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Suppose I have a canonical transformation on phase space, which is obtained by evolving a classical Hamiltonian system from time $t=0$ to $t=T$, with some arbitrary time-dependent Hamiltonian $H(t)$. That is, it's a finite canonical transformation which is a sum of infinitesimal canonical transformations. Then, can I always find some time-independent effective Hamiltonian $H_{eff}$ such that if I evolve the system from $t=0$ to $t=T$ under $H_{eff}$, this evolution gives me the original canonical transformation? Note that I don't require that the dynamics under $H(t)$ and $H_{eff}$ produce the same canonical transformation for all $t$, but only for some fixed $t=T$.

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This can be trivially achieved, if you are willing to go to extended phase space, that is add an extra degree of freedom to your system, represented by the momentum - position pair $(p_e, q_e)$.

Then the dynamic system evolves under the effective Hamiltonian $$ H_\text{eff} = p_e + H(\mathbf{p}, \mathbf{q}, q_e) $$

The effective Hamiltonian $H_{\text{eff}}$ is time independent, i.e. autonomous, and an integral of motion.

If this does not satisfy you, and your question is about effective Hamiltonians in the original, restricted phase space, for arbitrarily large times $T\rightarrow 0$, the answer is that, in general there can be no such effective Hamiltonian. This can be demonstrated by a simple example.

Consider an autonomous Hamiltonian system with one degree of freedom evolving under the Hamiltonian $H_0(p,q)$. The Hamiltonian is conserved, therefore the system is integrable. The orbits lie on contours given by $H_0 = \text{const.}$. Suppose now that this system is perturbed by a time dependent perturbation $H_1(p,q,t)$, so that $$ H(p, q, t) = H_0(p,q) + H_1(p,q,t). $$

The question then becomes, whether there can be an autonomous Hamiltonian of one degree of freedom $H_\text{eff}(p,q)$ that describes the motion under the non autonomous Hamiltonian $H(p, q, t)$.

It is a well known fact that in general, non autonomous systems exhibit chaotic behaviour, when
(see Poincare-Melnikov theory, for example). The perturbed Hamiltonian is no longer conserved and in general there are orbits that cover densely finite areas of phase space. Such orbits cannot be described as contour levels of some well behaved function $H_\text{eff}$, therefore no autonomous effective Hamiltonian exists.

You may weaken your requirements and ask for an effective Hamiltonian that exactly describes the motion only for a finite, possibly small, time $T$, after which it becomes invalid. In other words, the question becomes whether the non autonomous Hamitonian $H$ can be expressed as a series $$ H(p, q, t) = \sum_n H_{\text{eff}, n}(p, q) U_n(t) $$ where $$ U_n = \begin{cases} &1,\quad \text{for}\ n T \leq t < (n+1) T \\ &0,\quad \text{otherwise}. \end{cases} $$ I am not aware of any theorem that says that this is impossible, but even if it is not, in my opinion it would be impractical to calculate, in most cases.

On the other hand, if you are interested in finding effective Hamiltonians that approximately describe the motion for finite, possibly small, times $T$, for perturbed autonomous Hamiltonians, then you should delve into the field of Hamiltonian mappings. This is still an active field of research. Unfortunately, no general techniques are known and even the most common ones would be hard to summarise here.

For more information, see

Abdullaev, S. S. Construction of mappings for Hamiltonian systems and their applications Springer, 2006

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    $\begingroup$ I would be very interested to know if it is possible for one fixed value of $T$. It is intuitively clear that we can not have one autonomous Hamilton function giving the correct time evolution at all times $t$. But in the question, it is only required that the final system state at $t=T$ agrees with the final system state of the original time evolution. $\endgroup$
    – Noiralef
    Feb 17, 2020 at 12:17
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    $\begingroup$ In other words, in Quantum Mechanics, every time evolution is a unitary operator $U$. As shown in the answer by Clara Diaz Sanchez, we can find a constant generator that yields the correct evolution over the time $T$ simply by taking the logarithm of $U$. Is there an analogous way to "take the logarithm" of a canonical transformation in Classical Mechanics? $\endgroup$
    – Noiralef
    Feb 17, 2020 at 12:23
  • $\begingroup$ Thank you @zap for the detailed answer! I'm asking for a weaker version of what you've described, along the lines of Noiralef's comment. That is, I don't even mind if the evolution due to the effective Hamiltonian diverges from the exact driven evolution for times between t=0 and t=T, as long as t=T the two finite canonical transformations are equal to one another. $\endgroup$ Feb 17, 2020 at 17:57
  • $\begingroup$ Oh! I am sorry I misunderstood. This question is even more interesting than I thought at first. I am not sure about the answer and the more I think of it the more my intuition vacillates between "probably yes" and "definately not". When I find the time, I will give it some serious thought $\endgroup$
    – zap
    Feb 18, 2020 at 16:02
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Maybe this is too naive a picture, but under your time-dependent Hamiltonian you can define a unitary time-evolution operator:

$U(T,0) = {\cal T} \left( e^{-i \int_0^T H(t') dt'} \right)$

where $\cal T$ signifies the time-ordering operator. As $U$ is a unitary operator, you can express it in terms of a hermitian operator $H_{\mathrm eff}$ as

$U(T,0) = e^{-i T H_{\mathrm eff}}$.

This effective Hamiltonian will not be unique as you can add arbitrary multiples of $2 \pi/T$ to its spectrum, but it should exist. If your Hamiltonian were $T$-periodic, this construction would correspond to employing the Floquet formalism, with the eigenvlaues of $H_{\mathrm eff}$ being the Floquet quasienergies.

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  • $\begingroup$ Ah, I should've clarified that I'm thinking of classical systems. I've edited the question to specify that. But my question is inspired by the Floquet formalism! Essentially, I'm asking about the classical version of the Floquet formalism, where the time evolution operator U evolves probability distributions on phase space, instead of vectors in Hilbert space. $\endgroup$ Feb 8, 2020 at 17:18
  • $\begingroup$ $U(T,0)$ is a unitary operator, and so I can write it as $U = e^{i P}$ where $P$ is hermitian. I think there is nothing to stop me taking $P$ to be an effective Hamiltonian, although it may indeed have odd non-linear properties. This is rather tangential to the OP's question though, as he is just considering classical systems (as he made clear after I made my first comment). $\endgroup$ Feb 17, 2020 at 16:22
  • $\begingroup$ Can't help wondering why this was downvoted. $\endgroup$ Mar 29, 2020 at 14:05

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