In Griffith's Intro to QM Problem 1.5, I'm asked to normalize the wave function:
$\psi ( x,t) =Ae^{-\lambda |x|} e^{-i\omega t}$ (where $A$, $\lambda$ and $\omega$ are positive, real constants)
I know that this means solving for $A$ where:
$1=\int |\psi |^{2} dx$
According to the solutions this reduces to:
$2|A|^{2}\int ^{\infty }_{0} e^{-2\lambda x}dx$
...but I don't understand why!
1) Why does the $e^{-i\omega t}$ term disappear? Is it because $t$ is (treated as) constant? (If it was considered to be $0$, that would explain it, but that seems unreasonable given the definition of $\psi$?)
2) Is the 0 - $\infty$ limit of the integral due to the $|x|$ in the exponent?
$$\int_{-\infty}^{\infty} |\psi|^2 \, \text{d}x = \int_{-\infty}^{\infty} |Ae^{- \lambda |x|}|^2\cdot|e^{-i \omega t}|^2 \, \text{d}x = \int_{-\infty}^{\infty} |A|^2e^{-2\lambda |x|} \, \text{d}x$$ Since, $e^{-i \omega t}$ is a number with magnitude $1$(try Euler's formula). Now,
$$\int_{-\infty}^{\infty} |A|^2e^{-2\lambda |x|} \, \text{d}x = |A|^2 \int_{-\infty}^{\infty} e^{-2\lambda |x|} \, \text{d}x = 2|A|^2 \int_{0}^{\infty} e^{-2\lambda x} \, \text{d}x$$ By symmetry, replacing $|x| \mapsto x$ and considering the integral on $(0,\infty)$.