DIRAC DELTA FUNCTION - Physicspages

1 DIRAC DELTA FUNCTIONLink to: Physicspages home leave a comment or report an error, please use the auxiliary date: Original: 16 Feb 2011, 14 Nov 2011, 6 Feb 2017; Revised 26 Dec DIRACDELTAFUNCTION IN ONE DIMENSIONOne of the weirder bits of mathematics that the physics student will en-counter is the DIRAC DELTA FUNCTION (x). In one dimension, the FUNCTION (technically it s really not a FUNCTION at all, but a distribution) can be definedby saying (x) =0 for allx6=0 but (x) = atx=0. However, this def-inition isn t very satisfactory, and in fact doesn t define (x)uniquely.

2 It isbetter to define it using two conditions: (x) =0 ifx6=0(1) (x)dx=1(2)It is the integral condition which pins the DELTA FUNCTION down those that like to visualize functions, the DELTA FUNCTION can be thoughtof as a limit of a series of rectangular functions. If we define a rectangularfunction of width 1/wand heightw(so its area is 1) centred horizontallyon they-axis and sitting with its base on thex-axis, then we can visualizethe DELTA FUNCTION as the limit of this rectangle asw . Aswgets larger,the rectangle gets taller and thinner, and in the limit it is an infinitely highspike with only an infinitesimal width, but with an area always equal to consequence of the DELTA FUNCTION being zero everywhere except atx=0 is that if we multiply it by any FUNCTION , it doesn t matter what thatfunction s values are except atx=0.

3 That is, we can sayf(x) (x) =f(0) (x)(3)In terms of integrals, this means that1 DIRAC DELTA FUNCTION2 f(x) (x)dx= f(0) (x)dx(4)=f(0) (x)dx(5)=f(0)(6)using the second defining property 2 of (x)above. The effect of integratinga FUNCTION multiplied by the DELTA FUNCTION is to pick out the FUNCTION s valueatx= is easy enough to move the location of the DELTA FUNCTION s spike. If wewant the spike to appear atx=awe can use the FUNCTION (x a), since thespike occurs when the DELTA FUNCTION s argument is zero, that is, atx a= we can generalize the integral formula above to f(x) (x a)dx=f(a)(7)One of the trickier formulas that causes some consternation amongstphysics students is this formula: (kx) =1|k| (x)(8)wherekis a non-zero we discuss what this means, we can run through the proof.

4 Sincethe main use of DELTA functions is in integration, we can consider this for-mula in light of the integration condition. Suppose we define the variabletransformationy=kx. Thendy=k dx. We ll takek >0 first: f(x) (kx)dx= f(y/k) (y)1kdy(9)=1k f(y/k) (y)dy(10)=1kf(0)(11)This is the same result as what we would get if we evaluated:1k f(x) (x)dx=1kf(0)(12)so it seems reasonable to take (kx) = (x)/kin this DELTA FUNCTION3 Ifk <0,then the derivation is the same except that making the variablesubstitutiony=kxinverts the limits of integration, so we get f(x) (kx)dx= f(y/k) (y)1kdy(13)= f(y/k) (y)1kdy(14)= 1kf(0)(15)=1|k|f(0)(16)Therefore, saying (kx) = (x)/|k|covers both a problem many people have with this formula is this.

5 If the defini-tion of (x)is that it is zero everywhere, but infinite whenx=0, then sincethe same can be said of (kx), why can t we just say (kx) = (x)?The reason arises in the ambiguity of this non-integral definition of thedelta FUNCTION that I mentioned at the start. There are many way we candefine a FUNCTION that is zero everywhere but infinite atx=0. Look at itthis way. Instead of using the limit of the sequence of rectangles that I didat the start, suppose we use a sequence of rectangles of width 1/kwandheightw, so that their areas are all 1/k.

6 Now in the limit asw we getanother FUNCTION that is zero everywhere except atx=0 and has an infinitespike atx=0 but it is clearly not the same as (x)since the area of thespike is 1/kinstead of 1. This new FUNCTION which results from scaling thexaxis using the transformationx kxgives you a spike with an area of1/ktimes the generalization of this formula is (f(x) f(x0)) =1|f (x0)| (x x0)(17)The proof of this follows the same lines as above. We consider the casef (x0)>0 and do the integral g(x) (f(x) f(x0))dx(18)whereg(x)is any FUNCTION .

7 Now use the substitutionDIRAC DELTA FUNCTION4y=f(x) f(x0)(19)dy=f (x)dx(20)x=f 1(y+f(x0))(21)wheref 1denotes an inverse FUNCTION . Then we get g(x) (f(x) f(x0))dx= g(f 1(y+f(x0))) (y)f (x)dy(22)Sincey=0 whenx=x0we get g(f 1(y+f(x0))) (y)f (x)dy=g(f 1(f(x0)))1f (x0)(23)=g(x0)f (x0)(24)As above whenk <0, iff (x0)<0 we get g(x0)/f (x0)instead of 24so the general formula is as given in the even more general case wheref(x)has more than one zero, thisformula generalizes to (I won t bother with the proof here): (f(x)) = i (x zi)|f (zi)|(25)whereziare the zeroes off(x).

8 Note in particular thatf (zi)is thederiva-tiveoffevaluated at points where the original FUNCTION (notthe derivative!)is formula that can cause nightmares is the derivative of the stepfunction, that is of the functionH(x) ={0x 01x >0(26)Since the FUNCTION is constant everywhere except atx=0 its derivative iszero everywhere except atx=0. However, the step FUNCTION is discontin-uous at this point, and since it jumps a finite amount over a single point, itwould seem that its derivative is infinite at that point. To see what s goingon, suppose we try the integralDIRAC DELTA FUNCTION5 1 1f(x)dHdxdx=f(0) 1 1dHdxdx(27)=f(0)[H(1) H( 1)](28)=f(0)(29)(the limits on the integral could be any interval which includes 0).}

9 The firstline uses the fact thatdH/dx=0 everywhere exceptx=0, and the secondline is just the ordinary evaluation of an both conditions of the DELTA FUNCTION , so we can saydHdx= (x)(30)Note that if the step is a different size, sayk >0, so that we haveHk(x) ={0x 0k x >0(31)then the same analysis gives 1 1f(x)dHkdxdx=kf(0)(32)so from the earlier example, we getdHkdx= (x/k)(33)=k (x)(34)2. DIRAC DELTA FUNCTION AS LIMIT OF A GAUSSIAN INTEGRALYet another form of the DIRAC DELTA FUNCTION is as the limit of a Gaussianintegral.}

10 We start withg (x x )=1( 2)1/2e (x x )2/ 2(35)If 2is real and positive, we have1( 2)1/2 e (x x )2/ 2dx=1(36) DIRAC DELTA FUNCTION6[The Gaussian integral can be looked up in most tables of integrals, or eval-uated using Maple.]Thus the area under the curve is always 1, for any real value of 2. Nowas 2 0 the exponential becomes zero except whenx=x . The factor1/( 2)1/2tends to infinity as 2 0, but the exponential always tends tozero faster than any power of , sog (x x )tends to zero everywhere ex-cept atx=x.