DIRAC DELTA FUNCTION AS A DISTRIBUTION

1 MASSACHUSETTS INSTITUTE OF TECHNOLOGYP hysics : Relativistic Quantum Field Theory IProf .Alan GuthMarch 13, 2008 INFORMAL NOTESDIRAC DELTA FUNCTION AS A DISTRIBUTIONWhy the DIRAC DELTA FUNCTION is not a FUNCTION :The DIRAC DELTA FUNCTION (x) is often described by considering a FUNCTION thathas a narrow peak atx= 0, with unit total area under the peak .In the limit as thepeak becomes infinitely narrow, keeping fixed the area under the peak, the functionis sometimes said to approach a DIRAC DELTA FUNCTION .One example of such a limitisg(x) lim 0g (x),( )whereg (x) 1 2 e 12x2/ 2.( )The area underg (x) is 1, for any value of >0, andg (x) approaches 0 as 0for anyxother thanx= , it was pointed out long ago that the DELTA FUNCTION cannot be rigor-ously defined this way.

2 The functiong(x) is equal to zero for anyx =0,andisinfinite atx= 0; it can be shown that any such FUNCTION integrates to zero .To seethis, define the integral as the area under the curve, and consider the construction:In this picture the vertical axis is entirely encased in rectangles, each of which hasheight 1 .The width of the rectangles vary, with the lowest rectangle having width ,forsome >0, and each successive rectangle has half the width of the rectanglebelow .Note that the outline of the boxes is everywhere above the curveg(x), LECTURE NOTES 4, SPRING 2008: DIRAC DELTA FUNCTION as a Distributionp. 2the area underg(x) must be less than the total area of the boxes .But the totalarea of the boxes is given by a geometric series,A= +12 +14 +.

3 2 .( )Since can be chosen as small as one likes, the area under the limit functiong(x)must be result does not contradict the statement that the area underg (x)is1for any >0 .Rather, this is a case where the limit of an integral is not the sameas the integral of the limit of the integrand .The integral has the value 1 for every >0, so the limit of the integral as 0 is 1 .However, if one takes the limit ofthe integrand first, and then integrates, the answer is DELTA FUNCTION as a DISTRIBUTION :A DIRAC DELTA FUNCTION is defined to have the property that dx (x) (x a) (a).( )Butwehavejustseenthatthereisnofunction (x) which has this property, as longas integration is defined by the area under a curve .However, there is no problemdefining a DISTRIBUTION that behaves this way.

4 Using the notation we defined earlierfor distributions, we defineT (x a)[ ] (a).( )HereT (x a)is the name that we will use for the DISTRIBUTION that acts like integra-tion over the DELTA FUNCTION (x a), and (x) is the test FUNCTION .This is certainlya linear functional that is defined on all Schwartz functions , and therefore is a tem-pered DISTRIBUTION . We interpret Eqs. ( ) and ( ) as meaning the same thing,where the second form emphasizes the definition as a DISTRIBUTION , and the first formemphasizes that distributions can be viewed as a generalized kind of FUNCTION , witha notation that makes them look like functions .That is, we define dx (x) (x a) T (x a)[ ] (a).( )To put it differently, we must remember that an integral over a DELTA FUNCTION , suchas Eq.

5 (4 .4), is not defined as a standard integral instead it is symbolic integral,which is defined as a DISTRIBUTION which maps the FUNCTION that multiplies the deltafunction to its value at the point where the argument of the DELTA FUNCTION LECTURE NOTES 4, SPRING 2008: DIRAC DELTA FUNCTION as a Distributionp. 3 What about dpe ip(x a)?If we interpret this integral in the sense of Riemann or Lebesgue, it simply doesnot exist it diverges .As a DISTRIBUTION , however, we can interpretI(x) dpe ip(x a)( )as the Fourier transform off(p)=eipa.( )SinceI(x) is the Fourier transform of the distributionTf[ ], it must itself be awell-defined DISTRIBUTION .Using the notation of distributions, the DISTRIBUTION cor-responding toIis then given byTI[ ]= Tf[ ].

6 ( )To evaluate the distributionTI[ ], we first write the DISTRIBUTION correspondingto functionf(p) in the usual way:Tf[ ]= dpf(p) (p)= dpeipa (p).( )The Fourier transform of this DISTRIBUTION is then defined by applying the samedistribution to the Fourier transform of the test FUNCTION , so Tf[ ] Tf[ ]= dpeipa (p).( )But the inverse Fourier transform is given by (x)=12 dpeipx (p),( )so by comparing the two formulas above, one sees that Tf[ ]=2 (a).( )But this is exactly the definition of the DELTA FUNCTION DISTRIBUTION , as given inEq.( ),so Tf[ ]=2 T (x a)[ ].( ) LECTURE NOTES 4, SPRING 2008: DIRAC DELTA FUNCTION as a Distributionp. 4If we write these distributions as symbolic integrals, then the above equationbecomes dxI(x) (x)=2 dx (x a) (x),( )where we remembered from Eq.

7 (4 .9) that Tf[ ] is equal to the DISTRIBUTION corre-sponding toI(x) .Since Eq .(4 .15) holds for an arbitrary test FUNCTION (x), we canequate the distributions without showing their arguments:I(x)= dpe ip(x a)=2 (x a).( )The Derivative of a DELTA FUNCTION :If a DIRAC DELTA FUNCTION is a DISTRIBUTION , then the derivative of a DIRAC deltafunction is, not surprisingly, the derivative of a DISTRIBUTION .We have not yetdefined the derivative of a DISTRIBUTION , but it is defined in the obvious way .Wefirst consider a DISTRIBUTION corresponding to a FUNCTION , and ask what would be thedistribution corresponding to the derivative of the with a well-behaved ( , piecewise continuous and bounded by somepower oft) functionf(t), we defined the corresponding DISTRIBUTION byTf[ ] dtf(t) (t).

8 ( )Then if we write the DISTRIBUTION corresponding to df/dt,wegetTdf/dt[ ]= dtdfdt (t)( )Sincef(t) is bounded for large|t|by a power oft,and (t) falls off faster than anypower, we can integrate by parts without encountering a surface term:Tdf/dt[ ]= dtf(t)d dt= Tf[d dt].( )This result can then be taken as the general definition of the derivative of a distri-bution:T [ ] T[d dt].( )Applying this result to a DELTA FUNCTION in the notation of Eq .(4 .6), it looks exactlylike integration by parts: dtf(t)d (t a)dt= dfdt t=a.( )