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

Question

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?

Answer 1

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.

Answer 2

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).

Answer 3

There's several cases.

1. Maxwell daemon. The total energy of a dameon system is meaningless as the energy becomes information, ie you can build a perpetual motion machine with necessary information.

2. Virtual systems. Virtual particles are excluded from the hamiltonian.

3. Hawking radiation. If two black holes orbit each other they give off effectively infinite information while remaining in orbital equilibrium.

There's many, many more cases. Hamiltonians are a "good enough" formulation of a system.

How do these Hamiltonian systems have either rheonomic constraints for their generalised coordinates or a generalised potential? And if they don't, how do those cases in classical Hamiltonian mechanics relate to these? –

They do.

The general equation of a rheonomic constraint is the root of the squares of the coordinates minus a length value, with the equation equal to zero. In other words, a system is constrained as an oscillation. This can equivalently be defined as the change in one variable in polar coordinate terms plugged into the same equation- except the latter definition is rheonomic and not scleronomous.

In the examples I gave:

1. The length here could technically be infinite but could also be constrained to some value of L- which would prevent infinite energy and still differ from a formulation of a Hamiltonian.

2. Virtual particles can be arbitrary assigned and can be in any system.

3. The orbital part of the oscillation would be a Hamiltonian, hawking radiation is not.

To make any of these rheonomic they must be phrased as polar coordinates and not absolute coordinates, but the result is the same.

So all of these are rheonomic constrained oscillations- under some assumption of L.