Lecture notes on Mathematical Methods of Physics

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Mathematical Methods for Physics PHYS 30672 by Niels Walet with additions by Mike Godfrey, and based on work by Graham Shaw Spring 2015 edition Last changed on April 13, 2016iiContents 1 Introduction and Prerequisites 1 1.1 Prerequisites . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.2 Notation for scalar products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2 Linear vector spaces 3 2.1 Definition of a linear vector space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 2.1.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2.2 Linear independence and basis vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 2.2.1 The scalar product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 2.2.2 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.3 Function spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 2.3.1 Continuous basis functions: Fourier Transforms . . . . . . . . . . . . . . . . . . . . . 8 2.3.2 General orthogonality and completeness in function spaces . . . . . . . . . . . . . . 9 2.3.3 Example from Quantum Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 2.4 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 3 Operators, Eigenvectors and Eigenvalues 13 3.1 Linear operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3.1.1 Domain, Codomain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 3.1.2 Matrix representations of linear operators . . . . . . . . . . . . . . . . . . . . . . . . . 14 3.1.3 Adjoint operator and Hermitian operators . . . . . . . . . . . . . . . . . . . . . . . . 15 3.2 Eigenvalue equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 3.2.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 3.3 Sturm–Liouville equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 3.3.1 How to bring an equation to Sturm–Liouville form . . . . . . . . . . . . . . . . . . . 20 3.3.2 A useful result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3.3.3 Hermitian Sturm–Liouville operators . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3.3.4 Second solutions, singularities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3.3.5 Eigenvectors and eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 3.4 Series solutions and orthogonal polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 3.4.1 The quantum-mechanical oscillator and Hermite polynomials . . . . . . . . . . . . . 23 3.4.2 Legendre polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 3.4.3 Bessel functions and the circular drum . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 4 Green’s functions 31 4.1 General properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 4.1.1 First example: Electrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 4.1.2 The eigenstate method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 4.1.3 The continuity method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 4.2 Quantum mechanical scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 4.3 Time-dependent wave equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 iiiiv CONTENTS 4.3.1 Solution for the Green’s function by Fourier transforms . . . . . . . . . . . . . . . . 40 4.3.2 Wave equations in(2+ 1) dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 5 Integral Equations 45 5.1 Classification of linear integral equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 5.1.1 Examples equivalent to differential equations . . . . . . . . . . . . . . . . . . . . . . 45 5.1.2 Examples that are not equivalent to differential equations . . . . . . . . . . . . . . . . 46 5.2 Solution of integral equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 5.2.1 Equations soluble by simple methods . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 5.3 Neumann series solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 5.3.1 Convergence of Neumann series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52 5.3.2 Resolvent kernels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 5.4 Hilbert–Schmidt theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 5.4.1 Eigenfunction expansion of the kernel . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 5.4.2 Hilbert–Schmidt solution of integral equations . . . . . . . . . . . . . . . . . . . . . . 56 6 Variational calculus 59 6.1 Functionals and stationary points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 6.2 Stationary points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 6.3 Special cases with examples: first integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 6.3.1 Functional of first derivative only . