I Orthogonal Basis of Periodic Functions: Beyond Sines and Cosines

[QuantumCuriosity42]

@QuantumCuriosity42 : I have the impression that, 89 posts down the line, many of us are not clear on it. Maybe you can rephrase it, reframe it for us?

Alright, sorry for that, I'll try to explain it as clearly as possible.

My question arises from why there is such an intrinsic relationship with harmonics in physics, not only in terms of the mathematics we use to describe the world in a simple way for us, but it also seems that there are phenomena in nature that correspond with this frequency.

I'll start with the doubt related to mathematics, which might justify why nature inherently relates to harmonic frequencies.

Is there another basis of periodic functions besides sin(kx), cosine(kx), or e^ikx that can generate the entire space of functions? If the only one is that (harmonic waves), then here my doubts end, but if there are more, it doesn't make sense to me that the phenomena described next correspond with the harmonic frequency instead of frequencies from other bases. From what I've read so far, I understand that more bases can exist.

Regarding the physical phenomena I'm talking about, they include, among others:

• E=h*f, the energy of a photon depends on the frequency of the associated harmonic wave, rather than the frequency of any other base.
• In black body radiation, Planck discovered that there are oscillators of energy h*f (as appears in the black body radiation formula).
• The color interpreted by our eyes also depends on the <<harmonic>> frequency of the light we see, not on the frequencies of the wave's decomposition into another base of periodic waves of different frequencies.
• The energy of a harmonic oscillator, again, is quantified in quanta h*w (again, harmonic frequency).
• De Broglie wavelength refers to the length of a harmonic (sinusoidal) wave, not any other.
• For example, in electron microscopy, to distinguish something very small, a wavelength very close to the size of the object to be observed must be used, but this wavelength depends on the harmonic frequency of the wave. This is related to edge diffraction, where the same thing happens, or in a slit, where a wave's diffraction upon passing through it depends on its <<harmonic>> frequency. That is, the refraction of a wave depends on its frequency (again, its frequency in a harmonic decomposition, if there are several, its behavior will be the sum of the behaviors for each frequency).

And probably many more cases like this can be cited.

If you don't know why, please don't divert the question to other things. In fact, I would appreciate it if you could indicate that you don't know either, so maybe more people will see it, and we might get an answer from someone who does know. This is not meant to be confrontational, but that's why this thread has stretched so long. I've asked physics degree professors and they have not been able to answer my doubt. That's why I don't think it's a simple question at all, and strangely, it surprises me that no one else wonders the same as me, just assuming that it is so and that's it.

Thank you very much, let's see if we can get a satisfactory answer.

 

[berkeman]

The color interpreted by our eyes also depends on the <<harmonic>> frequency of the light we see, not on the frequencies of the wave's decomposition into another base of periodic waves of different frequencies.

Sorry, I don't understand this. Can you please elaborate? Thanks.

 

[QuantumCuriosity42]

Sorry, I don't understand this. Can you please elaborate? Thanks.

I mean, the electromagnetic wave that reaches our eyes can be decomposed using the Fourier transform into a sum of multiple sines/cosines of different frequencies.

Similarly, we could perform this decomposition not with a basis of multiple sines/cosines of different frequencies, but with one of any other periodic function that also forms a basis in the same way, with each element of the basis being that function at a different frequency. Analogous to how in the Fourier Transform a signal is decomposed by projecting it onto a basis formed by infinite sines/cosines, each with a different frequency. This is possible because sine or cosine is orthogonal to itself whenever the frequency is different. A property that I don't know if it is unique to sines and cosines or if there is another basis of periodic functions other than this.

If this basis exists, for example Walsh functions, we would have a decomposition in Walsh frequencies, instead of harmonic frequencies (of the sines or cosines). What I wanted to say is that the colors of the things we see correspond to the harmonic frequencies of the decomposition with the Fourier transform, not with the frequencies of another decomposition, such as Walsh's (which I am not clear if it forms a basis of periodic functions that are always the same and only change in frequency).

 

[berkeman]

for example Walsh functions,

Walsh functions are for discrete signals, not continuous signals...

Walsh functions form a complete orthogonal set of functions that can be used to represent any discrete function—just like trigonometric functions can be used to represent any continuous function in Fourier analysis.[1]

https://en.wikipedia.org/wiki/Walsh_function

 

[QuantumCuriosity42]

Walsh functions are for discrete signals, not continuous signals...https://en.wikipedia.org/wiki/Walsh_function

Perfect to know that, I wasn't sure as I said. Then there is no other base for continuos signals apart from cos, sin and e^ikx?

