Solving the Fourier Transform Issue Using Quantum Coherent States

Abstract

Data on frequency and time is not simultaneously available in a Fourier transform. Problems with the Fourier transform have led to the emergence of short-window Fourier analysis which itself has several limitations. The main problem with short-time Fourier transform is related to Heisenberg’s uncertainty principle. In the wavelet transform, the signal is analyzed on a set of wavelet functions (coherent states). In quantum mechanics, the coherent state is a type of quantum state which has the minimum value of uncertainty. In the present study, the concepts of quantum mechanics such as the uncertainty principle, Dirac’s notation and quantum coherent states are used to present a method for obtaining wavelet functions.

Appendix: Proof of Equations

By using the following identities:

$$ \langle\alpha|(a+a^{\dag})|\alpha\rangle=(\alpha+\bar{\alpha}), $$

(A.1)

$$ \langle\alpha|(a-a^{\dag})|\alpha\rangle=(\alpha-\bar{\alpha}), $$

(A.2)

$$ \langle\alpha|(a+a^{\dag})(a+a^{\dag})|\alpha\rangle=(\alpha+\bar{\alpha})^{2}+ 1, $$

(A.3)

$$ \langle\alpha|(a-a^{\dag})(a-a^{\dag})|\alpha\rangle=(\alpha-\bar{\alpha})^{2}-1, $$

(A.4)

therefore, 〈(Δx)²〉 and 〈(Δp)²〉 are obtained:

$$ \langle({\Delta} x)^{2}\rangle_{\alpha}=\langle x^{2}\rangle_{\alpha}-\langle x\rangle^{2}_{\alpha}=\frac{\hbar}{2m\omega} :x=\sqrt{\frac{\hbar}{2m\omega}}(a+a^{\dag}), $$

(A.5)

$$ \langle({\Delta} p)^{2}\rangle_{\alpha}=\langle p^{2}\rangle_{\alpha}-\langle p\rangle^{2}_{\alpha}=\frac{\hbar m\omega}{2} :p=i\sqrt{\frac{m\hbar\omega}{2}}(-a+a^{\dag}). $$

(A.6)

Now we want to consider completeness:

$$ \int {d^{2}}\alpha|\alpha\rangle\langle\alpha|=\int {d^{2}}\alpha e^{-|\alpha|^{2}}\sum\limits_{mn}\frac{\bar{\alpha}^{n}\alpha^{m}}{\sqrt{n!m!}}|m\rangle\langle n|, $$

(A.7)

{\break}α can be written using polar coordinates:

$$\alpha=re^{i\phi} {d^{2}}\alpha=d\phi dr r,$$

$$ {\int}_{0}^{2\pi} e^{i(n-m)\phi} d\phi= 2\pi\delta_{mn}, $$

(A.8)

$$ {\int}_{0}^{\infty} e^{-r^{2}} r^{2n + 1} dr=\frac{n!}{2}, $$

(A.9)

therefore

$$ \int {d^{2}}\alpha|\alpha\rangle\langle\alpha|=\pi\sum\limits_{n}|n\rangle\langle n|=\pi, $$

(A.10)