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 6.3.2 No explicit dependence on x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 6.4 Generalizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 6.4.1 Variable end points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 6.4.2 One endpoint free . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 6.4.3 More than one function: Hamilton’s principle . . . . . . . . . . . . . . . . . . . . . . 67 6.4.4 More dimensions: field equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 6.4.5 Higher derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 6.5 Constrained variational problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 6.5.1 Lagrange multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 6.5.2 Generalization to functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74 6.5.3 Extensions to the method of Lagrange multipliers . . . . . . . . . . . . . . . . . . . . 78 6.5.4 Eigenvalue problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 6.5.5 The Rayleigh–Ritz method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 A Contour Integration 83 A.1 The Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 A.2 Contour Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 A.3 Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 A.4 Example 1: Simplest case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 A.5 Example 2: Complex exponentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 A.6 Final case: poles on the real axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86Chapter 1 Introduction and Prerequisites This document is based on a summary of the main mathematical results of the course initially prepared by Graham Shaw. We hope they have turned into a reasonably complete (and accurate) guide to the material presented in the course. I would suggest you come back to this from time to time, as this is always very much a work in progress. Please let me know if you find any typos or slips, so I can fix them. This is not intended to be an absolutely complete set of notes, so do not be surprised if some deriva- tions and examples of applying the results are not given, or are very much abbreviated. Of course, this does not imply you do not need to be able to derive or apply these results. Nor need you necessarily mem- orise very complicated equations, just because they are included here. Common sense must be applied; use of good textbooks next to these notes is advisable. There are many different ways to remember mathematics and much of physics. One that is generally useful is to understand a number of the key principles underlying the work, so that you can derive most results quickly. Combined with practice from both the example sheets and additional material as can be found in the textbooks, should prepare you quite well for this course. 1.1 Prerequisites PHYS 20171, Mathematics of Waves and Fields, is a prerequisite for this course. Most students will also have taken PHYS 30201, Mathematical Fundamentals of Quantum Mechanics. There is some overlap between those courses and the introductory material in these notes. In addition, the Section on Green’s Functions requires basic knowledge of contour integration and the residue theorem. The latter material has been covered in PHYS 20672, but it is not essential for under- standing the lectures nor will it be tested in the exam. Some students who have not attended PHYS 20672 may still want to get the gist of the Green’s- function application of contour integration. They should read Appendix A (about 10 pages) and the first two or three pages of section 3.3 of Mathews and Walker, Mathematical Methods of Physics. (The later pages of section 3.3 involve integrating around cuts and branch points, which will not be required here.) There is also a Mathematica notebook (Contour.nb) available on the course web site, as well as a pdf file (Contour.pdf), and much of the material is also summarised in Appendix A. 1.2 Notation for scalar products There are currently two editions of the notes, to cater to different tastes:  In Notes.pdf and in the lectures I use (a, b) for the scalar product of vectors a and b. There is a small chance thatl(a, b) (meaning the product ofl with(a, b)) could be mistaken for a functionl with two arguments a and b, but the correct reading can always be determined from the context. 12 CHAPTER1. INTRODUCTIONANDPREREQUISITES  InNotesBK.pdf the scalar product is represented byhajbi, which is the notation most often found in textbooks of quantum mechanics and the one that students sometimes ask for. Please let me know if the automatic translation from(a, b) tohajbi has missed any cases: I haven’t checked every line.  