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.

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

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.