What operational procedure is required to measure $i[\hat\phi(x),\hat\phi(y)]$ in an interacting (or non-interacting) QFT? [assume smearing by test-functions, or give an answer in Fourier space, for $i\bigl[\widetilde{\hat\phi}(k),\widetilde{\hat\phi}(k')\bigr]$, to your taste; give electromagnetic field indices to $\hat\phi(x)$ and transpose the question to QED if the scalar field case seems unreal in an essential way].

In a non-interacting QFT, $i[\hat\phi(x),\hat\phi(y)]$ is a c-number, but in an interacting QFT it may presumably be an observable independent of $\hat\phi(x)$ and $\hat\phi(y)$. How do we measure it, however? If we cannot measure it, is it effectively chosen by convention?

An answer that describes the (operational) difference between $A=\hat\phi(x_1)...\hat\phi(x_i)\hat\phi(x_{i+1})...\hat\phi(x_n)$ and $B=\hat\phi(x_1)...\hat\phi(x_{i+1})\hat\phi(x_i)...\hat\phi(x_n)$ might be equally good (though $A^\dagger\not=A$ and $B^\dagger\not=B$).

As always, a reference would be good.

EDIT: Following Luboš' answer, and more particularly his comments, for which again thanks, $i[\hat\phi(x),\hat\phi(y)]$ is zero if $x-y$ is space-like, both in non-interacting and interacting quantum fields. If a quantum field $\hat\phi(x)$ is the basis for an empirically adequate quantum theory, measurement operators at time-like separation from a given space-like hyperplane at time $t$ can in principal be expressed as a nontrivial function of the field and field momentum at time $t$, $\hat\phi(y)=\int[\hat\phi(t,\mathbf{x}')f_t(\mathbf{x}')+\hat\pi(t,\mathbf{x}')g_t(\mathbf{x}')]\mathrm{d}^3\mathbf{x}'$, for two functions $f_t$ and $g_t$, which for a given theory are determined by the Lagrangian. Given many implicit assumptions, measuring the commutator $i[\hat\phi(t,\mathbf{x}),\hat\phi(y)]=-g_t(\mathbf{x})$ (because $[\hat\phi(t,\mathbf{x}),\hat\pi(t,\mathbf{y})]=i\delta^3(\mathbf{x}-\mathbf{y})$) is equivalent to measuring $g_t(\mathbf{x})$. If we also measure $i[\hat\pi(t,\mathbf{x}),\hat\phi(y)]$, we can determine (or at least significantly constrain) the Lagrangian. If we do not know the Lagrangian, therefore, it would seem to be desirable to have an operational way to measure $i[\hat\phi(x),\hat\phi(y)]$ in order to measure the Lagrangian (instead of the usual approach of guessing the Lagrangian and seeing how close the consequent predictions are to the measurements; the haphazardness of this process is something of a crisis in HEP). I would welcome comments on this or an answer that takes this EDIT into account. Alternatively, an operational approach to measuring the Lagrangian would implicitly be a measurement, in principle, of $i[\hat\phi(x),\hat\phi(y)]$; is there such a thing?

  • $\begingroup$ great question - $\endgroup$ – lurscher Apr 11 '12 at 16:13

I don't think that there will be any nontrivial reference about the hypothetical procedure because the misunderstanding seems to be hidden in the very question.

In a quantum field theory, $[\phi(x),\phi(y)]=0$ for spacelike separated $x,y$, due to the Lorentz invariance. This identity is an operator identity. So you're really asking what is the operational procedure to measure the value of an operator that is equal to zero. The right way to measure it is to be brave enough, to do nothing, and to say that the result is zero because it's the only eigenvalue of this operator.

If $x,y$ are timelike (or light-like, which is a marginal case) separated, the commutator is nonzero. But it's really meaningless to talk about the measurement of differently ordered operators at different times. Measurement in quantum mechanics is meant to be a measurement of a particular operator at a fixed time and when we consider a sequence of measurements, the ordering from the left to the right is correlated with their ordering in time because the measurements of operators at an increasing $t$ are acting on ket vectors one after another. That's why, for example, the $S$ matrix may be written as the time-ordered exponential of the integrated Hamiltonian.

Physically speaking, it only makes sense to place operators at points in the past to the right side of a product and operators from the future points to the left, assuming we are evaluating some evolution operator ready to act on ket vectors (for bra vectors it's the other way around, as seen by a simple Hermitian conjugation). Algebraically speaking, the oppositely ordered products are also meaningful but they don't correspond to any procedure.

One exception could be said: if you actually didn't want to measure the eigenvalue of the commutator but just the expectation value of the commutator of the elementary fields, a form of a propagator, then the "retarded commutator" (retarded means "with the theta-function imposing a right time-ordering") may be measured as a response function. One perturbs the operator at the initial time, and measures how much it changes the value of the operator at the final slice. Only the change of the expectation value is evaluated, and in this way, one may obtain the expectation value of the retarded commutator.

If $x,y$ were spacelike-separated but the commutator were nonzero, e.g. because the Lorentz symmetry would be broken, the commutator would still be equal to some particular operator that may be expressed as a functional of some "basic" operators. All Hermitian operators in a quantum mechanical theory are legitimate observables that may be measured. If you want me to describe a particular device and procedure to measure the value of the operator, what the gadget has to do, and so on, you must tell me exactly what theory you consider because devices to measure something do depend on the actual theory (i.e. its Hamiltonian).

