Canonical transformation from canonical to kinetic momentum

[DrDu]

My classical mechanics is very rusty. Recently, I wondered if it is possible to replace the canonical momentum with the kinetic momentum using a canonical transformation (so this isn't homework). I tried to work this out, but somehow, the Lorenz force got lost. Maybe some of you has an idea what goes wrong:
Consider the classical hamiltonian for a particle of mass $m$ and charge $e$ in an electromagnetic field.
Its position is $q(t)$ and it's canonical momentum $p(t)$.
$$
H=\frac{1}{2m}(p-eA(q,t))^2+e\Phi.
$$
The kinetic momentum is $P=p-eA$.
Can we introduce it as new canonical momentum?
We need to find a generating function $G(q,P,t)$ so that
$$
p=\frac{\partial{G}}{\partial q},
$$
$$
Q=\frac{\partial{G}}{\partial P},
$$
and
$$
H'(Q,P)=H+\frac{\partial{G}}{\partial t}
$$

Integrating the first equation, we obtain
$$
G(q,P,t)=\int_0^q (P+eA)\; dq=Pq+e\int_0^q A(q,t)\;dq + f(P,t).
$$
We consider a transformation with $f(P,t)=0$.
From the second equation, the canonical variable to $P$ is
$$
Q = q,
$$
so I will use $q$ furtheron instead of $Q$.
With
$$
\frac{\partial A}{\partial t} =-E -\nabla \Phi.
$$
we obtain from the third equation
$$
H'= \frac{1}{2m}P^2- e\int_0^q E(q,t)dq.
$$
So the transformed hamiltonian is completely free of potentials. The integration paths seems to be arbitrary.

We get the equations of motion
$$
\dot{q}= \frac{\partial H}{\partial P}= \frac{P}{m}
$$
and
$$
\dot{P} = -\frac{\partial H}{\partial q} = eE(q)
$$
somehow, the Lorenz force got lost!

 

[Demystifier]

You made a typo and a big conceptual error. The typo is that the last term in the new Hamiltonian should be $\partial G/\partial t$, not $\partial G/\partial P$. The big conceptual error is that you work with only 1 spatial dimension, while Lorentz force due to magnetic field does not exist in 1 spatial dimension.

Or to say the same thing in a positive way, what you have really shown is that in 1 spatial dimension the potential $A$ can be eliminated by a canonical transformation, which is a correct result.

Indeed, if you start with your initial Hamiltonian and never perform the canonical transformation, you get
$$\dot{q}=\frac{\partial H}{\partial p}=\frac{p-eA}{m}$$
$$\dot{p}=-\frac{\partial H}{\partial q}=\frac{p-eA}{m}e\frac{\partial A}{\partial q} -e\frac{\partial \Phi}{\partial q}$$
Inserting the first equation into the second, and using
$$\frac{\partial \Phi}{\partial q}=-E-\frac{\partial A}{\partial t}$$
you obtain
$$\dot{p}=eE +e\left( \frac{\partial A}{\partial t} + \frac{\partial A}{\partial q} \dot{q} \right) = eE +e\frac{dA}{dt}$$
Hence, taking the time derivative of the first equation you get
$$m\ddot{q}=\dot{p}-e\frac{dA}{dt} = eE$$
showing that there is no magnetic Lorentz force in 1 spatial dimension.

 

[DrDu]

Thank you for your feedback! I corrected the typo. Of course p and q etc. are 3d vectors. But you are right insofar as I naively assumed the derivative of the line integral to be the integrand at the upper end. This is not correct in more than 1 dimension.

 

[Demystifier]

In 3 spatial dimensions you have an integral of the form $\int {\bf A}\cdot d{\bf q}$, which depends on the path of integration. If you choose a path along which electric and magnetic fields are zero, perhaps in this way you can even obtain something that resembles the Aharonov-Bohm effect.

 

[DrDu]

Of course I also had the AB effect in mind. The derivation should be correct for a 1d particle on a ring where $q=\phi$ which describes correctly the acceleration by a changing magnetic flux in the center of the ring. In QM, one would have to clarify how to cope with the multivaluedness of the integral.
Also $\phi$ is not representable as a Hermitian operator due to the same reason.

 

[DrDu]

In QM, one could formally implement this transformation with the unitary $U(q,t)=\exp(ie\int Adq)$. Again, this is hard to interpret if the integral over $q$ is multi-valued.

 

[Demystifier]

In QM, one could formally implement this transformation with the unitary $U(q,t)=\exp(ie\int Adq)$. Again, this is hard to interpret if the integral over $q$ is multi-valued.

Perhaps it helps to integrate over closed loops only, then the canonical variable is not $q$, but an integral over the loop. Something similar is done in loop quantum gravity.

 

[DrDu]

You were right pointing out path dependence. Integration of $p=\nabla_q G$ is only possible if $p$ is rotation free. $p$ is essentially $A$, so we can at best transform away the rotation free part of $A$. This corresponds to a change to Coulomb gauge.