A What separates Hilbert space from other spaces?

SemM

Gold Member
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!

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fresh_42

Mentor
2018 Award
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

Gold Member
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

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2018 Award
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

Gold Member
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!

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fresh_42

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2018 Award
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.

SemM

Gold Member
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

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2018 Award
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
• 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

Gold Member
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
• 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

Gold Member
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

Gold Member
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

Gold Member
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

Gold Member
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

Homework Helper
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.]

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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

Gold Member
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

Mentor
2018 Award
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

Gold Member
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

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2018 Award
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

Gold Member
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

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2018 Award
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

Gold Member
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

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2018 Award
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.