I found an excerpt on page 171 of "The variational principles of mechanics" written by Cornelius Lanczos stated that

If, however, the conjugate variables $q_k$, $p_k$ are replaced by the complex variables $$\frac{q_k+ip_k}{\sqrt{2}}=u_k$$ and $$\frac{q_k-ip_k}{\sqrt{2}}=u_k^*,$$ then the double set of canonical equations $$ \dot q_i=\frac{\partial H}{\partial p_i}, \qquad\dot p_i=-\frac{\partial H}{\partial q_i}$$ can be replace by the following single set of complex equations $$\frac{du_k}{idt}=-\frac{d H}{d u_k^*}.$$

I did not understand that why shall we introduce the complex conjugate of $u_k$ and the coefficient $\sqrt 2$. It seems to me that the equation $$\frac{du_k}{idt}=-\frac{d H}{d u_k}$$ with $$u_k=q_k+ip_k$$ can already fulfill the task. Could you please tell me why my thinking is wrong?

P.S. I found another link (Relationship between complex analysis and Hamilton's canonical equation) related to this problem but not exactly the same.

  1. We need the complex conjugated variable $\bar{u}$ because the Hamiltonian $H(u,\bar{u})$ is typically not holomorphic in $u$.

  2. The normalization of the CCR$^1$
    $$\{u,\bar{u}\} ~=~-i \tag{1}$$ explains the $\sqrt{2}$-normalization in $$ u=\frac{q+ip}{\sqrt{2}}.\tag{2}$$

  3. Recall that Hamilton's equations can be written in terms of the Poisson bracket $$\dot{u}~=~\{ u, H\}~=~\{ u, u\}\partial H + \{ u, \bar{u}\}\bar{\partial} H. \tag{3}$$ In the last equality in eq. (3) we used that the Poisson bracket $\{ \cdot, \cdot\}$ is a first-order differential operator wrt. its second entry.

  4. We can now derive the complex form of Hamilton's equations $$\dot{u} ~\stackrel{(1)+(3)}{=}~-i\bar{\partial}H.\tag{4}$$


$^1$ Quantum mechanically, the Poisson bracket (1) should be replaced with $(i\hbar)^{-1}$ times the corresponding commutator
$$[\hat{u},\hat{u}^{\dagger}] ~=~\hbar \hat{\mathbb{1}}, \tag{5}$$ in accordance with the correspondence principle between classical and quantum physics. Here the operator $\hat{u}$ and the Hermitian adjoint operator $\hat{u}^{\dagger}$ are an annihilation and a creation operator, respectively.

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  • $\begingroup$ Is there really an interest in writing the Hamilton equations in complex form? What gives? $\endgroup$ – Cham Aug 8 '19 at 2:13
  • $\begingroup$ Yes. The complex Hamiltonian formulation is used in e.g. coherent state path integrals. $\endgroup$ – Qmechanic Aug 8 '19 at 5:48

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