Quantum Field Theory Homework 3 Solution

Transcription

1 Quantum Field Theory Homework 3 Solution 1 Complex Gaußian Integrals Consider the integral exp [ ix 2 /2 ] dx (1) First show that it is not absolutely convergent. Then we should define it as Next, show that the answer is (1 + i) π. lim exp [ ix 2 /2 ] dx. (2) R Now let s try to understand that result by making a few plots. Plot the real part of the integrand as a function of x for x [ 10, 10]. Then plot the incomplete integral... as a function of R for R [0, 10]. Repeat for the imaginary part of the integrand. takes. Use these plots to explain in words, how the integral manages to take the value that it 1.1 Solution The absolute value of exp(ix 2 /2) is 1. Therefore the absolute integral is which clearly diverges as R. exp [ ix 2 /2 ] dx Now let s do the integral. Using contour integration we can write f(x)dx (1+i)R (1+i)R f(z)dz ir 1 dx 2R (3) f(z)dz +ir R f(z)dz (4) for any f(z) an entire function (or free from singularities in the region bounded by the above contours). Applying to our problem, e ix2 /2 dx e z2 (1 + i)dz 0 e ir2 /2 e z iz2 /2 dz 0 e ir2 /2 e Rz iz2 /2 dz (5) and the first integral is (1+i) π for R 1, while the second and third integrals are bounded by 1/R. So in the large R limit we get what we want. But let s see how the pictures look: left/right are the functions / integrals. 1

2 2 Asymptotic series Consider the baby or toy version of scalar φ 4 theory, where it is just a single integral; ( ) φ 2 Z dφ exp 2 + λφ4. (6) 24 This is what the path integral for scalar φ 4 theory would look like if there were only one point in spacetime (and after rotating the contour for φ so the i s go away). Consider Z as a function of λ. 2.1 Values Evaluate Z(0) Z(0.01) Z(0.1) Z(0.4) Z(1) Z(5) numerically to 20 digits, for instance, with Mathematica. 2

3 2.1.1 Solution It turns out there is a closed form solution! Thank you Mathematica! The numerical values are Z(λ) 3e 3/4λ K 1 (3/4λ) 4. (7) λ 1/2 Z(0) Z(0.01) Z(0.1) Z(0.4) Z(1) Z(5) (8) 2.2 Series expansion Replace exp( λφ 4 /24) with its series expansion in λ (or equivalently, in φ). Find explicitly the λ 0 and λ 1 terms in the series. Evaluate the λ 0 and the sum of λ 0 and λ 1 terms (first and second partial sums) numerically, for each of the examples you did above. For which cases does the λ 1 term help improve the accuracy? Solution Replace e λx4 /24 1 λx 4 /24. The x 4 integrates into 3, so we get 2π(1 λ/8). Without the added term, all of them equal 2π Z(0). The new 20 digit results with the first correction are Z(0) Z(0.01) Z(0.1) Z(0.4) Z(1) Z(5) (9) The correction is an improvement out to λ 1 but it isn t much of an improvement there and it makes things worse at λ 5. It always underestimates the answer, for a reason we will see. 3

4 2.3 Asymptotic series Find the complete series expansion in λ in closed form, that is, write Z(λ) c n λ n, (10) and find an explicit expression for c n. (Do this by expanding the exponent, exchanging orders of summation and integration, and doing the integral for each term in the series.) Show that the radius of convergence (in λ) of this series is zero Solution Simple enough: we just series expand the exponential, interchange sum and integral, and perform all integrations; dxe x2 /2 e λx4 /24 dxe x2 /2 ( λ) n ( λ) n 2π 2π ( λ) n x 4n dxe x2 /2 x 4n (4n)! 2 2n (2n)! ( λ) n (4n)! 24 n 2 2n (2n)!n!. (11) In other words, Z(λ) c n λ n, c n 2π ( 1)n (4n)! 24 n 2 2n (2n)!n!. (12) The act of interchanging the summation and integration requires a uniform convergence property which is not satisfied; this is the origin of the trouble below. Note that (2n)!/2 n n! is often written (2n 1)!!. The radius of convergence can be found by considering the ratio of neighboring terms and examining the large n limit; c n+1 (4n + 4)(4n + 3)(4n + 2)(4n + 1) c n 96(2n + 2)(2n + 1)(n + 1) (4n + 3)(4n + 1) 24(n + 1) 2n 3. (13) In the large n limit this diverges absolutely; no nonzero λ, however small, can keep the series from diverging absolutely. Therefore the radius of convergence is zero. Note that the series is alternating. 4

