What are the reasons why we usually treat Quantum Field Theory in momentum space instead of position space? Are the computations (e.g. of Feynman diagrams) generally easier and are there other advantages of this formulation?
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2$\begingroup$ I believe it's just because we can take the $\partial_\mu$ of spacetime and replace it with a momentum $p_\mu$, no? In other words it's the same reason that anyone does anything in $k$-space. $\endgroup$– CR DrostCommented Dec 20, 2016 at 15:18
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$\begingroup$ There are plenty of resources at least briefly discussing Feynman diagrams in momentum space. Have you tried googling for e.g. "position space Feynman diagram"? $\endgroup$– ACuriousMind ♦Commented Dec 20, 2016 at 15:18
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$\begingroup$ I think it should be made clearer by specific examples, because I saw about four different QFT lecture styles and all of them accented the use of momentum space at different points and for different purposes. $\endgroup$– VoidCommented Dec 20, 2016 at 15:32
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$\begingroup$ You should still be mindful of position space diagrams. Think about boundaries etc ways translation gets destroyed. $\endgroup$– AHusainCommented Dec 21, 2016 at 12:08
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2 Answers
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The most important reasons we use momentum space Feynman rules are:
In position space, the Feynman rules generate convolutions of propagators. Because of the convolution theorem, the momentum space rules generate products of propagators, which are clearly easier to handle.
Moreover, in position space you have an integral for each vertex, while on momentum space you have one integral per loop, and in a general diagram there are many more vertices than loops, thus making the momentum space rules easier to use.
What's more, the LSZ theorem in momentum space is trivial to implement: we just drop the propagators on the external lines; in position you'd have to evaluate some exponential integrals (which are straightforward, but cumbersome).
Finally, the renormalisation conditions are naturally imposed in momentum space, and therefore you want the diagrams in momentum space.
answered Dec 20, 2016 at 15:19
AccidentalFourierTransformAccidentalFourierTransform
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$\begingroup$ Nice answer (+1) Is it fair to state that QFT is constructed in momentum space purely just to make analytical calculations easier? Would it be equally valid to construct a QFT in position space if one is not concerned at all with calculations (or have other means than analytical calculations to get at results)? $\endgroup$– JamesCommented Aug 26 at 0:09
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1$\begingroup$ @James Yes. Typically it is best to have both options at the same time, in order to use whichever is most convenient, not only at the level of simplifying calculations but also to facilitate conceptual reasoning. But it is a Fourier transform after all, which is invertible, so both representations contain the exact same information, and you can use whichever you like the most. $\endgroup$ Commented Aug 26 at 12:57
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1$\begingroup$ thank you. I realized I have subconsciously been mixing up QFT with second quantization... Did everything you say apply to both classical fields as well as second quantization fields equally? $\endgroup$– JamesCommented Aug 26 at 14:09
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I may add that the expressions for propagators $G(x,x^{\prime})\propto \int \frac{\mathrm{d}^D k}{k^2+m^2} \mathrm{e}^{-\mathrm{i}k (x-x^{\prime})}$ are quite cumbersome in the position space, and have a plenty of singularities. See, for example, this Wiki article.
answered Dec 20, 2016 at 15:55