# Is there a general prescription for commutators?

1. Sep 17, 2010

### pellman

Let a QM system be described in the Heisenberg picture by position variables $$q_j$$ with corresponding conjugate momenta $$p_j$$. We have the equal-time commutators

$$[q_j(t),p_k(t)]=i\hbar \delta_{jk}$$

In quantum field theory, for the Dirac spinor field we have the equal-time commutator

$$\left\{\psi_j(\vec{x},t),\psi^\dagger_k (\vec{x}',t)\right\}=\delta_{jk}\delta^3(\vec{x}-\vec{x}')$$

(subscripts refer to spinor components) Depending on how we scale the Lagrangian, the conjugate momentum to psi is

$$\pi_j=\frac{\partial \mathcal{L}}{\partial \dot{\psi_j}}=i\hbar\psi^\dagger_j$$

So the commutator in terms of the conjugate momentum has a similar form to that its QM counterpart:

$$\left\{\psi_j(\vec{x},t),\pi_k (\vec{x}',t)\right\}=i\hbar \delta_{jk}\delta^3(\vec{x}-\vec{x}')$$

A similar result holds (I think) for the real scalar field:

$$\left[\phi(\vec{x},t),\pi(\vec{x}',t)\right]=i\hbar\delta^3(\vec{x}-\vec{x}')$$

Is this a general rule? Do we always have commutators between a quantity and its conjugate momentum proportional to i*hbar (times the delta function if continuous)? If so, what is the deeper meaning? Is there a principle this can be derived from?

2. Sep 17, 2010

### tom.stoer

The principle you are looking for is "quantization". Take a classical Lagrangian, derive the conjugate momenta p via differentiation w.r.t. to the velocities dq/dt, use the Poisson brackets {q,p} = 1 of classical phase space variables q and p and translate them into commutators |q, p] = i of operators in a Hilbert space.

The last step is the "quantization" which can be motivated but not derived.

3. Sep 17, 2010

### meopemuk

Hi pellman,

There is no direct connection between commutators [position,momentum] and (anti)commutators of quantum fields.

Particle position and momentum are observable quantities, so their commutators must have a physical meaning. The operator of momentum is *defined* as the generator of space translations. This means that the following equations must be true (sorry if I screwed up the signs, but they are not so important)

$$e^{ip_xa}q_xe^{-ip_xa} = q_x + a$$
$$e^{ip_xa}q_ye^{-ip_xa} = q_y$$

...etc.

From these equations it is easy to derive the usual commutators.

On the other hand, quantum fields do not correspond to any observable quantity. They are artificial constructs, whose purpose is to facilitate construction of Hamiltonians for systems with variable number of particles. It appears that if we *choose* our field operators to be (anti)commuting at space-like intervals, then a relativistic Hamiltonian can be easily constructed out of them. Field commutators themselves do not have any particular physical meaning. The best place to read about it is section 5.1 in S. Weinberg "The quantum theory of fields", vol. 1.

Eugene.

4. Sep 17, 2010

### tom.stoer

I think you can motivate the commutation relations of quantum fields similar to ordinary (point particle) quantum mechanics.

If you put the field theory on a compact space and go to momentum space you find something like (infinitly many coupled) particles. So the canonical quantization of quantum fields can be motivated by quantizing N point particles (ordinary quantum mechanics) and then send N to infinity. Doing that you automatically get the commutation relations for quantum fields.

5. Sep 19, 2010

### pellman

Thanks. This looks promising. Last time I read it, it was beyond me but I think I'm up to it now.

6. Sep 20, 2010

### dx

Field commutators have physical meaning just like 'ordinary' commutators. The measurability of the electric and magnetic fields is restricted by quantum electrodynamical commutators just like the measurability of position and momentum of a particle is restricted by [x, p] = ih.

7. Sep 20, 2010

### meopemuk

One can take the position that we never measure electric and magnetic fields themselves. We measure only particle properties, e.g., accelerations. From these data we can calculate classical field values at points, but this does not mean that fields are measurable physical entities.

How would you measure electron's Dirac field, which is even non-Hermitian?

On page 198 of his book Weinberg writes:

"However we will be dealing here with fields like the Dirac field of the electron that do not seem in any sense measurable. The point of view take here is that [field commutators are] needed for the Lorentz invariance of the S-matrix, without any ancillary assumptions about measurability and causality."

Denying the physical relevance of quantum fields is the crucial point of Weinberg's method.

Eugene.

8. Sep 20, 2010

### dx

I don't know what Weinberg's method is, but atleast in the case of quantum electrodynamics, the commutation relation for the averaged field operators implies uncertainty relations for the classical field quantities E and B, and this uncertainty relation corresponds exactly to the experimental limitations of measurement, just like the case of p and q.

9. Sep 20, 2010

### meopemuk

The double-slit experiment demonstrates the simultaneous non-measurability (=nonzero commutator =uncertainty relations) of particle observables x and p.

Is there a similar experiment, which demonstrates the uncertainty relations for the fields E and B?

Eugene.

10. Sep 21, 2010

### tom.stoer

I don't think that experimental setups provide a good guidance towards commutation relations of fields.

As I said: one can show that the derivation of commutation relations of field operators follows from the very same principles as used in commutation relations for quantum mechanical operators, namely from translation of a canonical structure in classical phase space to QM operators in Hilbert spaces. Defining fields on a compact space discretizes the momentum space and automatically leads to harmonic-oscillator like creation and annihilation operators in Fock-space. It looks like QM with N degrees of freedom.

11. Sep 21, 2010

### dx

Yes, though the treatment is considerably more involved than in the case of p and q. See the following papers:

"On the Question of the Measurability of Electromagnetic Field Quantities"
"Field and Charge Measurements in Quantum Electrodynamics"

Both by Bohr and Rosenfeld.

12. Sep 21, 2010

### meopemuk

There are no experiments in these papers. Just theoretical speculations. So, the applicability of the uncertainty principle to the "measurements of electromagnetic fields" remains ... well ... uncertain.

Eugene.

13. Sep 21, 2010

### dx

I don't know what your objections are to the arguments presented in the papers, but you should know that these results are considered to be established.