MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department 8.323: Relativistic Quantum Field Theory I

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Departent 8.33: Relativistic Quantu Field Theory I INFORMAL NOTES DISTRIBUTIONS AND THE FOURIER TRANSFORM Basic idea: In QFT it is coon to encounter integrals that are not well-defined.last week we talked about the two point function φ(x)φ(y) for spacelike separations (x y) = r, which is given forally by D(r) = i dp (π) r p +. If this integral is defined in the usual way as Λ li dp Λ Λ p +, then it does not exist.the integral can be defined by putting in a convergence factor e ɛ p : li dp peipr e ɛ p p +. But how does one know whether a different convergence factor would get the sae result? One way to resolve these issues is to treat the abiguous quantity as a distribution, rather than a function.all tepered distributions (to be defined below) have Fourier transfors, which are also tepered distributions.furtherore, we can show that the ɛ-prescription used above is equivalent to the tepered-distribution definition of the Fourier transfor. Distribution: A distribution is a linear apping fro a space of test functions to real or coplex nubers.(an operator-valued distribution aps test functions into operators.) Test Functions: The space of test functions {ϕ(t)} deterines what type of distribution one is discussing.the test functions for tepered distributions belong to Schwartz space, the space of functions which are infinitely differentiable, and the function and each of its derivatives fall off faster than any power for large t.the Gaussian is a good exaple of a Schwartz function.any function in Schwartz space has a Fourier transfor in Schwartz space.(the Fourier transfor of a Gaussian is a Gaussian.)

2 8.33 NOTES, DISTRIBUTIONS AND THE FOURIER TRANSFORM, SPRING 3 p. Functions as Distributions: Given any function f(t) which is piecewise continuous and bounded by soe power of t for large t, one can define a distribution T f by T f [ϕ] dtf(t)ϕ(t). Since ϕ(t) falls off faster than any power, this integral will converge.note that because the class of ϕ(t) s is very restricted, the class of possible f(t) s is very large. Fourier Transfor: For any function f(t) which is integrable, eaning that dt f(t) converges, define f(ω) dte iωt f(t). Fourier Transfor of a Distribution: To otivate the definition, suppose f(t) is integrable, and consider T f [ϕ] = = = f(ω)ϕ(ω) dω dt = T f [ ϕ]. dtf(t) ϕ(t) e iωt f(t) ϕ(ω) dω Note that these integrals are absolutely convergent, so there is no proble about interchanging the order of integration.so, for any distribution T, define its Fourier transfor by T [ϕ] T [ ϕ]. Note that any function f(t) which is piecewise continuous and bounded by soe power of t for large t can define a distribution, and can therefore be Fourier transfored as a distribution.

3 8.33 NOTES, DISTRIBUTIONS AND THE FOURIER TRANSFORM, SPRING 3 p. 3 Relation to ɛ convergence factor: Suppose f(t) is not integrable, and so does not have a Fourier transfor.suppose, however, that there exists a continuous sequence of regulated functions f ɛ (t) which are integrable for ɛ>, which satisfy f ɛ (t) < f(t), and which for each t satisfy li f ɛ(t) =f(t). Proof: Exaple: f ɛ (t) =f(t)e ɛ t.note that the regulator that we used for the two-point function at spacelike separations has this property.to show: if we Fourier transfor f ɛ (t) and take the liit ɛ at the end, it is the sae as the distribution-theory definition of the Fourier transfor. The distribution-theory definition of the Fourier transfor is T f [ϕ] T f [ ϕ] = dtf(t) ϕ(t). The ɛ prescription is to use T f [ϕ] li T f ɛ [ϕ]. We need to show these are equivalent.use Tf [ϕ] = li dω f ɛ (ω)ϕ(ω) = li dω = li dte iωt f ɛ (t)ϕ(ω) dtf ɛ (t) ϕ(t). If we can take the liit inside the integral, we are done!

4 8.33 NOTES, DISTRIBUTIONS AND THE FOURIER TRANSFORM, SPRING 3 p. 4 Last step is proven with Lebesgue s Doinated Convergence Theore: If h ɛ (t) isa sequence of functions for which li h ɛ(t) =h(t) for all t, and if there exists a function g(t) forwhich dtg(t) converges, and for which g(t) h ɛ (t) for all t and all ɛ, then li dth ɛ (t) = dth(t). Note, by the way, that the existence of the integrable bounding function g(x) is absolutely necessary.a siple exaple of a function h ɛ (t) for which one CANNOT bring the liit through the integral sign would be a function that looks soething like: Analytically, this function can be written as h ɛ (t) = { 1 if 1 ɛ <t< 1 ɛ +1 otherwise. Note that the square well oves infinitely far to the right as ɛ, so h ɛ (t) for any t.but the integral of the curve is 1 for any ɛ, and hence it is 1 in the liit.the Lebesgue Doinated Convergence theore excludes functions like this, because any bounding function g(t) ust be 1 for all t, so g(t) cannot be integrable. The theore does apply, however, to li dtf ɛ (t) ϕ(t).

