Now make your choice. Your question sounds like: Why do we consider $\mathbb{Q}$ as a field and not as a vector space?

It appears to me, from what you write here, that $\mathbb{Q}$ is regarded as the integral area, among other properties, and thus can be attributed to the probability density. In Hilbert space, $\mathbb{Q}$ does not exist, but in R (or C) it does. So , unless I am confusing things here, the probability density of $\psi$ (where $\psi$ "lives" in Hilbert space) is actually in R. So, if $\mathbb{Q}$ is a scalar in R, or a field, like you say, I can see that Hilbert space, which is a vector space, differs from Banach space by being a vector space?

 

[SemM]

If this product exists, we have a Hilbert space. In a Banach space, there is in general no inner product, for otherwise we would call it a Hilbert space. It is not forbidden for a Banach space to have an inner product, it is simply not required.

This sounds like a daring question I am posting, but doesn't the inner product ALWAYS exist, as long as there are two vector elements ?

 

[Infrared]

This sounds like a daring question I am posting, but doesn't the inner product ALWAYS exist, as long as there are two vector elements ?

No, given a normed vector space $(V,|\cdot|)$, there is not always an inner product on $V$ that induces the norm (in the sense that $|v|=\sqrt{(v,v)}$ ). A norm $|\cdot|$ induced from an inner product must satisfy $|a+b|^2+|a-b|^2=2(|a|^2+|b|^2)$, but there are plenty of normed vector spaces that don't satisfy this identity, like $L^p$ for $p\neq 2$.

 

[SemM]

No, given a normed vector space $(V,|\cdot|)$, there is not always an inner product on $V$ that induces the norm (in the sense that $|v|=\sqrt{(v,v)}$ ). A norm $|\cdot|$ induced from an inner product must satisfy $|a+b|^2+|a-b|^2=2(|a|^2+|b|^2)$, but there are plenty of normed vector spaces that don't satisfy this identity, like $L^p$ for $p\neq 2$.

Thanks!

Is an example of this the case of the vector pair v1 and the null vector?

 

[Infrared]

Thanks!

Is an example of this the case of the vector pair v1 and the null vector?

I'm not sure what you mean- what is your chosen normed vector space? In any event, you will never get a counterexample by letting one of the vectors be $0$ (in my notation, $(a,b)=(v_1,0)$), because then both sides will equal $2|v_1|^2$, so I think the answer to your question, to the best I understand it, is "no".

 

[mathwonk]

try my favorite banach space, the space of continuous functions on the closed unit interval, with norm f equal to the supremum (largest value) of |f| on the interval. then try out some different functions and see what they give you for this desired equality.

[spoiler alert: try f = a linear function going down from 1 to 0 along this interval, and another linear function g going up fom 0 to 1.]

 

[chiro]

Hey SemM.

Hilbert spaces have to be continuous, they have to converge and they have to maintain consistency with C and not R which is not easy because of the i^2 = -1 property.

The continuity is hard because of infinite dimensions since you can find many ways to make matrix multiplication with a vector infinity (+ and -) and you have to find all of the conditions that make A*x a sensible vector. When that vector has to be in Hilbert space then it means that the integral on functional needs to converge [i.e. be a proper number] and finding all the constraints that make this possible is what is studied in operator algebras.

The infinite dimensional operators make this hard since there are so many ways you can make each section/coefficient of the resulting vector [think Ax = b] an infinity term and even if they are all real values, you also have to consider that <a,b> [inner product of two vectors] might be + or - infinity as well which screws up everything with consistency.

If one finds the constraints on the operator itself that allows everything to be finite number [real and complex] then the space of all operators have been found [infinite dimensional] and one can use that to make sure that if you use one [like in quantum mechanics] then all of the calculus and linear algebra for infinite dimensions will always work.

 

[SemM]

Hey SemM.

Hilbert spaces have to be continuous, they have to converge and they have to maintain consistency with C and not R which is not easy because of the i^2 = -1 property.

