Absolute value sign when normalizing a wave function 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. 
 A: $$\int_{-\infty}^{\infty} \psi^{\dagger}\psi dx = \int_{-\infty}^{\infty} (A e^{-\lambda |x|} e^{-i\omega t})^{\dagger}(A e^{-\lambda |x|} e^{-i\omega t})dx$$
$$= A^{\dagger}A\int_{-\infty}^{\infty} (e^{-\lambda |x|} e^{-i\omega t})^{\dagger}(e^{-\lambda |x|} e^{-i\omega t})dx$$
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.
A: Introduction to Quantum Mechanics -Second edition - David J. Griffiths 
Problem 1.5
Consider the wave function
$$\Psi(x,t)=Ae^{-\lambda|x|}e^{-i\omega t}$$
where $A$, $\lambda$, and $\omega$ are positive real constants.
(a) Normalize $\Psi$.

$$\begin{align*}
1&=\int\limits_{-\infty}^\infty|\Psi(x,t)|^2\,\mathrm dx\\
&=\int\limits_{-\infty}^\infty\left|Ae^{-\lambda|x|}e^{-i\omega t}\right|^2\,\mathrm dx\\
&=\int\limits_{-\infty}^\infty|A|\left(e^{-\lambda|x|}e^{-i\omega t}\right)^*\ |A|\left(e^{-\lambda|x|}e^{-i\omega t}\right)\,\mathrm dx\\
&=\int\limits_{-\infty}^\infty|A|e^{-\lambda|x|}e^{+i\omega t}\ |A|e^{-\lambda|x|}e^{-i\omega t}\,\mathrm dx\\
&=\int\limits_{-\infty}^\infty |A|^2(e^{-\lambda|x|-\lambda|x|}e^{+i\omega t-i\omega t})\mathrm dx\\
&=|A|^2\int\limits_{-\infty}^\infty e^{-2\lambda|x|}e^0\,\mathrm dx\\
&=2|A|^2\int\limits_{0}^\infty e^{-2\lambda|x|}\,\mathrm dx\qquad\text{(even integrand)}\\
&=2|A|^2\left.\left(\frac{e^{-2\lambda|x|}}{-2\lambda}\right)\right|_0^\infty\\
&=2|A|^2\left(0-\frac1{-2\lambda}\right)\\
&=\frac{|A|^2}{\lambda}\\
\\
\frac1{|A|^2}&=\frac1\lambda\\
\\
|A|^2&=\lambda\\
A^*A&=\lambda\\
A^2&=\lambda\qquad\text{($A$ is real )}\\
\\A&=\sqrt\lambda\\
\\
\therefore \boxed{\Psi(x,t)=\sqrt\lambda e^{-\lambda|x|}e^{-i\omega t}}
\end{align*}$$
A: Well, $\left|\Psi(x,t)\right|^2=\Psi^*(x,t)\Psi(x,t)=|A|^2e^{-2\lambda \left|x\right|}$, isn't it?
