Hamiltonian mechanics and the electric field

[snoopies622]

The dimensions of action divided by the dimensions of electric field strength are distance x time x charge.

Does this mean that distance x time x charge - whatever one might call that - is the "conjugate momentum" of an electric field?

If so - is there any physical significance to this? Does it manifest itself in quantum field theory?

 

[snoopies622]

Does every measurable quantity have a corresponding "conjugate momentum"?

 

[snoopies622]

What got me wondering about this was entry #5 of this thread

https://www.physicsforums.com/printthread.php?t=237356

which includes the following sentence,

"This means, for example, that the value of the electrostatic field in the vicinity of an electron can no longer be specified with complete certainty, as suggested by merely using Coulomb's law to infer it (of course, one could do this at the expense of completely losing information of the field's velocity)."

(corrections and emphasis added)

Distance x time x charge doesn't seem like the velocity of anything to me, yet it is the other half of that uncertainty relation, at least according to dimensional analysis.

Any thoughts about this?

 

[TriTertButoxy]

Every time-dependent dynamical quantity would have a corresponding conjugate momentum (no need to put in quotations, as this is exactly true). For the electromagnetic field, the dynamical quantity is the vector potential, which certainly has a conjugate momentum.

When passing over to quantum field theory, there we get an uncertainty principle among the dynamical quantity and its momentum conjugate.

 

[snoopies622]

Is the conjugate momentum of the electromagnetic vector potential expressible in terms of E and B? I can't make the dimensions work out using [itex] \epsilon_o [/itex] and [itex] \mu_o [/itex].

 

[snoopies622]

I'm still confused about the quotation I cited in entry #3. It suggests that electrical field velocity is the conjugate momentum corresponding to electrical field strength. But isn't the product of any dynamical quantity and its corresponding conjugate momentum equal to some quantity of action? In this case they make

( force / charge ) x (force / (charge . time))

which not only isn't action (force x distance x time), but - unless I made a math error - isn't action multiplied by some mix of [itex] \epsilon_o [/itex]'s and [itex] \mu_o[/itex]'s, either.

 

[snoopies622]

electric field conjugate

I asked a question very similar to this one in the classical section a few weeks ago

https://www.physicsforums.com/printthread.php?p=3060252#

but it didn't take, so I'm putting a quantum version of it here.

How does the uncertainty principle apply to electric field strength? That is, the more precisely one can know the magnitude of an electric field in a certain direction at a certain place and time, the less precisely one can know...what?

I have a couple follow-ups in mind, but I'd rather wait to get an answer to this one first.

Thanks.

 

[ZapperZ]

Threads merged. Please do not create multiple threads on similar topics.

Zz.

 

[snoopies622]

Anybody? Are these bad questions?

 

[tom.stoer]

In order to study (Q)ED in the canonical framework you have to extend this framework to treat constrained systems (a la Dirac).

First of all a canonical momentum is always derived via derivation w.r.t. to a time derivative. Setting A°=0 (which is called Weyl gauge or temporal gauge, allowed by gauge symmetry) one finds that the canonical momentum to A(x) is just E(x) (spatial components) - and that there is no canonical momentum for A° as there is no time derivative acting on A°. That's why A° is not a dynamical variable but a Lagrange multiplier generating an equation of constraint, the so-called Gauss law, which acts as a generator of a canonical transformation (classically) or a unitary transformation (in QED) for gauge time-independent gauge transformation respecting A°=A'°=0.

The treatment of contsrained systems whith local gauge invariance is rather complicated in quantum field theory. In QED I guess Bjorken-Drell explain how to treat QED in a so-called physical gauge.

 

[snoopies622]

Thanks Tom, that was helpful.

Part of my confusion comes from the fact that I've seen the term "canonical conjugate" defined in different ways, and I don't know if they're all equivalent.

One is from classical mechanics, where p is the conjugate of q because

[tex]

p = \frac {\partial L}{\partial \dot {q} }

[/tex]

This evidently requires the Lagrangian to be expressible in terms of q and [itex] \dot {q} [/itex], and the product of the p and q is always action.

Another is a purely mathematical definition, where one variable results from a Fourier transform of another, like time and frequency.

