Mathematics for Physicists:Introductory Concepts and Methods

Solutions to even-numbered problems

ALEXANDER ALTLAND & JAN VON DELFT

January 9, 2019

Contents

S Even-numbered solutions 3

SL Problems: Linear Algebra 5S.L1 Mathematics before numbers 5

Sets and Maps 5Groups 5Fields 7

S.L2 Vector spaces 8Vector spaces: examples 8Basis and dimension 11

S.L3 Euclidean geometry 12Normalization and orthogonality 12Inner product spaces 12

S.L4 Vector product 15Algebraic formulation 15Further properties of the vector product 16

S.L5 Linear Maps 17Linear maps 17Matrix multiplication 18The inverse of a matrix 19General linear maps and matrices 22Matrices describing coordinate changes 24

S.L6 Determinants 27Computing determinants 27

S.L7 Matrix diagonalization 29Matrix diagonalization 29Functions of matrices 34

S.L8 Unitarity and Hermiticity 35Unitarity and orthogonality 35Hermiticity and symmetry 36Relation between Hermitian and unitary matrices 41

S.L10 Multilinear algebra 42Direct sum and direct product of vector spaces 42Dual space 43Tensors 44

i

ii Contents

Alternating forms 45Wedge product 45Inner derivative 46Pullback 47Metric structures 47

SC Solutions: Calculus 49S.C1 Differentiation of one-dimensional functions 49

Definition of differentiability 49Differentiation rules 49Derivatives of selected functions 49

S.C2 Integration of one-dimensional functions 51One-dimensional integration 51Integration rules 52Practical remarks on one-dimensional integration 56

S.C3 Partial differentiation 59Partial derivative 59Multiple partial derivatives 59Chain rule for functions of several variables 60

S.C4 Multi-dimensional integration 60Cartesian area and volume integrals 60Curvilinear area integrals 63Curvilinear volume integrals 64Curvilinear integration in arbitrary dimensions 66Changes of variables in higher-dimensional integration 68

S.C5 Taylor series 70Complex Taylor series 70Finite-order expansion 71Solving equations by Taylor expansion 72Higher-dimensional Taylor series 73

S.C6 Fourier calculus 74The δ-Function 74Fourier series 79Fourier transform 81

Case study: Frequency comb for high-precision measurements 85S.C7 Differential equations 87

Separable differential equations 87Linear first-order differential equations 90Systems of linear first-order differential equations 91Linear higher-order differential equations 98General higher-order differential equations 99Linearizing differential equations 100

S.C8 Functional calculus 101Euler-Lagrange equations 101

iii

S.C9 Calculus of complex functions 103Holomorphic functions 103Complex integration 104Singularities 105Residue theorem 106

SV Solutions: Vector Calculus 112S.V1 Curves 112

Curve velocity 112Curve length 112Line integral 113

S.V2 Curvilinear Coordinates 114Cylindrical and spherical coordinates 114Local coordinate bases and linear algebra 117

S.V3 Fields 118Definition of fields 118Scalar fields 119Extrema of functions with constraints 120Gradient fields 122Sources of vector fields 123Circulation of vector fields 125Practical aspects of three-dimensional vector calculus 127

S.V4 Introductory concepts of differential geometry 134Differentiable manifolds 134Tangent space 134

S.V5 Alternating differential forms 135Cotangent space and differential one-forms 135Pushforward and Pullback 136Forms of higher degree 137Integration of forms 140

S.V6 Riemannian differential geometry 141Definition of the metric on a manifold 141Volume form and Hodge star 142

S.V7 Differential forms and electrodynamics 143Laws of electrodynamics II: Maxwell equations 143Invariant formulation 144

SL Problems: Linear Algebra

S.L1 Mathematics before numbers

S.L1.1 Sets and Maps

PL1.1.2 Composition of maps

(a) Since 02 = 0, (±1)2 = 1, (±2)2 = 4, the image of S under A is T = A(S) = {0, 1, 4} .

(b) Since√

0 = 0,√

1 = 1,√

4 = 2, the image of T under B is U = B(T ) = {0, 1, 2} .

(c) The composite map C = B ◦A is given by C : S → U, n 7→ C(n) =√n2 = |n| .

