Generating a Hilbert space representation of a wavefunction

[SeM]

Hello, I Have a particle with wavefunction Psi(x) = e^ix

and would like to find its Hilbert space representation for a period of 0-2pi. Which steps should I follow?

Thanks!

 

[strangerep]

Hmm. Is this homework? (If so, you should be posting this over in one of the HW forums.)

Anyway,... the answer should follow immediately if you list the defining properties of "Hilbert space" (which are easily obtainable from Wikipedia, or by googling).

 

[bhobba]

Hello, I Have a particle with wavefunction Psi(x) = e^ix

Do you know what a Hilbert space is? If so does what you wrote belong to it?

And if you don't know what it is then its a rather hard question for us to answer - the answer is it isn't even a member of a Hilbert Space - but any detail beyond that requires you to know what one is.

Thanks
Bill

 

[strangerep]

[...] the answer is it isn't even a member of a Hilbert Space [...]

Heh, that depends on whether the OP meant $0$ to $2\pi$ as a domain...

 

[bhobba]

Heh, that depends on whether the OP meant $0$ to $2\pi$ as a domain...

True - we really need more info.

Thanks
Bill

 

[SeM]

I think this explains what I am wondering about:
https://www.researchgate.net/deref/https%3A%2F%2Fphysics.stackexchange.com%2Fquestions%2F246647%2Fwhy-do-we-need-both-hamiltonian-and-hilbert-space-to-specify-a-quantum-system%2F246650
However, I am not sure which steps one should follow to prove the solutions to a Hamiltonian in a Hilbert space. So far, I have only used a Cartesian system with 2 dimensions, and got a solution to the Hamiltonian, the eigenstate f(x) = e^ikx. However, this is for a cartesian system. How can I analyse this solution and the Hamiltonian in a N-finite Hilbert space for the interval of 0 to 2pi? Are there 3-4 steps one can stick to in order to interpret the existing Hamiltonian and its Cartesian solution (for one particle) in a Hilbert space for N particles?

 

[bhobba]

Take the Hamiltonian of a free particle, Solve it. Are they in a Hilbert space? What you wrote is one such solution (p=1) in certain units. Is it in a Hilbert space?

Again I mention if you do not know what a Hilbert space is (all functions in a Hilbert Space are Lebesque square integrable as you should either know or look up in your investigations). Is the function you wrote square integrable - multiply it by its conjugate - that's the squaring bit, then integrate from -∞ to ∞. What do you get? Does that imply its in a Hilbert space? Indeed how do you apply the Born Rule to such a function?

Please post at least an attempt to answer these questions - its very very important and led no less a person than the great Von-Neumann to exasperation and to also no less a person than the equally great mathematician, Grothendieck, along with others such as Schwartz and Gelfland to sort out - and it took time as well - it didn't happen overnight.

Now I could tell you the answer but will not for three reasons. One - what you figure out for yourself you understand better. And two it will ensure you fully understand concepts like Hilbert space etc. Many beginning students see it mentioned in texts but don't grasp what it is. Its wise to rectify that. And thirdly you develop self reliance - really your question should have been what's going on with the solution you wrote and states form a Hilbert space, detailing the exact issue..

Thanks
Bill

 

[SeM]

Dear Bill, thanks for this extensive explanation. Answer to your question, the Hamiltonian its solved for a Cartesian space, not Hilbert space.

I will first try your advice, and then get back to you here or via PM, is that OK? It will be next week.

First I check if it is Hermitian, then I do a square integration over - to + infinity with its complex conjugate and normalize it to 1. At that stage I apply the Born rule to it, then I get back to you!

Thanks!

Have a good weekend

 

[bhobba]

Dear Bill, thanks for this extensive explanation. Answer to your question, the Hamiltonian its solved for a Cartesian space, not Hilbert space.

Nope - the solution is an element of a Hilbert space, or to be absolutely precise it is the representation of a member of a Hilbert space, but we will not be that carefull to start with.

I will first try your advice, and then get back to you here or via PM, is that OK?

I would rather you post in this thread.

First though can you tell us what textbook you are using?

