A Brief Look at Gaussian Integrals - weylmann.com

1 A Brief look at Gaussian IntegralsWilliam O. Straub, PhDPasadena, CaliforniaJanuary 11, 2009 Gaussian Integrals appear frequently in mathematics and physics. Perhaps the oddest thing about these integralsis that they cannot be evaluated in closed form over finite limits but are exactly integrable over[ , ]. A typicalGaussian expression has the familiar bell curve shape and is described by the equationy= exp( 12ax2)whereais a the path integral approach to quantum field theory, these Integrals typically involve exponential terms withquadratic and linear terms in the variables.

2 They are always multi-dimensional with limits at[ , ], and aretherefore difficult to solve. For example, the probability amplitude for a scalar field (x)to transition from onespacetime point to another can be expressed byZ= D exp[i/~ d4x[ 12 ( 2+m2) +J ]where the first integral is of dimensionnandD =d 1(x)d 2(x)..d n(x), wherengoes to infinity. Needless tosay, it helps to have a few formulas handy to calculate such quantum field theory, Gaussian Integrals come in two types. The first involves ordinary real or complexvariables, and the other involves Grassmann variables. The second type is used in the path integral description offermions, which are particles having half-integral spin.]

3 Grassmann variables are highly non-intuitive, but calculatingGaussian Integrals with them is very easy. Here, we ll briefly consider both Integrals with Ordinary VariablesThe simplest Gaussian integral isI= exp( x2)dx(1)which can solved solved quite easily. Squaring the quantityI, we getI2= exp( x2)dx exp( y2)dy= exp[ (x2+y2)]dxdyGoing over to polar coordinates, we haver2=x2+y2,dxdy=rdrd , andI2= 0 2 0exp( r2)rdrd Integration of this double integral is now straightforward, and we haveI= exp( x2)dx= Most problems of this sort will have the constant factora/2in the exponential term, but by substitution you shouldbe able to easily show thatI= exp( 12ax2)dx= 2 a(2)The presence of the constantacomes in handy when encountering Integrals of the kind1 xmexp( 12ax2)dx(3)wheremis some positive integer.

4 All we have to do is take the differentialdI/daof both sides, gettingdIda= 12 x2exp( 12ax2)dx= 12 2 a 3/2from which we easily find that x2exp( 12ax2)dx= 2 a 3/2By repeated differentiation, it is easy to show that x2nexp( 12ax2)dx= (2n 1)!! 2 a (2n+1)/2=(2n 1)!!an 2 a=(2n)!an2nn! 2 a(4)where(2n 1)!! = 1 3 is easy to see that (3) will always be zero whenmis odd, because the integrand is then odd inxwhile theintegration limits are even. The fact that the integral exists only for even powers ofxis important, because inthe path integral approach to quantum field theory the factorx2nis related to the creation and annihilation ofparticles, which always occurs next most complicated Gaussian integral involves a linear term in the exponential, as inI= exp( 12ax2+Jx)dx(5)whereJis a evaluate this integral, we use the old high school algebra trick of completing the 12ax2+Jx= 12a(x Ja)2+J22awe can use the substitutionx x J/ato see that this is essentially the same as (2)

5 With a constant term, so that exp( 12ax2+Jx)dx= exp[J22a] 2 a(6)We can evaluate Integrals likeI= xmexp( 12ax2+Jx)dxusing the same differentiation trick we used earlier, but now we have the choice of differentiating with respect toeitheraorJ. It really makes no difference, but differentiating withJis a tad easier. In that case we havedIdJ= xexp( 12ax2+Jx)dx=Jaexp[J22a] 2 aNote that whenJ= 0, we get the same odd/even situation as before, and the integral is zero. But now a seconddifferentiation gives2d2 IdJ2= x2exp( 12ax2+Jx)dx=1aexp[J22a] 2 a(1 +J2a)which is the same as (4) if we setJ= 0.

