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physic · May 30, 2012, 09:17 AM

this is simple of wave function:
$\Psi{(\vec r, t)}=Ae^{i(\phi+\vec k .\vec r - \omega t)}$

$\Psi{(\vec r, t)}=A|\cos{(\phi+\vec k .\vec r - \omega t)}+i\sin{(\phi+\vec k .\vec r - \omega t)}|$
i want to know
why wave function $\Psi{(\vec r, t)}$ shown with $\vec r$ and $t$ ?

what is $A$ Amplitude?

what is phase angle $(\phi+\vec k .\vec r - \omega t)$ ?

and what means cos and sin of it $\cos{(\phi+\vec k .\vec r - \omega t)}$ ?

ebaines · May 30, 2012, 11:21 AM

Originally Posted by physic

i want to know why wave function $\Psi{(\vec r, t)}$ shown with $\vec r$ and $t$ ?

Because the $\Psi$ is a function both of position $\vec r$ and time 't'.

Originally Posted by physic

what is $A$ Amplitude?

Yes , 'A' typically is used to mean "amplitude."

Originally Posted by physic

what is phase angle $(\phi+\vec k .\vec r - \omega t)$ ?

The phase angle defines how the magnitude of the wave is shifted relative to $\sin (\omega t)$ . So here the phase angle is $\phi + \vec k \cdot \vec r$ .

Originally Posted by physic

and what means cos and sin of it $\cos{(\phi+\vec k .\vec r - \omega t)}$ ?

This comes from:

$<br /> e^{i \theta} = \cos(\theta) + i \sin (\theta)<br />$