Why is $\langle x \rangle =0$ for the ground state hydrogen atom?

Question

From Griffiths, Introduction to Quantum Mechanics, 2nd ed:

I found $\langle r \rangle =\frac{3a}{2}$ and $\langle r^2 \rangle =3a^2$. Now I need to find the expectation value of x. However, I don't understand their hint. I understand that $r$ is the radius and $r^2=x^2+y^2+z^2$ is the definition for a sphere in cartesian coordinates. I didn't know how to use the part "exploit the symmetry of the ground state".

Edit: I should mention that the answer is $\langle x \rangle =0$ and $\langle x^2 \rangle = \frac{1}{3}\langle r^2 \rangle = a^2$ as given by the solution set. However, I have no clue why.

Answer 1

In spherical coordinates $r$ is the distance from the origin and can only ever be positive.
So the expectation value $\langle r \rangle$ has to be positive.
However in rectangular Cartesian coordinates $x$ can be both positive and negative.
In the case that you are considering with a spherically symmetric wave function you would expect the electron to spend as much time when $x<0$ as when $x>0$ so the expectation value of $x$, $\langle x \rangle =0$.

Since $x^2$ is always positive you would expect its expectation value to be positive.
To find $\langle x^2 \rangle$ you reason that because there is spherical symmetry $\langle x^2 \rangle =\langle y^2 \rangle =\langle z^2 \rangle$ and then use $\langle r^2 \rangle = \langle x^2 \rangle +\langle y^2 \rangle +\langle z^2 \rangle$ to get your answer.