Hilbert Book Model Project/Quaternionic Field Equations/Fourier Transform

= Fourier spaces = In an infinite dimensional Hilbert space, a Fourier transform accomplishes a complete transform of an old orthonormal base $$\{|q\rangle\}$$ to another orthonormal base $$\{|\tilde{q}\rangle\}$$, such that none of the new base vectors can be written as a linear combination that does not include all the old base vectors.

The base vector $$|q\rangle$$ is eigenvector of a normal operator $$|q\rangle\,q \,\langle q|$$ with eigenvalues $$\{q\}$$. Base $$\{|q\rangle\}$$ is orthonormal.

$$\langle q|q' \rangle=\delta(q-q')$$

Similarly, the base vector $$|\tilde{q}\rangle$$ is eigenvector of a normal operator $$|\tilde{q}\rangle\,\tilde{q} \,\langle \tilde{q}|$$ with eigenvalues $$\{\tilde{q}\}$$.

$$\langle \tilde{q}|\tilde{q}' \rangle=\delta(\tilde{q}-\tilde{q}')$$

The inner product $$\langle q|\tilde{q} \rangle$$ is a function of both $$q$$ and $$\tilde{q}$$ coordinates.

Remember that function $$f(q)$$ can be represented with respect to an orthonormal base $$\{|q\rangle\}$$ and operator $$f$$ as

$$f(q)=\langle q|f\, q\rangle$$

$$\tilde{f}(\tilde{q})=\langle \tilde{q}|\tilde{f}\, \tilde{q}\rangle$$

$$f(q) \Leftarrow \mathfrak{F}\Rightarrow \tilde{f}(\tilde{q})$$

$$f=|q\rangle\,f(q) \,\langle q|$$

$$\tilde{f}=|\tilde{q}\rangle\,\tilde{f}(\tilde{q}) \,\langle \tilde{q}|$$

$$\mathfrak{R}=|q\rangle\,q \,\langle q|$$

$$\tilde{\mathfrak{R}}=|\tilde{q}\rangle\,\tilde{q} \,\langle \tilde{q}|=|q\rangle\,\tilde{\mathfrak{R}}(q) \,\langle q|$$

These equations describe Fourier transform pairs $$\{f,\tilde{f}\}$$ and the same continuum $$\breve{f}$$. That continuum $$\breve{f}$$ is represented by $$f(q)$$ as well as by $$\tilde{f}(\tilde{q})$$ and these functions correspond respectively to the operators

$$f$$ and $$\tilde{f}$$. So $$f(q)$$ and $$\tilde{f}(\tilde{q})$$ describe the same thing, which is the continuum $$\breve{f}$$.

The inner product $$\langle q|\tilde{q} \rangle$$ is a function that fulfills the following corollaries.
 * Convolution of functions in the old base $$\{|q\rangle\}$$ representation becomes multiplication in the new base $$\{|\tilde{q}\rangle\}$$ representation.
 * Similarly, convolution of functions in the new base $$\{|\tilde{q}\rangle\}$$ representation becomes multiplication in the old base $$\{|q\rangle\}$$ representation.
 * Differentiation in the old base representation becomes multiplication by the new coordinate in the new base representation.
 * Similarly, differentiation in the new base representation becomes multiplication by the old coordinate in the old base representation.

Inner products
Remember that

$$\langle \alpha x | y\rangle=\alpha^* \langle x | y\rangle

$$

$$\langle x | \beta \,y\rangle=\langle x | y\rangle\,\beta

$$

Complex Fourier transform
Fourier transformation is well established for complex functions. We will apply that knowledge by establishing complex parameter spaces inside the quaternionic background parameter space.

If an $$x$$ axis along the normalized vector $$\vec{n}$$ is drawn through the quaternionic background parameter space, then

$$\tilde{f}(\xi) = \int_{-\infty}^{\infty} f(x)\ e^{- 2\pi \vec{n} x \xi}\,dx$$

$$f(x) = \int_{-\infty}^{\infty} \tilde f(\xi)\ e^{2 \pi \vec{n} \xi x}\,d\xi$$

Here $$x$$ plays the role of parameter $$q$$ along direction $$\vec{n}$$ and $$\xi$$ plays the role of parameter $$\tilde{q}$$ along direction $$\vec{n}$$. $$\vec{n}$$ can be taken in an arbitrary direction and can start at an arbitrary location in the quaternionic background parameter space..

The inner product $$\langle q|\tilde{q} \rangle$$ relates to a two parametric function that along the direction $$\vec{n}$$ corresponds to $$\langle x\vec{n}|\xi\vec{n}\rangle=\exp(2\pi x \vec{n}\xi)$$

Here $$f(x)$$ and $$\tilde f(\xi)$$ are complex functions with complex imaginary base number $$\vec{n}$$.

