# Orthogonal Basis of Periodic Functions: Beyond Sines and Cosines

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• QuantumCuriosity42
vanhees71 said:
The physical laws look as they look predominantly due to the underlying symmetries of the mathematical model describing it. First of all in Q(F)T one exploits the space-time symmetries of the spacetime model under consideration. In both Newtonian and special-relativistic physics the space-time translations are a symmetry. So any QT model must have the space-time translations as a symmetry, and the corresponding generators of these symmetry transformations are energy and momentum. For a single particle wave function thus you get
$$\psi(x,\alpha)=\exp(-\mathrm{i} \hat{p} \cdot x) \psi(0,\alpha),$$
where ##x=(t,\vec{x})## is the space-time four-vector (which you can also use in non-relativistic physics; I also use natural units ##c=\hbar=1## for convenience).

This suggests to work with energy eigenstates, which have the time dependence ##\propto \exp(-\mathrm{i} E t)##. That's why mode decompositions lead to harmonic time behavior naturally in theories with time-translation invariance.

The same holds for momentum-space representation, where naturally plane-wave solutions ##\propto \exp(+\mathrm{i} \vec{x} \cdot \vec{p}## come up.

For free particles you have both time-translation and space-translation invariance, and you get the above mentioned mode decomposition in terms of energy-momentum eigenvalues.

In addition, if you consider elementary particles, defined by irreducible representations of the space-time symmetry group, you have a energy-momentum relation. In SR it's simply the Casimir operator ##\hat{p} \cdot \hat{p}=\hat{M}^2##. In Newtonian physics it's a bit more complicated since there mass occurs as the non-trivial central charge of the Galilei group's Lie algebra.
Your response, while insightful, seems to presuppose the use of exponential functions (like the complex exponential in the wave function) rather than addressing the possibility of other bases. This feels like a circular argument, as it defines wave functions using these exponentials without considering why this representation is preferred or whether other legitimate alternatives exist. Am I correct in this observation?
I seriously think my questions still remain unanswered, with all due respect.

QuantumCuriosity42 said:
This feels like a circular argument, as it defines wave functions using these exponentials

You are in the solipsist vortex. There are no answers that will get you out. The descriptions we choose are the descriptions we choose. Listen to the Feynman on Why and p

DaveE
Of course, you can also just solve the eigenvalue equation. In the position representation we have ##\hat{p}=-\mathrm{i} \partial_x##. Then
$$\hat{p} u_p(x)=p u_p(x) \; \Rightarrow \; \partial_x u_p(x)=\mathrm{i} p u_p(x) \; \Rightarrow \; u_p(x)=N_p \exp(\mathrm{i} p x).$$
The exponential function is just the unique solution of the eigenvalue equation for the momentum operator.

QuantumCuriosity42
vanhees71 said:
Of course, you can also just solve the eigenvalue equation. In the position representation we have ##\hat{p}=-\mathrm{i} \partial_x##. Then
$$\hat{p} u_p(x)=p u_p(x) \; \Rightarrow \; \partial_x u_p(x)=\mathrm{i} p u_p(x) \; \Rightarrow \; u_p(x)=N_p \exp(\mathrm{i} p x).$$
The exponential function is just the unique solution of the eigenvalue equation for the momentum operator.
Where is the proof that it is the only solution?
Maybe that solution is the only <<plane-wave>> solution? Like here in the wave equation?

"Known as the Helmholtz equation, it has the well-known plane-wave solutions e^(+-ikx) with wave number k = ω/c." It doesn't specify if those are the only plane-wave solutions of they are just one example solution.

hutchphd said:
You are in the solipsist vortex. There are no answers that will get you out. The descriptions we choose are the descriptions we choose. Listen to the Feynman on Why and p
Thanks, but I already saw that video multiple times. And I keep believing that there is an answer to my question (it is well specified, and on the framework of physics and math (in reference to what he says in that video)), even if only few people know it.

