I. Are the statements in quotation marks true or false?
1. “The function sin (1/z) does not have a Taylor series expansion in the neighbor- hood of z = 0.”
2. A function f(z) is defined by the power series ∞
n=0
z2n+1/[n! (n+ 1)!] about the origin.
“f(z) is an entire function of z.”
3. “The only singularity of 1/Γ(z) is a simple pole at z= 0.”
4. “The Mittag-Leffler expansion of the gamma function is given by Γ(z) = ∞
n=0
(−1)n/[(z+
n)n!].”
5. Let f(z) =z+z3+z9+z27+· · · ad infinitum, for |z|<1.
“f(z) cannot be analytically continued outside the unit circle.”
6. “The power series ∞
n=1
(ln n)zn/n converges at all points on its circle of conver- gence.”
7. “The function f(z) = 1/(ez−1) is a meromorphic function of z.”
8. “Dispersion relations for the real and imaginary parts of a generalized suscepti- bilityχ(ω) can be derived only if the corresponding response function φ(t) decays to zero faster than any negative power of t, ast → ∞.”
9. “The derivative of the gamma function, Γ(z), has zero residue at each of its poles.”
10. “The Legendre function of the second kind, Qν(z), has branch points in the z- plane even when ν is a positive integer.”
11. “The Laplace transform of the functionf(t) = cosh πthas no singularities in the region Res > π.”
12. Bessel’s differential equation is
z2 d2
dz2 +z d
dz + (z2−ν2)
f(z) = 0, where ν is a parameter.
“If φν(z) is any solution of this equation, then φ−ν(z) must be equal to φν(z), apart from a possible multiplicative constant.”
13. “The group M¨ob (2,C) of M¨obius transformations of the complex plane has con- tinuous subgroups, but no discrete subgroups.”
14. “The group M¨ob (2,C) of M¨obius transformations of the complex plane is iso- morphic to the group SO(3,1) of homogeneous proper Lorentz transformstions in (3 + 1)-dimensional spacetime.”
15. “The Riemann surface of the function f(z) =z1/2(z−1)−1/3 has 6 sheets.”
16. “It is possible to find a contour integral representation of the beta function B(z, w) that is valid for all complex values of both z and w.”
17. “The Riemann zeta function ζ(z) cannot be continued analytically to the left of the line Rez = 12, because it has an infinite number of zeroes on that line.”
18. “The Fourier transform operator inL2(−∞, ∞) has a finite number of eigenval- ues, each of which is infinitely degenerate.”
19. LetG(x, x) denote the Green function of the differential operator d2/dx2 where x∈[−1,1].
“As a function of x, G is continuous at x = x, but its derivative ∂G/∂x has a finite discontinuity at x=x.”
20. “The fundamental Green function of the Laplacian operator∇2in four-dimensional Euclidean space is G(r,r) = −1/[4π2(r−r)2].”
21. Consider the diffusion equation in d-dimensional space, ∂f /∂t = D∇2f with boundary condition f(r, t)→0 as r → ∞and initial condition f(r,0) =δ(d)(r).
“The fundamental solution to this equation is a Gaussian in each Cartesian com- ponent of r, for all positive integer values of the dimension d.”
22. The scattering amplitude for the scattering of a nonrelativistic particle of mass m in a central potential λV(r) is given by
f(k, θ) =− mλ 2π2
d3r e−ik·rV(r)ψ(r), where k is the scattered wave vector.
“This formula is valid only if the potential V(r) decays to zero as r → ∞more rapidly than any inverse power of r.”
23. Continuation: “In the Born approximation, the scattering amplitude in the for- ward direction (θ= 0) vanishes identically.”
24. Continuation: “In the Born approximation, the imaginary part of the scattering amplitude vanishes identically.”
25. Consider the Helmholtz operator ∇2+k2 in three-dimensional space.
“The fundamental Green function of this operator, corresponding to outgoing spherical waves, is G(r−r) =−eik·(r−r)/
4π|r−r| .”
26. Consider the wave operator (1/c2)∂2/∂t2− ∇2 in (d+ 1)-dimensional spacetime, where c is the speed of light in a vacuum. Let G(d)(R, τ) denote the causal re- tarded Green function of the operator.