The continuity is hard because of infinite dimensions since you can find many ways to make matrix multiplication with a vector infinity (+ and -) and you have to find all of the conditions that make A*x a sensible vector. When that vector has to be in Hilbert space then it means that the integral on functional needs to converge [i.e. be a proper number] and finding all the constraints that make this possible is what is studied in operator algebras.

The infinite dimensional operators make this hard since there are so many ways you can make each section/coefficient of the resulting vector [think Ax = b] an infinity term and even if they are all real values, you also have to consider that <a,b> [inner product of two vectors] might be + or - infinity as well which screws up everything with consistency.

If one finds the constraints on the operator itself that allows everything to be finite number [real and complex] then the space of all operators have been found [infinite dimensional] and one can use that to make sure that if you use one [like in quantum mechanics] then all of the calculus and linear algebra for infinite dimensions will always work.

Thanks Chiro! I read also your other reply on the other thread, and they are both deep. I will need to get back to this after reading some more. Thanks!

 

[fresh_42]

Hilbert spaces have to be continuous, they have to converge and they have to maintain consistency with C and not R which is not easy because of the i^2 = -1 property.

The continuity is hard because of infinite dimensions

What is a continuous space? You probably meant complete!
And Hilbert spaces can also be real, and of finite dimension!

 

[SemM]

What is a continuous space? You probably meant complete!
And Hilbert spaces can also be real, and of finite dimension!

I think he meant that the each element in a Hilbert space must be continuous?

 

[fresh_42]

I think he meant that the each element in a Hilbert space must be continuous?

This doesn't make sense either, as at prior elements of a Hilbert space are just vectors. E.g. $\begin{bmatrix}1\\2\end{bmatrix}$ is an element of a Hilbert space, $\mathbb{R}^2$. Now what makes it continuous? And he said the space would be continuous, a term which is defined for functions on topological spaces, not the spaces themselves.

 

[SemM]

This doesn't make sense either, as at prior elements of a Hilbert space are just vectors. E.g. $\begin{bmatrix}1\\2\end{bmatrix}$ is an element of a Hilbert space, $\mathbb{R}^2$. Now what makes it continuous? And he said the space would be continuous, a term which is defined for functions on topological spaces, not the spaces themselves.

Yes I know, but what about the point of that a wavefunction, which is element of Hilbert space, can be represented by vectors?

 

[fresh_42]

Yes I know, but what about the point of that a wavefunction, which is element of Hilbert space, can be represented by vectors?

Wavefunctions can be added, stretched and compressed, so they are vectors in the sense of elements in a Hilbert space. Whether these are continuous as functions on topological vector spaces is another question and attributed to the fiunctions, not the space. Of course one can start as Let $\mathcal{H}$ be the Hilbert space of continuous functions ... but this has nothing to do with the definition of a Hilbert space, only with the selection of a specific one where the elements happen to be continuous functions.

 

[SemM]

Wavefunctions can be added, stretched and compressed, so they are vectors in the sense of elements in a Hilbert space. Whether these are continuous as functions on topological vector spaces is another question and attributed to the fiunctions, not the space. Of course one can start as Let $\mathcal{H}$ be the Hilbert space of continuous functions ... but this has nothing to do with the definition of a Hilbert space, only with the selection of a specific one where the elements happen to be continuous functions.

I see your point, its a definition question and it would be like saying a football field is composed of footbal-trajectories instead of being composed of a field to kick the footballs on.

 

[fresh_42]

It's more like saying a football field is green grass. Firstly, not the football field is green grass, instead of it consists of green grass, which is a difference. And secondly, not all football fields are made of green grass. There are also ones made of clay, artificial grass or theoretically one could even imagine grass of other colors. So the field with green grass is an example, not a requirement.

However, as with every comparison, it's not 100% accurate to put it this way.

 

[mathman]

Hilbert spaces are defined as vector spaces, which have an inner product and is complete. The inner product then defines the norm as ||x||^2=(x,x). Complete means every Cauchy sequence has a limit in the space. The most common examples are l^2, which are infinite series, where the squares of the absolute values have a finite sum, and L^2 which are functions where the integrals of the absolute value squared over the specified domain are finite.

