Would $[\hat{Q},\hat{H}]$ correspond to an observable? Where $\hat{Q}$ is an observable and $\hat{H}$ is the Hamiltonian.

Surely that would just mean that $[\hat{Q},\hat{H}]$ would commute i.e. = 0?:

$[\hat{Q},\hat{H}]\phi_{n} = \hat{Q}\hat{H}\phi_{n} - \hat{H}\hat{Q}\phi_{n} = \hat{Q}E_{n}\phi_{n} - \hat{H}q_{n}\phi_{n} = q_{n}E_{n}\phi_{n} - E_{n}q_{n}\phi_{n} = 0 $ ? Hence the commutation does NOT correspond to an observable?


closed as unclear what you're asking by ACuriousMind, innisfree, Danu, John Rennie, Kyle Kanos Jun 5 '15 at 13:22

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    $\begingroup$ An observable is just a self-adjoint operator. Why do you suppose it commutes with $H$, or rather, why do you assume $\hat{Q}\phi_n = q_n \phi_n$? What's the actual question here? $\endgroup$ – ACuriousMind Jun 4 '15 at 20:18
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    $\begingroup$ Not all observables are conserved/commute with the Hamiltonian $\endgroup$ – innisfree Jun 4 '15 at 20:18
  • $\begingroup$ And furthermore, why should 0 not be an observable? (Although a boring one) $\endgroup$ – Sebastian Riese Jun 4 '15 at 20:19
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    $\begingroup$ You've a hidden assumption that they can be diagonalized simultaneously, which is equivalent to assuming they commute. $\endgroup$ – Omry Jun 4 '15 at 20:42

$[A, B]$ for two observables $A$ and $B$ is an observable if, and only if, $A$ and $B$ commute.

Proof: $$ [A, B]^\dagger = (AB)^\dagger - (BA)^\dagger = B^\dagger A^\dagger - A^\dagger B^\dagger = BA - AB = -[A, B].$$

Note: An observable is any Hermitian operator. The commutator of two Hermitian operators is anti-Hermitian, as the proof shows. $0$ is an observable, but a "boring" one. (Actually any real number multiplied by the identity is an observable).

  • $\begingroup$ I like this +1, but in what sense is $0$ an observable? $\endgroup$ – innisfree Jun 4 '15 at 21:34
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    $\begingroup$ It is a corner case, admittedly. But it is a Hermitian operator, in all states we can give the probability ($p = 1$) of measuring the eigenvalue 0, all states are eigenstates with the eigenvalue 0. $\endgroup$ – Sebastian Riese Jun 4 '15 at 21:55

As noted by Sebastian, $[Q,H]$ will be anti-Hermitian and therefore generally not an observable (except in the trivial case).

However $i[Q,H]$ is an important observable. This corresponds to the classical Poisson bracket which can be see in the following formula,

$$ \frac{d \langle Q\rangle}{d t} = \frac{i}{\hbar} \langle [H,Q]\rangle + \langle \frac{\partial Q}{\partial t} \rangle.$$


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