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What is the meaning of the imaginary part of the plane wave function

by Chuck88
Tags: signal, signal analysis, wave
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Chuck88
#1
Mar6-12, 05:10 AM
P: 37
The plane wave function sometimes could be represented as:

[tex]
U(\mathbf{r} ,t ) = A_{0} e^{i(\mathbf{k} \cdot \mathbf{r} - \omega t + \phi)}
[/tex]

and we could separate the expression above into:

[tex]
U(\mathbf{r} ,t = \cos(\mathbf{k} \cdot \mathbf{r} - \omega t + \phi) + i \sin(\mathbf{k} \cdot \mathbf{r} - \omega t + \phi)
[/tex]

Then what is the practical meaning of the imaginary part, ##i \sin(\mathbf{k} \cdot \mathbf{r} - \omega t + \phi)##?
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PhilDSP
#2
Mar6-12, 11:37 AM
P: 603
Hi Chuck88,

The imaginary part is how much amplitude (of a copy of each frequency component that is shifted 90 degrees in phase) that must be added to the real part to end up with the correct phase and amplitude of the original signal [itex] U(\mathbf{r} ,t ) [/itex]. (Normally, a wave would be decomposed to a weighted amount of the real and imaginary components, such as when you assign the initial conditions to a solution of the wave equation)

If only the real part existed, the signal would have a fixed phase alignment. Adding the imaginary part effectively corrects the phase to match the actual experimental signal or wavefront.
Chuck88
#3
Mar6-12, 07:30 PM
P: 37
Quote Quote by PhilDSP View Post
Hi Chuck88,

The imaginary part is how much amplitude (of a copy of each frequency component that is shifted 90 degrees in phase) that must be added to the real part to end up with the correct phase and amplitude of the original signal [itex] U(\mathbf{r} ,t ) [/itex]. (Normally, a wave would be decomposed to a weighted amount of the real and imaginary components, such as when you assign the initial conditions to a solution of the wave equation)

If only the real part existed, the signal would have a fixed phase alignment. Adding the imaginary part effectively corrects the phase to match the actual experimental signal or wavefront.
Thanks for your reply. But ##\phi## in the real part ##\cos(\mathbf{k} \cdot \mathbf{r} - \omega t + \phi)## could represent the phase of the wave. And ##A_{0}## could represent the amplitude of the wave. It seems like that if we neglect the imaginary part of the wave, the amplitude and phase can still be presented only with the parts in the real part.

PhilDSP
#4
Mar7-12, 02:15 AM
P: 603
What is the meaning of the imaginary part of the plane wave function

Most likely ##\phi## is going to be used to align the phase globally. Likewise with ##A_0## for the amplitude. To model a realistic wave, each frequency component ##\omega_n## will have a unique phase offset that is accommodated by the combined amplitudes of the real and imaginary components for that frequency.

The wave function with only a single valued ##\omega## is a monochromatic wave. Most waves encountered experimentally are more complex.
baiber
#5
Mar7-12, 07:14 AM
P: 2
It seems like that if we neglect the imaginary part of the wave, the amplitude and phase can still be presented only with the parts in the real part.
PhilDSP
#6
Mar9-12, 10:07 AM
P: 603
Quote Quote by baiber View Post
It seems like that if we neglect the imaginary part of the wave, the amplitude and phase can still be presented only with the parts in the real part.
That's true in the simple case of a monochromatic wave, but not in general.
Chuck88
#7
Mar9-12, 10:31 AM
P: 37
Quote Quote by PhilDSP View Post
That's true in the simple case of a monochromatic wave, but not in general.
Can you explain that in detail?
PhilDSP
#8
Mar11-12, 11:57 AM
P: 603
A wave that is monochromatic consists of only a single sinusoid. Lasers tend toward producing monochromatic waves but almost all other light sources produce a broadband of frequencies (white light for example). The waveform or wave packet can be decomposed to containing a collection of frequencies. Sound waves are similar though the term monochromatic isn't really used to describe them.

Look up Fourier Analysis and wave superposition for more details.


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