QuantumCuriosity42 said:
This feels like a circular argument, as it defines wave functions using these exponentials

The descriptions we choose are the descriptions we choose. Listen to the Feynman on Why and perhaps reread my previous post and give it some thought.

hutchphd said:
The descriptions we choose are the descriptions we choose. Listen to the Feynman on Why and perhaps reread my previous post and give it some thought.
I am not against having chosen that. I am asking if there are more basis to choose, and why the phenomena I listed is related to that frequency and not the frequency of other basis. As simply as that.
I have a feeling photons don't really exist and we defined them as to have the relation E=hf, and then we built a theory around that asumption? Idk

I'm now convinced that there are three possible futures:
1) This thread goes on forever, without giving you an answer you are satisfied with.
2) No one responds anymore.
3) A mentor shuts it down as a waste of bandwidth.

I'll offer my help in implementing option 2. This is tiresome, and I'm done. I wish you well.

berkeman
QuantumCuriosity42 said:
Your response, while insightful, seems to presuppose the use of exponential functions (like the complex exponential in the wave function) rather than addressing the possibility of other bases. This feels like a circular argument, as it defines wave functions using these exponentials without considering why this representation is preferred or whether other legitimate alternatives exist. Am I correct in this observation?
No, you are wrong. There is no presupposition of exponential functions, they are derived. The quantum-mechanical state function ##\left|\Psi\left(t\right)\right\rangle## is a solution of the Schrödinger equation:$$i\hbar\frac{d}{dt}\left|\Psi\left(t\right)\right\rangle =\hat{H}\left|\Psi\left(t\right)\right\rangle \tag{1}$$where ##\hat{H}## is the Hamiltonian operator. If the state has constant energy ##E##, then ##\left|\Psi\left(t\right)\right\rangle## satisfies the eigenvalue condition:$$\hat{H}\left|\Psi\left(t\right)\right\rangle =E\left|\Psi\left(t\right)\right\rangle \tag{2}$$Substituting this into (1) gives:$$i\hbar\frac{d}{dt}\left|\Psi\left(t\right)\right\rangle =E\left|\Psi\left(t\right)\right\rangle \tag{3}$$which has the unique solution:$$\left|\Psi\left(t\right)\right\rangle =e^{-i\left(E/\hbar\right)t}\Psi_{0}\tag{4}$$where ##\Psi_{0}## is the time-independent quantum state. So unless you are prepared to re-invent quantum mechanics and replace the Schrödinger equation with something else, there are no other possible "bases" and you must accept that a state function of constant energy always varies with time sinusoidally.

hutchphd, QuantumCuriosity42 and berkeman
renormalize said:
No, you are wrong. There is no presupposition of exponential functions, they are derived. The quantum-mechanical state function ##\left|\Psi\left(t\right)\right\rangle## is a solution of the Schrödinger equation:$$i\hbar\frac{d}{dt}\left|\Psi\left(t\right)\right\rangle =\hat{H}\left|\Psi\left(t\right)\right\rangle \tag{1}$$where ##\hat{H}## is the Hamiltonian operator. If the state has constant energy ##E##, then ##\left|\Psi\left(t\right)\right\rangle## satisfies the eigenvalue condition:$$\hat{H}\left|\Psi\left(t\right)\right\rangle =E\left|\Psi\left(t\right)\right\rangle \tag{2}$$Substituting this into (1) gives:$$i\hbar\frac{d}{dt}\left|\Psi\left(t\right)\right\rangle =E\left|\Psi\left(t\right)\right\rangle \tag{3}$$which has the unique solution:$$\left|\Psi\left(t\right)\right\rangle =e^{-i\left(E/\hbar\right)t}\Psi_{0}\tag{4}$$where ##\Psi_{0}## is the time-independent quantum state. So unless you are prepared to re-invent quantum mechanics and replace the Schrödinger equation with something else, there are no other possible "bases" and you must accept that a state function of constant energy always varies with time sinusoidally.
Thanks, is there a proof that it is the only solution anywhere?
Also, is it the same problem as the one I said in #109? Is e^+-ikx the only plane-wave solution for the wave equation?

DaveE said:
I'm now convinced that there are three possible futures:
1) This thread goes on forever, without giving you an answer you are satisfied with.
2) No one responds anymore.
3) A mentor shuts it down as a waste of bandwidth.

