I thought that Hamilton's equations could be written like this:


By $q$, I mean the complex function $x+ip$

By $\hat{H}$, I mean the complex function $$\frac{\partial H}{\partial x}+i\frac{\partial H}{\partial p}$$ , where $H$ is the Hamiltonian.

Also, we can have two state functions $j(x,p)$ and $k(x,p)$, such that the function $c(x+ip)=c(q)=j(x,p)+ik(x,p)$ is complex differentiable

Then, $\frac{dc}{dt}=\frac{dc}{dq}\frac{dq}{dt}=-i\frac{dc}{dq}\hat{H}$

The first equation that I wrote looks a lot like the Schrodinger's equation. Is there any connection?

EDIT- Instead of defining $\hat{H}$ to be a sort-of complex gradient of $H$, we can also define a function $F(q)=H(q)+iG(q)$, where $G(q)$ is the Harmonic conjugate of $H(q)$. Then the equation would be $i\frac{dq}{dt}=conjugate(F'(q))$

  • $\begingroup$ The problem with comparing to the TDSE is your RHS is a gradient rather than a Laplacian. The fact $q$ is dependent on the LHS but independent on the RHS adds a further dissimilarity. $\endgroup$
    – J.G.
    Oct 30, 2021 at 14:07
  • $\begingroup$ Related: physics.stackexchange.com/q/495709/2451 $\endgroup$
    – Qmechanic
    Oct 30, 2021 at 17:16
  • $\begingroup$ @J.G. We could change the gradient into a complex derivative of $F(q)$ wrt $q$ by defining a complex differentiable function $F(q)=H(x,p)+iG(x,p)$, where $H(x,p)$ is the Hamiltonian $\endgroup$
    – Rain Deer
    Nov 1, 2021 at 3:10
  • $\begingroup$ @RainDeer That doesn't address my points. The equations you want to compare are so dissimilar it makes no sense to ask whether they have a connection. $\endgroup$
    – J.G.
    Nov 1, 2021 at 7:34
  • $\begingroup$ @J.G. Why are you so rude? Isn't the gradient just the square root of the Laplacian? And it's well-known that Hamilton's equations have a deep connection with QM. I was just wondering if writing the equations this way makes the connection with QM more straightforward. $\endgroup$
    – Rain Deer
    Nov 1, 2021 at 8:42

2 Answers 2


In my opinion, your “complexification” of Hamilton’s equations seems to make cotact with the symplectic geometry formulation of Hamiltonian mechanics. The phase space of a particle on a one-dimensional line, coordinatized by $(x,p)$, can be described in terms of complex variables $a:=(cx+ip)/K$ and $\bar{a}:=(cx-ip)/K$, where $c$ and $K$ are real constants introduced for matching dimensions. In this complexified description, multiplication by $i$ corresponds to “Fourier transform”—changing your Lagrange submanifold. The complex structure is rotating the phase space by $90^\circ$, which is the “unit rotation” generated by the Poisson bivector. So, roughly speaking, your $i$ is a “representation” of the Poisson bivector: $\omega^{\mu\rho}\omega^{\rho\nu} = -\delta^{\mu\nu} \rightsquigarrow ii=-1$. (To be more accurate, we should introduce metric and compose a $(^1_1)$-tensor by combining it with the symplectic form $dp\wedge dx$: $J^\mu{}_\nu = \delta^{\mu\rho} \omega_{\rho\nu} \implies J^\mu{}_\rho J^\rho{}_\nu = -\delta^\mu{}_\nu$.) Also, your “complex gradient” is called the “symplectic gradient” or “Hamiltonian vector fields” in symplectic geometry.

Most notably, this “holomorphic” description is extremely useful and insightful in studying the classical mechanics and the quantum Hilbert space (e.g., Weinberg or Polchinski) of one-dimensional simple harmonic oscillator, namely $c=m\omega$ and $K=\sqrt{2m\hbar\omega}$. In this case, the “$i$” corresponds to $90^\circ$ rotation on the phase space generated by the Hamiltonian flow $\{\,\,\,\,,\hbar\omega\bar{a}a\}$. In this sense, your “complex Hamilton’s equations” do have connections with quantum mechanics, namely Fourier transforming quantum states, quantization in the holomorphic language, coherent states, exploring the classical-quantum correspondence, etc.

If you want more direct connections to Schrodinger equation, I wonder if we can try to associate the $i$ entering in your complex version of $\{\,\,\,\,,\hbar\omega\bar{a}a\}$ and the $i \partial/\partial t$ entering in the time-evolution equation of the wavefunction.

There are also more interesting stories about this. If you are interested, you may want to take a deeper look at “geometric quantization” or quantization on Kahler manifolds.


In your complex rewrite of Hamilton's equations, the LHS has dependent $q$ and independent $t$, while the RHS has dependent $F$ and independent $q$. In the Schrödinger equation, $\psi$ is dependent on both sides. Comparing these is like comparing $y_t=z_y$ to $y_t=y_{xx}$.

Although Hamilton's equations and the TDSE are both equations of motion, they're tricky to compare because the former's phase space coordinates play a different role in $\psi$'s dynamics. To appreciate how and why, it helps to compare the actions $\int L(x,\,\dot{x})dt$ and $\int\left[\frac{i\hbar}{2}(\psi^\ast\psi_t-\psi^\ast_t\psi)-\frac{\hbar^2}{2m}\nabla\psi^\ast\cdot\nabla\psi-\psi^\ast V\psi\right]dtd^3x$; in the latter, $x$ becomes a variable over which we integrate a Lagrangian density.

What we can do, however, is note the classical $iq_t=(F_q)^\ast$ has quantum counterpart $i\langle q\rangle_t=\langle(F_q)^\ast\rangle$. This follows from the TDSE. Similarly, the TDSE implies Newton's second law becomes $\langle p_j\rangle_t=-\langle V_j\rangle$ in Cartesian coordinates.


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