(d) A, B and C are all surjective. B is injective and hence also bijective. A and C arenot injective, since, e.g., the elements +2 and −2 have the same image under A, withA(±2) = 4, and similarly C(±2) = 2. Therefore, A and C are also not bijective.

S.L1.2 Groups

PL1.2.2 The groups of addition modulo 5 and rotations by multiples of 72 deg

(a) 0 1 2 3 40 0 1 2 3 41 1 2 3 4 02 2 3 4 0 13 3 4 0 1 24 4 0 1 2 3

The neutral element is 0.The inverse element ofn ∈ Zq is 5− n.

(b) • r(0) r(72) r(144) r(216) r(288)

r(0) r(0) r(72) r(144) r(216) r(288)

r(72) r(72) r(144) r(216) r(288) r(0)

r(144) r(144) r(216) r(288) r(0) r(72)

r(216) r(216) r(288) r(0) r(72) r(144)

r(288) r(288) r(0) r(72) r(144) r(216)

The neutral element is r(0).The inverse element of r(φ) is r(360− φ).

(c) The groups (Z5, ) and (R72, • ) are isomorphic because their group compositiontables are identical if we identify the element n of Z5 with the element r(72n) of R72.

(d) The group (R360/n, • ) of rotations by multiples of 360/n deg is isomorphic to thegroup (Zn, ) of integer addition modulo n.

5

6 S.L1 Mathematics before numbers

PL1.2.4 Group of discrete translations on a ring

(a) Consider the group axioms:(i) Closure: by definition a, b ∈ Z ⇒ (a + b)modN ∈ ZmodN . Thus: x, y ∈ G ⇒

∃nx, ny ∈ ZmodN : x = λnx, y = λny. It follows that T (x, y) = λ · (nx +ny)modN ∈ λ · ZmodN = G. X

(ii) Associativity: The usual addition of integers is associative, m,n, l ∈ Z ⇒ (m+n) + l = m + (n + l), and this property remains true for addition modulo N .For x, y, z ∈ G we therefore have: T (T (x, y), z) = λ · ((nx +ny) +nz)(modN) =λ · (nx + (ny + nz))(modN) = T (x, T (y, z)). X

(iii) Neutral element: The neutral element is 0 = λ · 0 ∈ G: For all x ∈ G we have:T (x, 0) = λ · (nx + 0)(modN) = x. X

(iv) Inverse element: The inverse element of n ∈ ZmodN is [N + (−n)]modN ∈ZmodN . Therefore the inverse element of x = λ·n ∈ G is given by −x ≡ λ·(N+(−n)) ∈ G, since T (x,−x) = λ ·(n+(N+(−n)))(modN) = λ ·0(modN) = 0. X

(v) Commutativity (for the group to be abelian): For all x, y ∈ G we have T (x, y) =λ · (nx + ny)modN = λ · (ny + nx)modN = T (y, x), since the usual additionof real numbers is commutative, and this property remains true for additionmodulo N . X

Since (G, T ) satisfies properties (i)-(v), it is an abelian group. X

(b) The group axioms of (T, ) follow directly from those of (G, T ):(i) Closure: Tx, Ty ∈ T ⇒ Tx Ty = TT (x,y) ∈ T, since x, y ∈ G ⇒ T (x, y) ∈ G

[see (a)]. X

(ii) Associativity: For Tx, Ty, Tz ∈ T we have: (Tx Ty) Tz = TT (x,y) Tz =TT (T (x,y),z)

(a)= TT (x,T (y,z)) = Tx TT (y,z) = Tx (Ty Tz). X

(iii) Neutral element: The neutral element is T0 ∈ T: For all Tx ∈ T we have:Tx T0 = TT (x,0) = Tx+0 = Tx. X

(iv) Inverse element: The inverse element of Tx ∈ T is T−x ∈ T, where −x is theinverse element of x ∈ G with respect to T , since Tx T−x = TT (x,−x) =Tx+(−x) = T0. X

(v) Commutativity (for the group to be abelian): For all x, y ∈ G we have Tx Ty =TT (x,y) = TT (y,x) = Ty Tx, since the composition rule T inG is commutative. X

Since (T, ) satisfies properties (i)-(v), it is an abelian group. X

PL1.2.6 Decomposing permutations into sequences of pair permutations

(a) The permutation [132] is itself a pair permutation, as only the elements 2 and 3 areexchanged, hence its parity is odd.