And the other is a quantum mechanical one, where if the commutator of two operators is [itex] i \hbar [/tex], they are canonical conjugates.

Are these really all the same thing?

 

[tom.stoer]

They are related.

In field theory the coordinate x is no longer the canonical variable, but simply an index. So the canonical conjugate variables in QED are

[tex]A_\mu(x),\;\Pi^\mu(x) = \frac{\delta S}{\delta (\partial_0 A_\mu(x))}[/tex]

In the Weyl gauge you directly find the electric field for the i-component and you find Zero for A° as there is no time derivative of A° b/c the field strength tensor is by definition antisymmetric and has vanishing diagonal elements.

Now the relation between canonical variables is defined in terms of Poisson brackets in the classical Hamiltonian formulation:

[tex]q,\;p=\frac{\partial L}{\partial (\partial_0 q}[/tex]

and

[tex]\left\{q, p\right\} = 1[/tex]

The Poisson bracket for arbitrary functions on phase space depending on p and q is defined as

[tex]\left\{f(q,p), g(q,p)\right\} = \frac{\partial f(q,p)}{\partial q}\frac{\partial g(q,p)}{\partial p} - \frac{\partial g(q,p)}{\partial q}\frac{\partial f(q,p)}{\partial p}[/tex]

Using f=q and g=p you immediatey find the relation {q,p}=1.

Quantizing this system means replacing q,p with the corresponding operators acting on a Hilbert space, replacing the Poisson bracket with the commutator and inserting an "i". Strictly speaking this step cannot be derived but has to be postulated and justified afterwards (justification is simply: QM is self-consistent and described nature rather well :-)

Once you have a pair of canonical variables you try to find new variables and expressions between them. In field theory one often uses creation and annihilaton operators which are related to the simply operators for the harmonic oscillator. The construction uses a Fourier decomposition of the basic field operators for which the canonical commutations are already known.

The new commutation relations for the new operators can be derived exactly. Let's make a simple example.

Assume we have

[tex]\phi(x),\;\pi(y)[/tex]

and

[tex][\phi(x),\;\pi(y)] = i\delta(x-y)[/tex]

Now we define

[tex]f(k) = \int dx\,\e^{-ikx}\phi(x)[/tex]

[tex]g(k) = \int dx\,\e^{ikx}\pi(x)[/tex]

Now we can calculate the commutation relations for these new operators f and g rather easily

[tex]\left[f(k), g(k^\prime)\right] = \int dx\,e^{-ikx} \int dy\,e^{ik^\prime y}\left[ \phi(x), \pi(y)\right] = \int dx\,dy\,e^{-i(kx - k^\prime y)}\delta(x-y) = \int dx\,e^{-i(k -k^\prime)x} = 2\pi\delta(k - k^\prime)[/tex]

Therefore we have derived the commutation relation for f and g. In the same sense one derives the commutation relations for creation and annihilation operators which requires a little more care - but we are nearly there.

 

[snoopies622]

Wow - thank you!

 

[tom.stoer]

I tried to correct some LaTeX typos

 

[snoopies622]

In field theory the coordinate x is no longer the canonical variable, but simply an index. So the canonical conjugate variables in QED are

[tex]A_\mu(x),\;\Pi^\mu(x) = \frac{\delta S}{ \delta ( \partial_0 A_\mu(x) ) } [/tex]

Goldstein's a good source for going from a discrete set of particles to a mechanical field, no?

edit: I wonder why LaTeX won't make that second right parenthesis in the denominator.
edit 2: ah, there it is.

 

[tom.stoer]

There was a question regarding S in

[tex]A_\mu(x),\;\Pi^\mu(x) = \frac{\delta S}{\delta (\partial_0 A_\mu(x))}[/tex]

S is the action. Formally the functional derivative is defined as

[tex]\frac{\delta (\partial_0 A_\mu(y))}{\delta (\partial_0 A_\nu(y))} = \delta_\mu^\nu \delta(x-y)[/tex]

[tex]\frac{\delta A_\mu(x)}{\delta A_\nu(y)} = \delta_\mu^\nu \delta(x-y)[/tex]

therefore one needs the integral of the action to get rid of the delta functions. The expression using the Lagrangian density instead of S is a sloppy notation.