PHYS 3900 Homework Set #03

Transcription

1 PHYS 3900 Homework Set #03 Part = HWP Due: Mon. Feb. 2, 208, 4:00pm Part 2 = HWP 3.05, Due: Mon. Feb. 9, 208, 4:00pm All textbook problems assigned, unless otherwise stated, are from the textbook by M. Boas Mathematical Methods in the Physical Sciences, 3rd ed. Textbook sections are identified as h.cc.ss for textbook hapter cc, Section ss. omplete all HWPs assigned: only two of them will be graded; and you don t know which ones! Read all Hints before you proceed! Make use of the latest version of the PHYS3900 Homework Toolbox, posted on the course web site. Do not use the calculator, unless so instructed! All arithmetic, to the extent required, is either elementary or given in the problem statement. State all your answers in terms of real-valued elementary functions (+,, /,, power, root,, exp, ln, sin, cos, tan, cot, arcsin, arccos, arctan, arcot,...) of integer numbers, e and π; in terms of i where needed; and in terms of specific input variables, as stated in each problem. So, for example, if the result is, say, ln(7/2) + (e 5 π/3), or 7 9/2, or (6 4π) 0, then just state that as your final answer: no need to evaluate it as decimal number by calculator! Simplify final results to the largest extent possible; e.g., reduce fractions of integers to the smallest denominator etc..

2 HWP 03.0: Do the integrals in textbook problems h. 4.3, No. 6 and No., by parameterization of the integration curves. Hints: () All the integrals must be done by finding a parameterization z(τ) for the integration curve where z(τ) is defined on a real parameter interval a τ b. Then convert the complex line integral into definite integrals over [a, b]: f(z)dz = b a f(z(τ)) d z(τ) dτ dτ Also, for a curve consisting of two or more connected straight or circular segments [in No. 6(b) or ], the integral along should first be broken up into the sum of integrals along each segment and then a parameterization should be applied to integrate over each segment. (2) Use the parameterizations of straight line and circular arc segments that were discussed in the lecture, as follows: Straight line from z a to z b : z(τ) = z a + (z b z a ) τ a b a with a < b, τ [a, b] The choice of the real parameter interval bounds a and b is arbitrary, as long as you choose a < b. (You may want to try different ones to see this explicitly!) A convenient choice is sometimes a = 0 and b = ; or a = and b =. Another convenient choice is very often: a = 0 and b = L where L is the length of the straight line segment. ircular arc segment on a circle of radius R, centered at an arbitrary complex number c, starting at an angle θ a and ending at θ b : or z(τ) = c + Re iτ, τ [a, b] with a θ a, b θ b, if θ a < θ b ; z(τ) = c + Re iτ, τ [a, b] with a θ a, b θ b, if θ b < θ a. The case θ a < θ b represents a counterclockwise arc with parameter interval bounds a θ a and b θ b ; and the case θ b < θ a represents a a clockwise arc, with parameter interval bounds a θ a and b θ b ; Thus, setting θ a = 0 and θ b = 2π gives a full counterclockwise circle; setting θ a = 0 and θ b = 2π gives a full clockwise circle. Make sure to verify all the foregoing statements yourself, e.g. by plugging in some numbers for c, R, θ a and θ b and plotting out the corresponding curves. HWP 03.02: Do textbook problem h. 4.6, No. 2. That is, show that the residue of a complex differentiable (D) function f(z) with an n-th order pole at z = z o can be calculated as follows: onstruct the function F (z) := (z z o ) m f(z) with an integer m n. Evaluate its (m )-th derivative, F (m ) (z), at z o, and divide by (m )! to obtain the residue of f(z) at z o : Resf(z o ) = (m )! F (m ) (z o ) 2