(b) To obtain 123 [231]7−→ 231 via pair permutations, we bring the 2 to the first slot, thenthe 3 to the second slot: 123 [213]7−→ 213 [321]7−→ 231, thus [231] = [321] ◦ [213], with evenparity.

Below we proceed similarly: we map the naturally-ordered string into the desired orderone pair permutation at a time, moving from front to back:

7 S.L1.3 Fields

(c) 1234 [3214]7−→ 3214 [1432]7−→ 3412 ⇒ [3412] = [1432] ◦ [3214] even

(d) 1234 [3214]7−→ 3214 [1432]7−→ 3412 [2134]7−→ 3421 ⇒ [3421] = [2134] ◦ [1432] ◦ [3214] odd

(e) 12345 [15342]7−→ 15342 [13245]7−→ 15243 [12435]7−→ 15234 ⇒ [15234] = [12435] ◦ [13245] ◦ [15342]odd

(f) 12345 [32145]7−→ 32145 [21345]7−→ 31245 [15342]7−→ 31542 ⇒ [31542] = [15342] ◦ [21345] ◦ [32145]odd

S.L1.3 Fields

PL1.3.2 Complex numbers – elementary computations

For z1 = 3 + ai and z2 = b− 2i (a, b ∈ R), we find [brackets give results for a = 2, b = 3]:

(a) z1 = 3− ai [3− 2i](b) z1 − z2 = 3− b+ (a+ 2)i [4i](c) z1z2 = (3 + ai) · (b+ 2i) = −2a+ 3b+ i(6 + ab) [5 + 12i]

(d) z1

z2= 3− aib− 2i = (3− ai)(b+ 2i)

(b− 2i)(b+ 2i) = 2a+ 3b+ i(6− ab)b2 + 4 [1]

(e) |z1| =√

9 + a2 [√

13](f) |bz1 − 3z2| = |i(ab+ 6)| = |ab+ 6| [12]

PL1.3.4 Algebraic manipulations with complex numbers

(a) (z + i)2 = (x+ i(y + 1))2 = x2 − (y + 1)2 + i2x(y + 1) ,

(b) z

z + 1 = z

z + 1 ·z + 1z + 1 = (x+ iy)

(x+ 1 + iy) ·(x+ 1− iy)(x+ 1− iy)

= x(x+ 1) + y2 + i(y(x+ 1)− xy)(x+ 1)2 + y2 = x(x+ 1) + y2 + iy

(x+ 1)2 + y2 ,

(c) z

z − i = z

z − i ·z + iz + i = (x− iy)

(x+ i(y − 1)) ·(x+ i(1− y))(x− i(y − 1))

= x2 + y(1− y) + i(x(1− y)− yx)x2 + (y − 1)2 = x2 + y(1− y) + ix(1− 2y)

x2 + (y − 1)2 .

PL1.3.6 Multiplying complex numbers – geometrical interpretation

z1 = 1√8 + 1√

8 i 7→ ( 1√8 ,

1√8 ) ρ1 =

√18 + 1

8 = 12 φ1 = arctan

(1)

= π4

z2 =√

3− i 7→ (√

3,−1) ρ2 =√

3+1 = 2 φ2 = arctan(− 1√

3

)= 11π

6

z3 = z1z2 = ( 1√8 + 1√

8 i)(√

3−i) ρ3 =√

38 + 1

8 + 38 + 1

8 = 1 φ3 = arctan(√

3−1√3+1

)= π

12

8 S.L2 Vector spaces

=√

38 +

√18 + (

√38 −

√18 )i 7→ (

√38 +

√18 ,

√38 −

√18 )

z4 = 1z1

=√

81+i =

√8(1−i)

(1+i)(1−i) ρ4 =√

2 + 2 = 2 φ4 = arctan(−1)

= 7π4

=√

2−√

2i 7→ (√

2,−√

2)

z5 = z1 = 1√8 −

1√8 i 7→ ( 1√

8 ,−1√8 ) ρ5 =

√18 + 1

8 = 12 φ5 = arctan

(−1)

