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Let $|\phi(r)\rangle$ be a spherically symmetric ground state with $\langle\phi|\phi\rangle=1$, e.g. the ground state of the Schrödinger equation for the hydrogen atom. My professor claimed today that for an operator $Q^{ij}$ we have for such a state \begin{equation} \langle\phi|Q^{ij}|\phi\rangle=\frac{\delta^{ij}}{3}\langle\phi|Q^{kk}|\phi\rangle, \end{equation} where $i,j=1,2,3$ are the indices of the spatial coordinates. Whereas for, let's say, $Q^{ij}=r^ir^j$ this can be easily shown by explicit integration, I wonder if there is a way to prove the relation general. $Q^{ij}$ might for example be an operator appearing in second order perturbation theory and thus contain the reduced Green's function, which makes it non-trivial.

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The assertion as you have phrased it is patently incorrect. With overwhelming probability, you misunderstood your instructor's explanation, particularly regarding the conditions that $Q^{ij}$ needs to satisfy for the result to be true.

It should be obvious that the result as you have stated it cannot be true: "for any operator $Q^{ij}$..." - what if I choose $Q^{ij}=1$ for all $i$ and $j$? Without additional qualifiers on $i$ and $j$ and their role inside $Q^{ij}$, why even give it superscripts?

That said, if you suitably qualify what the $Q^{ij}$ does, then yes, the result can probably be proved for a pretty broad class of operators. As mentioned in the comments, the main tool for this job is the Wigner-Eckart theorem, which tells you that if $Q^{ij}$ has some special interaction with rotations, then its dependence on the directional indices $i$ and $j$ can be tightly constrained.

In this specific case, you obviously want to generalize the case of $Q^{ij}=r^ir^j$, so the natural condition on $Q$ is to require that if under some rotation you have $r^i\mapsto r'^i=R^{ik}r^k$, then the $Q$ transforms as $$ Q^{ij}\mapsto Q'^{ij}=R^{ik}R^{jl}Q^{kl}. $$ This includes $Q^{ij}=r^ir^j$ but also plenty of nontrivially different operators, like e.g. $Q^{ij}=\frac12(r^ip^j+p^jr^i)$, or any operator of the form $Q^{ij}=f(|r|^2)r^ir^j$, which can have vastly different matrix elements. Under this transformation rule, however, the Wigner-Eckart theorem means that the expectation values of the $Q^{ij}$ under a spherically symmetric state must vanish. This is because $r^ir^j$ fills out a representation of the rotation group which includes the scalar $T^{(0)}$ and the quadrupole $T^{(2)}$ irreducible representations; the latter have all-zero Clebsch-Gordan coefficients between $s$ states, and you're left with only the scalar representation, which is what your result shows.

However, I will leave it to you to flesh out the details of that argument; hopefully it will be a good lesson on the fact that when someone says "if $x$ then $y$" you can't just drop the "if $x$" and pretend that the "then $y$" still holds.

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  • $\begingroup$ Thank you for the quick answer. I agree, the statement was a bit unfortunate. I of course only thought of meaningful operators, like the examples in your second to last paragraph. $\endgroup$ – Dominik Jan 11 '17 at 20:37
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    $\begingroup$ @Dominik No, that completely misses the point. "Meaningful" is such a loose and undefined qualifier that it is essentially meaningless. Would you consider e.g. $Q^{ij}=r^1r^ir^j$ to be "meaningful"? ...because the result is false for it. You do actually need to pay attention to the "if $x$" parts. $\endgroup$ – Emilio Pisanty Jan 11 '17 at 20:48
  • $\begingroup$ By "meaningful" I mean operators that fulfill the condition $Q^{ij}\mapsto Q'^{ij}=R^{ik}R^{jl}Q^{kl}$, which seems to be the case for the operators I have to deal with. $\endgroup$ – Dominik Jan 11 '17 at 21:34
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    $\begingroup$ Then you don't say 'meaningful', you say 'tensor operator of rank two'. $\endgroup$ – Emilio Pisanty Jan 12 '17 at 6:18

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