Some necessary math (part I) Note: This is a partial collection of several formulae that we will need throughout the first part of the course. For convenience, we will review them now quickly. The ones for which I do not give proofs here are either very well known, or will be parts of your first assignment – if needed, you can find the missing proofs in the solutions of the first assignment. 1 Euler integrals (a) The Gamma function is defined for any a> 0 as: ∞ Γ(a)= dxxa−1e−x (1) Z0 Properties: 1 1. Γ(1) = 1;Γ( 2 )= √π. 2. Γ(a +1) = aΓ(a) 3. Γ(n +1) = n! for any integer n 1. We will use this a lot! ≥ (b) We will use this second function just once, to compute the volume of a N-dimensional sphere, a bit later on. This type of integral appears quite often all over physics, so it’s good to know it in general. For any a> 0,b> 0, we define: 1 B(a, b)= dxxa−1(1 x)b−1 (2) Z0 − Then, Γ(a)Γ(b) B(a, b)= Γ(a + b) 2 Gaussian integrals (a) one-dimensional: (1) For any a such that Re(a) > 0, we define: ∞ 2 π I(a)= dxe−ax = (3) Z−∞ r a This is an integral that we will use many many times, so let us prove that it is true. The trick is to compute [I(a)]2 using polar coordinates: ∞ ∞ 2 2 [I(a)]2 = dxe−ax dye−ay Z−∞ Z−∞ Now, change to polar coordinates x = r cos(φ), y = r sin(φ) dxdy = rdrdφ, x2 + y2 = r2, and of course the integral is over the whole 2D plane, φ [0, 2π], r →[0, ) ∈ ∈ ∞ → ∞ 2π ∞ 2 du π [I(a)]2 = drr dφe−ar = 2π e−au = Z0 Z0 Z0 2 a where we changed the variable u = r2. The last step is a trivial integral. 1 (2) For any b and a with Re(a) > 0, we define: ∞ 2 −ax2+bx π b J(a, b)= dxe = e 4a (4) Z−∞ r a ′ b The proof of this is straightforward, by changing the variable of integration to x = x 2a . (3) For any even integer n, we define: − n ∞ Γ +1 n −ax2 2 In(a)= dxx e = ³n+1 ´ (5) Z−∞ a 2 If n is odd, In(a) = 0 because the integrand is an odd function. 2 n n−1 n+1 Proof: change variables ax = y xdx = dy/(2a) x dx = dyy 2 /(2a 2 ) → → → n ∞ ∞ Γ +1 n −ax2 1 n−1 −y 2 2 In(a) = 2 dxx e = n+1 dyy e = ³n+1 ´ Z0 a 2 Z0 a 2 by definition of the Γ function. (b) m-dimensional Gaussian integral: leta ˆ be a m m matrix which is symmetric (aij = aji), non- × singular (deta = 0) and positive defined (all its eigenvalues, let’s call them λ , ..., λm, are larger than 6 1 zero: λi > 0, i = 1, ..., n). Then: ∞ ∞ ∞ m − m m a x x π i j ij i j I(ˆa)= dx1 dx2 . dxme =1 =1 = (6) −∞ −∞ −∞ sdeta Z Z Z P P The demonstration is fairly straightforward: perform a change of variables to new variables for m m m 2 which the exponent is diagonalized: aijxixj λiy . The integral now factorizes i=1 j=1 → i=1 i in m independent, simple gaussian integrals.P P Since deta =P λ1λ2 ...λm, the final results follows immediately. Finally, if ˆb =(b1, . , bm) is a collection of m numbers, then: ∞ ∞ ∞ m − m m a x x m b x π 1 m m a−1 b b i j ij i j + i i i 4 i j ( )ij i j J(ˆa, ˆb)= dx1 dx2 . dxme =1 =1 =1 = e =1 =1 Z−∞ Z−∞ Z−∞ sdeta P P P P P (7) where a−1 is the inverse of the matrix a. So these last two are just direct generalizations of the one-dimensional Gaussian integrals. I don’t think we will need them – generally we will only encounter the one-dimensional gaussian integrals. Those, you absolutely have to be able to deal with, they will appear over and over. 3 Saddle-point approximation and Stirling formula The saddle-point (or gaussian) approximation is a very convenient way to obtain approximations for integrals of the general form: ∞ I = dxeF (x) (8) Z0 when F (x) is a function which has a single maximum at a value xM 1 (see sketch below). In this ≫ case, the integrand exp(F (x)) is a function that is highly-peaked at xM , and quickly becomes very small for values x far from xM . The idea, then, is to approximate the function F (x) accurately in the vicinity of xM , since that’s