# Is there a quantum mechanical analog to classical rheonomic constraints wherein the Hamiltonian is not the total energy?

The Wikipedia article on the Hamiltonian operator in QM says that the Hamiltonian corresponds to the total energy of the system, but qualifies that statement with a "in most cases" tacked on the end. In classical mechanics, the Hamiltonian of a system with rheonomic constraints will generally not (never?) equal to the total energy of the system. Is there an analogous situation in QM in which the Hamiltonian operator doesn't correspond to the total energy of the system? What are the exceptions the article is alluding to?

• As a simple example, the Hamiltonian of a particle of mass $m$ and charge $q$ in an electromagnetic potential $(\phi, \textbf{A})$ is $H = \frac{1}{2m} (\textbf{p}-q\textbf{A})^2 + q\phi(\textbf{r})$ Commented Jul 27, 2015 at 21:50
• @Ultima that's not a good example, you may want to follow the discussion on this answer to understand there subtlety there, see qmechanic's reference of Goldstein. physics.stackexchange.com/questions/11905/… Commented Dec 28, 2023 at 16:01

You might like the time dependent Hamiltonian $$H=\frac{p^2}{2m}\, e^{-\lambda t}+\frac{m\omega^2 x^2}{2} \, e^{\lambda t}, \qquad \lambda >0, \quad[x,p]=i \hbar \mathbf{1},$$ describing a harmonic oscillator in the presence of a velocity-dependent damping term with the equation of motion $$\ddot{x}+ \omega^2 x +\lambda \dot{x}=0.$$ To see this, we employ the Heisenberg equation of motion $$\dot{A} \equiv \frac{dA}{dt}=\frac{i}{\hbar} \left[H,A\right] +\frac{\partial A}{\partial t}$$ on $$x$$ and the canonical momentum $$p$$: $$\dot{x}=\frac{i}{\hbar}\left[ \frac{p^2}{2m} e^{-\lambda t}, x\right]=\frac{i e^{-\lambda t}}{2m \hbar}[p^2,x]=\frac{p}{m} e^{-\lambda t},\\ \dot{p}=\frac{i}{\hbar}[H,p]=\frac{i}{\hbar}\left[ \frac{m \omega^2 x^2}{2}e^{\lambda t}, p \right]= \frac{i m \omega^2e^{\lambda t}}{2 \hbar}[x^2,p]=-m\omega^2 e^{\lambda t} x.$$ Obviously, the canonical momentum $$p=m\dot{x}e^{\lambda t}$$ differs from the physical momentum $$m\dot{x}$$ by the time dependent factor $$e^{\lambda t}$$. In the next step, one computes $$\ddot{x}=-\omega^2 x-\lambda \dot{x}$$, verifying the equation of motion for $$x$$ mentioned above.
Clearly, the total energy $$E=\frac{m \dot{x}^2}{2}+\frac{m \omega^2 x^2}{2}= e^{-\lambda t} H$$ of the harmonic oscillator differs from the Hamiltonian $$H$$. Its time derivative $$\dot{E}=-\lambda m \dot{x}^2$$ describes the energy loss of the system due to the friction term.
• So, in this case, are both $E$ and $H$ Hermitian operators? I understand that they are, as $E = \mathrm{e}^{-\lambda t} H$ and $H$ should be one. Commented Dec 30, 2023 at 15:14
That really heavily depends on how you define energy. Usually (I think) the Hamilton operator is defined as the "energy operator", so then there obviously aren't any exceptions. But if you define $$E= T+V$$ (which you can do in a quantum-mechanical sense), then $$H \ne E$$ in general (for example if a magnetic field is present). Also, and I think that was what Wikipedia alludes to, the energy of a system changes over time if the hamiltonian is time-dependent (i. e. there is no eigenstate of the hamiltonian satisfying Schrödinger's equation).