First I check if it is Hermitian, then I do a square integration over - to + infinity with its complex conjugate and normalize it to 1. At that stage I apply the Born rule to it, then I get back to you!

Try that and see what happens - but I will this time outline it first - multiply e^ix by its conjugate and what you get is 1. Take the integral from -∞ to ∞ and what do you get? That depends on your definition of such improper integrals - in some texts you take the limit as l → ∞ of the integral from -l to l in which case you get zero. In others you break it into two bits and you get -∞ + ∞. What that means again depends on how you interpret such things. Some may say its undefined while others again say it's zero - we will take zero again. To normalize your wave-function divide it by zero - well again depending on your view of such things that is either undefined or infinite. Either way you are up the creek without a paddle.

Whats going on? That solution to the Schrödinger equation is NOT a member of a Hilbert space, its not normalizeable - as you would expect physically anyway since its cos x + i sine x for x -∞ to ∞. It's not zero at infinity which is physical nonsense - in fact the limit at infinity of cos x or sine x, since its an oscillating function, is undefined - but it does not go to zero. You have obtained an nonphysical solution.

Once you understand that I can explain the resolution which depends on your mathematical sophistication.

Thanks
Bill

 

[SeM]

Dear Bill, the wavefunction is actually quite more complex than this, I used this simple version to see what to do in the first place. Now I see that:

1. Check hermiticiy
2. Integrate the complex conjugate with the wavefunction over infinite space, normalize to 1, then check for it towards the Born Rule.
3. Get back to you.

If there is something else I could do, was thinking of generating the Product of the WFN with N, so

Psi1* Psi2*...Psi

I will do this on Monday, because its too important to be done on a half way on a Sunday.

Sergio

 

[bhobba]

Dear Bill, the wavefunction is actually quite more complex than this,

Yes the actual solution for Schrödinger's equation is more complex and like you said you should really do separation of variables to solve it.

Do all those things - but the same issue will arise - for the Schrödinger equation with no potential term you get terms of the form you wrote and exactly the same issues arise when you try to find its norm then calculate the normalized solution - you will not be able to do it.

Get back here when you have done it and I will explain the solution.

It will be greatly facilitated if you do a bit of study before into what's called distribution theory:
http://orion.lnu.se/pub/education/course/electrical/4ED044/Distributions.pdf

Thanks
Bill

 

[SeM]

Dear Bill, that is all correct which you wrote. The potential term is not included because there is no proper potential form, as these are quite free particles without the conventional potential form. I have therefore omitted that term, and look only at the kinetic term. I will look at all you have wrote, and get back in some days. This may take some more time.

Thanks for your help!

 

[bhobba]

The potential term is not included because there is no proper potential form, as these are quite free particles without the conventional potential form.

The wave-functions of free particles do not belong to a Hilbert space - they are non-physical as I explained.

However one can, in a sense I will not go into (called weak convergence) find a sequence of functions, fn, that are zero outside a certain interval (called functions of compact support) such that fn → f where f is a function like you got. Since the functions are of compact support they are of course square integergrable. So all you do is consider the solution you got simply as an approximation to fn when n is so large for all practical purposes its the same as f - the function you got. The interval where its not zero could be the length of the universe - just as long as its way outside what can be measured. That's all that required to resolve it - physically that is. Mathematically - the solution is much harder and requires Rigged Hilbert Spaces:
https://arxiv.org/pdf/quant-ph/0502053.pdf

Thanks
Bill

 

[vanhees71]

There are plenty of wave functions for free particles that are square integrable. You can start, e.g., with a Gaussian wave packet, and it's easy to calculate that with time it stays a Gaussian wave packet with its "center" moving with constant velocity and its "width" getting broader and broader with time. It's a good exercise to calculate the corresponding solution of the Schrödinger equation from scratch!

 

[SeM]

Dear Bill, the function does however satisfy the given condition:

\begin{equation}
\int_{0}^{2\pi}|\psi|^2 dx < \infty,
\end{equation}

which is satisfied, one can define the inner product of two such functions by the formula

\begin{equation}
\langle f,g \rangle = \int_{0}^{2\pi} f g dx ,
\end{equation}

How does this NOT coincide with a Hilbert "Geometry" ?

