The expectation value of $r=\sqrt{x^2 + y^2 + z^2}$ for the electron in the ground state in hydrogen is $\frac{3a}{2}$ where a is the bohr radius.
I can easily see from the integration that the expectation values of $x$ , $y$, $z$ individually is $0$ because of the factor $\cos\phi$, $\sin\phi$ and $\cos\theta$(.$\sin\theta$) respectively for $x$ $y$, $z$ integrating out to $0$.
What does this mean? Is it an artifact of the spherically symmetric potential? Since $r$ has a non zero value but what makes up $r$ ie $x$, $y$, $z$ (or rather the squareroot of the sum of their squares) each has individually $0$ and i can simultaneously determine their values since they commute with each other, which means $x,y,z$ are all simultaneously $0$ and $r$ is not?!
Best Answer
$x$, $y$ and $z$ can take any value from $-\infty$ to $\infty$, and for all three the wavefuntion is symmetric about the origin. So it's (hopefully) obvious that the expectation value of all three is zero.
By contrast $r \ge 0$ i.e. $r$ cannot be negative. So it's (hopefully) obvious that the expectation value of $r$ will lie somewhere in the range $0 \le r \le \infty$.
You would get a similar result if you calculated the expectation value of $|x|$ rather than $x$.