6
$\begingroup$

Context: I'm not a physicist.

I've come across the Hamiltonian in classical physics and in quantum physics, and I can't recognise why they have the same name. They seem very different. So I probably am missing something.

Classical. My understanding of the classical Hamiltonian is: it is a function $H(p,q,t)$ of the (generalized) coordinates of the system, letting $q$ be the generalized "position" vector, and $p$ be the generalized momentum vector. The time evolution of the system is obtained as:

$$\dot q=\frac {\partial H} {\partial p}\quad\quad\quad \dot p=-\frac {\partial H} {\partial q}$$ Where the dots denote time derivatives.

Quantum. My understanding of the quantum Hamiltonian, is that it is a Hermitian operator $H$ on the quantum state $|\Psi \rangle $, that is used to calculate the time derivative (generalized Schrodinger equation) :

$$|\dot \Psi\rangle = \alpha H|\Psi \rangle$$

(Where $\alpha=-\frac i {2\pi h}$, which is not important for now).

How are these related? They seem totally different objects:

  • Type signature. The classical hamiltonian is a function from state-space to $\mathbb R$. The quantum Hamiltonian is an operator from state space to state space. (both possibly time dependent, in the classical case I mean by state space, $(q,p)$ space, in the quantum case I mean the quantum state space).

  • Derivatives. In the classical case, we take the derivative of the Hamiltonian. In the quantum case, we don't, we just use the hamiltonian itself, and use it to operate on the state space.

  • Analog of Newton's law?. In classical physics, the Hamiltonian is based on a variational principle (the Hamiltonian is derived from minimizing the Action functional). The "local" laws of motion in classical physics are Newton's laws: $\dot p = F(q,p)$. It seems to me that the Schrodinger equation is the quantum analog of Newton's law (insofar as there is one), not the analog of Hamilton's equations in classical mechanics: The "force operator" also is an operator from state space to state space. Using nonstandard notation by letting the classical state $s=(q,p)$ to emphasize the analogy with the Schrodinger equation, Newton's laws can be written as: $$\dot s = F s$$ Where $F_q(q,p) = \frac 1 m p$ and $F_p(q,p)$ is the force. It is surprising to me that the Schrodinger equation, which has the form of a local law of motion, contains the energy of the system (the Hamiltonian), whereas the classical law of motion contain forces rather than energy.

Am I wrong that the Schrodinger equation is more analogous to Newton's laws than to the classical Hamiltonian equations? Why do they have the same name, if they seem like very different types of objects?

$\endgroup$
9
  • $\begingroup$ You are seriously overthinking this. Ask yourself: what is a Hamiltonian? What does it quantify? $\endgroup$
    – Gert
    Feb 9, 2020 at 16:44
  • $\begingroup$ @Gert, A hamiltonian is an operator on the quantum space, and its expectation quantifies the "expected energy" of the system. $\endgroup$
    – user56834
    Feb 9, 2020 at 17:06
  • $\begingroup$ Why the quote marks? $\endgroup$
    – Gert
    Feb 9, 2020 at 17:15
  • 1
    $\begingroup$ The quantum and classical formalisms are different of course, but in both cases $\hat{H}$ is the total energy of the system (quantum OR Newtonian). $\endgroup$
    – Gert
    Feb 9, 2020 at 17:31
  • 1
    $\begingroup$ The quantum mechanical Hamiltonian plays an analogous role in the quantum mechanical formalism as does the classical Hamiltonian in the classical formalism, in particular, they both tell you the energy of a system in the language the formalism is speaking: the quantum Hamiltonian spits out eigenvalues and expectation values, the classical Hamiltonian simply spits out some number. Furthermore, one can usually 'convert classical and quantum Hamiltonians into each other' by quantizing and taking a classical limit. $\endgroup$
    – Stijn B.
    Feb 9, 2020 at 19:33