 

[berkeman]

Then there is no other base for continuos signals apart from cos, sin and e^ikx?

I can't say that for sure (I'm no expert), but that is my understanding. Have you tried doing a general internet search to see if you can find any scholarly articles about any alternatives for continuous functions?

 

[hutchphd]

What I wanted to say is that the colors of the things we see correspond to the harmonic frequencies of the decomposition with the Fourier transform, not with the frequencies of another decomposition, such as Walsh's (which I am not clear if it forms a basis of periodic functions that are always the same and only change in frequency).

There is difficulty here separating "nature" from how are brains have historically analyzed nature. I mentioned Kepler's Harmonice Mundi earlier. Why do we humans find harmonic music pleasing to the ear?

Further we have assumed it will please God when we castrate boys to add a few more high harmonics to the choir. There is certainly a human proclivity to harmonic series so its ubiquity in our natural philosophy is not surprising.
So there is a question in my mind of cause and effect here. Your quandary may be
with our attempts to understand nature and not by nature itself. I do not know, but I do know this road rapidly leads to a solipsist bog of no return.

The analysis of color you present is filled with harmonic assumptions having little to do with our actual perceptions. Yes we have agreed upon a set of analytic steps between Sol and brain, but they are largely mechanisms of our own creation. Do we require Maxwell's equation's to see color? A rose by any other name.......

 

[QuantumCuriosity42]

I can't say that for sure (I'm no expert), but that is my understanding. Have you tried doing a general internet search to see if you can find any scholarly articles about any alternatives for continuous functions?

Yes I did, but I did not find much literature on the topic. Much less a formal proof. To me it seems a very relevant thing to have been asked on the Internet or proved in the past by physicists.

 

[DaveE]

why nature inherently relates to harmonic frequencies.

An interesting philosophical question. But 'why?' questions seldom have good answers. Nature is what it is. I definitely don't know why things are the way they are. I doubt that anyone does if you dig deep enough. I doubt you will get a satisfactory answer from us, certainly not from me. The best I can do is relate it to the fact that many basic phenomena are well described by simple differential equations, which create these simple solutions. Many famous physicist have wondered about this sort of question, which I would lump together as "why is nature so well described by simple mathematics".

However, it's a good sensible query which doesn't require the rest of your elaborations, which frankly are a bit confusing for us to sort through. The answer to many of those is to study them more until you have reduced them to this basic query.

You may enjoy this answer from Feynman, if you haven't seen it before:

 

[QuantumCuriosity42]

You have already asked this question and it has been clearly answered by several people. I would suggest studying Linear Algebra before asking it (yet) again.

The answers given before said Walsh functions were valid, but in #94 @berkeman said they are not valid for continuous functions, only discrete.

Yes. So why did you just ask us that again?

Because as I said, I receive contradictory answers, now it looks like they are not valid for continuous signals.

Vision in mammals is way to complex of system to be a good example for these basic questions. Let's pretend we're not ready for that yet. In any case it is well understood by people who study it. It's not a mystery, it's just complex.Again, not the best example, classical optics is well defined. Mostly based on Huygens principle. Goodman - Fourier Optics is a great place to learn this. In any case, while fourier/hartley transforms are easy, I imagine you could use wavelets or other transforms too (although no one actually would).

The thing is that wavelets are not periodic. And as of yet I did not find a basis of periodic functions (Walsh functions are not as said by @berkeman ).

 

[DaveE]

The answers given before said Walsh functions were valid, but in #94 @berkeman said they are not valid for continuous functions, only discrete.

Because as I said, I receive contradictory answers, now it looks like they are not valid for continuous signals.

The thing is that wavelets are not periodic. And as of yet I did not find a basis of periodic functions (Walsh functions are not as said by @berkeman ).

Yes, good questions. I'm not sure, that's why I deleted these comments from my post, perhaps as you were replying? I need to think more about this.

 

[anuttarasammyak]

Walsh functions are for discrete signals, not continuous signals...

https://en.wikipedia.org/wiki/Walsh_function

It would be natural and easy to apply Fourier decomposition to continuous functions and Walsh decomposition to discrete functions, however, both continuous and discrete functions CAN be decomposed to Fourier or Walsh. Is this my understanding wrong ?

 

[berkeman]

It would be natural and easy to apply Fourier decomposition to continuous functions and Walsh decomposition to discrete functions, however, both continuous and discrete functions CAN be decomposed to Fourier or Walsh. Is this my understanding wrong ?