= 7π4

As expected, we find:ρ3 = ρ1ρ2

φ3 = φ1 + φ2

ρ4 = 1/ρ1

φ4 = −φ1

ρ5 = ρ1

φ5 = −φ1

1z

2z

)zRe(

)zIm(

2z1z=3z

6π/

4π/

2

1z1=4z

1z= ¯5z

12π/

6π/11

4π/7

2

4π/

121

S.L2 Vector spaces

S.L2.3 Vector spaces: examples

PL2.3.2 Vector space axioms: complex numbers

First, we show that (C,+) forms an abelian group. Notation: zj = xj + iyj , j = 1, 2, 3.(i) Closure holds by definition: z1 + z2 ≡ (x1 + x2) + i(y1 + y2) ∈ C . X(ii) Associativity: (z1 + z2) + z3 = ((x1 + x2) + i(y1 + y2)) + (x3 + iy3)

= ((x1 + x2) + x3) + i((y1 + y2) + y3)= (x1 + (x2 + x3)) + i(y1 + (y2 + y3))= z1 + (z2 + z3) . X

(iii) Neutral element: z + 0 = (x+ 0) + i(y + 0) = x+ iy = z . X

(iv) Additive inverse: with − z = (−x) + i(−y)is z + (−z) = (x− x) + i(y − y) = 0 . X

(v) Commutativity: z1 + z2 = (x1 + x2) + i(y1 + y2)= (x2 + x1) + i(y2 + y1) = z2 + z1 . X

Second, we show that multiplication by a real number, · , likewise has the propertiesrequired for (C,+, ·) to form a vector space. For λ, µ ∈ R, we have

(vi) Closure: λz = λ(x+ iy) = (λx) + i(λy) ∈ C, since λx, λy ∈ R

9 S.L2.3 Vector spaces: examples

(vi) Multiplication of a sum of scalars and a complex number is distributive:

(λ+ µ) · z = (λ+ µ)(x+ iy) = (λ+ µ)x+ i(λ+ µ)y= (λx+ µx) + i(λy + µy) = λz + µz . X

(vii) Multiplication of a scalar and a sum of complex numbers is distributive:

λ(z1 + z2) = λ((x1 + x2) + i(y1 + y2)) = λ(x1 + x2) + iλ(y1 + y2)= (λx1 + λx2) + i(λy1 + λy2) = λz1 + λz2 . X

(viii) Multiplication of a product of scalars and a complex number is associative:

λ · (µ · z) = λ · (µx+ iµy) = (λµ)x+ i(λµ)y = λµ(x+ iy) = (λµ) · z . X

(ix) Neutral element: 1 · z = 1 · (x+ iy) = z . X

Therefore, the triple (C,+, ·) represents an R-vector space.Remark: This R-vector space is two-dimensional (i.e. isomorphic to R2), since each com-plex number z = x+ iy is represented by two real numbers, x and y. This fact is utilizedwhen representing complex numbers as points in the complex plane, with coordinates xand y in the horizontal and vertical directions, respectively.

PL2.3.4 Vector space of polynomials of degree n

(a) The definition of addition of polynomials and the usual addition rule in R yield

pa(x) + pb(x) = a0x0 + a1x

1 + . . . anxn + b0x

0 + b1x1 + . . . bnx

n

= (a0 + b0)x0 + (a1 + b1)x1 + . . . (an + bn)xn = pa+b(x) ,

since a + b = (a0 + b0, . . . , an + bn)T ∈ Rn+1. Therefore pa pb = pa+b . X

The definition for the multiplication of a polynomial with a scalar and the usualmultiplication rule in R yield

cpa(x) = c(a0x0 + a1x

1 + . . . anxn) = ca0x

0 + ca1x1 + . . . canx

n = pca(x) ,

since ca = (ca0, . . . , can)T ∈ Rn+1. Therefore c • pa = pca . X

(b) We have to verify that all the axioms for a vector space are satisfied. First, (Pn, )indeed has all the properties of an abelian group:

(i) Closure: adding two polynomials of degree n again yields a polynomial of degreeat most n. X

(ii,v) Associativity and commutativity follow trivially from the corresponding prop-erties of Rn+1. For example, consider associativity:

pa (pb pc) = pa pb+c = pa+(b+c) = p(a+b)+c = pa+b pc = (pa pb) pc.X

(iii) The neutral element is the null polynomial p0, i.e. the polynomial whose coef-ficients are all equal to 0. X