Followers of fashion should note thatha, bi is yet another notation for the scalar product; it is often found in the recent mathematical literature. Any appropriate notation is acceptable in answers to exam questions.  The scalar product is denoted by(a, b) in the edition of the notes that you are currently reading.Chapter 2 Linear vector spaces 2.1 Definition of a linear vector space A linear vector space V over a scalar set S (we shall typically consider sets S =R orC) is a set of objects (called vectors) a, b, c, . . . with two operations: 1. Addition of any two vectors, c= a+ b; 2. Multiplication by a scalarl2 S, b= la. These must satisfy the following conditions 1. V is closed under addition,8a, b2 V : a+ b2 V. 2. Addition is commutative: 8a, b2 V : a+ b= b+ a and associative 8a, b, c2 V :(a+ b)+ c= a+(b+ c). 3. There exists a null vector 02 V,8a2 V : a+ 0= a. 4. Every element a2 V has an inversea2 V such that a+(a)= 0. 5. The set V is closed under multiplication by a scalar,8a2 V,l2 S : la2 V. 6. The multiplication is distributive for addition of both vectors and scalars, 8a, b2 V,l2 S : l(a+ b)= la+lb, 8a2 V,l,m2 S : (l+m)a= la+ma, and associative, 8a2 V,l,m2 S :l(ma)=(lm)a. 7. There is a unit element 1 in S, such that 1a= a. Note that we have not defined subtraction; it is a derived operation, and is defined through the addition of an inverse element. 34 CHAPTER2. LINEARVECTORSPACES Example 2.1: 0 1 x 3 A The spaceR of vectors r= y = xi+ yj+ zk is a vector space over the set S=R. z Example 2.2: The space of two-dimensional complex spinors       a 1 0 = a +b , b 0 1 a,b2C, is a vector space. Note: If we look at the space of up and down spins, we must require that the length of the 2 2 vectors (the probability),jaj +jbj , is 1. This is not a vector space, since     2 a a 1 2 2 2 2 2 2 2   + =ja +aj +jb +bj =jaj +jbj +jaj +jbj + 2(a a +b b ), 1 2 1 2 1 1 2 2 2 2 1 1 b b 1 2 which is not necessarily equal to 1. Example 2.3: R 2 The space of all square integrable (i.e., all functions f with dxj f(x)j ¥) complex func- tions f of a real variable, f :R7C is a vector space, for S=C. The space defined above is of crucial importance in Quantum Mechanics. These wave func- tions are normalisable (i.e., we can define one with total probability 1). R 2 2 The space of all functions f , f :R7C with dxj f(x)j ¥ is denoted asL (R). 2.1.1 Problems 1. Show that the zero vector 0 is unique, and that for each a there is only one inversea. 2.2 Linear independence and basis vectors A set of vectors a, b, . . . , u is said to be linearly independent provided the equation la+mb+ . . .+su= 0 has no solution exceptl= m= . . .= s = 0. This can be used to show that when you pick one of the vectors a, b, . . . , u, it can not be expressed as a sum over the rest. In cases where there is a largest number of independent vectors:2.2. LINEARINDEPENDENCEANDBASISVECTORS 5 The dimension n of a space is the largest possible number of linearly independent vectors which can be found in the space. Any set of n linearly independent vectors e , e , . . . , e in an n-dimensional space is said to form a 1 2 n complete set of basis vectors, since one can show that any vector x in the space can be expanded in the form x= x e + x e + ...+ x e , (2.1) n n 1 1 2 2 where the numbers x are called the components of x in the basis e , e , . . . , e . 2 n i 1 Example 2.4: Show that the vectors(1, 1, 0),(1, 0, 1) and(0, 1, 1) are linearly independent. Find the compo- nent of a general vector(x, y, z) in this basis. Solution: We get the three coupled equations x = x + x , 1 2 y= x + x , 1 3 z= x + x . 2 3 These have a unique solution if the determinant is not zero, 0 1 1 1 0 A det 1 0 1 6= 0. 0 1 1 This is true, since the determinant equals2. The components are found by solving the equa- tions (Gaussian elimination is quite easy in this case): 1 1 1 x = (x+ y z), x = (x+ z y), x = (y+ z x). 1 2 3 2 2 2 Theorem 2.1. The decomposition (2.1) is unique. Proof. Suppose that there is a second decomposition, x = y e + y e + ...+ y e . Subtract the left- and 1 1 2 2 n n right-hand sides of the two decompositions, collecting terms: 0=(x y )e +(x y )e + ...+(x y )e . n n n 1 1 1 2 1 2 Linear independence of the vectorsfeg implies that x = y , which contradicts our assumption of a i i i second decomposition, and thus it is unique. Let us look at an example in an infinite dimensional space: Example 2.5: The Fourier decomposition of a function defined only on the intervalp,p is given by ¥ f(x)= a + (a cos nx+ b sin nx) . 