I have a question about difference between Hamiltonian function (the description of system in classical physics) and the Hamiltonian operator (quantum mechanics).

I think that there two different points of view: a physical one and mathematical (more technical) one.

In classical Hamiltonian mechanics state of the system (just for sake of simplicity let's consider one dimensional case) is determined by the variables $p, q$. It actually means that if one has defined the initial values of $p$ and $ q$ in arbitrary point of time $t$ then one can find their values in subsequent moment of time $t + \Delta t$ $$ q(t + \Delta t) = q(t) + \dot q(t) \Delta t $$ $$ p(t + \Delta t) = p(t) + \dot p(t) \Delta t $$ using canonical equations: $$\dot q = \partial H / \partial p$$ $$\dot p = - \partial H / \partial q$$ where $H$ classical Hamiltonian function.

In case of quantum mechanics. The state of the system is defined by $\Psi(q, t)$. And if we know $\Psi$ at given moment of time $t$ we can calculate it at subsequent moment $t + \Delta t$: $$\Psi(q, t + \Delta t) = \Psi(q, t) + \dot \Psi(q, t) \Delta t$$ where $i \hbar \dot \Psi = \hat H \Psi$ and $\hat H$ is the Hamiltonian operator.

As for me this leads to the following consequences

  • In classical physics Hamiltonian defines canonical variables, but in QM Hamiltonian operator defines only one quantity (psi function)
  • Classical motion is defined by canonical equation (principle of the least action), QM Hamiltonian constructed in such a way to satisfy Schrodinger equation (it is not derived from principle of least action)
  • Mathematically Hamiltonian in CM is just a function of $q,p$ variables but in quantum mechanics it is a Hermitian operator

Please could you tell me if I am right or if I have something missed here? I am actually interested in what is the difference between quantum and classical Hamiltonian? I will be very pleased because it is very interesting topic for me.


2 Answers 2


Your question is misbegotten, starting from your 3rd point:

Mathematically, the Hamiltonian in CM is just a function of q,p variables, but in quantum mechanics it is a Hermitian operator

This is one of the most enduringly popular untruths in the trade, associated with the peculiar particular development of the field. In fact, Quantum mechanics can be described perfectly well in phase space, and classical mechanics in Hilbert space. The reason is the existence of the invertible Wigner-Weyl map bridging phase space and Hilbert space in complete and practical equivalence.

This, then, obviates the false contrast of your first point:

In classical physics, the Hamiltonian determines canonical variables, but in QM the Hamiltonian operator determines only one quantity, ψ.

Misleading. In both cases the Hamiltonian does the same job. The classical Liouville evolution equation (for which deterministic trajectories can be defined through δ-function Liouville densities) can be extended in QM by Moyal's deterministic equation (performing the function of the Schroedinger equation), which describes the evolution of a Wigner quasibrobability distribution, instead. Because this formulation automatically encodes the uncertainty principle, complete δ-function localization in phase space is impossible, and so trajectories, strictly speaking, do not exist, and the probability fluid diffuses.

Nevertheless, this does not quite obviate your second point, which is factually wrong on its own:

Classical motion is determined by canonical equations (principle of least action), while the QM Hamiltonian is constructed in such a way as to satisfy Schroedinger's equation (it is not derived from the principle of least action).

Indeed, Liouville's equations follow directly from the canonical equations of motion resulting from an action extremizing principle, now understood to be a classical limit of QM, after the 1933 insight of Dirac, p 69, and codified in Feynman's path integral formulation, the celebrated small $\hbar$ limit.

Strangely, and rather disconcertingly, Schroedinger's equation can also be derived from a well-known extremization of $\int dt dx ~\psi^*(x,t) (i\hbar \partial_t -H) \psi(x,t)$, even though no such destructive interference limit is involved!(?) I have wondered about it for a long time, and concluded, possibly wrongly, that it is a "fluke", namely the trivial fact that any linear equation amounts to the extremum of a quadratic Hermitian form. But I really can't be sure. Can you?


$\bullet$ The wave function has information about both position and momentum, so in a sense you're right. But it's not particularly useful to count quantities in the way you do here: Write $X_\Psi:=(q,p),\ \nabla:=(\tfrac{\partial}{\partial q},\tfrac{\partial}{\partial p})$ and $\omega:=\begin{pmatrix} 0&1\\ -1&0 \end{pmatrix}$. Now your classical equation is $\dot X_\Psi=\omega \nabla H$, where $H$ and therefore also the vector $\omega \nabla H$ is a function of $X_\Psi$, and this is also just one equation for one quantity.

$\bullet$ Yeah, the wave function is viewed as a weighted deviation for the action principle for classical point particles. But for the one-particle wave function, the Schrödinger equation is just a field equation and this also has a Lagrangian, $\frac{\partial \mathcal{L}}{\partial \psi^{*}} - \left(\frac{\partial}{\partial t} \frac{\partial \mathcal{L}}{\partial\frac{\partial \psi^{*}}{\partial t}} + \sum_{j=1}^3 \frac{\partial}{\partial x_j} \frac{\partial \mathcal{L}}{\partial\frac{\partial \psi^{*}}{\partial x_j}}\right) = 0$ with $\mathcal{L}\left(\psi, \mathbf{\nabla}\psi, \dot{\psi}\right) := \mathrm i\hbar\, \frac{1}{2} (\psi^{*}\dot{\psi}-\dot{\psi^{*}}\psi) - \frac{\hbar^2}{2m} \mathbf{\nabla}\psi^{*} \mathbf{\nabla}\psi - V( \mathbf{r},t)\,\psi^{*}\psi$.

$\bullet$ These are the definitions, right. But both provide both, energy function an operator generating time developement. The function for the Hermitian operator maps $\psi$ to $\left\langle\phi\right|H\left|\phi\right\rangle$, see here, and the equations above provide a flow mapping a state at a time $t_i$ to a state at a time $t_f$.

  • $\begingroup$ Thank you very much for your comments on my suggestions. Now it's clear why first point is not a good answer. Second point i am not fully agree with you. I think that Schrödinger equation is not derived from principle of least action, it just satisfies this principle (i am not sure in this statement). Third comment requires some time to think for me. What about your thoughts? How do you answer the question about difference classical and quantum Hamiltonian? $\endgroup$ Sep 6, 2014 at 9:06
  • 1
    $\begingroup$ @Wingonafly: Regarding your response to the second point, what you derive from what is a matter of choice. Usually the Schrödinger equation isn't taken to be derived from anything but taking as a starting point. I don't know what you mean by "How do you answer the question about..." because your initial post didn't even include specific questions. A main difference is that classical observables always commute. $\endgroup$
    – Nikolaj-K
    Sep 6, 2014 at 9:17

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