Quaternionic Fourier transform
More generally the specification of the quaternionic Fourier must cope with the non-commuting multiplication of quaternionic functions.

$$\langle x|{\color{Blue}{\tilde{f}}} y \rangle= \int\limits_{\tilde{q}} \langle x\color{Red}{|\tilde{q}\rangle }{\color{Blue}{\left\{\int\limits_{q}\langle \tilde{q}|q\rangle f(q) \langle q|\tilde{q}\rangle \, dq\right\}}} {\color{Red}{\langle \tilde{q}|}}\color{Black}{y\rangle\, d \tilde{q}} = \int\limits_{\tilde{q}} \langle x\color{Red}{|\tilde{q}\rangle } {\color{Blue}{\tilde{f}(\tilde{q})}}\langle \tilde{q}|\color{Black}{y\rangle\, d \tilde{q}}$$

$$\tilde{f}(\tilde{q})=\int\limits_{q}\langle \tilde{q}|q\rangle f(q) \langle q|\tilde{q}\rangle \, dq$$

We see in the formulas that this method merely achieves a rotation of parameter spaces and functions. In the complex number based Hilbert space, it would achieve no change at all.

The Fourier transform installs only a partial rotation. This results in left and right oriented Fourier transforms.

Left oriented Fourier transform
The left oriented Fourier transform $$\mathfrak{F}_L$$ has an inverse $$\mathfrak{F}_L^{-1}$$.

$$\tilde{f}=\mathfrak{F}_L(f)=|\tilde{q}\rangle\,\mathfrak{F}_L(f)(\tilde{q}) \,\langle \tilde{q}|$$

$$f=\mathfrak{F}_L^{-1}(\tilde{f})$$

The left oriented Fourier transform is defined by:

$$\tilde{f}_L(\tilde{q}_L)=\int\limits_{q} \langle \tilde{q}_L|q\rangle f(q)\, dq =\int\limits_{q} \langle \tilde{q}_L| f(q)\,q\rangle\, dq$$

For two members $$|q\rangle$$ and $$|q'\rangle$$ of an orthonormal base $$\{|q\rangle\}$$ holds

$$\langle q|q'\rangle=\delta(q-q')$$

For two members $$|\tilde{q}_L\rangle$$ and $$|\tilde{q}'_L\rangle$$ of an orthonormal base $$\{|\tilde{q}_L\rangle\}$$ holds

$$\langle \tilde{q}_L|\tilde{q}_L'\rangle=\delta(\tilde{q}_L-\tilde{q}_L')$$

$$\int_{\tilde{q}_L} \langle q'|\tilde{q}_L\rangle\langle\tilde{q}_L|q\rangle \,d\tilde{q}_L=\delta(q-q')$$

$$\int_{q} \langle \tilde{q}'_L|q\rangle\langle q|\tilde{q}_L\rangle \,dq=\delta(\tilde{q}_L-\tilde{q}'_L)$$

The reverse transform is given by

$$f(q)=\int_{\tilde{q}} \langle q|\tilde{q}_L\rangle \tilde{f}_L(\tilde{q}_L)\, d\tilde{q}_L= \int_{\tilde{q}}\langle q|\tilde{q}_L\rangle\int_{q'}\langle\tilde{q}_L|q'\rangle f(q')\, dq'\, d\tilde{q}_L = \int_{q'}f(q')\int_{\tilde{q}}\langle q|\tilde{q}_L\rangle\langle\tilde{q}_L|q'\rangle \, d\tilde{q}_L\, dq' =\int_{q'}f(q')\,\delta(q-q')d q'$$

$$f(q)=\int_{\tilde{q}} \langle q|\tilde{q}_L\rangle \tilde{f}_L(\tilde{q}_L)\, d\tilde{q}_L= \int_{\tilde{q}} \langle q|\tilde{f}_L(\tilde{q}_L)\,\tilde{q}_L\rangle \, d\tilde{q}_L$$

Right oriented Fourier transform
Similarly for the right oriented Fourier transform

$$\tilde{f}_R(\tilde{q}_R)=\int\limits_{q} f(q)\langle q|\tilde{q}_R\rangle\, dq= \int\limits_{q} \langle f(q)^*\, q|\tilde{q}_R\rangle\, dq$$

$$f(q)= \int_{\tilde{q}} \tilde{f}_R(\tilde{q}_R)\langle \tilde{q}_R|q\rangle \, d\tilde{q}_L= \int_{\tilde{q}} \int_{q'} f(q')(\tilde{q}_R)\langle q'|\tilde{q}_R\rangle\langle\tilde{q}_R|q\rangle\,d\tilde{q}_L\, dq'=\int_{q'}f(q')\,\delta(q-q')d q'

$$

Conclusion
The extra value of the right oriented and left oriented Fourier transforms is low. The complex number based Fourier transform has much greater value for the spectral analysis of continuums. However that analysys then restricts to a single direction per case,

Important is the fact that Fourier transform pairs $$\{f,\tilde{f}\}$$ describe the same continuum $$\breve{f}$$.