These additional data you would have to add would of course describe a hypothetical situation because in our world, the commutator of spacelike-separated operators does vanish. A theory in which it wouldn't vanish would be a nonlocal one so the intuition in the Universe described by that other hypothetical theory would differ from the right intuition in our Universe, especially when it comes to things like the description of positions of objects (which is not a terribly useful concept if the theory is nonlocal, anyway).

You asked about the ways how to measure the value of a particular operator $0$ which was trivial. But maybe you wanted to ask a different question: how do we empirically prove that the commutator is zero, i.e. how do we prove this particular corollary of locality. Well, we could prove that the commutator is nonzero just like for the operators $x,p$ which appear in the uncertainty principle: the measurement of one of them inevitably disturbs the other. If the measurement of $\phi(x)$ disturbed the distibution of results obtained by measuring $\phi(y)$, the commutator would be nonzero.

However, to measure the actual value of the operator equal to the commutator is subtle. In general, the operator $[\phi(x),\phi(y)]$ commutes neither with $\phi(x)$ nor with $\phi(y)$. The case of $xp-px=i\hbar$ is kind of special because the commutator is a $c$-number. Because the commutator is a completely independent operator from $\phi(x)$ and $\phi(y)$, its measurement procedure is of course completely independent from the measurement of $\phi(x)$ and $\phi(y)$. In particular, if we want to measure $i(xp-px)$, the right way is once again to do nothing because the right value is always $-\hbar$.

That's another important misconception that is implicitly contained in your question. You seem to believe that the measurement of the commutator or product of several operators may be obtained by measurements of the individual factors in the commutator or the product. But this is not true at all. Take a simple example of $-i[J_x,J_y]/\hbar$. How do you measure the value of this operator equal to the commutator? Measuring $J_x$ and $J_y$ won't be any useful because the commutator is actually equal to $J_z$, the third and completely independent component! The classical observable corresponding to the commutator is not a function of the observables appearing in the commutator! In general, it's a completely different operator that may only be measured by a completely different procedure.

There's one more thing you may wanted to address: the situation in which $x,y$ are equal or very close to each other so that some nonzero term may arise from the delta-function-like commutator (or perhaps from some quantum loop corrections or renormalization, as seen in the effective action, etc.). Linking fields to actual apparata is always subtle because the identity of the fields in a field theory (or any theory) is only given up to field redefinitions (which may be nonlocal), and so on. So without having a particular convention about what you mean by the fields, which also implies that you may exactly calculate what the commutator is, it is meaningless to ask about such subtle questions. When you talk about $\phi(x)$, you have to use some particular scheme of conventions etc. so you're no longer talking about operationally doable things. If you want an operational answer, you need to formulate the question operationally as well.

At any rate, even in this case of a nonzero commutator from nearby $x,y$, the main points of the text above still hold. The measurement of the commutator is completely independent from the measurement of the operators whose commutator is being computed. And if the commutator is a $c$-number (which is always the case e.g. when there are only non-derivative interactions), the measurement procedure is to do "nothing".

Last comment. It also seems to me that you have a wrong idea about the actual value of the commutator in common interacting QFTs. In the Standard Model, all renormalizable interaction terms are non-derivative – they are polynomial of the fields but don't depend on the field derivatives. So if you compute the canonical momenta as the derivative of the Lagrangian with respect to the time-derivatives of the fields, the interaction terms will contribute nothing at all (because there are no time derivatives in the interaction Lagrangian)! So the equal-time commutators in the Standard Model are exactly the same as they are in the non-interacting limit of the Standard Model.

The timelike-separated (non-equal-time) commutators are affected by the interactions, of course, but the products that are not properly time-ordered don't correspond to any operational procedure.

  • $\begingroup$ That's the first time I've come up against Luboš full force, I think. I don't understand how it can be that "it's really meaningless to talk about the measurement of differently ordered operators at different times", given that HEP is about signals that change over time, which are then subjected to various signal analysis algorithms (albeit ad hoc ways are used to extract event timings from time-varying CCD currents, etc.). "Measurement in quantum mechanics is meant to be a measurement of a particular operator at a fixed time": so QM cannot model measurements at different times? $\endgroup$ – Peter Morgan Apr 11 '12 at 17:17
  • $\begingroup$ Dear Peter, "so QM cannot model measurements at different times?" It can. One can measure operators at arbitrary times, but one may only do so chronologically. One first measures the operator at an earlier time and then he measures another operator at a later time. The earlier operator acts on the ket as the first one, so it's on the right side, and the later operator acts later, so it's on the left side of the product. It makes no sense to ask how you make the two measurements in the opposite order. The order of the measurements is unambiguously determined by the ordering of arguments $t_i$. $\endgroup$ – Luboš Motl Apr 11 '12 at 17:20
  • $\begingroup$ You suggest in your last paragraph that I "believe that the measurement of the commutator or product of several operators may be obtained by measurements of the individual factors in the commutator or the product." I take my "in an interacting QFT [the commutator] may presumably be an observable independent of $\hat\phi(x)$ and $\hat\phi(y)$" to say otherwise. $\endgroup$ – Peter Morgan Apr 11 '12 at 17:21
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    $\begingroup$ "Measurement in quantum mechanics is meant to be a measurement of a particular operator at a fixed time" Practically, is there a measurement other than a yes/no observation, which is not a time average? $\endgroup$ – Nikolaj-K Apr 11 '12 at 17:52
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    $\begingroup$ "operators at one moment are functionals of operators at any other moment" -- right, that's an answer, albeit encoded, that is a fundamental aspect of how QM is structured, in terms of the classical non-relativistic concept of a phase space. It leaves quite a few questions that seem very delicate, but thanks Luboš. +1 for the comments. $\endgroup$ – Peter Morgan Apr 11 '12 at 18:08

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