Undergrad Orthogonal Basis of Periodic Functions: Beyond Sines and Cosines

[hutchphd]

But like, why look at the radiation spectra of a black body in that basis of sines (Fourier transform)!?

Because the simplest model was solving Maxwell's eqations for a closed -wall box at temperature T. This naturally leads to standing-wave solutions. The multiplicity of such solutions was thereby most easily counted (facilitating the calculation of the energy distribution using thermodynamics).
If you are curious about quantum , this is in many books.

Also how did people originally decompose black body radiation on its armonic spectra? They applied Fourier Transform?.

No they used the physics of light: Maxwell's equations. Fourier transform is a mathematical formalism. Planck made the physical Ansatz that energy could be exchanged with atoms only in discrete amounts There is a wonderfull learning tool called Wikipediia: see Planck spectrum . Doh.
It turns out that quantum electrodynamics demands this rule.

 

[QuantumCuriosity42]

Because the simplest model was solving Maxwell's eqations for a closed -wall box at temperature T. This naturally leads to standing-wave solutions. The multiplicity of such solutions was thereby most easily counted (facilitating the calculation of the energy distribution using thermodynamics).
If you are curious about quantum , this is in many books.

No they used the physics of light: Maxwell's equations. Fourier transform is a mathematical formalism. Planck made the physical Ansatz that energy could be exchanged with atoms only in discrete amounts There is a wonderfull learning tool called Wikipediia: see Planck spectrum . Doh.

But only sine and cosines are standing-wave solutions? There can be more right?
I read Wikipedia and much more but there my questions don't have answer sadly.

 

[hutchphd]

I read Wikipedia and much more but there my questions don't have answer sadly.

We are glad to help.
There are many ways to solve for the standing waves. The easiest to solve is a cubic box ( largely because of the simplicity of the Fourier decomposition) but the detailed shape of the box does not really affect the result. You could solve for a spherical bax if you had smart slaves (i.e. grad students) to do the calculations but the resulting physics is the same.

 

[Haborix]

Mathieu functions Hard to discern what exactly is being sought here, but have a look at Mathieu functions. A kind of generalization of sine and cosine.

 

[marcusl]

The extension from sines and cosines is a stretch; Mathieu functions are the radial wave functions in elliptic cylinder coordinates.

 

[Haborix]

The extension from sines and cosines is a stretch; Mathieu functions are the radial wave functions in elliptic cylinder coordinates.

They can be recovered by taking the limit of one of their parameters. I don’t see that your objection amounts to more than saying they aren’t sine and cosine, which seems to be what OP wants.

 

[Svein]

How about this set as an exasmple ?
View attachment 334783

These are called Walsh functions. They are another example of an orthogonal basis function set.

 

[QuantumCuriosity42]

These are called Walsh functions. They are another example of an orthogonal basis function set.

Those are not orthogonal, for example the dot product between the first and third is < 0.

 

[vanhees71]

An example for periodic complete eigenfunctions are the energy eigensolutions for a non-relativistic simple pendulum, the Mathieu functions,

https://arxiv.org/abs/quant-ph/0307079

 

[marcusl]

They can be recovered by taking the limit of one of their parameters.

I didn't realize that. Which parameter do you set to reduce to sines and cosines?

These are called Walsh functions. They are another example of an orthogonal basis function set.

They should be, but they aren't quite right; see below.

Those are not orthogonal, for example the dot product between the first and third is < 0.

Actually, waveform 3 in the diagram in post #11 is incorrect. The center lobe should be longer (as long as one side of waveform 2) and the two side lobes half as long. Then they are orthogonal Walsh codes.

 

[Haborix]

I didn't realize that. Which parameter do you set to reduce to sines and cosines?

See the wiki page. Setting $q=0$ gives sine and cosine, which makes sense given the differential equation the Mathieu functions solve reduces to one for a harmonic oscillator when $q=0$.

 

[pasmith]

I don't see what that has to do with my question about a basis formed by infinite versions of the same periodic function, each one with a different frequencies (like cos(kx)).

Any piecewise smooth periodic function can be written as a series of sines and cosines (ie. a Fourier Series). For a function f with period 2\pi we have f(x) = \sum_{n=0}^\infty a_n\cos(nx) + b_n \sin(nx). Then we can define a transform <br /> \hat{g}(k) = \int_{-\infty}^\infty g(x)f(kx)\,dx which can be expresed in terms of the Fourier sine and cosine transforms of g (at least provided a_0 = 0).

