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Writing the relationship between canonical momenta $\pi _i$ and canonical coordinates $x_i$

$$\pi _i =\text{ }\frac{\partial \mathcal{L}}{\partial \left(\frac{\partial x_i}{\partial t}\right)}$$

Then using the Lagrangian density for classical electrodynamics

$$\mathcal{L} = \frac{1}{2}\left(\epsilon _0E^2- \frac{1}{\mu _0}B^2\right)- \phi \rho _{\text{free}} + A\cdot J_{\text{free}} + \mathbb{E}\cdot \mathbb{P} + \mathbb{B}\cdot \mathbb{M}$$

Q1: Does it make sense to substitute the canonical scalar field $\phi$ (or electric scalar potential) for $\pi _i$ ; and $\phi \rho _{\text{free}}$ for canonical coordinates $x_i$. Such that these satisfy Heisenberg's Uncertainty relation?

Q2: Is there a generalized method for taking a Lagrangian and deriving conjugate variables, which then satisfy Heisenberg's Uncertainty relation?

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3 Answers 3

The fields in the electromagnetic Lagrangian satisfy the uncertainty principle with their conjugate momenta, but the result is a distributional uncertainty principle as appropriate for the distributional quantum fields.

From the Lagrangian, the canonical conjugate momentum to the field $A$ is the electric field $E$. If you take a gauge choice where A is purely spatial ($\phi=0$ gauge), you get the Bohr-Rosenfeld uncertainty relation between a measurement of A and a measurement of E in a region.

The commutation relation for the quantum field is

$$ [A_i(x),E_j(y)] = \delta(x-y)\delta_{ij}$$

And the resulting uncertainty principle comes from smearing both fields using test functions. Define

$$ A^f_i = \int f(x) A_i(x) dx $$ $$ E^g_j = \int g(y) E_j(y) dy$$

Where f,g are positive $C^\infty$ bump functions, then

$$ [A^f, E^g] = \delta_{ij} \int dx f(x)g(x)$$

This leads to a field theoretic uncertainty principle

$$ \delta A_i^f \delta E_i^g \ge (\int dx f(x)g(x))$$

This uncertainty principle demands that the uncertainty products diverge in a particular way as the region becomes small, and Bohr and Roesenfeld verify that the physical uncertainty must be present when you use small charged quantum pith-balls to measure the value of E and A as best you can. This is automatic from a Lagrangian formulation of quantum mechanics, since you derive the commutation relation from the Lagrangian in a bosonic path integral. So this is one of those things that became trivial after Feynman's Lagrangian formulation became standard. It was controversial in Bohr and Rosenfeld's time, because field quantization made photons.

For a general bosonic Lagrangian, the canonical commutators

$$ [\phi(x),\pi(y)] = \delta(x-y) $$

Where $\pi(x) = {\partial L\over \partial \dot{\phi} }$ leads to an uncertainty relation between $\phi$ and $\pi$ in the exact same way. These field uncertainty relations are in every way analogous to the usual momentum-position uncertainties in ordinary quantum mechanics.

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I think this is the right method, but I'm still tracking down the Bohr-Rosenfeld uncertainty relation and trying to understand. Thanks Ron. Just so you know, I'm looking for a generalized method for taking a Lagrangian and deriving conjugate variables, which then satisfy Heisenberg's Uncertainty relation? –  metzgeer Apr 28 '12 at 12:01
    
@metzgeer: When the Lagrangian is quadratic in momentum, these are the path integration variables and the classical conjugate momenta. The Bohr-Rosenfeld paper is from 1933, and is a perfectly clear classic. –  Ron Maimon Apr 28 '12 at 14:27

It doesn't really make sense to do this, for a few reasons: first, classical field theory has no concept of an uncertainty. It's not even clear what $\sigma_\phi$ and $\sigma_\pi$ would mean in that case. Besides, these quantities are necessarily going to commute. The uncertainty principle relates the product of the variances to the commutator, so the only result you'll get is that $\sigma_\phi\sigma_\pi \ge 0$. It's completely trivial. (If you're dealing with matrix-valued classical fields, then I guess you could get a nonzero commutator, but you still have the issue of why you're computing the $\sigma$s.)

