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.


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ Then you don't say 'meaningful', you say 'tensor operator of rank two'. $\endgroup$ – Emilio Pisanty Jan 12 '17 at 6:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.