# Does it mean anything if the commutator of an operator with the Hamiltonian is equal to the Hamiltonian?

Question says it all, really. I have $[\hat{H},\hat{O}]=-2i\hbar\hat{H}$. Does this mean that the operator $\hat{O}$ (an observable) is special in some way?

Does this mean that the operator $\hat O$ (an observable) is special in some way?

I believe it means there is no such $\hat O$.

If $\hat O$ corresponds to an observable, we require the eigenvalues to be real.

Let $|o\rangle$ be an eigenket of $\hat O$ with real eigenvalue $o$:

$$\hat O |o\rangle = o |o\rangle$$

Now consider the following

$$\hat O \hat H |o\rangle = (\hat H \hat O - [\hat H, \hat O])|o\rangle = \hat H o|o\rangle + 2i\hbar \hat H |o\rangle = (o + 2i\hbar)\hat H |o\rangle$$

Thus, $\hat H |o\rangle$ is an eigenket of $\hat O$ with complex eigenvalue $(o + 2i\hbar)$ in contradiction with the requirement that the eigenvalues of $\hat O$ are real.

• Could this be seen as an extension of what @gonenc said? Since only if $|o\rangle$'s energy eigenfunctions are zero is your proof contradicted? – quantum_loser Apr 12 '15 at 17:19
• @quantum_loser: Alfred's conclusion only holds if $H\neq 0$. Both answers show that no $O$ with the property you propose can exist for non-zero $H$. – ACuriousMind Apr 12 '15 at 17:22
• However $\hat O$ might not correspond to an observable. In this case it is inevitable that you find a contradiction. – Gonenc Apr 12 '15 at 18:43


It means basically that all of the energy eigenstates has zero energy eigenvalue. Ups...

Let $\left| \psi \right>$ be a normalized energy eigenstate with energy eigenvalue $E_\psi$.

$$\bok{\psi}{[H,O]}{\psi}=\bok{\psi}{HO-OH}{\psi}=E_\psi \left\{ \bok{\psi}{O}{\psi} - \bok{\psi}{O}{\psi}\right\}=0$$ On the other hand:

$$\bok{\psi}{[H,O]}{\psi} = \bok{\psi}{2i\hbar H}{\psi} = 2i\hbar E_\psi = 0 \quad \forall \left| \psi \right>$$

$$\implies E_\psi=0 \quad \forall \left| \psi \right>$$

• You said the commutator was $(-)2i\hbar H$ but that hardly plays any role. – Gonenc Apr 12 '15 at 16:36
• Interesting, but you never use the property of the action of $\hat{O}$ on $|\psi\rangle$. That must play a role somehow or else all energy eigenstates would have energy zero. Clearly I'm missing something obvious... – quantum_loser Apr 12 '15 at 16:56
• Note that "all of the energy eigenstates have zero eigenvalue" means that $H=0$, so such an $O$ cannot exist for $H\neq0$. – ACuriousMind Apr 12 '15 at 17:16
• @quantum_loser they explicitly use that $\left[H,0\right]\propto H$ in the first equality of the second line of maths. – or1426 Apr 12 '15 at 22:51

Suppose the space-time group includes dilatations which expand or contract space. Points in space $x^{i}\in V_{3}$ transform under a small dilatation $\epsilon$ near the identity as, $$x'^{i}=x^{i}+\epsilon x^{i} \ .$$ The change in the coords is, $$\frac{d x^{i}}{d\epsilon}=x^{i}$$ In the Hamiltonian formulation, the generator of dilatations will be some phase-space function $O$ such that the PB, $$\frac{dx^{i}}{d\epsilon}=[x^{i},O]_{PB}=\frac{\partial x^{i}}{\partial x^{k}}\frac{\partial O}{\partial p^{k}}-\frac{\partial x^{i}}{\partial p^{k}}\frac{\partial O}{\partial x^{k}}=\frac{\partial O}{\partial p^{i}}=x^{i}$$ Integrating, gives the phase space function as, $$O=p^{i}x^{i} \ .$$ If the spacetime group is Galilean relativity, the Hamiltonian is, $$H=\frac{p^{i}p^{i}}{2m} \ .$$ The PB of interest is then, $$[H,O]_{PB}=-\frac{\partial H}{\partial p^{k}}\frac{\partial O}{\partial x^{k}}=-\frac{p^{k}p^{k}}{m}=-2H \ .$$ Now go over to quantum mechanics by replacing the phase space functions with operators, $$[\hat{H},\hat{O}]=-2i\hat{H}$$ This recovers the commutator in the question and shows it has the meaning of a dilatation of Galilean space-time.

The other answers claim that $\hat{O}$ is not Hermitian or that it does not exist. However, $\hat{O}$ must exist and be Hermitian because it's the generator of dilatations in the affine space-time and all affine spaces - those with a notion of parallelism - have dilatations in addition to translations (see chapter 13 of Coxeter's "Introduction to Geometry"). The dilatations are unfamiliar, but one can set up a similar commutator for the boost $\hat{K}$ and the arguments in the other answers would run again and say the boosts are not Hermitian or don't exist. So, the algebra for a boost is, $$[\hat{K},\hat{P}]=i\hat{H}$$ $$[\hat{K},\hat{H}]=i\hat{P}$$ Subtracting, $$[\hat{P}-\hat{H},\hat{K}]=i(\hat{P}-\hat{H}) \ .$$ This is the same as $[\hat{H},\hat{O}]=-2i\hat{H}$, modulo a numerical factor, with $\hat{H}\rightarrow\hat{P}-\hat{H}$ and $\hat{O}\rightarrow \hat{K}$.

• This $O$ is not Hermitian after quantization, and so not an observable as the other answers assume (because OP does say in a parenthesis that $O$ is meant to be an observable). I think compatibility with the other answers arises if $O$ is a symmetry of the theory, because then $O$ needs to be implemented Hermitian as the generator of a unitary operator, but the Hamiltonian vanishes (weakly) in theories with reparametrization invariance, and the dilatation is a reparametrisation of the radial coordinate. – ACuriousMind Apr 12 '15 at 22:49