# I Generating a Hilbert space representation of a wavefunction

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1. Nov 7, 2017

### Staff: Mentor

I don't know what you mean by that.

Here is a very brief overview of whats 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

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Last edited: Nov 7, 2017
2. Nov 7, 2017

### 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?

3. Nov 7, 2017

### Staff: Mentor

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

4. Nov 7, 2017

### SeM

Maybe I misunderstand. The overlap integral of the wavefunction and its complex conjugate :

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

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

5. Nov 7, 2017

### Staff: Mentor

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 cant 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 cant 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

Last edited: Nov 7, 2017
6. Nov 7, 2017

### strangerep

No, you're (so far) discussing some incoherent nonsense, and making it hard for others to help you. Bhobba is persevering and has made the effort to write several nontrivial posts to try and help you, and yet you can't even be bothered to write out your actual wavefunction.

That's wrong. That's not what "orthogonal" means in the context of Hilbert spaces. If $S$ is finite, then $\psi$ is "normalizable", else (if $S\to\infty$) it is non-normalizable and hence not a member of the Hilbert space of square-integrable functions on the domain $(-\infty,+\infty)$.

7. Nov 7, 2017

### Mark Harder

When we examine your function over all x space, we see that it increases without bounds as x → ∞, therefore it is not square integrable over the entire range of x. How do we fix this so that ei x is square integrable per the rigorous definition? By multiplying it by a "hat" function, which is zero everywhere but the interval [0, 2π], where it equals one. The integral of the product over [-∞ ,+∞ ] can be broken up by adding the integrals over [- , 0] and [2 π,+∞ ], which are each 0, and the integral of |ψ|2 over the range on which your new wavefunction is non-zero. The latter equals 2 π. So your wavefunction thus modified is square integrable as far as the math is concerned. But what does such a wavefunction describe in the physical world? As a student of physics, you ought to consider that question, I think, because multiplying by the hat function must represent something physical. As physicists, I don't think we can just ignore it.

8. Nov 7, 2017

### Staff: Mentor

Exactly. A wave-function is the expansion of the eigenvalues of position from -∞ to ∞ so cant be just taken from 0-2π - it makes no physical sense.

What you can do is take it over some large - but finite domain and have it zero outside that domain then at the end of your calculation take the length to infinity and see if that helps - but except as an approximation it cant be restricted.

It will not work too well with e^ix though - try it and see. So you need some other strategy.

Really you need to post exactly what you are trying to do, why your paper was rejected, and then people can help you. For me, as Strangerep has said, I have basically poured out my brains with some pretty non-trivial stuff that is not usually mentioned in beginner, or even intermediate texts to see if it can help you. If it hasn't then I am at wits end and someone else needs to help you with it.

As I said it should really be part of the general training of applied mathematicians - but usually isn't and is why I persevered. But what I have done is pretty much my best effort and I can go no further.

Thanks
Bill

Last edited: Nov 8, 2017
9. Nov 8, 2017

### vanhees71

It doesn't make any sense to say a wave function is orthogonal or non-orthogonal. It also doesn't make sense to say so about the geometrical vectors of the usual Euclidean space. Two non-zero vectors are called orthogonal if $\langle \psi_1|\psi_2 \rangle=0$. If $\langle \psi|\psi \rangle=0$ necessarily you have $|\psi \rangle=0$ (positive definiteness of the scalar product of a Hilbert space).

10. Nov 8, 2017

### SeM

Vanhees, how can it not make any sense? Many text books, such as MQM , almost start with that definition of orthogonality as part of the solutions to the Schrödinger equation. Why is the overlap integral "not sensable"? It is straight out from the textbook!

Thanks Bill, I am fully aware of your help, as I have also written in PMs. I am checking out the Dirac function in Bohms Quantum theory (again).

Mark, your suggestion is something I have not tried. So far I have developed a chapter based on the wavefunction in a cyclic boundary interval, 0 to 2pi, which is also similar to the MQM approach.

What I am trying to do here is to show some properties of a non-orthogonal function (solution to a PDE) . If I write it out, as it is part of a paper, I can as well publish the paper here. This function is as non.orthogonal as e^ikx, so there is really not reason to spell it out, as I am looking for a general principle of study for a solution to a PDE in Hilbert space, which is not orthogonal and therefore does not satisfy the overlap integral condition. The Born rule, as advised by Bill, is precisely the overlap integral within a finite domain such as 0 to 2 pi (citing MQM), so why is also that wrong?

11. Nov 8, 2017

### vanhees71

Only two vectors can be orthogonal or non-orthogonal, but not a single vector. A single function (which is just a representation of the vector in the position representation) can also consequently not be orthogonal or non-orthogonal.

The functions
$$u_k(x)\exp(\mathrm{i} k x)/\sqrt{2 \pi},$$
where $k \in \mathbb{Z}$, form a complete orthonormal set, since
$$\langle k'|k \rangle=\int_0^{2 \pi} \mathrm{d} x u_{k'}^*(x) u_k(x)=\frac{1}{2 \pi} \int_0^{2 \pi} \mathrm{d} x \exp[\mathrm{i} x(k-k')]=\delta_{k k'}.$$

12. Nov 8, 2017

### SeM

I am not sure we are talking about the same here. In MQM the non-orthogonality of a function and its complex conjugate is clearly described by that integral above.

13. Nov 8, 2017

### vanhees71

Ok, I give up. You should first read a good textbook on quantum theory or functional analysis.

14. Nov 8, 2017

### SeM

Mark, the hat function approach seems similar to the Delta function approach advised by Bill. I take the Dirac function approach further for the moment,

Thanks

15. Nov 8, 2017

### vanhees71

I've no clue what MQM means. Quantum theory is quantum theory. There's only one such theory, and it is based on clear definitions, and you should use these clear definitions. It doesn't make sense to say "one vector is orthogonal"!!! First of all one must learn proper and concise definitions (and, of course, understand them)!

16. Nov 8, 2017

### SeM

Vanhees, it's Molecular Quantum Mechanics

Mark, I read your suggestion with care now, and I noticed I already did what you wrote, however, I called it the hat function a "testing function", as derived from one of Bills links. The testing function is 1, and the integral is for 1 * Psi over the interval 0 to 2pi, which makes it a tempered distribution, allowed for Fourier Transform.

Whether it makes any physical sense is not the scope of this paper, however, I have done a Hilbert transform of it, to see its behaviour as a signal in a physical world.

Bill, I will look into the Dirac function alternative.

Thanks all

17. Nov 8, 2017

### SeM

Vanhees:

I didnt say a vector is orthogonal, I said the wavefunction is non-ortogonal to its complex conjugate, which is the reason for the overlap integral being a non-zero value. However, the very function is square integrable over a finite region, and it is normalizable.

18. Nov 8, 2017

### vanhees71

Please read the passage you quoted again carefully word by word! Nowhere it's said a wave function is orthogonal or non-orthogonal to its complex conjugate. This wouldn't make any sense! It doesn't matter whether you work in molecular quantum mechanics or any other application, the math of QM is unique.

19. Nov 8, 2017