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.

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You are assuming A is real. –  Ron Maimon Jul 14 '12 at 8:57

Well, $\left|\Psi(x,t)\right|^2=\Psi^*(x,t)\Psi(x,t)=|A|^2e^{-2\lambda \left|x\right|}$, isn't it?

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I still don't see where the | | came from. It is true that $A^2 = |A|^2$ but I'm not that satisfied with placing an absolute value sign where there needn't be one. I get $\left|\Psi(x,t)\right|^2=\Psi^*(x,t)\Psi(x,t)=Ae^{-\lambda \left|x\right|}Ae^{-\lambda \left|x\right|}=A^2e^{-2\lambda \left|x\right|}$. Note that I'm assuming A is a real constant, which seems reasonable. –  Joebevo Jul 14 '12 at 8:41
I think there's my problem. If A can be complex then your expression seems to make sense. –  Joebevo Jul 14 '12 at 8:47
It is not true that $A^2=|A|^2$ and it is not true that $A$ is real. A general amplitude isn't real; and your particular $A$ clearly isn't real, either. The probabilities are always expressed as squared absolute values of the amplitudes. A formula without the absolute value would clearly be wrong as probabilities are real. –  Luboš Motl Jul 14 '12 at 8:48

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|Ae^{-\lambda|x|}e^{-i\omega t}|^2\,\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|}\,\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*}

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It is not generally true that the amplitude ($A$) is real. –  dmckee Dec 30 '12 at 22:05

$$\int_{-\infty}^{\infty} \psi^{\dagger}\psi dx = \int_{-\infty}^{\infty} (A e^{-\lambda |x|} e^{-i\omega t})^{\dagger}(A e^{-\lambda |x|} e^{-i\omega t})dx$$ $$= A^{\dagger}A\int_{-\infty}^{\infty} (e^{-\lambda |x|} e^{-i\omega t})^{\dagger}(e^{-\lambda |x|} e^{-i\omega t})dx$$

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.

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