I Canonical transformation from canonical to kinetic momentum

AI Thread Summary
The discussion centers on the feasibility of replacing canonical momentum with kinetic momentum through a canonical transformation in a Hamiltonian framework involving electromagnetic fields. The original Hamiltonian is analyzed, revealing that the transformation leads to a loss of the Lorentz force, attributed to a conceptual error regarding dimensionality. The participants clarify that in one spatial dimension, the potential can be eliminated, but this does not hold in three dimensions where path dependence of the vector potential A is crucial. The conversation also touches on implications for quantum mechanics, particularly regarding the multivaluedness of integrals and the Aharonov-Bohm effect. The conclusion emphasizes the importance of considering path integrals and gauge choices in these transformations.
DrDu
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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!
 
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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.
 
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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.
 
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.
 
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.
 
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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.
 
DrDu said:
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.
 
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.
 
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