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I would like to ask some questions to confirm if what I have acquired so far are correct, please point out my faults if any.
If all the below statements are too bulky to read, my question is actually as short as:
\psi (x) = A{e^{-j(kx - \omega t)}}
What is the propagation direction of this wave? Positive x or negative x? Why(which terms tell this)?
Regarding the propagation direction of wave equations, to the best of my understanding, in the following wave equation:
\psi (x) = A{e^{ - j(kx - \omega t)}}, let B = j(kx - \omega t)
The direction of the wave propagation is towards the positive x direction (+ve), for the following reasons:
1. when we consider the exponential, we are actually taking only the real part of it, which is \cos (kx - \omega t).
As \cos x = \cos ( - x), therefore no matter the sign before B is + or - (i.e. \psi (x) = A{e^{j(kx - \omega t)}}{\rm{ }}or{\rm{ }}\psi (x) = A{e^{ - j(kx - \omega t)}}), the propagation direction will not change, remains as positive x direction.
However, from a lecture note I recently read, I saw something kind of contradicting to the above:
\psi (x) = A{e^{j(kx - \omega t)}}{\rm{ + }}B{e^{ - j(kx - \omega t)}}
"Consider a free particle that travels only in the positive direction of x. Let the arbitrary constant B be zero."
It seems the negative sign on the latter term told us that it's propagating towards negative x direction and that is omitted, which is not consistent with my previous understanding of wave equations.
I am sorry if I have put too much redundancy here, simply trying to make it as clearly as I can.
Thanks for reading.
If all the below statements are too bulky to read, my question is actually as short as:
\psi (x) = A{e^{-j(kx - \omega t)}}
What is the propagation direction of this wave? Positive x or negative x? Why(which terms tell this)?
Regarding the propagation direction of wave equations, to the best of my understanding, in the following wave equation:
\psi (x) = A{e^{ - j(kx - \omega t)}}, let B = j(kx - \omega t)
The direction of the wave propagation is towards the positive x direction (+ve), for the following reasons:
1. when we consider the exponential, we are actually taking only the real part of it, which is \cos (kx - \omega t).
As \cos x = \cos ( - x), therefore no matter the sign before B is + or - (i.e. \psi (x) = A{e^{j(kx - \omega t)}}{\rm{ }}or{\rm{ }}\psi (x) = A{e^{ - j(kx - \omega t)}}), the propagation direction will not change, remains as positive x direction.
However, from a lecture note I recently read, I saw something kind of contradicting to the above:
\psi (x) = A{e^{j(kx - \omega t)}}{\rm{ + }}B{e^{ - j(kx - \omega t)}}
"Consider a free particle that travels only in the positive direction of x. Let the arbitrary constant B be zero."
It seems the negative sign on the latter term told us that it's propagating towards negative x direction and that is omitted, which is not consistent with my previous understanding of wave equations.
I am sorry if I have put too much redundancy here, simply trying to make it as clearly as I can.
Thanks for reading.