Why equal-time commutators?

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Main Question or Discussion Point

Why are the commutators in QFT equal-time commutators? I am talking about things like

[tex][\phi(x,t),\pi(x',t)]=i\delta (x-x')[/tex]

where pi is the canonically-conjugate momentum density to phi.

Shouldn't a relativistic approach treat time and space more equivalently? Something like

[tex][\phi(x,t),\pi(x',t')]=i\delta (x-x')\delta(t-t')[/tex]
 

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Avodyne
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You have to pick a frame to work in before you know what x and t are. Then, it's just standard QM that, in the Heisenberg picture, the simple commutators are at equal times. Eg, in QM, [itex][x,p]=i\hbar[/itex] in the Schrodinger picture becomes [itex][x(t),p(t)]=i\hbar[/itex] (that is, the equal-time commutator) in the Heisenberg picture.
 
666
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Eg, in QM, [itex][x,p]=i\hbar[/itex] in the Schrodinger picture becomes [itex][x(t),p(t)]=i\hbar[/itex] (that is, the equal-time commutator) in the Heisenberg picture.
thanks. But that's my point. It's just a leftover from non-relativistic QM. What am I missing?
 
nrqed
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thanks. But that's my point. It's just a leftover from non-relativistic QM. What am I missing?
It depends the approach you follow to build the quantum field theory.


Most people like to postulate that the right approach is to quantize a classical field theory. They then postulate that the amplitude of the field is analoguous to a position and then the above ETCR follows from the generalization of QM to QFT.
So yes, in that sense, it follows by analogy with QM.

I personally don't like this approach (because there are two huge leaps of faith involved). I prefer much more the approach followed by Weinberg in his first QFT book which could be summarized as "particles first, then fields" instead of the above "fields first then particles". But most people on this borad (and most physicists in general I think) prefer the fields first approach.
 
666
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I like the idea of "particles first, then fields". But still, if it leads to the same commutators, then it must have some justification for the break in covariance. I will look through Weinberg.
 
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Are you sure it breaks covariance? Recall that
[tex]\pi = \frac{\delta S}{\delta \partial_0 \phi}[/tex],
so it is not a scalar field like [tex]\phi[/tex], but rather 1 component of a tensor [tex]\pi^\mu[/tex].

I think you can derive the ETCR from the time-ordered two-point function. So perhaps you want to figure out why that object manages to stay Lorentz invariant.
 
Last edited:
strangerep
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Why are the commutators in QFT equal-time commutators? I am talking about things like

[tex][\phi(x,t),\pi(x',t)]=i\delta (x-x')[/tex]

where pi is the canonically-conjugate momentum density to phi.
You need the covariant form of the commutators to understand what's
really going on here...

Shouldn't a relativistic approach treat time and space more equivalently?
Something like

[tex][\phi(x,t),\pi(x',t')]=i\delta (x-x')\delta(t-t')[/tex]
The covariant commutators don't involve [itex]\delta(t-t')[/itex].
Rather, the RHS involves a propagator specific to the type of
field being considered. The relativistically-sensible propagators
reduce to something like [itex]\delta^{(3)}(x-x')[/itex] for
spacelike-separated x,x' and for that case it is possible to find
a frame of reference in which t = t', leading to the familiar
equal-time form of the CCRs.
 
666
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Recall that
[tex]\pi = \frac{\delta S}{\delta \partial_0 \phi}[/tex],
so it is not a scalar field like [tex]\phi[/tex], but rather 1 component of a tensor [tex]\pi^\mu[/tex].

That's it. Thanks, lbrits!

and thanks also to strangerep. I actually typed up a lengthier response to your post but then I realized that lbrits explained where I was going wrong.
 
666
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From PCT, Spin and Statistics, and All That

In old-fashioned** field theory, this requirement [of local commutativity] was often met by assuming that the fields provided an irreducible set of operators satisfying the canonical commutation relations at a given time

[tex][\phi(x,t),\pi(x',t)]=i\delta (x-x')[/tex] (3-7)

However, (3-7) requires that the fields make sense as operators when smeared in x only, and this is an additional strong assumption which goes beyond our axioms. Furthermore there are hints from examples that, in general, [tex][\phi(x,t),(\partial\phi/\partial t')(x',t')][/tex] has singularities at [tex]t-t'=0[/tex], even after being smeared in x and x'. In this case it is difficult to given (3-7) a meaning. Thus, one is reluctant to accept canonical commutation relations as an indispensable requirement on a field theory.



** This was published in 1964!

when the authors talk about being smeared in x, they mean that according to their strict approach, these sorts of restrictions cannot be applied to [tex]\phi(x,t)[/tex] but only to

[tex]\phi(f)=\int f(x)\phi(x)d^4x[/tex]

where f is an appropriate test function.

In other words, i don't really understand what they are saying. But its nice to know someone else is bothered by it besides me.
 
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Some people are bothered by "functions" that approach delta functions in some limit. Personally I always fall back to some box normalization picture or regularization picture when I'm bothered by this. It's always the final answer that has to make sense physically, not the individual steps =)

Thing is, these singularities are where the fun hides in QM and QFT. Ever wonder about [tex]\operatorname{tr}\,(XP - PX) \neq 0[/tex]? These contact terms come from short-distance behaviour and our attempts to apply finite-dimensonal rules to infinite dimensional systems. For instance, for an infinite dimensional matrix you have to be careful how to define [tex]\det A[/tex], and most of the time it'll be the case that [tex]\det A \det B \neq \det AB[/tex]. This gives rise to various anomalies. Finally there is normal ordering which is even more fun!
 
reilly
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For free fields, given basic creation and destruction operators, you can find the (anti)commutators for arbitrary coordinates, in each field involved. Weinberg's treatment is based on creation and destruction operators( see p173 and on in Vol 1 of his Field Theory treatise, and then go to p201 and following.) Note, also, that the Jacob and Wick helicity formalism preceded Weinberg's treatment.

So, really, Weinberg uses canonical commutation rules, but in a somewhat disguised fashion. My sense is that the key for Weinberg is the Cluster Decomposition Principle.

One of the major players in what might be called practical formal field theory is Gunner Kallen. He was writing in the 1960s about smeared out fields, distributions,commutation rules, propagators, vertex functions and the like. If you want to get a better handle on such matters, find his work, maybe on Google -- my source is the lectures from the 1960 Les Houches summer school in a volume called Dispersion Relations and Elementary Particles, edited by De Witt and Omnes. Yes, it is very old stuff, but still highly relevant and insightful.

And remember, physicists are often sloppy about mathematics -- cf. Goldberger and Watson's Collision Theory. They go into infinite detail about the formal difficulties of the Lippman Schwinger eq., the Heitler Integral Eq, and all of formal scattering theory, which directly connects with field theory. Good stuff to know if you want to understand QM dynamics, whether relativistic or non-relativistic.
Regards,
Reilly
 

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