5 8.33 NOTES, DISTRIBUTIONS AND THE FOURIER TRANSFORM, SPRING 3 p. 5 Take and h ɛ (t) =f ɛ (t) ϕ(t), h(t) =f(t) ϕ(t), g(t) = f(t) ϕ(t). Botto Line: The ɛ prescription used by physicists is equivalent to the unabiguous definition of the Fourier transfor in tepered-distribution theory.that is, if the function to be Fourier-transfored f(t) is not integrable, one can proceed as long as one can find an integrable regulator f ɛ (t) such that f ɛ (t) < f(t), and for each t, li f ɛ(t) =f(t). One can then Fourier transfor f ɛ (t) instead.in the general case one cannot take the liit ɛ iediately, but one ust leave ɛ in the expression for the distribution. Only after the distribution is evaluated for a particular test function can the liit ɛ be taken.reeber, for exaple, that we wrote the Fourier transfor of the Feynan propagator as i p + iɛ. With the ɛ in place one can carry out integrals involving the propagator, and then one can take the liit ɛ at the end.if one tried to set the ɛ ter to zero iediately, then the poles in the propagator would lead to ill-defined integrations. The two-point function at spacelike separation: When we applied the regulator e ɛ p to the integral for the two-point function at spacelike separation, we found that we could obtain a definite function, D(r) = li i e ɛ p (π) r p + = 4π r K 1(r). The distribution theory analysis, however, did not justify the last step of taking the liit ɛ, but instead indicated that we should keep ɛ in place until after the distribution has been applied to a test function.in the previous lecture I showed that the integral shown above can be evaluated exactly with finite nonzero ɛ, and

6 8.33 NOTES, DISTRIBUTIONS AND THE FOURIER TRANSFORM, SPRING 3 p. 6 the result involved K 1 (r + iɛ), and also the iaginary part of a Struve function H 1 (i(r + iɛ)) and the real part of a Bessel function J 1 (i(r + iɛ)).if one defines i e ɛ p (π) r p + D(r, ɛ), then the distribution theory approach iplies that we should evaluate this distribution on a test function ϕ(r) by coputing li dr D(r, ɛ) ϕ(r). We, however, would like to siplify this further by taking the liit through the integral sign and using li D(r, ɛ) = 4π r K 1(r). Since ϕ(r) is required to be very well behaved, one needs only a oderate aount of unifority in the liit above to justify bringing the liit through the integral sign, using again the Lebesgue doinated convergence theore.a unifor liit would ean that for every δ>thereexistsanɛ such that D(r, ɛ) 4π r K 1(r) <δ. To bring the liit through the integral sign, it would be enough to show that for every δ>thereexistsanɛ such that D(r, ɛ) 4π r K 1(r) <g(r) δ, where g(r) is soe fixed function which blows up no faster than a power as r, so that drg(r) ϕ(r) would be guaranteed to converge.i assue that this can be ade to work, but I will not pursue it further. There is, however, an interesting point to look at concerning the use of different regulators.the distribution analysis shows that any regulator should be equivalent to any other, provided only that f ɛ (t) < f(t), and for each t, li f ɛ(t) =f(t).

7 8.33 NOTES, DISTRIBUTIONS AND THE FOURIER TRANSFORM, SPRING 3 p. 7 This includes the possibility of a sharp cutoff at Λ = 1/ɛ.So 1/ɛ dp p + should be an acceptable regulator.but earlier we rejected this definition, because it looked like the liit did not exist.to understand what is happening, it is easiest to integrate by parts: 1/ɛ dp p + = 1 ir eir/ɛ 1 1+ ɛ ir 1/ɛ dp e ipr (p + ) 3/. The second ter is well-behaved, since it falls off at large p as 1/p 3.This integration by parts is just the recipe that I recoended for a purely nuerical evaluation of the two-point function.for the exponential regulator the surface ter vanished, but this tie we have the probleatic ter proportional to e ir/ɛ.this factor has no liit as ɛ if one thinks of it as a function, but as a distribution it vanishes: li dre ir/ɛ ϕ(r) is defined by first Fourier transforing ϕ(r) at p = 1/ɛ, and then taking the liit ɛ, or equivalently p.but the Fourier transfor of a Schwartz function is also a function of Schwartz type, and therefore it falls off faster than any power as p. In the above arguent I ignored the factor of 1/r that arose in the Fourier transfor. For the spacelike separation proble r ust be positive, so we restrict the ϕ(r) to vanish if r.but we have not yet specified how the test functions ϕ(r) should behave as r.following the general philosophy of choosing test functions to be extreely well-behaved, we can require each test functions to vanish in an open neighborhood about r =.We do not fix the size of these open neighborhoods, but just insist that for each acceptable ϕ(r), there is soe δ> such that ϕ(r) = if r < δ.then ultiplication by 1/r is perfectly well-defined operation on test functions, and we are justified in setting to zero. 1 1 ir eir/ɛ 1+ ɛ