6 Takingsuccessivederivatives leads to more complicated expressions, butthey all reduce to (4) whenJ= 0. The business of settingJ= 0after differentiation is a fundamental technique inquantum field theory, as it is used to generate particles from the source termJ. And again, it always only appliesto even powers in the variablex(which becomes the field (x)in the path integral).Up to this point we have dealt only with Gaussian Integrals having the single variablex. But in quantumfield theory there can be an infinite number of variables, and so we need to investigate how the Gaussian integralsbehave when the variablexbecomes then-dimensional vectorx, where the dimensionnmay be infinite.

7 So tobegin, let s look at the generalization of (2) inndimensions, which looks like .. exp( 12xTAx) (7)wherexTis a row vector,xis a column vector, andAis a symmetric, non-singularn nmatrix. The situationat first glance appears hopeless, but we can simplify things by transforming the vectorxto some other vectorywith an orthogonal matrixS(ST=S 1) with a determinant of unity:x=Sydx= |S| |S|= 1. The integral now looks like exp( 12yTS 1 ASy) I am now using just one integral sign for compactness. We now demand that the matrixSbe a diagonalizingmatrix, which means thatS 1AS= d100 0d20 00d3.

8 AndS 1A 1S= 1/d100 01/d20 001/d3 .. This changes the integral to exp[ 12(d1y21+d2y22+..dny2n)] the variables thus separated, we can use (2) to write the answer as exp( 12xTAx) 2 d1 2 d2 2 dn=(2 )n/2|A|1/2(8)which reduces to (2) for the casen= 1(remember that|A|=d1d2 dn). In quantum field theory the numeratorblows up asn , but this term ends up in an overall normalization constant, so we don t worry about it we have to tackle then-dimensional version of (5), which is a bit tricky. This integral looks like exp( 12xTAx+JTx) (9)whereJis a vector whose elements are all constants [in quantum field theory, theJ s become source functions forthe field (x), and thus are very important].

9 We proceed as before, using the diagonalizing matrixS, with thetransformation 12xTAx+JTx 12yTDy+JTSyExpanded, this looks like 12(d1y21+d2y22+.. dny2n) +JaSa1y1+JaSa2y2+.. JaSanynwhere summation over the indexais assumed. We now have to complete the square for everyyiterm. For example,the first set of terms will be 12d1y21+JaSa1y1= 12d1(y1 JaSa1d1)2+(JaSa1)22d1which can be expressed with matrix summation notation as 12di(yi JaSaidi)2+(JaSai)22diNote that division bydi[= (S 1AS)ii]looks a tad odd but, becausediis a diagonal element, the summationnotation is preserved. Now, by again making the substitutionyi yi JaSai/di, we have (2) all over again, sothe integral becomes simply exp( 12xTAx+JTx)dnx=(2 )n/2|A|1/2exp[(JaSai)22di]While correct, the exponential term can be put into a more recognizable form.

10 First, we have to ask ourselves: justwhat exactly is(JaSai)2/2di? It has to be a scalar, and the only way to write this as such isJaSai(JbSbi)T2di=JaSaiS 1ibJb2di(remember thatST=S 1). It would appear now thatSaiS 1ib= ab, but this would giveJTJ/2di, with anindexed denominator that has nowhere to go. So instead we put the1/diin the middleand writeJaSaiS 1ibJb2di=12 JaSai 1di S 1ibJb(10)Now recall that matrix theory guarantees that the diagonal matrix defined byS 1AS=Dallows a similarexpression for the inverse ofA, which isS 1A 1S=D 1. In matrix summation notation, this isS 1ikA 1kbSbi=D 1ii=1diPlugging this into (10), theS 1 Sterms cancel at last, and we have12 JaSai 1di S 1ibJb=12 JaSaiS 1ikA 1kbSbiS 1icJc=12 JkA 1kcJc=12 JTA 1J4 The Gaussian integral is thus exp( 12xTAx+JTx) (2 )n/2|A|1/2exp[12 JTA 1J](11)In quantum field theory, the matrixA 1is related to what is known as apropagator, and it s akin to the 2+m2term in the probability amplitude we presented earlier.