I'll offer my help in implementing option 2. This is tiresome, and I'm done. I wish you well.
If you don't know it I don't think how your comment helps. I think you should also pursue the answer. Or make constructive criticism if my question is somewhat an unanswerable question.

QuantumCuriosity42 said:
This feels like a circular argument, as it defines wave functions using these exponentials
QuantumCuriosity42 said:
If you don't know it I don't think how your comment helps. I think you should also pursue the answer. Or make constructive criticism if my question is somewhat an unanswerable question.
Constructive criticism: Take a slightly larger view.
The descriptions we choose are the descriptions we choose. Listen to the Feynman on Why and perhaps reread my previous post and give it some thought.

hutchphd said:
Constructive criticism: Take a slightly larger view.
The descriptions we choose are the descriptions we choose. Listen to the Feynman on Why and perhaps reread my previous post and give it some thought.
The descriptions we choose are not the descriptions we choose. The descriptions we choose are true descriptions, if not they are discarded.
I already saw your comment and the Feynman On Why video like I told you previously, but I think that is unrelated to my question.
In fact right now I am starting to undestand it more because of what vanhees71 and renormalize told me. It looks like that is the only possible solution in math, not is simply one convenient to work with.

But it is we humans who have devised (or chosen to particularly discover) the math. What you say is self evidently true if you choose to not pursue anything else. Do you really think that we have discovered all of mathematics and that the process of such discovery does not depend upon the mind of the discoverer? Larger view.

Of course what was told to you previously (@vanhees71 et al) was exactly correct but not the only (or even necessarilly fundamental) answer to your question. It is probably the only appropriate answer for Physics Forums, however!

DaveE said:
3) A mentor shuts it down as a waste of bandwidth.
Long thead is closed for Moderation...

hutchphd
QuantumCuriosity42 said:
Thanks, is there a proof that it is the only solution anywhere?
Also, is it the same problem as the one I said in #109? Is e^+-ikx the only plane-wave solution for the wave equation?
May I ask the level of your knowledge about differential equations and their solutions? In post #114 I stated that the solution ##\exp\left(-i\left(E/\hbar\right)t\right)\Psi_{0}## to the time-dependent Schrödinger equation for constant E is unique. I know this because mathematicians long ago proved the existence and uniqueness of solutions to linear ODE initial-value problems. For example, see this page from the lecture notes at https://personalpages.manchester.ac...tYearODEs/Material/ExistenceAndUniqueness.pdf:

The substitutions:$$t\rightarrow x,\:\left|\Psi\left(t\right)\right\rangle \rightarrow y\left(x\right),\:i\frac{E}{\hbar}\rightarrow p\left(x\right),\:q\left(x\right)\rightarrow0$$along with the initial-value statement:$$\left|\Psi\left(0\right)\right\rangle =\Psi_{0}\rightarrow y\left(0\right)=\Psi_{0}$$ maps the Schrödinger equation precisely into the form of eqs.(3),(4), so the theorem applies and the unique solution is the exponential one found in post #114. And same is true for your plane-wave solution to the 2nd-order ODE it satisfies. The solutions must be exponentials and only exponentials. It's proven mathematics!

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vanhees71, QuantumCuriosity42, DaveE and 1 other person
As the titled mathematical question of "Orthogonal Basis of Periodic Functions: Beyond Sines and Cosines" there is nothing more to to talk with, I think. No periodic function case and continuity/discontinuity of base were also considered.

Beyond the titled question, @QuantumCuriosity42 seems to extend the question how to deduct photon energy of ##n\hbar\omega## from the mathematics. Photons have parameter of ##(\omega,\mathbf{k})## with relation of ##\omega=c|\mathbf{k}|##. If we need another parameter for quantization, it would be the physics of another world.

My short attempt to reply the question.
Q: Why parameter is ##\omega## of vivbrational character ##e^{i\omega t}## ?
A: It comes from harmonic dynamics in quantization of em field. Harminics are common in light, sound, etc. But it is not universal. Non harmonic or non linear phenomena are also popular.

