# Can one canonical conjugate variable be considered to be the “frequency” of the other one? (which could be a “wavelength”)?

So, from http://en.wikipedia.org/wiki/Conjugate_variables#Derivatives_of_action, we have...

• The energy of a particle at a certain event is the negative of the derivative of the action along a trajectory of that particle ending at that event with respect to the time of the event.

• The linear momentum of a particle is the derivative of its action with respect to its position.

• The angular momentum of a particle is the derivative of its action with respect to its angle (angular position).

• The electric potential (φ, voltage) at an event is the negative of the derivative of the action of the electromagnetic field with respect to the density of (free) electric charge at that event.

• The magnetic potential (A) at an event is the derivative of the action of the electromagnetic field with respect to the density of (free) electric current at that event. The electric field (E) at an event is the derivative of the action of the electromagnetic field with respect to the electric polarization density at that event.

• The magnetic induction (B) at an event is the derivative of the action of the electromagnetic field with respect to the magnetization at that event.

• The Newtonian gravitational potential at an event is the negative of the derivative of the action of the Newtonian gravitation field with respect to the mass density at that event.

We know that one canonical conjugate variable can be Fourier transformed into its dual. So all properties of Fourier transforms apply.

It also appears that "momentum" can be considered to be the "frequency" of position. Does this analogy also apply to each and every once of these cases above?

In that the energy of a particle is the "frequency" of its trajectory? Or that the electric potential is the "frequency" of the density of free electric charge?

Or that the electric field is the "frequency" of the electric polarization density? And that the magnetic induction is the "frequency" of the magnetization?

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## 1 Answer

The answer for all the cases mentioned in the question is positive: There is a Fourier transform which intertwines the conjugate variables, but this is not true in general.

To be specific I'll elaborate the particle (in one dimension) case, but this elaboration can be generalized for the other cases as well.

In the particle case, we know how to construct a canonical transformation that exchanges the position and the momentum.

$F(q, Q) = qQ$

The momenta are given by:

$p = \frac{\partial F}{\partial q}=Q$

$P = -\frac{\partial F}{\partial Q}=-q$

The time independent Schrödinger equation in the first set of variables corresponding to a Hamiltonian \$H(p,q) is:

$H(i \hbar\frac{\partial}{\partial q}, q) \psi(q) = E \psi(q)$

while in the second set of variables, it is:

$H(Q, -i \hbar\frac{\partial}{\partial Q}) \phi(Q) = E \phi(Q)$

Now, it is easy to verify that the Fourier transform:

$\phi(Q) = \int exp(\frac{i}{\hbar} F(q, Q)) \psi(q) dq$

intertwines between the energy eigenfunction of the two Hamiltonians, i.e., if $\psi(q)$ is an eigenfunction of the first Hamiltonian with an energe $E$, then $\phi(Q)$ will be an eigenfunction of the second Hamiltonian with the same energy.

Now, is the above description general for any two classicaly conjugate variables and any canonical transformation, the answer is no. In general we will have

$\phi(Q) = \int exp(\frac{i}{\hbar} F_{quant}(q, Q)) \psi(q) dq$,

where $F_{quant}(q, Q)$ ia a "Quantum canonical transformation", which can be expanded in powers of $\hbar$, such that the leading term is the "classical" canonical transformation:

$F_{quant}(q, Q) = F(q, Q) + \hbar F_1(q, Q) +\hbar^2 F_2(q, Q) + . . .$

This equation essentially constitutes the basis of deformation quantization.

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