The most useful property of complex fourier transforms is that they turn differentiation with respect to x into multiplication of the transform by ik; other choices of basis won't necessarily do that and are therefore less useful.

 

[QuantumCuriosity42]

Any piecewise smooth periodic function can be written as a series of sines and cosines (ie. a Fourier Series). For a function f with period 2\pi we have f(x) = \sum_{n=0}^\infty a_n\cos(nx) + b_n \sin(nx). Then we can define a transform <br /> \hat{g}(k) = \int_{-\infty}^\infty g(x)f(kx)\,dx which can be expresed in terms of the Fourier sine and cosine transforms of g (at least provided a_0 = 0).

The most useful property of complex fourier transforms is that they turn differentiation with respect to x into multiplication of the transform by ik; other choices of basis won't necessarily do that and are therefore less useful.

But what other transform is which uses a basis of periodic and orthogonal functions, each with same shape but different frequencies.

 

[anuttarasammyak]

How about this set as an exasmple ?
View attachment 334783

I would like to propose the below instead. Orthogonalyty and normalization i.e.
\int_{-L/2}^{L/2}|f|^2 dx =1 are obvious. Completeness has not been checked.

[EDIT]
From Wiki cited below I now know it is Radamacher function which is incomplete subsystem of Walsh function.
---
Notice that �2�

is precisely the Rademacher function rm. Thus, the Rademacher system is a subsystem of the Walsh system. Moreover, every Walsh function is a product of Rademacher functions:
---

 

[QuantumCuriosity42]

I would like to propose the below instead. Orthogonalyty and normalization i.e.
\int_{-L/2}^{L/2}|f|^2 dx =1 are obvious. Completeness has not been checked.

View attachment 335147

I think that set would work. I am going to call it f(kx), where k is the associated frequency (0 for the first, and increasing on the next ones).
If we can also decompose EM waves in that orthogonal basis of different frequencies formed by functions f(kx), we are going to arrive at an espectral decomposition of unique frequencies k1, k2, k3...
Why is it then that photon energy depends on the frequency on the armonic decomposition and not on this decomposition for example?
Was the photon a consequence of decomposing EM waves in armonics, and by choosing this new basis we will arrive at another formula relating its energy with this new k? Or really there is a strange connection between armonics and photon frequency in nature (darkest mistery ever? Why nobody asks this question?).

 

[vanhees71]

It's, because you want to write down a solution of the free em. field equation in terms of energy eigenstates. The most simple choice is to work with momentum eigenstates, and this inevitably leads to the expansion in plane waves. An alternative choice is to use angular-momentum eigenstates, leading to a decomposition into (vector) spherical harmonics.

For details, see Landau and Lifshitz, vol. 4.

 

[WWGD]

My personal interest here is why do we mostly study Fourier series on $L^2[a,b]$; mostly $[a,b]=[-\pi, \pi]$

 

[vanhees71]

It's because it naturally occurs in many problems. E.g., for any problem of a particle in an external central potential or a two-particle problem with a central interacting potential (e.g., the hydrogen atom) a complete set of compatible observables are energy (Hamiltonian) and orbital angular momentum, i.e., $\vec{L}^2$ and $L_z$. With the spherical harmonics as a basis for the angular part of the corresponding eigenfunctions, which are of the form
$$u_{E,\ell,m}(\vec{x})=R_{E\ell}(r) \text{Y}_{\ell m}(\vartheta,\varphi),$$
and $\text{Y}_{\ell m} \propto \exp(\mathrm{i} m \varphi)$, i.e., the expansion of an arbitrary wave function are Fourier series in $\varphi$.

 

[anuttarasammyak]

These are called Walsh functions. They are another example of an orthogonal basis function set.

Thanks for teaching. Wiki https://en.wikipedia.org/wiki/Walsh_function says what might be of OP's interest, i.e.
-----
Walsh functions and trigonometric functions are both systems that form a complete, orthonormal set of functions, an orthonormal basis in Hilbert space �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 �

. 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.
-----

 

[WWGD]

Thanks for teaching. Wiki https://en.wikipedia.org/wiki/Walsh_function says what might be of OP's interest, i.e.
-----
Walsh functions and trigonometric functions are both systems that form a complete, orthonormal set of functions, an orthonormal basis in Hilbert space �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 �

. 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.
-----

Nice. What do you mean by " digital counterparts " ?