where most of the contribution to the integral comes from. 2 F(x) We use a Taylor expansion, and stop to the first non-trivial term: 2 (x xM ) F (x) F (x )+(x x )F ′(x )+ − F ′′(x )+... M M M 2 M xM x ≈ − ′ ′′ exp[F(x)] But F (xM ) = 0,F (xM ) < 0 (conditions for a ′′ maximum). Let a = F (xM )/2 > 0. Then: − ∞ ∞ 2 2 I dxeF (xM )−a(x−xM ) = eF (xM ) dye−ay ≈ Z0 Z−xM −ax2 where y = x xM . Now, if xM 1 e M 1, and we may− as well extend the≫ lower→ limit all≪ the x x M way to , since we’re only adding an extremely Fig 1. Upper plot: some function F (x) with a maximum small contribution.−∞ In this case, the integral be- at xM . The dashed line shows the Taylor expansion to second order (a parabola). The long vertical lines show comes a simple gaussian (hence the name gaussian where is this Taylor expansion a good approximation. approximation) and we find: Lower plot: the exponential of F (x) (i.e., the integrand of interest to us) is generally large in the region where ∞ F (x) 2π F (x ) the Taylor expansion is valid. M I = dxe ′′ e 0 ≈ s F (xM ) Z | | The most famous example of such an approximation is the Stirling formula, which we will use very many times. This is an approximation for the Gamma function, for large values of a 1: ≫ ∞ ∞ Γ(a +1) = dxxae−x = dxea ln x−x Z0 Z0 ′ a Define F (x)= a ln x x. It has an extremum at F (x)= x 1 = 0 xM = a 1, by assumption. − ′′ a ′′ −1 → ≫ This is a maximum, because F (x)= x2 < 0 F (xM ) = a ,F (xM )= a ln a a. Then, applying the gaussian approximation, we find that:− → | | − 2π a a F (xM ) √ a ln a−a √ Γ(a + 1) ′′ e = 2πae = 2πa (9) ≈ s F (xM ) e | | µ ¶ We will use this most often for integers a = n 1, in which case we have: ≫ n n Γ(n +1) = n! √2πn (10) ≈ µ e ¶ From this, it follows that: 1 ln(n!) n ln n n + ln(2πn)+ . (11) ≈ − 2 Remember that n is a very large number; as a result, the first term here is the largest one, the second one is smaller, the third is even smaller, and neglected corrections are even smaller. Depending on how large n is, we can even stop at the first term. Stirling’s formula keeps just the first two terms: n ln(n!) n ln( ) (12) ≈ e 3 1 0.995 ratio 0.99 0.985 60 80 100 120 140 x x Figure 1: Checking the accuracy of Stirling’s formula. The red (lower) curve shows the ratio [x ln( e )]/ ln(Γ(x + 1)), x 1 while the green (upper curve) shows the improved approximation [x ln( e )+ 2 ln(2πx)]/ ln(Γ(x+1)). Even for relatively small numbers x 100, the formulae work quite well. Even Stirling’s formula recovers over 99% of the actual value. For the sorts of numbers∼ of interest to us, n 1023, this approximation is REALLY good. ∼ Note: this plot was made with Maple. If you want to test the Stirling formula some more, all you need to know is that the Maple name of the Gamma function is GAMMA(x). 4 Volume of a n-dimensional hypersphere This result is demonstrated in the textbook (calculation begins on the bottom of page 129 and page 130); there it is done by a different method. While I obviously expect you to know the answers for n = 1, 2, 3, I will give you the general formula, if it is needed in an exam – so don’t worry about learning it by heart. Although it is sort of a cool result. Question: what is the “volume” enclosed in a n-dimensional hypersphere of radius R, defined by n 2 2 the equation i=1 xi = R ? By definition: P n 2 2 Vn(R)= dx1 . dxnΘ(R xi ) − i Z Z X=1 Θ(x) is called a Heaviside (or step) function and appears quite often in more advanced courses. It is defined by: 0 ,x< 0 Θ(x)= ( 1 ,x> 0 All it does is to tell us that we should “sum” the infinitesimal volumes dV = dx1dx2 ...dxn as long as R2 n x2, i.e. as long as we are inside the hypersphere. ≥ i=1 i First,P define yi = xi/R. It follows immediately that: 1 1 n n 2 n Vn(R)= R dy1 . dynΘ(1 yi )= R Vn(1) −1 −1 − i Z Z X=1 4 i.e.