Evaluating position vector between 2 hydrogen states I am trying to find the quantity:
$$\langle1,0,0|\vec r|2,0,0\rangle$$
Where $|n,l,m\rangle$ are the hydrogen states.  For this, can I just integrate r from? 0 to infinity?  Or do I have to break it into the x, y, z components and go over all space?
 A: In general, you must integrate over all space. When you work with spherically symmetric systems like the hydrogen atom, the eigenstates $|n,l,m\rangle$ separate into functions of $(r, \theta, \phi)$: $\psi(\vec{r}) = R(r)\Theta(\theta)\Phi(\phi)$. Because of this, it is vastly easier to do the integrals in the $(r,\theta,\phi)$ space: you can carry out the integral over each coordinate independently of the other coordinates.
In this particular case, because $l = m = 0$, you should find that the $\theta$ and $\phi$ integrations are particularly easy, leaving you just $r$. You should find that this particular case will allow you to derive some selection rules.
A: ITHER :
You write $\textbf{r} = x\textbf{i} + y\textbf{j} + z\textbf{k} = r\sin{\theta}\cos{\phi}\textbf{i} + r\sin{\theta} \sin{\phi}\textbf{j} + r\cos{\theta}\textbf{k} $, 
work out the values of $x$, $y$ and $z$ separately and from them work out the values of $\phi$ and $\theta$ 
using $ \theta = \arctan(\frac{z}{\sqrt{x^2 + y^2 + z^2} })$ and $ \phi = \arctan(\frac{y}{x }$) 
OR:
find the values of $r$, $\phi$ and $\theta$ using:
$ \langle r\rangle = \langle1,0,0|r|2,0,0\rangle $,
$ \langle \phi\rangle = \langle1,0,0|\phi|2,0,0\rangle $,
$ \langle \theta \rangle = \langle1,0,0|\theta|2,0,0\rangle $
which I think will be harder to compute algebraically because $\theta$ and $\phi$ appear both in linear terms and in trigonometric formulae.
A: I don't think this problem is as hard as it looks.  As others have pointed you will have an integral for each of the three cartesian directions: $r_x$, $r_y$, and $r_z$.  Each integral is a triple integral, best done in spherical coordinates.  First do the $\phi$ integral, then $\theta$, leaving $r$ for last.
A: Actually the result can be given with zero calculation, by just noting that the wave functions $\left|n,0,0\right\rangle$ are radially symmetric, that is, invariant under arbitrary rotations around the origin, and the integration range is the whole space, and therefore also invariant under rotations, and therefore the result also has to be invariant under arbitrary rotations around the origin. On the other hand, the result is a vector. But there exists only one rotationally invariant vector: The zero vector.
A: That depends on what basis do you use. Since $|l,m,n⟩$ means the Schrödinger-eq. was solved in spherical coordinates, than you get the following:
$$
⟨1,0,0|r⃗ |2,0,0⟩=\int_VdV \Psi_{100}^*(\vec r)\vec r \Psi_{200}(\vec r)=$$$$=\int_0^{2\pi}\int_0^{\pi}\int_0^{\infty} \frac{1}{\sqrt{\pi}a_0^{3/2}}e^{-r/a_0}r \frac{1}{4\sqrt{2\pi}a_0^{3/2}}(2-\frac{r}{a_0})e^{-r/2a_0} r^2sin\theta \,dr\,d\theta \,d\phi
$$
Since in spherical coordinates, after integrating as per solid angle, $r⃗ \rightarrow r$, nothing depends on $\theta$ or $\phi$, so you need to integrate from 0 to $\infty$ as per $r$.
Source of wave functions: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.html#c3
