In Cohen-Tannoudji's Quantum mechanics book, I was reading about an example of a wave packet. The wave function is a superposition of 3 waves with different wave number:
$k_0$, $k_0 + \frac{\Delta k}{2}$, $k_0 - \frac{\Delta k}{2}$
And Amplitudes:
1, $\frac 12$,$\frac 12$.
The wave function is:
$$\psi(x)= Const. [e^{ik_0x} + \frac 12 e^{i(k_0 + \frac{\Delta k}{2})x} +\frac 12e^{i(k_0 - \frac{\Delta k}{2})x}]$$
$$\psi(x)= Const.e^{ik_0x} [1+cos(\frac {\Delta k}{2}x)]$$
One way to find out the distance for a destructive interference is to equalize with zero the expression in the bracket with the cos() expression. With this method you simply write
$cos(\frac {\Delta k}{2}x)=-1=cos\pi$ and from here we get $x=\frac{2\pi} {\Delta k}$.
But in the book the following is said:
As one moves away from x=0,the waves become more and more out of phase, and $|\psi(x)|$decreases.The interference becomes completely destructive when the phase shift between $e^{ik_0x}$ and $e^{i(k_0 \pm \frac{\Delta k}{2})x}$ is equal to $\pm \pi$:$\psi(x)$ goes to zero when $x=\pm \frac{2\pi} {\Delta k}$.
How does this translates mathematically? How can one study the phase shift when the waves are given in a complex expression? As it can be clearly seen, here you get $x=\pm\frac{2\pi} {\Delta k}$ instead of $x=\frac{2\pi} {\Delta k}$. One can argue that, initially, I could also write:
$cos(\frac {\Delta k}{2}x)=-1=cos(-\pi)$ and I would get the $x=-\frac{2\pi} {\Delta k}$ but this seems forced, knowing that $cos(-x)=cosx$.
So, to sum it up, I am interested into how can I investigate the phase shift of 2 waves (3 in this case) when the waves are given in a complex expression. I want to be able to cleanly write the correct result