 

[bhobba]

How does this NOT coincide with a Hilbert "Geometry" ?

Because the axioms of QM say inner products etc are taken over infinity.

Maddening isn't it.

You can try what Vanhees said as well - but that is not an eigenfunction of momentum - still that maybe is all that you require - a solution that is a member of the Hilbert space - not necessarily an eigenfunction of momentum - I can't comment because I do not know your intent.

Added later
In light of Stragerep's correct comment the position observable X has eigenvalues from -∞ to +∞ which implies the expansion in terms of position eigenvectors |x> is a function f(x) such that |u> = ∫f(x)|x> where since the eigenvalues can go from -∞ to +∞ so does the integral and the inner product.

But in QM all that is required is a compete set of commuting observables, and one does not have to be the position observable.

Thanks
Bill

 

[SeM]

Yes, a solution member of the Hilbert space is indeed sufficient. I will generate the wavepackets, as this has also been advised from another person.

Thanks so far.

 

[strangerep]

Because the axioms of QM say inner products etc are taken over infinity.

Huh? Which axioms are you using?

(One can derive useful quantum features from a purely abstract Hilbert space if an algebra of observables is given. Cf. the derivation of the quantum angular momentum spectrum in Ballentine sect 7.1.)

 

[bhobba]

Huh? Which axioms are you using?

The ones in Von-Neumann based on the Hilbert Space of square integrateable functions.

But I take your point. This was in relation to normal wave-functions of position. All, of course, one needs is a complete set of commuting observables which is one of Dirac's axioms.

Thanks
Bill

 

[SeM]

Hi Bill, given that the wavefunction at hand here in this thread is non-orthogonal, are there are classifications of non-orthogonal wavefunctions that can be made? For instance, if it obeys some inequality or some integrals which separate it from other non-orthogonal functions? In other words, except for the following rigged Hilbert space definition, defined by

\begin{equation}
\int_{0}^{2\pi}|ψ|^2dx<\infty
\end{equation}

can I add some more classifications? I have generated a Hilbert Transform of it for instance, which is quite interesting. However, I am currently not aware of other "classifications" of non-orthogonal wavefunctions, before I go to wavepackets.

 

[vanhees71]

That doesn't make any sense. Position has dimension Length, but $2 \pi$ is a dimensionless number. Any physicist should have a bulit-in syntax checker, telling her or him that this formula can't be right already for dimensional reasons.

 

[strangerep]

given that the wavefunction(s)at hand here in this thread is non-orthogonal, [...]

Are you sure? IIUC, your "wavefunctions" are of the form $$\psi_k = e^{ikx} ~.$$ If that's what you meant, then try evaluating an arbitrary inner product:
$$\int_{0}^{2\pi}\psi_k^* \psi_n \, dx ~=~ ? ~~, $$where $k,n$ are integers.

 

[SeM]

No, its not for that wavefunction. The condition is:

\begin{equation}
\int_{0}^{2\pi}|\psi|^2 dx < \infty,
\end{equation}Its similar to the Born interpretation which is:

\begin{equation}
\int |\psi|^2 dx < \infty,
\end{equation}

which holds for the wavefunction ( can't write here) its too long.

 

[strangerep]

No, its not for that wavefunction.

Huh? You've been talking about simple exponentials for most of this thread.

[...] which holds for the wavefunction (cant write here) its too long.

Maybe you better try to write out your actual wavefunction here (or give a link to it). It's hard for anyone to help you sensibly if you don't disclose what you're actually trying to work with.

 

[SeM]

I am discussing a general rule here Strangerep.

 

[bhobba]

Hi Bill, given that the wavefunction at hand here in this thread is non-orthogonal

I don't know what you mean by that.

Here is a very brief overview of what's going on with RHS's.

What you do is you start with a space of functions with nice properties. A function f(x) is called rabidly decreasing if x^nf(x) is bounded for any n. If f(x) is continuously differentiable and all derivatives are also rapidly decreasing then it is called a good function. Now it seems a bit of a complex definition to make but it has the very nice property of the Fourier transform of a good function is also a good function.

