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Guillemet

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- Thread starter Guillemet
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In summary: Nope, Hartley doesn't. Each Fourier coefficient encodes the amplitude and phase of the specified frequency, whereas to determine phases in the Hartley transform you need to look at all the coefficients together (using cepstral methods for example). In other words, the transform does not directly give you the phase information.

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Guillemet

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friend

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Guillemet said:

Try this:

http://arxiv.org/abs/0907.0909

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Guillemet

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jtbell

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Guillemet said:Had Hartley analysis been invented first

Can you provide a reference about this? A Google search for "Hartley analysis" yields only pages about the British writer L. P. Hartley, as far as I can tell (clicking through five pages of search results). Nothing to do with mathematics.

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Guillemet

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Wikipedia: Hartley transformjtbell said:Can you provide a reference about this?

Weird, I get three math-related matches on the first page: http://ieeexplore.ieee.org/xpls/abs_all.jsp?arnumber=272142 http://dx.doi.org/10.1109/CIC.1989.130514 http://www-hsc.usc.edu/~jadvar/CinC-Cepstrum.pdfjtbell said:A Google search for "Hartley analysis" yields only pages about the British writer L. P. Hartley, as far as I can tell (clicking through five pages of search results). Nothing to do with mathematics.

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ajw1

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lugita15

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See this thread about writing the Schrodinger equation as two real differential equations.

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IttyBittyBit

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Bottom line is, you need a system of encoding 'amplitude' and 'phase'. Both are important. Phase is important because it enables interference to take place.

The Hartley transform does not give you the phase information, so no, I don't see any (obvious) way that quantum mechanics could be adapted to use Hartley analysis instead, though it might be possible through some non-obvious method.

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akhmeteli

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Guillemet said:

As I wrote several times in this forum, a lot more can be done in quantum theory using just real numbers (not pairs of real numbers) than people tend to think. For example, for each solution of the Klein-Gordon equation there is a physically equivalent solution (coinciding with the original solution up to a gauge transform) with a real matter field (E. Schroedinger, Nature (London) 169, 538 (1952) ). Furthermore, in a general case, the Dirac equation can be rewritten as an equation for just one real function (http://akhmeteli.org/wp-content/uploads/2011/08/JMAPAQ528082303_1.pdf , published in Journ. Math. Phys.).

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the_pulp

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"each sequence of measurement outcomes obtained in a given experiment is represented

by a pair of real numbers"

I have an obvious intuition but I am searching for a precise (mathematical) definition of what the word "represented" means. Something like "a sequence of measurement outcomes is said to be represented by a mathematical object when ..."

Any thoughts would be very welcome.

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Guillemet

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IttyBittyBit said:Bottom line is, you need a system of encoding 'amplitude' and 'phase'. Both are important. Phase is important because it enables interference to take place.

The Hartley transform does not give you the phase information, so no, I don't see any (obvious) way that quantum mechanics could be adapted to use Hartley analysis instead, though it might be possible through some non-obvious method.

Hartley and Fourier both give phase spectra.

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IttyBittyBit

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Guillemet said:Hartley and Fourier both give phase spectra.

Nope, Hartley doesn't. Each Fourier coefficient encodes the amplitude and phase of the specified frequency, whereas to determine phases in the Hartley transform you need to look at all the coefficients together (using cepstral methods for example). In other words, the transform does not directly give you the phase information.

As an example, you can consider a gabor function (sinusoid multiplied by gaussian). The Fourier transform would be a peak at the frequency of the sinusoid which falls off as a gaussian curve. The phase of the peak in the frequency domain tells you the offset of the function in the time domain. The magnitude of the peak tells you the amplitude of the signal. The Hartley transform of this function has a similar structure to the real part of the Fourier transform, BUT the height of the peak is no longer independent of phase.

As I said, there might be some convoluted method to get Hartley analysis to work as a substitute for complex numbers. There is no obvious or straightforward way, though.

Complex numbers are numbers that have both a real and imaginary component. In quantum theory, they are used to represent the probability amplitudes of quantum states, which are necessary for understanding the behavior of particles on a microscopic level.

Complex numbers are fundamental in quantum theory because they provide a mathematical framework for describing the probabilistic nature of quantum systems. They allow for the superposition of states and the calculation of probabilities, which are essential for understanding quantum phenomena.

No, quantum theory cannot be fully understood without using complex numbers. The equations and principles of quantum theory are based on the use of complex numbers, and they are crucial for making accurate predictions and explanations of quantum phenomena.

The uncertainty principle in quantum theory is based on the concept of wave-particle duality, which is described using complex numbers. The imaginary component of complex numbers represents the wave-like behavior of particles, which is a key factor in understanding the uncertainty principle.

Yes, complex numbers are used in various areas of physics, including electromagnetism, fluid dynamics, and signal processing. They are a powerful mathematical tool for describing and analyzing physical phenomena that involve oscillations, waves, and probabilities.

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