It might make more sense in quantum field theory, where you could try to apply the generalized uncertainty relation,

$$\sigma_{\phi_i}\sigma_{\pi^{j\mu}} \ge \left|\frac{1}{2i}\langle[\phi_i,\pi^{j\mu}]\rangle\right|$$

There's a problem, though: the thing on the right (the commutator) is a field, which has a potentially different value at every point in spacetime, whereas the thing on the left is just a number. So this doesn't really work. It's only in basic quantum mechanics (or for only special operators), where the operator fields $\phi$ and $\pi$ are constant over all of space, that you can ignore the position dependence on the right.

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Why are you assuming the question is about classical fields? The OP didn't say this. –  Ron Maimon Apr 28 '12 at 2:03
    
Yes, he did say it was about classical electrodynamics. –  David Z Apr 28 '12 at 2:31
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He just said he is using the Lagrangian for classical EM, this doesn't mean he isn't doing quantum mechanics, since the quantum Lagrangian is the same as the classical Lagrangian. The question makes no sense outside of QM. –  Ron Maimon Apr 28 '12 at 3:33
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@metzgeer: You can't avoid the vector potential in EM--- the E and B formulation doesn't work because the phase change for a particle path is by the integral of A along the path. The gauge invariance is a headache that must be dealt with. If you want a source-free E,B formulation, you can do that, but it requires changing the action for a different path integral variables. The path-integral variable for the standard action is A, and you need to gauge fix it to make it correct (but you can do this in several ways without learning too much). The path integral is necessary for doing this right. –  Ron Maimon Apr 28 '12 at 6:54
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@metzgeer: The "generalized uncertainty" is just canonical commutators for a general Lagrangian. For Path-Integral systems (Lagrangians at most quadratic in p) this is given by the classical fields (Path Integral variables) and their conjugate momenta. The derivation of the canonical commutation is on Wikipedia's Path Integral page, and it implies the uncertainty. The classical theory has unit Poisson brackets between any two canonical pair, but quantumly the commutators get corrections. –  Ron Maimon Apr 28 '12 at 14:31

It seems that OP's generalized Heisenberg uncertainty principle as formulated in the second subquestion(v5) is nothing but canonical quantization of an arbitrary classical Lagrangian field theory.

Formally, at the classical level, this involves performing a Dirac-Bergmann analysis to obtain a Hamiltonian formulation via a (not necessarily regular) Legendre transformation. This may$^{\dagger}$ introduce first class and second class constraints. In case of first class constraints, this indicates that the system possesses gauge symmetry. In case of second class constraints, the canonical Poisson bracket should be replace with the Dirac bracket.

Quantum mechanically, at the leading order in $\hbar$, the bracket should be replaced with $(i\hbar)^{-1}$ times the corresponding commutator, in accordance with the quantum mechanical correspondence principle, thereby leading to (generalized) Heisenberg uncertainty relations. At subleading orders in $\hbar$, the issue of operator ordering ambiguities arises.

Related topics are geometric and deformation quantization.

Reference:

  1. M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1994.

--

$^{\dagger}$ E.g., in the Hamiltonian formulation of classical electrodynamics, Gauss' law becomes a first class constraint.

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For the special case of EM (as opposed to non-Abelian gauge fields) Henneaux and Teitelboim is overkill. The methods there are a little annoying to learn, because they don't provide the path integral dictionary, you have to make it yourself, and translating first and second class constraints to Path Integral is something that isn't presented in the literature. The generalized uncertainty relation is the distributional one I presented, and it depends on the size of the smearing function, requiring distributional fields. This can be understood in a simple gauge, like Dirac's. –  Ron Maimon Apr 30 '12 at 4:59

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