5 2.4 So what good is it? The expansion e x m0 ( 1) m has the following property for x > 0: the partial sums f n (x) x m 1 x + x2 2 x (14) n m0 ( 1) m x m (15) are alternately strict over-estimates and strict under-estimates of the actual function; that is, for x > 0, f 0 (x) 1 > e x, f 1 (x) (1 x) < e x, f 2 (x) (1 x+x 2 /2) > e x, and so forth with the <, > alternating. (Extra credit: prove this.) Use this property to prove that the partial sums found above, Eq. (10) with n cut off at 0, 1, 2, 3,..., are alternately over-estimates and under-estimates. Therefore, the true answer always lies between neighboring terms in the series of partial sums. Use this property to find a bound for Z(λ) at λ 1, by evaluating alternating terms until they start to diverge. How tight is the bound? Repeat for Z(0.4) and Z(0.1). Argue that the bound becomes tighter and tighter as λ gets smaller, so at small λ, while the series does not converge, it gives us very good information about the value of Z(λ). A series with this property zero radius of convergence but the ability to give good information near the origin is called an Asymptotic Series Solution First, we prove the alternating nature of partial sums of e x. Note that 1 > e x > 0 for all positive x. Therefore the property holds for n 1, 0. Now suppose the property holds up to some integer n 1 and consider the function f n (x) e x. Note that df n (x) dx d dx n m0 ( 1) m x m n m1 ( 1) m x m 1 (m 1)! n 1 m0 ( 1) m x m f n 1 (x). Therefore d dx (f n(x) e x ) e x f n 1 (x). By our induction assumption f n 1 (x) e x satisfies the property. Furthermore f n (0) e 0 0. If a function starts at zero and has positive/negative derivative, then it is positive/negative at all positive values. Therefore if f n 1 (x) e x is positive then f n e x is negative and vice versa. The result then follows by induction. 5

6 sum Now we use this condition. For real x, x 4 is nonnegative and real. Hence, for odd N the N ( λ) n x 4n and similarly with > for even N. Therefore it is strictly true that e x2 /2 e λx4 /24 e x2 /2 < exp( λx 4 /24), (16) N ( λ) n x 4n, (17) for odd N, because e x2 /2 > 0. Integration preserves this inequality; if one integral has a smaller integrand, everywhere, it will be a smaller integral. Furthermore, since the summation is finite, there is no problem interchanging the sum and integration, so we arrive at N dxe x2 /2 e λx4 /24 c n λ n, N odd. (18) Similarly for N even, the partial sum is always larger. Therefore each partial sum is either an overestimate or an underestimate, and the series of partial sums alternates. By writing out the partial sums for N odd and finding the largest one, and writing out the partial sums for N even and finding the smallest one, we can find a bound on the integral. As soon as the increments start to grow, which occurs around n 3/2λ (by our determination of the ratio of c n ), we should terminate the partial summation. For λ 1, simply evaluating the sum term by term gives that the best limit available is < Z(1) < , (19) where the values N 1, 2 set the limits. For λ 0.4, the limit is < Z(0.4) < , (20) where N 3, 4 set the limits. For λ 0.1, the limits are < Z(0.1) < , (21) where N 15, 14 set the limits. Note that the 3/2λ rule is working. It is usually argued that the best estimate is the value halfway between the bounds; in this case it is accurate to 4 parts in If we do λ 1/20 the tightness of the bound is wide, and the midpoint estimate is good to a part in ; for λ 1/100 the series is good for around 50 digits. At small λ the asymptotic series is very accurate. At large values it is useless. 6

7 2.5 Negative λ What happens when λ < 0? Solution In this case, the original integral is not convergent. The partial sum can be shown to be a strict underestimate, for even or odd N; the fact that the series diverges reflects the fact that the original integral is not convergent. The function Z(λ) can be considered as a function of λ in the complex plane. For Re λ > 0, it is a finite and analytic function; but its definition by the integral given is not convergent in the negative real half plane. It is possible to analytically continue it; in fact I presented an analytic formula which is well defined at nonzero λ. However, the formula involves square roots, and therefore has a branch cut on the negative real axis which runs up to the origin. The origin is not a point of analyticity for the function, which is why the radius of convergence is zero. However, the asymptotic series about λ 0 gives partial sums which bound the function in the positive real half plane in a manner discussed; this turns out to prove that the full analytic function can be reconstructed from the series we have found a process called Borel resummation, which we will not discuss in this class. 7