2 Answers 2

2
$\begingroup$

The relationship between classical and quantum mechanics is expressed the clearest in the process of Dirac's canonical quantisation. This consists of promoting replacing all Poisson brackets $$ \{A, B\} = \frac{\partial A}{\partial q} \frac{\partial B}{\partial p} - \frac{\partial B}{\partial q} \frac{\partial A}{\partial p} $$ by the corresponding commutators $$ -\frac{i}{\hbar} [\hat{A}, \hat{B}] = -\frac{i}{\hbar} (\hat{A} \hat{B} - \hat{B} \hat{A}). $$ In, particular, a classical observable $A$ evolves according to $$ \dot{A} = \{A, H\} $$ and a quantum observable $\hat{A}$ in the Heisenberg picture evolves according to $$ \dot{\hat{A}} = -\frac{i}{\hbar} [\hat{A}, \hat{H}] $$ (an alternative formulation to the Schrödinger equation).

I don't claim to have a good intuition for what this correspondence "means", but at least I can comment on your point on the type signature. As QM observables are Hermitian, they are completely characterised by their eigenvalues and eigenvectors. So you could see it as a mapping assigning each eigenvector a value, the eigenvalue. Representing this function as an operator is nice as it allows you to apply the function to all parts of a superposition simultaneously.

For the mathematics of how canonical quantisation works (and why the naive version doesn't work), you should probably look into deformation quantisation or the like. Seems interesting but I don't know any details. It might illuminate your question of where the derivatives come from. (Let me know if you figure it out! I am also puzzled.)

$\endgroup$
2
  • 1
    $\begingroup$ "As QM observables are Hermitian, they are completely characterised by their eigenvalues and eigenvectors. So you could see it as a mapping assigning each eigenvector a value, the eigenvalue" THIS is starting to clarify things for me. $\endgroup$
    – user56834
    Feb 10, 2020 at 6:48
  • 1
    $\begingroup$ "Let me know if ...". The derivatives, indeed, come from a very technical place, the Wigner map. They induce translation operations on the quantum states and the quantum operators. I could be wrong, but I cannot imagine there is a glib summary bypassing the math adequately. $\endgroup$ Feb 10, 2020 at 15:36
2
$\begingroup$

As far as I know, the classical and quantum Hamiltonian are related because of the Hamilton-Jacobi equation. The Hamilton Jacobi equation is a classical equation of motion that uses the Hamiltonian itself, and it reads $$\frac{\partial S}{\partial t}+H=0$$, and $S$ is termed Hamilton's principal function, together with the equation $p_j=\frac{\partial S}{\partial q_j}$, and there is even an analogue of the time independent Schödinger equation, by inserting $S=-Et+W$, where $W$ is called Hamilton's characteristic function, one gets $H=E$. And in the Schrödinger equation of the form, $$i\hbar\frac{\partial \psi}{\partial t}=-\frac{\hbar^2}{2m} \nabla \cdot \nabla \psi +U\psi$$, if one inserts $\psi=\sqrt{\rho}e^{\frac{iS}{\hbar}}$, and substitutes $\hbar \rightarrow 0$ it becomes the following form of the Hamilton-Jacobi equation: $$\frac{\partial S}{\partial t}+\frac{|\nabla S|^2}{2m}+U=0 $$ Another reason why they have the same name is because of their relation with energy. The QM Hamiltonian is of the form $$\hat{H}=-\frac{\hbar^2}{2m}\nabla\cdot\nabla+U$$, while the CM Hamiltonian is of the form $$H=\frac{|\mathbf{p}|^2}{2m}+U$$

Also, it is worth mentioning that one can derive the QM Hamiltonian from CM Hamiltonian by substituting dynamical variables with operators.

In short, their name is because of analogy between them. Hope this helps.

$\endgroup$
2
  • $\begingroup$ But isn't there still a big difference in that $S$ is a variational object (in the sense that it is an Action), whereas the quantum state is not? $\endgroup$
    – user56834
    Feb 10, 2020 at 9:05
  • $\begingroup$ @user56834 Well, there's a difference between the $S$ of the principle of least action and the $S$ in the HJE. The main difference is that the former is a functional, whereas the latter is a function. One can be developed completely without the other. They're of course related via $\mathcal{L}=\frac{\mathrm{d} S}{\mathrm{d} t}$. Unfortunately, they have the same customary symbol. $\endgroup$
    – Don Al
    Feb 10, 2020 at 22:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.