A symmetrized operator for the operators $\hat{H}$ and $\hat{N}$ is given by


When $\hat{H}$ is the Hamiltonian and $\hat{N}$ is the first moment of energy, the symmetrized operator $\hat{R}$ is the center of energy.

The author of the paper I'm reading (1310.6570) then says "There is, however, one disconcerting property of the center of energy operator: its components do not commute. This follows directly from the commutator of energy densities..." which is non-zero at two different positions.

He then goes on to say that "... the noncommutativity of the components of $\hat{R}$ is a direct consequence of the noncommutativity of Lorentz transformations. This non-commutativity does not preclude the use of $\hat{R}$ to characterize the spatial extension of the system."

I understand why the components of the operator do not commute. But I don't see why this is so "troubling" that he feels the need to point it out and then to justify using it. Is there, generally speaking, some potential problems if the components of an operator do not commute - especially if you wish to interpret that operator as an observable?

For further clarity the definitions of $\hat{H}$ and $\hat{N}$ are: $$ \hat{H}=\int dr \hat{E}(r), $$ $$ \hat{N}=\int dr \, r\hat{E}(r), $$ and $$ \left[\hat{E}(r),\hat{E}(r')\right]\neq 0. $$

  • $\begingroup$ What precisely do you call the "components" here? $H$ and $N$? $\endgroup$ – ACuriousMind Apr 20 '15 at 16:28
  • 2
    $\begingroup$ A simpler vector operator whose components do not commute is the angular momentum operator. This feature is not trivial - it is the basis for the entirety of angular momentum theory within QM, and it would be weird if a textbook author failed to remark on it. Whether it's "troubling" - well, that depends on what you're troubled by. $\endgroup$ – Emilio Pisanty Apr 20 '15 at 16:32
  • $\begingroup$ @ACuriousMind Yes. I've edited the original question. Hope that makes it more clear. $\endgroup$ – quantum_loser Apr 20 '15 at 16:35

The disconcerting thing is that a scalar Hermitian operator assigns a real number to each element of a basis, so you might naively think that a vector Hermitian operator assigns a vector to each element of a basis. Therefore, a generic state can be written as a superposition of states, each of which has a definite vector value for the vector operator.

This is wrong because the components of the vector operator don't commute, but many people naively think this way. For example, the "rotating vector model" of spin presented in some chemistry textbooks makes precisely this simplification/mistake.


Agreed, I don't see what is so troubling. If two operators don't commute, so be it, there's nothing you can do about that. It may make the problem at hand much more difficult however. For example, in path integral formulations of QM we encounter operators of the form $e^{\hat{T}+\hat{V}}$, which is not equal to $e^{\hat{T}}e^{\hat{V}}$ if $[\hat{T},\hat{V}]\neq0$, and so certain simplifications can't be made.


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.