(iv) The additive inverse of pa is p−a. X

10 S.L2 Vector spaces

Moreover, multiplication of any polynomial with a scalar also has all the propertiesrequired for (Pn, , •) to be a vector space. Multiplication with a scalar c ∈ R satisfiesclosure, since c • pa = pca again yields a polynomial of degree n. X All the rules formultiplication by scalars follow directly from the corresponding properties of Rn+1.XEach element pa ∈ Pn is uniquely identified by the element a ∈ Rn+1 – this identi-fication is a bijection between Pn and Rn+1, hence (Pn, , •) is isomorphic to Rn+1

and has dimension n+ 1. X

(c) The bijection between Pn and Rn+1 associates the standard basis vectors in Rn+1,namely ek = (0, . . . 1, . . . , 0)T (with a 1 at position k and 0 ≤ k ≤ n), with a basis inthe vector space (Pn, , •), namely {pe0 , . . . , pen}, corresponding to the monomials{1, x, x2, . . . , xn}, since pek (x) = xk. This statement corresponds to the obvious factthat every polynomial of degree n can be written as linear combination of monomialsof degree ≤ n.

PL2.3.6 Vector space with unusual composition rule – multiplication

(a) First, we show that (Va, ) forms an abelian group.

(i) Closure holds by definition. X(ii) Associativity: (vx vy) vz = vx+y−a vz = v(x+y−a)+z−a = vx+y+z−2a

= vx+(y+z−a)−a = vx vy+z−a = vx (vy vz).X

(iii) Neutral element: vx va = vx+a−a = vx, ⇒ 0 = va .X

(iv) Additive inverse: vx v−x+2a = vx+(−x+2a)−a = va = 0, ⇒ −vx = v−x+2a .X

(v) Commutativity : vx vy = vx+y−a = vy+x−a = vy vx . X

(b) To ensure that the triple (Va, , ·) forms an R-vector space, scalar multiplication,· , which by definition satisfies closure, also has to have the following four properties,each of which amounts to a condition on the form of f :

(vi) Multiplication of a sum (γ + λ) · vx = γ · vx λ · vx

of scalars and a vector v(γ+λ)x+f(a,γ+λ) = vγx+f(a,γ)+λx+f(a,λ)−a

is distributive: f(a, γ + λ) = f(a, γ) + f(a, λ)− a (1a)

(vii) Multiplication of a scalar λ · (vx + vy) = λ · vx λ · vy

and a sum of vectors vλ(x+y−a)+f(a,λ) = vλx+f(a,λ)+λy+f(a,λ)−a

is distributive: −λa + f(a, λ) = f(a, λ) + f(a, λ)− a (1b)

(viii) Multiplication of a product (γλ) · vx = γ · (λ · vx)of scalars and a vector v(γλ)x+f(a,γλ) = vγ(λx+f(a,λ))+f(a,γ)

is associative: f(a, γλ) = γf(a, λ) + f(a, γ) (1c)

(ix) Neutral element: 1 · vx = vx

vx+f(a,1) = vx

f(a, 1) = 0 (1d)

Evidently, the form of f is fully determined by the distributivity condition (vii), since

11 S.L2.4 Basis and dimension

(1b) yields f(a, λ) = a(1− λ) . It is easy to check that this form also satisfies theequations (1a), (1c) and (1d) resulting from the other three conditions (vi), (viii) and(ix).

(c) The above arguments hold, too, if a and x are elements of Rn, for any positiveinterger n. In other words, there is nothing special about the case n = 2 consideredabove.

S.L2.4 Basis and dimension

PL2.4.2 Linear independence

(a) The three vectors are linearly independent if and only if the only solution to theequation

0 = a1v1 + a2v2 + a3v3 = a1(1

23

)+ a2

(246

)+ a3

(−1−1

0

), with aj ∈ R, (1)

is the trivial one, a1 = a2 = a3 = 0. The vector equation (1) yields a system of threeequations, (i)-(iii), one for each of the three components of (1), which we solve asfollows:

(i) 1a1 + 2a2 − 1a3 = 0 (ii)-(i)⇒ (iv) a1 = −2a2

(ii) 2a1 + 4a2 − 1a3 = 0 (iv) in (i)⇒ (v) a3 = 0

(iii) 3a1 + 6a2 + 0a3 = 0 (iv) in (iii)⇒ (vi) 0 = 0

(ii) minus (i) yields (iv): a1 = −2a2. Inserting (iv) into (i) yields a3 = 0. (iv) into(iii) yields no new information. There are thus infinitely many non-trivial solutions(one for every value of a1 ∈ R), hence v1, v2 and v3 are not linearly independent.

