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MIT 8 323 - DISTRIBUTIONS AND THE FOURIER TRANSFORM

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MIT OpenCourseWare http://ocw.mit.edu 8.323 Relativistic Quantum Field Theory I Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department 8.323: Relativistic Quantum Field Theory I Prof. Alan Guth March 3, 2008 INFORMAL NOTES DISTRIBUTIONS AND THE FOURIER TRANSFORM Basic idea: In QFT it is common to encounter integrals that are not well-defined. Peskin and Schroeder, for example, give the following formula (p. 27, after Eq. (2.51)) for the two point function 0 |φ(x)φ(y)| 0 for spacelike separations (x − y)2 = −r2: −i � ∞ peipr D(r)= 2(2π)2r −∞ dp � p2 + m2 . If this integral is defined in the usual way as � Λ peipr lim dp � , Λ→∞ −Λ p2 + m2 then it does not exist. The integral can be defined by putting in a convergence factor e−|p|: ∞ peipr e−|p|lim dp � . →0 −∞ p2 + m2 But how does one know whether a different convergence factor would get the same result? One way to resolve these issues is to treat the ambiguous quantity as a distribution, rather than a function. All tempered distributions (to be defined below) have Fourier transforms, which are also tempered distributions. Furthermore, we can show that the -prescription used above is equivalent to the tempered-distribution definition of the Fourier transform. Distribution: A distribution is a linear mapping from a space of test functions to real or complex numbers. (An operator-valued distribution maps test functions into operators.) Test Functions: The space of test functions {ϕ(t)} determines what type of distribution one is discussing. The test functions for tempered distributions belong to “Schwartz space,” the space of functions which are infinitely differentiable, and the func-tion and each of its derivatives fall off faster than any power for large t. The Gaussian is a good example of a Schwartz function. Any function in Schwartz� 8.323 LECTURE NOTES 3, SPRING 2008: Distributions and the Fourier Transform p. 2 space has a Fourier transform in Schwartz space. (The Fourier transform of a Gaussian is a Gaussian.) Functions as Distributions: Distributions are sometimes called generalized functions, which suggests that a function is also a distribution. This is not quite true, but a wide range of functions can also be thought of as distributions. Given any function f (t)which is piecewise continuous and bounded by some power of t for large t,one can define a corresponding distribution Tf by � ∞ Tf [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)’sisvery large. Fourier Transform: For any function f (t) which is integrable, meaning that � ∞ dt |f (t)|−∞ converges, define � ∞ f˜(ω) ≡ dte−iωtf(t) . −∞ Fourier Transform of a Distribution: To motivate the definition, suppose f (t) is integrable, and consider � ∞ Tf˜[ϕ]= f˜(ω)ϕ(ω) dω −∞ � ∞ ∞ = dt e −iωt f(t) ϕ(ω) dω −∞ −∞ � ∞ =dtf(t)˜ϕ(t) −∞ = Tf [˜ϕ] . Note that these integrals are absolutely convergent, so there is no problem about interchanging the order of integration. So, for any distribution T , define its Fourier transform by T˜[ϕ] ≡ T [˜ϕ] .8.323 LECTURE NOTES 3, SPRING 2008: Distributions and the Fourier Transform p. 3 Note that any function f(t) which is piecewise continuous and bounded by some power of t for large t can define a distribution, and can therefore be Fourier transformed as a distribution. Relation to � convergence factor: Suppose f (t) is not integrable, and so does not have a Fourier transform. Sup-pose, however, that there exists a continuous sequence of “regulated functions” f(t) which are integrable for >0, which satisfy |f(t)| < |f (t)| , and which for each t satisfy lim f(t)= f(t) . →0 Example: f(t)= f(t)e−|t|. Note that the regulator that we used for the two-point function at spacelike separations has this property. To show: if we Fourier transform f(t) and take the limit  → 0 at the end, it is the same as the distribution-theory definition of the Fourier transform. Proof: The distribution-theory definition of the Fourier transform is T˜f [ϕ] ≡ Tf [˜ϕ] � ∞ = dtf(t)˜ϕ(t) . −∞ The  prescription is to use Tf ∗[ϕ] ≡ lim Tf˜� [ϕ] . →0 We need to show these are equivalent. Use � ∞ Tf ∗[ϕ] = lim dω f˜(ω)ϕ(ω)→0 −∞ � ∞ � ∞ = lim dω dte−iωt f(t)ϕ(ω)→0 −∞ −∞ � ∞ lim dtf(t)˜ϕ(t) . →0 −∞ =� �   8.323 LECTURE NOTES 3, SPRING 2008: Distributions and the Fourier Transform p. 4 If we can take the limit inside the integral, we are done! Last step is proven with Lebesgue’s Dominated Convergence Theorem: If h(t) is a sequence of functions for which lim h(t)= h(t) for all t, →0 and if there exists a function g(t)for which dtg(t) converges, and for which g(t) ≥|h(t)| for all t and all , then � � lim dth(t)= dth(t) . →0 Note, by the way, that the existence of the integrable bounding function g(x) is absolutely necessary. A simple example of a function h(t)for which one CANNOT bring the limit through the integral sign would be a function that looks something like: Analytically, this function can be written as 1if 1 <t< 1 +1h(t)= 0otherwise . Note that the square well moves infinitely far to the right as  → 0, so h(t) → 0 for any t. But the integral of the curve is 1 for any , and hence it is 1 in the limit. The Lebesgue Dominated Convergence theorem excludes functions like this, because any bounding function g(t)must be ≥ 1 for all t,so g(t) cannot be integrable. The theorem does apply, however, to � ∞ lim dtf(t)˜ϕ(t) . →0 −∞8.323 LECTURE NOTES 3, SPRING 2008: Distributions and the Fourier Transform p. 5 Take h(t)= f(t)˜ϕ(t) , h(t)= f(t)˜ϕ(t) , and g(t)= |f (t)˜ϕ(t)| . Bottom Line: The  prescription used by physicists is equivalent to the unambiguous defini-tion of the Fourier transform in tempered-distribution theory. That is, if the function to be Fourier-transformed f (t) is not integrable, one can proceed as long as one can find an integrable regulator


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