Normalizing the Complex Planewave

A wavefunction must always satisfy

$\int_{-\infty}^{\infty} |\Psi(x, t)|^{2} dx = 1$

(In other words, the probability of finding a particle somewhere in space must be 100%.)

Since $|\Psi(x, t)|^{2} = |Ae^{-2\pi|x|}e^{i\pi t}|^{2} = A^{2}e^{-4\pi|x|}$ and (because $\int a^{x} dx = \frac{a^{x}}{\ln a}$ )

$\int A^{2}e^{-4\pi x} dx = A^{2}\frac{e^{-4\pi x}}{-4\pi}$

(note the absence of the absolute value), hence

$\int_{-\infty}^{\infty} A^{2}e^{-4\pi|x|} dx = 2\int_{0}^{\infty} A^{2}e^{-4\pi x} dx = A^{2}\frac{1}{2\pi} = 1$

From this, we can then know that

$A = \sqrt{2\pi} \approx \boxed{2.507}$

Le math

So in the picture I skipped a few steps for brevity, here are the steps I skipped:

The wave function times its own complex conjugate, since it’s exponential, the iπt cancels out. This happens a lot with exponential. This leaves us with -2π|x|, but since we squared it, it becomes the -4π|x| in the first equation.

Knowing that an absolute value function is really just piecewise, I add two intervals from negative infinity to zero and zero to positive infinity. Since on negative numbers, |x| = -x, I sub that in and cancel negatives.

For x >= 0, |x| = x. This gives me line 2.

Simple integration takes over equation 3, and the rest is self explanatory.