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renormalize said:
May I ask the level of your knowledge about differential equations and their solutions? In post #114 I stated that the solution ##\exp\left(-i\left(E/\hbar\right)t\right)\Psi_{0}## to the time-dependent Schrödinger equation for constant E is unique. I know this because mathematicians long ago proved the existence and uniqueness of solutions to linear ODE initial-value problems. For example, see this page from the lecture notes at https://personalpages.manchester.ac...tYearODEs/Material/ExistenceAndUniqueness.pdf:
View attachment 335710
The substitutions:$$t\rightarrow x,\:\left|\Psi\left(t\right)\right\rangle \rightarrow y\left(x\right),\:i\frac{E}{\hbar}\rightarrow p\left(x\right),\:q\left(x\right)\rightarrow0$$along with the initial-value statement:$$\left|\Psi\left(0\right)\right\rangle =\Psi_{0}\rightarrow y\left(0\right)=\Psi_{0}$$ maps the Schrödinger equation precisely into the form of eqs.(3),(4), so the theorem applies and the unique solution is the exponential one found in post #114. And same is true for your plane-wave solution to the 2nd-order ODE it satisfies. The solutions must be exponentials and only exponentials. It's proven mathematics!
Thank you so much! Your post was insightful. (At least if Schrodinger equation was derived without supposing the premise you arrive at using it for a state of constant energy. If not it would be underwhelming.)
But then, what really means an "state of constant energy"? Is that a photon?
Also, the reasoning we use now to derive E=h*f as @PeterDonis told me in another thread (see screenshot) is that one you showed me @renormalize ?

anuttarasammyak said:
As the titled mathematical question of "Orthogonal Basis of Periodic Functions: Beyond Sines and Cosines" there is nothing left to to talk, I think. No periodic function case and continuity-iscontinuity of base were also considered.

Beyond the title @QuantumCuriosity42 seems to extend the question how to deduct photon energy of ##n\hbar\omega## from the mathematics. Photons have parameter of ##(\omega,\mathbf{k})## with relation of ##\omega=c|\mathbf{k}|##. If we need another parameter for quantization, it would be the physics of another world.

My short attempt to reply the question.
Q: Why parameter is ##\omega## of vivbrational character ##e^{i\omega t}## ?
A: It comes from harmonic dynamics of quantization of em field. Harminics are common in light, sound, etc. But it is not universal. Non harmonic or non linear phenomena are also popular.
What do you mean by: "No periodic function case and continuity of base were already considered."?
I think that as of yet nobody told another valid basis of periodic orthogonal functions with different frequencies.
And what do you mean by: "If we need another parameter for quantization, it is the physics of another world."?
How do you know that?

anuttarasammyak said:
As the titled mathematical question of "Orthogonal Basis of Periodic Functions: Beyond Sines and Cosines" there is nothing more to to talk with, I think. No periodic function case and continuity/discontinuity of base were also considered.

For function f of finete |f(x)|^2 in ##(-\infty,+\infty)##, ##\{\delta(x-t)\} ## is a complete set of basis.
Integral tarnsform
$$f(x)=\int f(t)\delta(t-x) dt$$
This identical transformation is obvious and not so interesting. Anyway it is a transform formally.
Orthogonality ＆Normalization
$$\int \delta(t-a)\delta(t-b) dt = \delta(a-b)$$
It shows also "normalization" for contimuous variable also to ##\delta(0)## as well as Fourier transform case.
Continuity
It can be regarded continuous in Schwartz method of continuous function sequence. Even derivatives exist which works as
$$\int f(t) \delta'(t-x) dt = - f'(x)$$
Periodicity
Not periodic.

PS Walsh-Fourier tranform basis is periodic.
ref. https://www.stat.pitt.edu/stoffer/dss_files/walshapps.pdf
It is discrete in popular sense, but we can regard them continuous as well as ##\delta## funciton is.
Actually derivative of Walsh functions are sum of periodical chain of ##\delta## functions.

In the last century wavelet transform of both discrete and continuous was found ref. https://en.wikipedia.org/wiki/Wavelet_transform
Scaling and shifting of a chosen mother wavelet seems a new idea apart from FT.
FT basis has infinite length which does not decay at all. It is mathematical ideal body but many physical phenomena take place locally. Wavelet transform with farther decaying mother wavelet might be convenient in such cases, I suspect.

