How does one know if two variables are conjugate pairs?

Question

First of all, I am having a hard time finding any good definition of what a conjugate pair actually is in terms of physical variables, and yet I have read a number of different things which use the fact that two variables are a conjugate pair to justify things.

For instance, every once in a while I'll see something which will say something along the line of, "we know the commutator of these variables is non-zero because they are a conjugate pair."

Also, I have gathered that position and momentum are a conjugate pair as are energy and time. I have seen once that the imaginary and real components of an electric field are a conjugate pair.

So, it seems that if I took two variables and their quantum mechanical operators do not commute, they are a conjugate pair. This doesn't feel quite right to me though... Or at least that's a lame definition cause it doesn't speak to classical anything.

So, that brings me to two questions:

1. How does one define a conjugate pair of variables?

2. What important roles do conjugate pairs play in both classical and quantum mechanics?

Answer 1

Let there be given a symplectic manifold $(M,\omega)$ and two functions $f,g: M \to \mathbb{R}$.

Definition. The two functions $f$ and $g$ constitute a canonical pair if there exists an atlas of local canonical/Darboux coordinate functions $$(q^1, \ldots, q^n, p_1, \ldots p_n):U\to \mathbb{R}^{2n}, $$ where $U\subseteq M$ is an open subset, such that the local restrictions satisfy $$ f|_U~=~q^1 \quad \text{and}\quad g|_U~=~p_1 .$$

Answer 2

In terms of physical variables, conjugate pairs come from Lagrangian mechanics. If you have a coordinate x, you can get the associated momentum, p from the Lagrangian function, L, which is the difference of the Kinetic energy, T, minus the potential energy V. e.g. L = T-V. Then the momentum is given by the partial derivative of the Lagrangian with respect to the velocity, dx/dt. If one makes a "canonical" transformation from the coordinates x, and p to different coordinate Q and P, then Q and P are similarly related to the new Lagrangian.

Such canonical pairs are often referenced when constructing quantum hamiltonians using the correspondence principle to argue that the same pairs from Lagrangian mechanics must also be canonically conjugate in quantum mechanics, and therefore should have the same commutation relations as x and p do.

A good discussion is here: https://en.wikipedia.org/wiki/Canonical_transformation

Answer 3

From the wikipedia definition, two variables are Conjugate if one is the fourier transform of the other. What does this mean? well, you will probaly know that the fourier transform of a gaussian function with $\sigma$ is another gaussian with standart deviation $1/\sigma$.

This means that if we try to measure one of the variables in a system which give us a high precision in one of them, the other will have great uncertainty. A great example of wikipedia is time and frequency of a sound wave. Lets say you wanna measure the timing center of the sound wave of the sound and the frecuency. You can obtain the frecuency of a very long wave easily, but the error at obtining the time will be huge. The oposite goes arround for a very short and sharp sound. This happens a lot in many other classical systems. The drill is that in QM, any two variable which are conjugated will show a similar property. If you have two not commutative operators, such as $p=-i\hbar\delta x$ and $x$, that dont commute, you could derive the Uncertainty principle: $\sigma_x\sigma_p>\hbar/2$, and again if you are able to measure one of them with very high precision the other will have a huge error. It's all about gaussians

Answer 4

I believe a common usage of the term "conjugate pair" just means that two dynamical variables have a Poisson bracket of $\pm 1$.