Generating a Hilbert space representation of a wavefunction

[strangerep]

I am discussing a general rule here 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.

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.

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 $(-\infty,+\infty)$.

 

[Mark Harder]

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!

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 e^{i 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.

 

[bhobba]

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.

Exactly. A wave-function is the expansion of the eigenvalues of position from -∞ to ∞ so can't 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 can't 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

 

[vanhees71]

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?

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).

 

[SeM]

Vanhees, how can it not make any sense? Many textbooks, 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?

 

[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
$$\langle k'|k \rangle=\int_0^{2 \pi} \mathrm{d} x u_{k'}^*(x) u_k(x)=\frac{1}{2 \pi} \int_0^{2 \pi} \mathrm{d} x \exp[\mathrm{i} x(k-k')]=\delta_{k k'}.$$

 

[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.

 

[vanhees71]

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

 

[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

 

[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)!

 

[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

 

[SeM]

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)!

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.

 

[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.

 

[PeterDonis]

Thread closed for moderation.