“G(d)(R, τ) vanishes identically when (cτ,R) is a time-like four-vector.”
27. Continuation: “G(d)(R, τ) is singular when (cτ,R) is a light-like four-vector.”
28. Let J = (Ji, J2, J3) be the generators of rotations in three-dimensional space, satisfying the Lie algebra [Jj, Jk] =ijklJl.
“The lowest-dimensional, non-trivial, unitary representation of the generators is in terms of (2×2) matrices with complex elements.”
29. “The parameter space of the group SU(n) is doubly connected.”
30. “The first homotopy group of the parameter space of the special orthogonal group SO(n), for every n≥3, is Z2.”
II. Fill in the blanks in the following.
1. Given that the imaginary part of an entire functionf(z) is v(x, y) =e(x2−y2) sin (2xy), the function is f(z) =· · ·.
2. The meridian of longitude ϕ on the Riemann sphere is mapped into a straight line in the complex plane. The equation of this straight line isy =mx+c, where m =· · · and c=· · ·.
3. The region of absolute convergence in the complex z-plane of the power series ∞
n=0
%(n+ 1)/(n2+ 1)&
(12z)n is · · ·.
4. The residue at infinity of the function f(z) = (z−z−1)3 is Res
z=∞f(z) =· · ·. 5. Let [z1, z2; z3, z4] denote the cross-ratio of the four points z1, z2, z3 and z4 in the
complex plane. Then [z1, z2; z3, z4] + [z1, z3;z2, z4] =· · ·.
6. The M¨obius transformation z → w such that three given points z1, z2, z3 are mapped respectively into three other given points w1, w2, w3 is expressed by a relation between w and z that reads · · ·.
7. Under the M¨obius transformation z →w= (z+ 1)/(z+ 2), an infinitesimal area element δA centered at the point z = −3/2 is mapped to an element of area λ δA, where the value of λ is · · ·.
8. The Bernoulli numbersBnare defined via the expansionz/(ez−1) = ∞
n=0
Bnzn/n!.
Therefore Bn is given by the contour integral Bn=· · ·. (You must specify both the integrand and the contour.)
9. The Chebyshev polynomial of the second kind, Un(cos θ), has the generating function
1
1−2t cos θ+t2 = ∞ n=0
Un(cos θ)tn,
where θ ∈ [0, π]. Therefore Un(cos θ) can be expressed as a contour integral in the t-plane given byUn(cos θ) =· · ·. (You must specify both the integrand and the contour.)
10. Continuation: Evaluating the contour integral and simplifying the result, the final expression forUn(cos θ) isUn(cos θ) =· · ·. (You must express your answer in terms of trigonometric functions of θ.)
11. Continuation: Hence the polynomialU1(cos θ) reduces toU1(cos θ) =· · ·. 12. The function f(z) = (z2+ 2)1/3 has branch points atz =· · ·.
13. Express your answer in terms of a Bessel function:
The residue of f(z) = exp (z−z−1) at z = 0 is Res
z=0 f(z) =· · ·.
14. The inverse Laplace transform of f(s) = 1/(s 2−2s+ 1) is f(t) =· · ·. 15. Let λ be a positive constant. The Laplace transform of the function
f(t) = t
0
dtn tn
0
dtn−1· · · t2
0
dt1e−λ(t−t1) is f(s) =· · ·.
16. Let
f(x) =
1− |x|, |x| ≤1 0, |x|>1.
If f(k) denotes the Fourier transform of f(x), the value of the integral ∞
−∞
dk |f(k)|2 =· · · .
17. The positional probability distribution at any time t≥0 of a random walker on a square lattice with sites labelled by the integers (, m) is given by
P(, m, t) =e−λt(p1/q1)/2(p2/q2)m/2I 2λt√
p1q1 Im
2λt√ p2q2
,
where λ is the mean jump rate and pi, qi are directional probabilities such that p1+q1+p2+q2 = 1. The leading asymptotic behavior ofP(, m, t) at very long times (λt1) is given by P(, m, t)∼ · · ·.
18. The diffusion equation for the positional probability density of a particle diffusing on the x-axis in the region−∞ < x ≤ a, in the presence of a constant force, is given by
∂p(x, t)
∂t =−c∂p(x, t)
∂x +D∂2p(x, t)
∂x2 .
