# What are $a$ and $a^*$ called in the context of a classical harmonic oscillator?

Consider a harmonic oscillator defined by the coupled differential equations \begin{align} \begin{split} \dot{X} &= \omega Y \\ \dot{Y} &= - \omega X \, . \end{split} \tag{1} \end{align} Defining new variables $$a = X + i Y$$ and $$a^* = X - i Y$$, produces a new uncoupled system of equations \begin{align} \begin{split} \dot{a} &= - i \omega \, a \\ \dot{a}^* &= i \omega \, a^* \, . \end{split} \tag{2} \end{align}

In classical physics [1] (or just in the mathematical context of this transformation used to solve a pair of coupled differential equations) what are the variables a and $$a^*$$ called?

[1]: In the context of quantum mechanics, the variables $$a$$ an $$a^*$$ would in fact be operators and would be called the "raising" and "lowering" operators.

• Aren't they canonical variables; at least this is the term I was familiar with when decoupling differential equations. Those, though, were generally not complex, so I wonder if the usage of $i$ here changes that? Dec 30, 2019 at 17:41
• I think your equations 3 & 4 are incorrect. Shouldn't they have an $i$ out front with the $\omega$? Dec 30, 2019 at 18:08
• @Geoffrey In the course of getting to the starting form Daniel suggests you multiply the position and momentum by constants so that $X$ and $Y$ have the same units (otherwise the initial expressions would also be dimensionally inconsistent). Alas that confuses the issue of interpretation. Dec 30, 2019 at 21:13
• Daniel, it might help the discussion to exhibit the transformation for some specific system (mass on a spring?), though I admit that risks contaminating the resulting interpreation with the specifics of that particular system. Dec 30, 2019 at 21:15
• A few months ago I learned how to number groups of equations, and I thought it might be useful here. Dec 30, 2019 at 21:30

I would call $$a$$ and $$a^*$$ the complex amplitude of the oscillator. Or I guess $$a$$ is the complex amplitude itself and $$a^*$$ is the complex conjugate of the amplitude but the distinction is unimportant as they carry the same information (just like in the quantum case).
• There is a simple way to figure out what $a$ and $a^*$ are: express the observable things for the oscillator via these "complex amplitudes". To be rigorous - this is the only right way of "seeing" their true meaning. Dec 30, 2019 at 18:07