- #1

Sandbo

- 18

- 0

If all the below statements are too bulky to read, my question is actually as short as:

[tex]\psi (x) = A{e^{-j(kx - \omega t)}}[/tex]

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:

[tex]\psi (x) = A{e^{ - j(kx - \omega t)}}[/tex], let B = [tex]j(kx - \omega t)[/tex]

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 [tex]\cos (kx - \omega t)[/tex].

As [tex]\cos x = \cos ( - x)[/tex], therefore no matter the sign before B is + or - (i.e. [tex]\psi (x) = A{e^{j(kx - \omega t)}}{\rm{ }}or{\rm{ }}\psi (x) = A{e^{ - j(kx - \omega t)}}[/tex]), 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:

[tex]\psi (x) = A{e^{j(kx - \omega t)}}{\rm{ + }}B{e^{ - j(kx - \omega t)}}[/tex]

"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.