 

[bhobba]

For those interested in this stuff they need to study functional analysis. My personal favorite starting book on that is the following:
http://matrixeditions.com/FunctionalAnalysisVol1.html

Unfortunately not available now as a hard copy - which is a pity - but still my favorite book.

You are then prepared to look into the issues of its relation to QM:
https://www.amazon.com/dp/0821846302/?tag=pfamazon01-20

Thanks
Bill

 

[FactChecker]

Please forgive me if I repeat things that have been said before.
A Banach space is complete and has a norm, so distance and convergence is defined. Every Hilbert space is a Banach space that additionally has a dot product which defines the norm and angles. So the extra things that you get with a Hilbert space are angles, orthogonal decomposition, Pythagorean Theorem, parallelogram law, etc. If you are working in QM with a space that is a Hilbert space, then you would want to take advantage of those extra things that give you so much more geometric intuition.

 

[chiro]

I actually meant continuous. Everything has to have continuity when it comes to analysis of these operators and continuous things are also measurable [no infinity terms or divergence].

Continuity is easy to think about when you have to have every mapping being draw with a pencil [or filled in if it is a region] by taking a pencil and putting it on some paper and doing the mapping without taking it off the page.

Also - finite dimensional spaces are not studied because they don't need to be. We assume in a hilbert space that the coefficients of the vector in l^2 are real or complex numbers of any kind and in L^2 we assume the integral to always be finite when it comes to norms and metrics.

It's easy to prove because summing and multiplying numbers always gives finite result if all are finite. The infinite part is what makes hilbert space and operator algebra theory hard.

You might want to consider how hard it is to find a way to get multiple infinite series to converge all at the same time to know why the generalization to infinite dimensions is not trivial.

 

[fresh_42]

I actually meant continuous.

... which is still a property of functions and not space.

Everything has to have continuity when it comes to analysis

... which is wrong in this generality. E.g. step functions are frequently used in analysis or with the integrals in functional analysis. If we consider linear operators, we have continuity for them, but these aren't the only ones which are studied and says nothing about the elements these operator act on. It is not part of the definition of a Hilbert or Banach space. It doesn't even have to be functions or sequences! The square integrable functions which are often used in physics are simply an example, not the necessity.

Also - finite dimensional spaces are not studied because they don't need to be.

... which doesn't change the fact that $\mathbb{R}^n$ is a Hilbert space and thus also a Banach space: real and finite dimensional.

 

[PeroK]

Continuity is easy to think about when you have to have every mapping being draw with a pencil [or filled in if it is a region] by taking a pencil and putting it on some paper and doing the mapping without taking it off the page.

Also - finite dimensional spaces are not studied because they don't need to be. We assume in a hilbert space that the coefficients of the vector in l^2 are real or complex numbers of any kind and in L^2 we assume the integral to always be finite when it comes to norms and metrics.

This is an A level thread. It's hard to accept "drawing a line without lifting a pencil off the page" as an adequate description of continuity of linear operators on function spaces!

The theory of quantum spin utilises finite dimensional Hilbert spaces.

 

[chiro]

There is no point for a finite dimensional space since all the theory is worked out in R^n or C^n.

You do need some geometric intuition for continuity in a hilbert space since you have infinite dimensions.

How are you going to get that intuition with an infinite dimensional matrix and an infinite dimensional vector?

 

[fresh_42]

There is no point for a finite dimensional space since all the theory is worked out in R^n or C^n.

This is a debatable point of view. In any case these examples have still to hold, if theorems on arbitrary Hilbert spaces are proven. They are a good test object.

You do need some geometric intuition for continuity in a hilbert space since you have infinite dimensions.

... since you can have infinite dimension. Whether a geometric intuition is needed is a very personal decision. I think no intuition and merely formulas are still better than a wrong intuition. And to draw a line with a pencil without interruption is wrong since the days of Cantor.

