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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 $\langle r^{s} \rangle$?

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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.

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  • $\begingroup$ Can you please bit elaborate? Is there any way that I can simplify this expression for $<x>$ and $<y>$ and $<z>$. $\endgroup$ – user135580 Apr 16 at 7:14
  • $\begingroup$ Does it vanish for all the states $\psi_{nlm}$? $\endgroup$ – user135580 Apr 16 at 7:14
  • $\begingroup$ Yes, it can be simplified: $$\langle x\rangle = \langle y \rangle = \langle z\rangle = 0,$$ for all $\psi_{nlm} $, on symmetry grounds. $\endgroup$ – Emilio Pisanty Apr 16 at 7:50
  • $\begingroup$ If you absolutely want to do things via methods which are utterly unnecessarily complicated, you can reduce the integrals in your question to the triple products of harmonics in e.g. this question (being careful with the conjugate), and then using the properties of the Wigner 3j symbols (specifically, that they vanish if the bottom row dies not add to zero). But that route is the wrong thing to do - unless you really understand the physical content, you're just obscuring the reason instead of clarifying it. $\endgroup$ – Emilio Pisanty Apr 16 at 8:02
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$\def\mxelm#1#2#3{\langle#1|\,#2\,|#3\rangle}$ 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?

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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.

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  • $\begingroup$ This works for x and y, but not for z. $\endgroup$ – Emilio Pisanty Apr 16 at 8:58
  • $\begingroup$ For $z$ one can rotate the coordinate system so that $z$ lies in the plane $\theta = \pi/2$ and the same argument holds, since the location of your "northpole" is completely arbitrary. $\endgroup$ – ohneVal Apr 16 at 9:16
  • $\begingroup$ No, that argument doesn't work - your state is already specified, and rotating the system would change the state. The hamiltonian is symmetric, but the eigenstates do not share its full symmetry. $\endgroup$ – Emilio Pisanty Apr 16 at 9:17
  • $\begingroup$ I can change coordinates within the integral if it serves you better, then use the property of rotations of the harmonics which will produce just an annoying combination of harmonics with the same $\ell$ and opposite sign $m$'s which will end up in the same sort of integrals as above. $\endgroup$ – ohneVal Apr 16 at 9:23
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    $\begingroup$ For $z$, why not just use the symmetry of $P^m_l(\cos(\theta))$.We have the volume element $dV = d\cos(\theta)d\phi dr$. $z = r \cos(\theta)$, thus the integrand becomes $$\propto (P^m_l(\cos(\theta)))^2\cos(\theta)d\cos(\theta)$$ where $\cos(\theta) \in (0,1)$. Since $P^m_l(x) = (-1)^{m+l}P^m_l(-x)$ it quickly follows $\langle z \rangle = 0$. $\endgroup$ – denklo Apr 16 at 10:15

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