# Why/Do Hamilton's equations Hold with Complex Variables?

I am investigating the problem of taking a hamiltonian of bilinear terms, and converting them into a bunch of uncoupled oscillators, such as in a periodic lattice. To do this, you have to introduce complex coordinates. My question is: why does the below still hold true?

Normally the Lagrange equations are derived using "real partial" derivatives, but of course the partial derivative with respect to a complex variable (like below) is defined as: 1/2( d/dx - i d/dy). I only have a very basic familiarity with the Hamiltonian-Lagrange formalism, but I guess my question is, is it true that the Hamiltonian equations can be extended to complex conjugate coordinates as seen below? If so, why? This seems to be a topic that is essentially nonexistent in the literature, and everywhere I go to investigate this topic of decoupled oscillators takes it as a given. • I think actually, as a potential answer to my own question: When you are taking a complex partial derivative, you are technically taking the derivative of both the real and imaginary part. So these are the actual variables that need to be minimized in the Lagrange equation. 71 is essentially two equations in 1. – Electric to be May 2 at 3:57

The Hamiltian formalism works with a set of N coordinates. Since complex-conjugate coordinates $$Q_s^*$$ are independent from the $$Q_s$$, the Hamiltonian formalism for N complex coordinates is the same as if you had 2N real coordinates. For example, the harmonic oscillator equation (71) can be written as two independent equation for $$Q_s$$ and $$Q_s^*$$.
• Well, if real and immaginary parts of $Q$ are independent, and $Q$ and $Q*$ are two different combination, they must be independent as well right? Regarding the condition $Q(-k) = Q^*(k)$ I think it is related to your particolar problem only. If it bothers you, you can remove the 1/2 and sum on positive $k$ only. This is just a relabelling of the indeces, and you obtain the independence you wanted ($\frac{d Q Q^*}{dQ*} = Q$) – Carlo Cepollaro May 2 at 8:47
• I have always intended the independence of $z$ and $z^*$ as the independence of (x,y) and (x,-y): it is true that given (x,y) the vector (x,-y) is fixed, nevertheless they are two independent vectors meaning that their linear combination is zero only if all the coefficients are zero. Their independence is on the real vector space with 2N components, and not on the N complex vector space. Can it be the source of the problem? – Carlo Cepollaro May 2 at 20:40