Here c and D are positive constants denoting the drift velocity and diffusion constant, respectively. p(x, t) is normalized to unity for all t ≥ 0. There is a reflecting boundary at the point x = a. The boundary condition satisfied by p(x, t) at x=a is then given by · · ·.
19. Continuation: As t → ∞, p(x, t) tends to the stationary probability density pst(x). This quantity satisfies the ordinary differential equation · · ·.
20. Continuation: The normalized solution for pst(x) ispst(x) =· · ·.
21. A quantum mechanical particle of mass m moving in one dimension has the Hamiltonian H = p2/(2m), where p is the momentum operator of the particle.
Its momentum-space wave function at t = 0 is given to be φ(p). Therefore its momentum-space wave function at any time t≥0 is given byψ(p, t) =· · ·. 22. The scattering amplitude for a nonrelativistic particle of mass m in a central
potential λV(r) is given, in the Born approximation, by fB(k, θ) =−2mλ
2Q ∞
0
dr r sin (Qr)V(r),
whereQis the magnitude of the momentum transfer vectorQ. The forward scat- tering amplitude in the Born approximation is therefore given by the expression fB(k,0) =· · ·.
23. Continuation: The backward scattering amplitude in the Born approximation is therefore given by the expression fB(k, π) = · · ·.
24. Let R = r−r and τ = t−t, as usual. Let G(d)(R, τ) denote the fundamen- tal Green function of the Klein-Gordon operator +µ2, where µ is a positive constant and = (1/c2)(∂2/∂t2)− ∇2, in (d+ 1)-dimensional spacetime. Then G(d)(R, τ) can be expressed in the form
G(d)(R, τ) = 1 (2π)d
ddkφ(k,R, τ), where φ(k,R, τ) = · · ·.
25. Let J = (J1, J2, J3) denote the generators of rotations in three-dimensional space, and let ψ be an arbitrary angle. The quantity e−iJ1ψJ2eiJ1ψ, expressed as a linear combination of the generators, is · · ·.
26. Continuation: Let n = (n1, n2, n3) be an arbitrary unit vector. Then the commutator [J·n, J2] = .
27. The number of generators of the orthogonal group O(n) and the special orthog- onal group SO(n) are, respectively, · · · and · · ·.
28. The number of generators of the unitary group U(n) and the special unitary group SU(n) are, respectively, · · · and · · ·.
29. Letxandpdenote the position and momentum operators of a quantum mechan- ical particle moving in one dimension, so that their commutator [x , p] = iI, where I is the unit operator. Let a be a real constant with the physical dimen- sions of length. Using Hadamard’s Lemma, the operator eiap/x e−iap/ simplifies to
eiap/x e−iap/ =· · · .
This result tells us why the momentum operatorpis the generator of translations in position space.
30. Continuation: Let b be a real constant with the physical dimensions of linear momentum. Once again, using Hadamard’s Lemma, the operator e−ibx/p eibx/
simplifies to
e−ibx/p eibx/ =· · · .
This result tells us why the position operator x is the generator of translations in momentum space.
Quiz 2: Solutions
I. True or false:
1. T 2. T 3. F 4. F 5. T 6. F 7. T 8. F 9. T 10. T 11. T 12. F 13. F 14. T 15. T 16. T 17. F 18. T 19. T 20. T 21. T 22. F 23. F 24. T
25. F 26. F 27. T 28. T 29. F 30. T
II.Fill in the blanks:
1. f(z) =ez2.
2. m = tanϕ and c= 0.
3. |z|<2.
4. Res
z=∞f(z) =−3.
5. 1.
6. (w−w2)(w1−w3)
(w−w3)(w1−w2) = (z−z2)(z1−z3) (z−z3)(z1−z2). 7. λ= 16.
8. Bn= n!
2πi
C
dz
zn(ez−1), where C encloses the origin once in the positive sense.
9. Un(cos θ) = 1 2πi
C
dt
tn+1(1−2t cos θ+t2), where C encloses the origin once in the positive sense.
10. Un(cos θ) = sin (n+ 1)θ sin θ . 11. U1(cos θ) = 2 cos θ.
12. i√
2, −i√ 2,∞. 13. −J1(2).
14. t et.
15. 1/[s(s+λ)n].
16. 4π/3.