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anuttarasammyak said:

For function f of finete |f(x)|^2 in ##(-\infty,+\infty)##, ##\{\delta(x-t)\} ## is a complete set of basis.
Integral tarnsform
$$f(x)=\int f(t)\delta(t-x) dt$$
This identical transformation is obvious and not so interesting. Anyway it is a transform formally.
Orthogonality ＆Normalization
$$\int \delta(t-a)\delta(t-b) dt = \delta(a-b)$$
It shows also "normalization" for contimuous variable also to ##\delta(0)## as well as Fourier transform case.
Continuity
It can be regarded continuous in Schwartz method of continuous function sequence. Even derivatives exist which works as
$$\int f(t) \delta'(t-x) dt = - f'(x)$$
Periodicity
Not periodic.

PS Walsh-Fourier tranform basis is periodic.
ref. https://www.stat.pitt.edu/stoffer/dss_files/walshapps.pdf
It is discrete in popular sense, but we can regard them continuous as well as ##\delta## funciton is.
Actually derivative of Walsh functions are sum of periodical chain of ##\delta## functions.

In the last century wavelet transform of both discrete and continuous was found ref. https://en.wikipedia.org/wiki/Wavelet_transform
Scaling and shifting of a chosen mother wavelet seems a new idea apart from FT.
FT basis has infinite length which does not decay at all. It is mathematical ideal body but many physical phenomena take place locally. Wavelet transform with farther decaying mother wavelet might be convenient in such cases, I suspect.
So the only other basis of periodic functions apart from sines/cosines, are walsh functions?
Wavelets are generic waves I think.

QuantumCuriosity42 said:
So the only other basis of periodic functions apart from sines/cosines, are walsh functions?
Yes, as far as I know now. I know I do not know very much.

marcusl
Or the Mathieu functions already mentioned in this thread. Or the various Jacobi elliptic functions... But I'm not sure what we're getting at anymore.

vanhees71
Haborix said:
Or the Mathieu functions already mentioned in this thread. Or the various Jacobi elliptic functions... But I'm not sure what we're getting at anymore.
Mathie functions are not periodic. I think Jacobi elliptic functions are not orthogonal.

QuantumCuriosity42 said:
Thank you so much! Your post was insightful. (At least if Schrodinger equation was derived without supposing the premise you arrive at using it for a state of constant energy. If not it would be underwhelming.)
But then, what really means an "state of constant energy"? Is that a photon?
Also, the reasoning we use now to derive E=h*f as @PeterDonis told me in another thread (see screenshot) is that one you showed me @renormalize ?
View attachment 335721
Does anyone know what is the other derivation "using different reasoning from the reasoning Placnk originally used" that he mentions on that screenshot?

The modern derivation of Planck's Law is of course to use QED at finite temperature. The result is the same.

QuantumCuriosity42 said:
Mathie functions are not periodic. I think Jacobi elliptic functions are not orthogonal.
An infinite subset of Mathieu functions are periodic. You may be right about elliptic functions not being orthogonal, but if I recall they are complete so a basis could be constructed from them.

vanhees71 said:
The modern derivation of Planck's Law is of course to use QED at finite temperature. The result is the same.
Could you point me to the derivation please?

Haborix said:
An infinite subset of Mathieu functions are periodic. You may be right about elliptic functions not being orthogonal, but if I recall they are complete so a basis could be constructed from them.
I want a periodic and orthogonal basis, because if the only one satisfying both conditions are sine/cosine then that would explain why they are the fundamental frequencies.

We have repeatedly quoted other periodic orthornomal function systems. Why should sine and cosine the only periodic orthonormal functions?

QuantumCuriosity42
vanhees71 said:
We have repeatedly quoted other periodic orthornomal function systems. Why should sine and cosine the only periodic orthonormal functions?
Ok, then that is set. And could you tell me where can I find the derivation of Planck law you told me before?

These two newly released videos (today) are the best I've ever seen on the topic.

In the first video he ignores any other light pulse and goes with a monochromatic one without taking into account any other or explaining why.
But yeah, at the end of the second video she says we still don't know why light slows down in water with a non monochromatic pulse. Looks like my question was unexpectedly at the limit of today's current physics knowledge (She's a theoretical physicst).