3 Hints: () Since f(z) has an n-th order pole at z o, you can expand it into a Laurent series around z o of the form n f(z) = a j (z z o ) j b j + (z z o ) j j=0 For an n-th order pole, the b j -coeffients terminate after j = n, i.e., b j = 0 for all j > n. (2) Multiply this Laurent series by (z z o ) m, to get the Taylor series expansion for F (z). (3) Take the (m )-th derivative of this Taylor series, set z = z o and divide by (m )!. HWP 03.03: Do textbook problems h. 4.6, No. 3. You are allowed to use (without proof) the result from HWP and/or the Residue Recipes, distributed with the Toolbox. HWP 03.04: This problem deals with a proof of the residue theorem for the simplest case of a single isolated singularity, enclosed by a simple closed curve. Therefore you are not allowed to use or assume validity of the residue theorem here: you re supposed to prove it! However, you are allowed to use without proof auchy s integral theorem and a theorem pertaining to the existence and convergence of a Laurent series around an isolated singularity, stated in Part (b) below. (a) Suppose K is a clockwise circle of radius R centered at a complex number z o. Use the parameterization function z(τ) for a clockwise circle centered at z o (see HWP 3.0 or Toolbox) to evaluate K m := dz(z z o ) m for any integer m = 0, ±, ±2,,... Show that K K m = 2πi if m = ; else : K m = 0 if m = 0, +, ±2, ±3, ±4,.... Footnote: Notice the minus sign here: K is clockwise! Hint: () Use the circle parameterization given in HWP03.0 to turn K m into a definite integral of the real integration variable τ over the interval [0, 2π]. (2) After you ve expressed K m as an integral over areal parameter variable τ, write out the τ-integrand explicitly for the special case m = and note how it differs from the case of all other integer values m with m. (3) To evaluate the definite τ-integral when m, use the fact that e inτ /(in) is an indefinite integral of e inτ, but only if n 0. Also use the fact that e 2πin = for any integer n. (4) What is the indefinite integral of e inτ in the special case n = 0? Use that to evaluate the definite τ-integral for the special case m =. (b) Assume, some complex differentiable function f(z), defined on some open set V, has an isolated singularity at z o. By the Laurent series convergence theorem, f(z) can then be expanded into a Laurent series around z o f(z) = a m (z z o ) m + m=0 j= m= b m (z z o ) m and this series will be absolutely convergent with a zero inner and a positive outer convergence radius R c, i.e., for all z with 0 < z z o < R c. 3

4 Use this and the result from Part (a), to prove the residue theorem for any clockwise circle K centered at z o inside the Laurent convergence disk: f(z)dz = 2πi b for 0 < R < R c. K Footnote: Here b is also referred to as the residue of f(z) at z o : b Res f(z o ). Note the minus sign here again: K is clockwise! (c) Generalize the result from Part (b) to any counter-clockwise simple closed curve which encloses z o, but no other singularities of f(z), as shown in Fig (A), i.e. prove that f(z)dz = +2πi b Note the plus sign here: is counter-clockwise! Hints: () Add to, and then subtract again from, the lollipop curve L shown in Fig (A). So, write: f(z)dz = f(z)dz + f(z)dz f(z)dz. As shown in Fig (A), L consists of three pieces: the clockwise circle K with radius R chosen so that 0 < R < R c and so that K is entirely in the interior of ; and two parallel straight line segments S δ and T δ connecting to K and K to, respectively: L L = S δ K T δ. (2) S δ and T δ are separated by a (small) distance δ. This δ > 0 is taken small enough so that the endpoints of both S δ and T δ fall on K, at z QS and z QT, and define a short curve segment Q δ that is part of K. Likewise, the endpoints of S δ and T δ on, z P S and z P T, define a short curve segment P δ that is part of. Also, let δ denote the curve from z P T to z P S along, obtained from by removing the short segment P δ. Likewise, let K δ denote the open path from z QS to z QT along K, obtained from K by removing the short segment Q δ. So, and K each have now been broken up into two pieces: = δ P δ and K = K δ Q δ. (3) Re-connect and re-arrange the pieces: combine δ, S δ, K δ and T δ into a new closed path to show that f(z)dz = N N := δ S δ K δ T δ. f(z)dz + f(z)dz + f(z)dz P δ Q δ L L f(z)dz (4) Where is z o located relative to N? Is z o inside or outside of N? Thus explain why auchy s integral theorem applies to the integral of f(z) over N and use it to get f(z)dz. N Hence show: f(z)dz = f(z)dz + f(z)dz f(z)dz f(z)dz f(z)dz P δ Q δ K S δ T δ 4