(b) The desired vector v′2 = (x, y, z)T should be linearly independent from v1 and v3,i.e. its components x, y and z should be chosen such that the equation 0 = a1v1 +a2v′2 + a3v3 has no non-trivial solution, i.e. that it implies a1 = a2 = a3 = 0:

(i) 1a1 + xa2 − 1a3 = 0 (ii)-(i)⇒ (vi) choose x = 1 , then a2 = 0.

(ii) 2a1 + ya2 − 1a3 = 0 (iv) in (i)⇒ (v) choose y = 0 , then a3 = 0.

(iii) 3a1 + za2 + 0a3 = 0 (iv) in (iii)⇒ (iv) choose z = 0 , then a1 = 0.

(iii) yields (iv): 3a1 = −za2; to enforce a1 = 0 we choose z = 0. (iv) inserted into(ii) yields (v): a3 = ya2; to enforce a3 = 0 we choose y = 0. (iv,v) inserted into (i)yields xa2 = 0; to enforce a2 = 0, we choose x = 1. Thus v′2 = (1, 0, 0)T is a choicefor which v1, v′2 are v3 linearly indepedent. This choice is not unique – there areinfinitely many alternatives; one of them, e.g. is v′2 = (2, 4, 1)T .

PL2.4.4 Einstein summation convention

(a) aibi = a1b

1 + a2b2 = 1 · (−1) + 2 · x = 2x− 1 .

12 S.L3 Euclidean geometry

(b) aiajbibj = (aibi)(ajbj)

(a)= (2x− 1)(2x− 1) = 4x2 − 4x+ 1 .

(c) a1ajb2bj = a1b

2ajbj (a)= 1 · x · (2x− 1) = 2x2 − x .

S.L3 Euclidean geometry

S.L3.2 Normalization and orthogonality

PL3.2.2 Angle, orthogonal decomposition

(a) cos(^(a,b)) = a · b‖a‖‖b‖ = 2 ·

√2 + 0 · 1 +

√2 · 1√

4 + 0 + 2 ·√

2 + 1 + 1=√

32 ⇒ ^(a,b) = π

6

(b) Vector from P to Q: c = q − p =(

32

)Vector from P to R: d = r− p =

(0

13a

)R

Q

S

P

d⊥d

‖d

c

Comp. of d parallel to c: d‖ = (d · c)c = (d · c)c‖c‖2 = (13a) · 2

9 + 4

(32

)= a

(64

)Comp. of d perp. to c: d⊥ = d− d‖ =

(0

13a

)− a(

64

)= a

(−6

9

)Consistency check: d⊥ · d‖ = a2(6 · (−6) + 4 · 9

)= 0. X

Coordinates of S: s = r− d⊥ =(

−1−1 + 13a

)− a(−6

9

)=(−1 + 6a−1 + 4a

)(c) Distance from R to S: RS = ‖d⊥‖ = a

√36 + 81 = a

√117

Distance from P to S: PS = ‖d‖‖ = a√

36 + 16 = a√

52

S.L3.3 Inner product spaces

PL3.3.2 Unconventional inner product

All the defining properties of an inner product are satisfied:(i) Symmetric:

〈x,y〉 = x1y1 + x1y2 + x2y1 + 3x2y2 = y1x1 + y1x2 + y2x1 + 3y2x2 = 〈y,x〉 . X

(ii,iii) Linear:

〈λx + y, z〉 = (λx1 + y1)z1 + (λx1 + y1)z2 + (λx2 + y2)z1 + 3(λx2 + y2)z2

= (λx1z1 + λx1z2 + λx2z1 + 3λx2z2) + (y1z1 + y1z2 + y2z1 + 3y2z2)