 

[anuttarasammyak]

That's what Wiki says not me. I assume it depicts that the value of the functions are 1 or -1 with periodic sudden changes. We may imagine sinuosidal waves inside these rectangles.

 

[WWGD]

That's what Wiki says not me. I assume it depicts that the value of the functions are 1 or -1 with periodic sudden changes. We may imagine sinuosidal waves inside these rectangles.

Maybe they mean discrete approximations?

 

[anuttarasammyak]

Maybe yes. Trigonometric
\sin kx
would correspond to Walsh
sgn(\sin kx)
where sgn is sign function. A kind of digitalization, isn't it ?

[EDIT]
Ref. https://www.stat.pitt.edu/stoffer/dss_files/walshapps.pdf

 

[hutchphd]

Maybe yes. Trigonometric
\sin kx
would correspond to Walsh
sgn(\sin kx)
where sgn is sign function. A kind of digitalization, isn't it ?

This historically was one route to the (FFT) Fast Fourier Transform for descrete sets of data if memory serves.
Harmonices mundi

 

[DaveE]

Nice. What do you mean by " digital counterparts " ?

They must mean as applied to sampled data systems (discrete time). Like the z-transform.
https://www.mathworks.com/help/signal/ug/discrete-walsh-hadamard-transform.html

 

[QuantumCuriosity42]

It's, because you want to write down a solution of the free em. field equation in terms of energy eigenstates. The most simple choice is to work with momentum eigenstates, and this inevitably leads to the expansion in plane waves. An alternative choice is to use angular-momentum eigenstates, leading to a decomposition into (vector) spherical harmonics.

For details, see Landau and Lifshitz, vol. 4.

But sure there are more plane waves apart from A*e^i(kx-wt), or not?

 

[vanhees71]

The general solution in the fully fixed radiation gauge for the vector potential is
$$\hat{\vec{A}}_{\mu}(t,\vec{x})=\sum_{\lambda \in \{-1,1\}} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 k}{\sqrt{(2 \pi \hbar)^3 2 \omega_k}} [\hat{a}_{\lambda}(\vec{k}) \exp(-\mathrm{i} k \cdot x) \epsilon_{\mu}(\vec{k},\lambda) + \text{h.c.}].$$
Here $\hat{a}_{\lambda}(\vec{k})$ is an annihilation operator for a photon with helicity $\lambda \in \{1,-1\}$, $\epsilon_{\mu}(\vec{k},\lambda)$ the corresponding transverse polarization vector (for $\vec{k}$ in $3$-direction, it's $\epsilon^0=0$ and $\vec{\epsilon}=(\vec{e}_x + \lambda \mathrm{i} \vec{e}_y)$. The four-momentum $k=(\omega_k,k)$ is "on shell", i.e., $\omega_k=|\vec{k}|=k$ (massless "particle").

 

[QuantumCuriosity42]

The general solution in the fully fixed radiation gauge for the vector potential is
$$\hat{\vec{A}}_{\mu}(t,\vec{x})=\sum_{\lambda \in \{-1,1\}} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 k}{\sqrt{(2 \pi \hbar)^3 2 \omega_k}} [\hat{a}_{\lambda}(\vec{k}) \exp(-\mathrm{i} k \cdot x) \epsilon_{\mu}(\vec{k},\lambda) + \text{h.c.}].$$
Here $\hat{a}_{\lambda}(\vec{k})$ is an annihilation operator for a photon with helicity $\lambda \in \{1,-1\}$, $\epsilon_{\mu}(\vec{k},\lambda)$ the corresponding transverse polarization vector (for $\vec{k}$ in $3$-direction, it's $\epsilon^0=0$ and $\vec{\epsilon}=(\vec{e}_x + \lambda \mathrm{i} \vec{e}_y)$. The four-momentum $k=(\omega_k,k)$ is "on shell", i.e., $\omega_k=|\vec{k}|=k$ (massless "particle").

I don't think that answers my question?

 

[vanhees71]

Then I misunderstood your question.

 

[WWGD]

@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?