5 (5) Get rid of the integrals over P δ, Q δ, S δ, and T δ by considering their combined curve R := P δ T δ Q δ S δ where T δ means T δ with direction reversed and S δ means S δ with direction reversed. As can be seen in Fig.03.04, R is the counter-clockwise closed loop R : z P S z P T z QT z QS z P S. The integrals over P δ, Q δ, S δ, and T δ in the -integral can therefore be combined into P δ f(z)dz T δ f(z)dz + Q δ f(z)dz S δ f(z)dz = f(z)dz. Also, R does R not enclose z o, i.e., f(z) is complex differentiable in the interior of R, without any singularity. Therefore, auchy s integral theorem, applied to f(z)dz, gives...? R (6) Now what s left over of f(z)dz? Use the result from Part (b) to complete the proof. Footnote: It should be quite obvious how to generalize the foregoing construction to prove the general residue theorem for a counter-clockwise simple closed curve which encloses in its interior more than one isolated singularity of f(z). You are not required to do this as part of your homework, but it is illustrated in Fig (B) which shows a enclosing 3 singularities: z, z 2 and z 3. For each singularity z k (k =, 2, 3), a lollipop L k is attached to. Each lollipop contains a clockwise circle K k which is centered at z k ; and straightline segments S k,δ and T k,δ which again cut out short P k,δ - and Q k,δ -segments, from and K k, respectively. [The Q k,δ -segments have been removed from each K k and are not labeled in Fig (B), just for clarity]. The integral of f(z) along each L k is then added and subtracted to/from f(z)dz. Just as in Fig (A), a new closed curve N is then constructed by removing all P k,δ a- segments from and the Q k,δ -segments from the respective K k. As in Fig (A), this new closed curve N thus comprises most of and most of each K k, except for the P k,δ - and Q k,δ -segments. Again, N excludes all singularities from its interior and we can thus apply auchy s integral theorem to N f(z)dz. One can then again combine each set of P k,δ -, T k,δ -, Q k,δ - and S k,δ -segments into a closed, singularity free loop R k which does not enclose any singularities of f(z). So, auchy s integral theorem again gives R k f(z)dz. One thus obtains f(z)dz = k K k f(z)dz. This gives the general residue theorem, after using the result from Part (b) to evaluate each K k f(z)dz. 5

6 Fig HWP 03.05: Using contour integration and the residue theorem, evaluate the following Fourier integrals: Γ sin(ωt) F (t) := dω Γ 2 + (ω Ω) 2 F 2 (t) := F 3 (t) := Γ 3 cos(ωt) dω [Γ 2 + (ω Ω) 2 ] 2 dω Γ3 sin(ωt) Γ 4 + (ω Ω) 4 with real-valued constants Γ > 0 and Ω. Express your answers in terms of t, Γ and Ω. Hints: Before starting the contour integral formulation: () Since cos(ωt) = Re(e iωt ) and sin(ωt) = Im(e iωt ), write F, F 2 and F 3 as either the real part or the imaginary part of complex-valued integrals where the cos- and sin-functions have been replaced by the e iωt - function, using, e.g., + dωre[f(ω)] = Re[ + dωf(ω)]. (2) Get rid of the Γ and Ω parameters inside the integrals, by the following substitutions: express ω by x := (ω Ω)/Γ, t by s := Γt, and Ω by ν := Ω/Γ. (3) Depending on the sign of t, then close the realaxis contour with a semi-circle of radius R either in the upper half or in the lower half of 6

7 the complex plane such that the contribution from the semi-circle vanishes in the limit of R. (4) To find the poles of the integrands in the complex plane, recall that the two 2nd roots of are +i and i (needed for F and F 2 ). and the four 4th roots of are ±(i + )/ 2 and ±(i )/ 2 (needed for F 3 ). (5) Use Residue Recipes from the Toolbox to find the residues and complete the calculation. HWP 03.06: In quantum field theory, one must sometimes calculate so-called Matsubara imaginary frequeny sums which might look like this, for example: S n = n k= n iπ(2k + ) a iπ(2k + ) b with real a and b and 0 a b 0, and the limit n to be taken. This problem illustrates how to evaluate such an infinite Matsubara sum, S := lim n S n. After completion of Part (a), you may choose one of two possible routes to solve this problem: Route I = Parts (b), (c), (d) or Route II = Parts (e), (f), (g). You do not have to complete both routes! Hint: Route II is easier. (a) For complex argument z, let f(z) := e z + and g(z) := z a z b and h(z) := f(z)g(z). Find all poles of g(z); and find and enumerate all poles of f(z). Use those to find and enumerate all poles of h(z) and their corresponding residues. Make a neat drawing of the complex plane to mark the location of each of these poles with a little : you ll need this treasure map to deal with the land mines below. Hints: () Assume, first without proof (then show it later after you ve found them all), that all poles of g(z) and f(z) are st order and the resp. sets of poles are disjoint [meaning: wherever g(z) has a pole f(z) does not, and vice versa]. (2) f(z) has a pole at some z o, if, and only if, /f(z) = 0 at z = z o (more precisely: lim z zo [/f(z)] = 0, since, strictly speaking, f(z) is not defined at a pole z o ). The same reasoning applies to the poles of g(z). (3) Note that e iπ = and that e 2iπk = if, and only if, k is an integer, k = 0, ±, ±2, ±3,... Hence show that e z = if, and only if, z is one of the following complex numbers, z k : z k := iπ(2k + ) with integer k = 0, ±, ±2, ±3... Use this to find all zeroes of /f(z), and hence all poles of f(z), and show that they are disjoint from the poles of g(z). 7

