Expectation value of $x,y,z$ for general $nlm$ state of hydrogen atom

Question

How to calculate expectation value of $\langle x\rangle, \langle y\rangle,\langle z\rangle$ for the general $\psi_{nlm}$ state? $x$ has $\sin(\theta)\cos(\phi)$ angular part which can be expressed as $\frac{1}{\sqrt{2}}(Y_{1}^{-1}+Y_{1}^{1})$, now the angular integration becomes $$\frac{1}{\sqrt{2}}\left(\int (Y_{1}^{-1}Y_{l}^{m*}Y_{l}^{m})\sin(\theta)d\theta d\phi+\int (Y_{1}^{1}Y_{l}^{m*}Y_{l}^{m})\sin(\theta)d\theta d\phi\right).$$ Here after I can apply Wigner-Eckart theorem and the problem can be solved. However, is there any other way of simplifying this expression to something simple general formula just like the Kramer relations for $\langle r^{s} \rangle$?

Answer 1

The probability densities for all of those states are symmetric under rotations around the $z$ axis and reflections in the $x,y$ plane. This then requires all of those expectation values to vanish.

Answer 2

$\def\mxelm#1#2#3{\langle#1|\,#2\,|#3\rangle}$ It's much easier to use parity, i.e. symmetry of the wavefunction wrt space inversion $$x \to -x \qquad y \to -y \qquad z \to -z.$$ It's known that $$\psi_{nlm}(-x,-y,-z) = (-1)^l\,\psi_{nlm}(x,y,z).$$ Then $|\psi_{nlm}|^2$ is even whereas $x$ is odd. You have $$\mxelm{nlm}x{nlm} = \int\!x\,|\psi|^2\>dx\,dy\,dz.$$ The integrand is odd under space inversion, so the integral vanishes. The same holds true for $y$ and $z$.

Note that Wigner-Eckart theorem if applied to rotation SO(3) group can't give the answer. Consider $L_z=x\,p_y-y\,p_x$. Under rotations it transforms as $z$ does, yet $$\mxelm{nlm}{L_z}{nlm} = m\,\hbar$$ and not 0. Of course this result doesn't contradict W-E theorem as it only says that $$\mxelm{nlm}x{nlm} = k\,\mxelm{nlm}{L_x}{nlm}$$ $$\mxelm{nlm}y{nlm} = k\,\mxelm{nlm}{L_y}{nlm}$$ $$\mxelm{nlm}z{nlm} = k\,\mxelm{nlm}{L_z}{nlm}$$ withe same $k$, but doesn't rule out $k=0$.

So $k=0$ has another cause: which?

Answer 3

It is easy to prove given the amount of symmetry as mentioned in the comments. Focus on the azimuthal angle, since standard spherical coordinates are taken such that the $z$-axis coincides with $\theta=0=\pi$, we expect the azimuthal integration already to be zero since the atom looks the same from every angle. So recall that the spherical harmonics have the form: $$Y_\ell^m(\theta,\phi) = K(\ell,m)P_\ell^m(\cos\theta) e^{im\phi}$$ where $K$ is a normalization coefficient and depends on $\ell$ and $m$. When one multiplies a spherical harmonic with its conjugate you will eliminate the $\phi$ part, $$Y^{*m}_\ell Y^m_\ell \propto P_\ell^{-m}P_\ell^m,$$ this holds for any $\ell$ or $m$. So the azimuthal part of the integrals in your question are reduced to $$\int d\phi\, e^{-i\phi} + \int d\phi\, e^{i\phi} = 0.$$ So the message is to have the physical intuition to say that it is zero and then prove it rigorously exploiting the physical observations.