Absolute value sign when normalizing a wave function

Question

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.

Answer 1

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

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.

Answer 2

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 $\Psi$.

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$$\begin{align*} 1&=\int\limits_{-\infty}^\infty|\Psi(x,t)|^2\,\mathrm dx\\ &=\int\limits_{-\infty}^\infty\left|Ae^{-\lambda|x|}e^{-i\omega t}\right|^2\,\mathrm dx\\ &=\int\limits_{-\infty}^\infty|A|\left(e^{-\lambda|x|}e^{-i\omega t}\right)^*\ |A|\left(e^{-\lambda|x|}e^{-i\omega t}\right)\,\mathrm dx\\ &=\int\limits_{-\infty}^\infty|A|e^{-\lambda|x|}e^{+i\omega t}\ |A|e^{-\lambda|x|}e^{-i\omega t}\,\mathrm dx\\ &=\int\limits_{-\infty}^\infty |A|^2(e^{-\lambda|x|-\lambda|x|}e^{+i\omega t-i\omega t})\mathrm dx\\ &=|A|^2\int\limits_{-\infty}^\infty e^{-2\lambda|x|}e^0\,\mathrm dx\\ &=2|A|^2\int\limits_{0}^\infty e^{-2\lambda|x|}\,\mathrm dx\qquad\text{(even integrand)}\\ &=2|A|^2\left.\left(\frac{e^{-2\lambda|x|}}{-2\lambda}\right)\right|_0^\infty\\ &=2|A|^2\left(0-\frac1{-2\lambda}\right)\\ &=\frac{|A|^2}{\lambda}\\ \\ \frac1{|A|^2}&=\frac1\lambda\\ \\ |A|^2&=\lambda\\ A^*A&=\lambda\\ A^2&=\lambda\qquad\text{($A$ is real )}\\ \\A&=\sqrt\lambda\\ \\ \therefore \boxed{\Psi(x,t)=\sqrt\lambda e^{-\lambda|x|}e^{-i\omega t}} \end{align*}$$

Answer 3

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