Can Quantum Mechanics be Studied in Banach Spaces or Other Non-Hilbert Spaces?

[SemM]

Hi, are there any models known in QM where the wavefunctions do not have to be infinitely differentiable, and thus can exist in other spaces than the Hilbert space? I assume Banach spaces allow elements that are not infinitely differentiable as subsets. Can therefore certain phenomena in QM be studied in Banach spaces or in other metric spaces where the elements are not inf. diff.?

Thanks

 

[fresh_42]

Wavefunctions are expressions in the complex exponential function and therewith infinitely differentiable - qua definitionem. Otherwise, they wouldn't be wavefunctions. Perhaps one can manipulate the amplitudes by some functions, which are not smooth, but this misses the point. And function spaces, which are Hilbert spaces, do not require to consist of smooth functions. The order is: functions to be considered $\longrightarrow$ common properties of these functions $\longrightarrow$ space they form, Hilbert, Banach or whatever - not the other way around. If you have a naked Hilbert space given, you know absolutely nothing about it's elements, apart from some convergence and arithmetic properties. They don't even have to be functions, e.g. directions will do. Important Hilbert spaces are $L^2-$spaces, which are defined by integrability, not differentiability.

It is as if you said, functions in a vector space have to be continuous. This statement doesn't make sense.

 

[martinbn]

SemM, what Hilbert space do you have in mind, where all the functions are smooth?

 

[SemM]

SemM, what Hilbert space do you have in mind, where all the functions are smooth?

Where all elements are continuous within a given boundary.

 

[fresh_42]

Where all elements are continuous within a given boundary.

But continuous doesn't imply even once differentiable. $x \longmapsto |x|$ is continuous and not differentiable at all.

 

[SemM]

But continuous doesn't imply even once differentiable. $x \longmapsto |x|$ is continuous and not differentiable at all.

I am mixing apparently two things. Differentiability and continuity. Sin(x) is continuous, and it is infinitely differentiable. Tan(x) is infinitely differentiable, but its not continuous.

 

[fresh_42]

Differentiability implies continuity. Where $\tan$ is defined, it is differentiable. It is not at the points, where it isn't defined, and where it is discontinuous. It's one of the advantages of the expression by exponential functions to see immediately all critical points.

 

[rubi]

Wave functions don't need to be continuous or differentiable. The Hilbert space of standard quantum mechanics is $L^2(\mathbb R)$, the space of (equivalence calsses of) square-integrable functions. It contains lots of non-differentiable functions such as $\chi_{[0,1]}(x)$. Wave functions don't even need to approach $0$ as $x\rightarrow\infty$, as the example $f(x)=\sum_{n=0}^\infty\chi_{[n,n+2^{-n}]}(x)$ shows.

 

[SemM]

Wave functions don't need to be continuous or differentiable. The Hilbert space of standard quantum mechanics is $L^2(\mathbb R)$, the space of (equivalence calsses of) square-integrable functions. It contains lots of non-differentiable functions such as $\chi_{[0,1]}(x)$. Wave functions don't even need to approach $0$ as $x\rightarrow\infty$, as the example $f(x)=\sum_{n=0}^\infty\chi_{[n,n+2^{-n}]}(x)$ shows.

Which part of QM is this in? Standard QM requires solutions to be square-integrable, finitely differentiable, continuous and single-valued.

 

[martinbn]

Where all elements are continuous within a given boundary.

I am asking about the space. You see the problem is that a typical space of continuous (or differentiable) functions will not be complete.

 

[rubi]

Which part of QM is this in? Standard QM requires solutions to be square-integrable, finitely differentiable, continuous and single-valued.

I'm talking about standard textbook QM. It requires square-integrability, but neither continuity nor differentiability. It's perfectly admissible to have discontinuous and non-differentiable wave-functions. The time evolution operator $e^{-\mathrm i t \hat H}$ is bounded and thus can be applied to any function in the Hilbert space. Of course, such states are physically unrealistic, but quantum mechanics as a mathematical theory has no problems dealing with them.