How are you going to get that intuition with an infinite dimensional matrix and an infinite dimensional vector?

I work with the definitions and finite dimensional examples. A couple of things are different in finite and infinite dimensional vector spaces, but this wan't the question here. The answer to the OP's question is: inner product and completeness, not dimensionality.

 

[FactChecker]

There is no point for a finite dimensional space since all the theory is worked out in R^n or C^n.

It may be better to say that our knowledge of Rⁿ or Cⁿ is abstracted by identifying their Banach and Hilbert properties and applying them to other spaces. Then we can have Rⁿ or Cⁿ motivated insight in those spaces.

You do need some geometric intuition for continuity in a hilbert space since you have infinite dimensions.

The Dirac delta function, associated comb functions and other generalized functions are very important in Fourier analysis. Continuity is not required for a function to be measurable and Lebesgue integration and measure can handle very strange functions.

How are you going to get that intuition with an infinite dimensional matrix and an infinite dimensional vector?

Integration or convolution with an appropriate kernel function is the infinite-dimensional equivalent of matrices.

 

[bhobba]

Just elaborating a bit on what FactChecker said.

Hilbert spaces were invented by David Hilbert and others to help in areas like Fourier Analysis, Partial Differential Equations and eventually QM.

Nowadays, especially for applications to QM and Fourier Transforms, IMHO the easiest way to naturally motivate infinite Hilbert spaces is via what's called Rigged Hilbert spaces of great value in understanding mathematically difficult/unusual things like the Dirac Delta Function - which is not part of a Hilbert space.

What you do is take the set of all complex (just to be definite row) vectors of finite length then consider its dual. Its in fact any infinite sequence at all. Well that leads you naturally to look at dual's of subspaces of that huge and very general space. What you notice is sometimes that dual contains the original space and sometimes its the other way around. There is one very special space whose dual is basically itself - that is called the Hilbert space. This is expressed in the so called Gelfland Triple:

S* ⊃ H ⊃ S - where H is a Hilbert space (ie the space such that the sum of the squares of the sequences absolute value exists) - S is some subspace of H and S* is dual.

This view is especially useful in studying distribution theory that makes Fourier transform's a snap - otherwise you quickly run into issues of convergence and functions that do not belong to a Hilbert space eg that dreaded Dirac Delta Function mentioned before. It bedeviled the greats like Von-Neumann who I believe actually invented the name - Hilbert Space - in honor of David Hilbert who he was assistant to (wow what a pair that would be - like Wheeler and Feynman).

That's just a way of motivating why its interesting to study it - you can of course treat it purely mathematically with no motivation at all.

Added Later:
In light of Marks response - note the original space of finite vectors. Finite vectors are Hilbert spaces ie they all can be placed in one to one correspondence with its dual. We see in that light infinite Hilbert spaces are simply a generalization of that interesting property to infinite spaces. The Dirac Delta function belongs to the dual of a subset of a Hilbert space - the space of what are called good functions is usually used because it has the interesting property of the Fourier transform of a good function is a good function. Its dual is called a Schwartz space - and its easy to define the Fourier Transform of any member of that space because of the Fourier Transform property of good functions. The Dirac Delta function is a member of the Schwartz space. It took quite a while for mathematicians to figure it all out and solve what the great Von-Neumann couldn't.

Thanks
Bill

 

[Mark44]

There is no point for a finite dimensional space since all the theory is worked out in R^n or C^n.

You do need some geometric intuition for continuity in a hilbert space since you have infinite dimensions.

A Hilbert space can be of infinite dimension, but doesn't have to be.

Emphasis added in both quotes.
From Wikipedia (https://en.wikipedia.org/wiki/Hilbert_space):

(A 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.

From Wolfram MathWorld (http://mathworld.wolfram.com/HilbertSpace.html)

Examples of finite-dimensional Hilbert spaces include
1. The real numbers $\mathbb R^n$ with <v, u> the dot product of v and u.
2. The complex numbers $\mathbb C^n$ with <v, u> the dot product of v and the complex conjugate of u.