Complex conjugate and expectation values in QM

Question

I'm currently trying to understand expectation values within Quantum Mechanics. I have a few questions that I could need a little bit of help with understanding how to interpret and how to further do calculations.

Notations and other important variables for this example:
A free particle has the wave function $\Psi(\vec{r}) = Ne^{-r\gamma}$ where $N$ is a normalisation constant, $\gamma$ is a positive real parameter and $$r = \sqrt{x^2+y^2+z^2}$$ is the distance from the origin of the Cartesian coordinate system.

(1) Since both r and $\gamma$ are positive and real, does this mean that $\Psi(\vec{r})^{\ast}\Psi(\vec{r})$ is just the square of $\Psi(\vec{r})$? I have this thought since the complex conjugate of a real function should be equal to the original function, correct?

(2) When calculating the expecation value $<\vec{r}>_{op}$, is it correct to assume that $$<\vec{r}>_{op} = (<x>_{op}, <y>_{op}, <z>_{op})~? $$ And if that is the case, can I then for example assume that when calculating $<x>_{op}$ that $y=z=0$ and therefore $r=x$ (see definition of r under Notations), and then rewrite my integral as: $$<x>_{op} = \int_{0}^{\infty} \Psi^\ast(\vec{r})x\Psi(\vec{r}) \,dx~?$$

This should in theory, if my assumptions are correct, correspond in the following representation of $<\vec{r}>_{op}$ as: $$[(\int_{0}^{\infty} \Psi^\ast(\vec{r})x\Psi(\vec{r}) \,dx), (\int_{0}^{\infty} \Psi^\ast(\vec{r})y\Psi(\vec{r}) \,dy), (\int_{0}^{\infty} \Psi^\ast(\vec{r})z\Psi(\vec{r}) \,dz)],$$ or have I understood this wrongly?

Thanks for your help and patience, please let me know if anything is unclear so that I can have a look at it and explain it in a better way!

Edit 1: changed wave function from $\Psi(\vec{r}) = Ne^{r\gamma}$ to $\Psi(\vec{r}) = Ne^{-r\gamma}$ due to writing out the wrong wave function by accident.

Answer 1

A free particle has the wave function $\Psi(\vec{r}) = Ne^{r\gamma}$ where N is a normalisation constant, $\gamma$ is a positive real parameter and $r = \sqrt{x^2+y^2+z^2}$ is the distance from the origin of the Cartesian coordinate system.

This function is clearly non-normalizable, and so is not a valid wavefunction for a quantum mechanical particle.

1. Yes
2. Yes - when we talk about $\langle \vec r \rangle$, we mean the list $\big(\langle x\rangle,\langle y\rangle,\langle z\rangle\big)$. No, you cannot make that assumption - the integrals over $x$ and $y$ must be performed as usual, i.e. $$\langle X \rangle_\psi = \int_{-\infty}^\infty\mathrm dx \int_{-\infty}^\infty \mathrm dy \int_{-\infty}^\infty \mathrm dz \ x |\psi(x)|^2$$ Of course, since your proposed wavefunction is not normalizable, you won't get finite answers for these anyway.