 

[PeroK]

I am mixing apparently two things. Differentiability and continuity. Sin(x) is continuous, and it is infinitely differentiable. Tan(x) is infinitely differentiable, but its not continuous.

$\tan x$ is a continuous function. It's not defined for $x = \pi/2$ etc.

 

[SemM]

$\tan x$ is a continuous function. It's not defined for $x = \pi/2$ etc.

I am asking about the space. You see the problem is that a typical space of continuous (or differentiable) functions will not be complete.

The metric space is pre-Hilbert if it's not complete. If it's complete , it is complete by that it's elements are square-integrable. However, this poses a question which is quite important: if square integration is a simple integration of the square of the modulus of a function f, then any function that is finite can be part of a subset of H, which itself is complete. But that does not make H complete. I am not sure. I take Fresh can specify this better than me.

PeroK:
Would tan(x) be an element outside the complete domain in H, composed of square integrable functions? It is apparently continuous but it's not square integrable.

Martinbn:
The space composed of discontinuous functions would be outside the complete sub-domain in H, so it would be an own domain in H,which is composed of non-square integrable functions i.e sin(x)?

 

[PeroK]

$\tan x$ is not square integrable.

 

[PeroK]

Hi, are there any models known in QM where the wavefunctions do not have to be infinitely differentiable, and thus can exist in other spaces than the Hilbert space? I assume Banach spaces allow elements that are not infinitely differentiable as subsets. Can therefore certain phenomena in QM be studied in Banach spaces or in other metric spaces where the elements are not inf. diff.?

Thanks

This post is such a confusion of ideas that it's practically unanswerable.

 

[SemM]

This post is such a confusion of ideas that it's practically unanswerable.

The answers here cleared up the confused ideas.

 

[SemM]

This post is such a confusion of ideas that it's practically unanswerable.

No this is not completely true. There are many papers on Hamiltonians for 2D particles for instance with non-standard wave-functions, which give complex eigenvalues to the observables. And these complex functions are actually related to the vortex lines of the wave amplitudes for these models, which are useful for emerging pinning algorithms. So I wouldn't dismiss this so bluntly, Perok.

 

[fresh_42]

The metric space is pre-Hilbert if it's not complete.

The other way around: A pre-Hilbert space is a metric space. A Hilbert space is a complete pre-Hilbert space, esp. also a pre-Hilbert space, since Hilbert spaces are a subset of pre-Hilbert spaces. Thus they are both metric spaces.

If it's complete , it is complete by that it's elements are square-integrable.

The other way around. See post #2. Completeness is a property about convergence, which has a priori nothing to do with the nature of the elements, which converge.

However, this poses a question which is quite important: if square integration is a simple integration of the square of the modulus of a function f, then any function that is finite ...

What is a finite function?

... can be part of a subset of H, which itself is complete.

Subsets are not automatically complete. The closed ones are. This is again primarily a topological property and not a property of the elements (functions) involved.

But that does not make H complete.

Completeness makes H complete, nothing else. Completeness means that all Cauchy sequences converge.

My recommendation is to read a book about the fundamental topological concepts followed by an introduction to functional analysis, which certainly covers the various aspects of pre-Hilbert, Hilbert and Banach spaces. Those books normally also contain a lot of examples. Mine has e.g. on pages 15ff: $C_\infty^0[0,1]$ (Banach), $C_2^0[1,0]$ (pre-Hilbert), $\mathbb{C}^n , \mathbb{R}^n,l_2\, , \,L_2(M;\rho)$ (Hilbert).

The following insight article and the discussion at the end might be helpful. At least it contains some books, which I can recommend.
https://www.physicsforums.com/insights/tell-operations-operators-functionals-representations-apart/

As this thread turned out to be a collection of facts about Hilbert spaces, rather than models in quantum mechanics, which all indicated some fundamental misconceptions about the topological vector spaces involved, and which are not necessary to discuss on an "A" level thread, I'll close this thread for now.

@SemM: In case you want to discuss certain models in quantum mechanics, please give us a source upon which we can debate about physics, rather than mathematics.