8 (4) Because of (), each pole of f(z) or g(z) is a st order pole of h(z). Those are the only poles of h(z). Use e.g. Residues Recipes (D) or (E) from the Toolbox to find the residue of h(z) for each of its poles. Route I: (b) Show that S n can be written as S n = M + n dz 2πi h(z) + M n dz 2πi h(z) where M ± n are two counter-clockwise closed rectangular loop contours, as shown in the Fig.03.06(A). The height r n, width 2δ, and real-axis off-set ±iɛ of these rectangular loops are chosen such that δ > 0 and 0 < ɛ < π and r n = π(2n + ). Hint: Show that neither loop, M ± n, encloses any poles of g(z); loop M + n encloses the first (n + ) poles of f(z) along the positive imaginary axis (nearest to the real axis in the upper complex half-plane); and loop M n encloses the first (n + ) poles of f(z) along the negative imaginary axis (nearest to the real axis in the lower complex half-plane). Then express the above two loop integrals, and, in terms of the respective sums of residues of M + n M n enclosed poles, using the residue theorem. (c) Deform M ± n into semi-circles n ±, plus closure by corresponding straight-line segments L ± n, as shown in Fig.03.06(B): the n ± have radius r n ; the L ± n are parallel to the real axis with real parts ranging from r n to +r n and off-sets of ±iɛ from the real axis, as shown in Fig.3.06(B). When combined with short vertical straight line segments P n and Q n (each of length 2ɛ), the straight-line segments L ± n form a clockwise(!) closed loop R n : R n = L + n + P n + L n + Q n. Using again the residue theorem or the auchy integral theorem, with appropriate concatenation of integration contours, show that dz S n = R n 2πi h(z) dz P n 2πi h(z) dz Q n 2πi h(z) + dz n + 2πi h(z) + dz n 2πi h(z) Assuming without proof (see Footnote below) that the last four integrals are vanishing in the limit n, thus show that the infinite-series limit of S n is given by: S := lim S n = lim n n R n dz 2πi h(z) Footnote: To prove that lim n is zero for each of the last four integrals in S n, one can use the upper bound theorem, valid for any complex integrand function h(z), integrated along any finite-length contour X : 0 dz h(z) L(X ) max( h(z) ) =: B(h, X ) X 8 z X

9 where L(X ) is the length of the contour X and max z X ( h(z) ) is the maximum absolute value of the integrand h(z) along the contour X. For each of the last four integrals in S n (i.e., for contour X n P n or Q n or + n or n ), one can show that the upper bound B to the integral absolute values vanishes: lim n B(h, X n ) = 0. And since each integral s absolute value, X n dz h(z) is squeezed between 0 and B(h, X n ), we also have lim n X n dz h(z) = 0. Q.E.D..0& Fig x - lj t,- " (d) Using the residue theorem and the results from (a), evaluate the integral S n dz := 2πi h(z) R n for n chosen suffiently large so that R n encloses a and b, i.e., for r n > max( a, b ). Note here that R n then encloses only the poles of g(z), but no poles of f(z), since ɛ < π. 9

10 Using the last result from (c), thus state your final result for the infinite Matsubara sum S k= iπ(2k + ) a iπ(2k + ) b, expressed in terms of elementary algebraic and exponential functions of the real parameters a and b. Route II: (e) Using the residue theorem and the results from (a), evaluate the integral I n = h(z) dz n 2πi, where n is a counter-clockwise circle, centered at 0, with radius s n = π(2n + 2) and n is a positive integer, chosen large enough so that s n > a and s n > b. Express I n as a sum over the residues of all isolated singularities of h(z) which n encloses. State clearly which singulaties of h(z) are enclosed and which ones are not enclosed by n if s n obeys the foregoing conditions: s n > a and s n > b. Hints: Recall from Part (a) that at every singularity of h(z) is either a singularity of f(z) or a singularity of g(z). Draw the circle n into the treasure map constructed in Part (a)! (f) Use the results from Part (a) for the residues of h(z) and from Part (e) to write I n in the form I n = S n + R where S n is the finite sum defined above in the problem statement. Show that the term R is independent of n, assuming again that s n > a and s n > b. Evaluate R, expressed in terms of elementary algebraic and exponential functions of the parameters a and b. Hint: The term R contains only contributions from the residues of h(z) at the singulaties of g(z), but none from the residues of h(z) at any singularities of f(z). (g) You may assume without proof that lim I n = 0 n [The proof of this can be obtained by application of the upper bound theorem, similiar to the Footnote given in Part (c) above. You are not required to provide this proof as part of your HW solution, but I highly recommnd you try to prove it to yourself!] 0

11 Use the foregoing n -limit of I n and the result from Part (f), to evaluate S k= iπ(2k + ) a iπ(2k + ) b, expressed in terms of elementary algebraic and exponential functions of the parameters a and b.