Here is the trick of RHS's - start with good functions and consider its dual. If g is any element of its dual you can define its Fourier transform F(g) in the following way - <g|F(f)> = <F(g)|f>. In this way you can get Fourier transforms of all sorts of functions you normally could not - one of them is e^ix - I will let you look up what it is. This dual is called a Schwartz space. Since the Fourier transform of a wave-function is the representation in momentum space this makes a Schwartz space useful in QM.

Now in a finite dimensional space the eigenvectors of a Hermitian operator |bi> are orthogonal ie <bi|bj> = δij - the Kronecker Delta. Also ∑|bi><bi| = 1.

Now in this Schrwartz Space you can have the |bi> as a continuum rather than discreet which I will label by |x>. These have infinite length and we have <x|y> = δ(x-y) where δ is the Dirac Delta Function and ∫|x><x|dx = 1.

There is no easy way to explain why this is, all I can do is refer you to the attached paper.

I don't know if this answers your question, but its the best I can do at this sort of level.

Thanks
Bill

 

[SeM]

Thanks for this excellent example Bill. I completely understand what you mean here. However, being non-orthogonal the wavefunction at hand, does this imply that one must transform it using some testing function in order to develop a Fourier integral and thus a momentum space?

 

[bhobba]

Thanks for this excellent example Bill. I completely understand what you mean here. However, being non-orthogonal the wavefunction at hand, does this imply that one must transform it using some testing function in order to develop a Fourier integral and thus a momentum space?

By non-orthogonal I presume you mean not square integrate.

If you are working with square integrable functions, as wave-packets are, there should be no problem.

I still do not understand what you mean by non-orthogonal. Orthogonality is a condition that depends on the inner product of two vectors. So far you have only talked about one.

Thanks
Bill

 

[SeM]

Maybe I misunderstand. The overlap integral of the wavefunction and its complex conjugate :\begin{equation}
S= \int_{-\infty}^{\infty} \psi \psi^{*}dx
\end{equation}

yields a non-zero value for S, for the infinite domain.'

This categorizes Psi as non-orthogonal. That is why I call it non-orthogonal.

Thanks

 

[bhobba]

\begin{equation}
S= \int_{-\infty}^{\infty} \psi \psi^{*}dx
\end{equation}

yields a non-zero value for S, for the infinite domain.'

No - it gives the length squared of psi considered as a member of a Hilbert space. You divide by root S to get a normalized psi, which is what you normally work with in QM because it allows you to directly use the Born Rule. The integral you gave from x1 to x2 then gives the probability of the particle, when observed, being found from x1 to x2. That's the precise reason the function you posted at the start is pathological in QM - it is non zero out to infinity - it not normalizeable and the Born rule will not work. You can transform it with a Fourier transform so psi now is in momentum space. You do that and you get the Dirac delta function at 1. Its not strictly normalizeable either (but being loose in use of squaring such a function, which actually is a rather advanced topic - it can't be done in distribution theory - but most applied mathematicians don't worry about it and just naively do it - you can try - I will explain later) but in probability theory it represents a probability of momentum 1 with a dead cert. Its one of the uses, applied math wise, of Distribution Theory, how one handles discreet probabilities like that of a dice in continuous probability functions:
https://www.probabilitycourse.com/chapter4/4_3_2_delta_function.php

Its why IMHO this stuff needs to be known by ALL applied mathematicians. It should be part of every physics, math, engineering, physical chemistry etc curriculum - but unfortunately isn't.

Here is the math in momentum space. You have as the momentum function δ (p-1). So you have ∫δ (p-1)δ (p-1) to normalize it. Now apply the delta function rule to δ (p-1) itself you get δ(0) which is undefined. You can't do that - its not legit but we will not be that careful. So its not normalizeable. What you do is consider it as a normal function that is non zero in a very small region about 1 - its not the Dirac Delta function but FAPP you take it as if it is. That can be normalized and you see the momentum is 1 from the Born Rule.

If you have access to a library have a look at Dirac - Principles of QM - Section 23 on the momentum representation. It makes mathematicians wince, and guys like Von-Neumann, get a bit upset - as you will see if you read his textbook.

Thanks
Bill