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.
Best Answer
$\bullet$ 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 $t_f$.