I have solved the following problem from Griffiths "Introduction to Quantum Mechanics".
Consider the wavefunction:
$\Psi (x,t) = A e^{-\lambda |x|} e^{-i\omega t} $
Normalize $\Psi$.
Now, we want $ \int_{-\infty}^\infty |\Psi (x,t)|^2 dx = 1$
It is fairly straightforward, where the modulus is $|\Psi (x,t)|= r = A e^{-\lambda |x|}$. Therefore I square $r$ and integrate. I deal with the absolute value sign by multiplying by $2$ and integrating from 0 to $\infty$, while dropping the absolute value sign, to get:
$ 2\int_0^\infty (A e^{-\lambda x})^2 dx$
This should give me a factor of $A^2$ which I can take outside the integral sign. However, instead of a simple $A^2$, the solution gives an $|A|^2$. I don't understand where the absolute value sign came from. After all, taking the above expression $r$ as being equal to $|\Psi(x,t)$|, the modulus has already been dealt with.
Best Answer
$$\int_{-\infty}^{\infty} \psi^{\dagger}\psi dx = \int_{-\infty}^{\infty} (A e^{-\lambda |x|} e^{-i\omega t})^{\dagger}(A e^{-\lambda |x|} e^{-i\omega t})dx$$ $$= A^{\dagger}A\int_{-\infty}^{\infty} (e^{-\lambda |x|} e^{-i\omega t})^{\dagger}(e^{-\lambda |x|} e^{-i\omega t})dx$$
Where $\dagger$ represents the Hermitian conjugate, or the complex conjugate in the case of $A$, so $$A^{\dagger}A = |A|^2$$
and that is where the $|A|^2$ comes from, regardless of whether $A$ is real or not.