I dunno; the Walsh functions sure seem discrete to me...

(from the Wikipedia link):

Walsh function is a product of Rademacher functions:

W k ( x ) = ∏ j = 0 ∞ r j ( x ) k j W_{k}(x)=\prod _{{j=0}}^{\infty }r_{j}(x)^{{k_{j}}}

Comparison between Walsh functions and trigonometric functions

Walsh functions and trigonometric functions are both systems that form a complete, orthonormal set of functions, an orthonormal basis in Hilbert space L 2 [ 0 , 1 ] L^{2}[0,1] of the square-integrable functions on the unit interval. Both are systems of bounded functions, unlike, say, the Haar system or the Franklin system.

Both trigonometric and Walsh systems admit natural extension by periodicity from the unit interval to the real line R {\mathbb R}. Furthermore, both Fourier analysis on the unit interval (Fourier series) and on the real line (Fourier transform) have their digital counterparts defined via Walsh system, the Walsh series analogous to the Fourier series, and the Hadamard transform analogous to the Fourier transform.

 

[anuttarasammyak]

We surely know that Fourier can be applied to discrete functions, e.g.
\delta(x)=\frac{1}{2\pi}\int e^{ikx} dk
and its integral, discrete
\theta(x)=\frac{1}{2\pi}\int \frac{e^{ikx}}{ik} dk

at least. We may expect vice versa.

[EDIT] Ref. 5. The Walsh-Fourier Transform and Spectrum in https://www.stat.pitt.edu/stoffer/dss_files/walshapps.pdf
It seems to be affirmative to my expectation.

 

[vanhees71]

Alright, sorry for that, I'll try to explain it as clearly as possible.

My question arises from why there is such an intrinsic relationship with harmonics in physics, not only in terms of the mathematics we use to describe the world in a simple way for us, but it also seems that there are phenomena in nature that correspond with this frequency.

I'll start with the doubt related to mathematics, which might justify why nature inherently relates to harmonic frequencies.

Is there another basis of periodic functions besides sin(kx), cosine(kx), or e^ikx that can generate the entire space of functions? If the only one is that (harmonic waves), then here my doubts end, but if there are more, it doesn't make sense to me that the phenomena described next correspond with the harmonic frequency instead of frequencies from other bases. From what I've read so far, I understand that more bases can exist.

Regarding the physical phenomena I'm talking about, they include, among others:

• E=h*f, the energy of a photon depends on the frequency of the associated harmonic wave, rather than the frequency of any other base.
• In black body radiation, Planck discovered that there are oscillators of energy h*f (as appears in the black body radiation formula).
• The color interpreted by our eyes also depends on the <<harmonic>> frequency of the light we see, not on the frequencies of the wave's decomposition into another base of periodic waves of different frequencies.
• The energy of a harmonic oscillator, again, is quantified in quanta h*w (again, harmonic frequency).
• De Broglie wavelength refers to the length of a harmonic (sinusoidal) wave, not any other.
• For example, in electron microscopy, to distinguish something very small, a wavelength very close to the size of the object to be observed must be used, but this wavelength depends on the harmonic frequency of the wave. This is related to edge diffraction, where the same thing happens, or in a slit, where a wave's diffraction upon passing through it depends on its <<harmonic>> frequency. That is, the refraction of a wave depends on its frequency (again, its frequency in a harmonic decomposition, if there are several, its behavior will be the sum of the behaviors for each frequency).

And probably many more cases like this can be cited.

If you don't know why, please don't divert the question to other things. In fact, I would appreciate it if you could indicate that you don't know either, so maybe more people will see it, and we might get an answer from someone who does know. This is not meant to be confrontational, but that's why this thread has stretched so long. I've asked physics degree professors and they have not been able to answer my doubt. That's why I don't think it's a simple question at all, and strangely, it surprises me that no one else wonders the same as me, just assuming that it is so and that's it.

Thank you very much, let's see if we can get a satisfactory answer.

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.

 

[QuantumCuriosity42]

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.

 

[hutchphd]

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

 

[vanhees71]

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]

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.

 

[QuantumCuriosity42]

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.

 

[hutchphd]

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.

 

[QuantumCuriosity42]

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

 

[DaveE]

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.

 

[renormalize]

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.

 

[QuantumCuriosity42]

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?

 

[QuantumCuriosity42]

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.

I will receive and answer when someone that knows the answer writes it.
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.

 

[hutchphd]

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

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.

 

[QuantumCuriosity42]

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.

 

[hutchphd]

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.

 

[hutchphd]

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!