0 å n n n=1 This means that for the Fourier series the basis functions are: 1,fsin nx, cos nxg, n= 1, . . . ,¥ . It is not at all easy to show that this basis is complete This is a general complication in infinite dimensional spaces.6 CHAPTER2. LINEARVECTORSPACES 2.2.1 The scalar product 1 For any two vectors a, b we can define a scalar product (a, b) a generalizable notation for a b, which satisfies:  (a, b)=(b, a) , (2.2) (a,lb+mc)= l(a, b)+m(a, c) , (2.3) together with (a, a) 0, (2.4) where the equality holds for a= 0 only. Note: The inner product is sometimes described as a mapping from V V7 S, which is another way of saying that it is a scalar function of a pair of vectors in V. We can use the inner product to define the norm (a more correct description of the length) of the vector a, 1/2 kak(a, a) . One can define a length without an inner product. A good example is the so-called “1-norm” of a vector,jjxjj = å jxj, which is used quite commonly in numerical analysis of linear 1 n n algebra. One of the important uses of an inner product is to test whether two vectors are at straight angles: The vectors a and b are said to be orthogonal if(a, b)= 0. Triangle and Cauchy–Schwartz inequality The triangle inequality is the ‘obvious’ statement that the length of the sum of two vectors is less than the sum of the lengths, ka+ bkkak+kbk. From the triangle inequality we could prove the Cauchy–Schwartz inequality, but we take a different approach here: Theorem 2.2. For any two vectors a, b, we have j(a, b)jkakkbk. (2.5) Proof. The proof is simple. We suppose that b6= 0, because if b = 0, Eq. (2.5) holds as a trivial equality, 0= 0. Consider the non-negative function f(x) defined by f(x)=(a+ xb, a+ xb) 0 , where x is a complex variable. By using the properties of the scalar product, this can expanded to give  2 f(x)=(a, a)+ x (b, a)+ x(a, b)+jxj (b, b) 0 . (2.6) 2 For the special value x =(b, a)/kbk , (2.6) becomes 2 2 2 kak j(a, b)j /kbk  0 , which is easily rearranged to give the Cauchy–Schwartz inequality, Eq. (2.5). 1 We shall also use the term inner product for this operation.2.2. LINEARINDEPENDENCEANDBASISVECTORS 7 You might want to show for yourself that the ‘special’ value of x that we needed in the proof is actually the value for which f(x) takes its smallest possible value. Note, if you try this, that f is not an analytic function of x (why not?), so a little care will be needed to find the minimum: you could, for example, minimize with respect to the real and imaginary parts of x separately. Orthogonalisation and orthonormalisation There are several ways to turn an arbitrary set of vectors into an orthogonal set—one where every pair of vectors is orthogonal—, or even better an orthonormal set: an orthogonal set where each vector has length one. We discuss only two of these methods. The traditional approach is the Gram-Schmidt procedure. This procedure is defined recursively. In the 0 mth step of the algorithm one defines the vector e that is orthonormal to the m 1 orthonormal vectors m defined in previous steps. Thus m1 00 0 0 1 :e = e (e , e )e ; m m m å i i i=1 0 00 00 2 :e = e /ke k. m m m The first line above removes all components parallel to the m 1 previous normalised and orthogonal 0 vectors (check), the second step normalises the result, so that e is normalised. m A more modern approach (using numerical linear algebra) is based on the construction of the “overlap matrix” N (also called norm matrix, which is why we use the symbol), with entries N = (e , e ). This ij i j 1/2 matrix is Hermitian (or symmetric if the basis is real), and we can now define a matrix N , such that 1/2 1/2 N NN = I. This can then be used to define the orthonormal basis 0 1/2 e =(N ) e . l k lk For a real symmetric matrix M (and similarly for an Hermitian one, but we shall concentrate on the first case here) we can define matrix powers in a simple and unique way by requiring that the powers are also symmetric matrices. The easiest way to get the result is first to diagonalise the matrix M, i.e., to find its eigenvalues (i) T l and eigenvectors e . We can then write M = O diag(l)O , with O the matrix with the i j (j) normalised eigenvectors as columns, O = e . The eigenvectors are orthonormal, and thus ij i T O O = I. The matrix diag(..) has the eigenvalues l on the diagonal, and is zero elsewhere. T T (Convince yourself that O O= I and O MO= diag(l).) We then define arbitrary powers of M by a a T M = O diag(l )O . (2.7) 2 Orthonormal basis functions: For discrete vector spaces one can always choose an orthonormal set of basis functions satisfying (e , e )= d . (2.8) i j ij Here we have introduced the Kronecker delta d , defined for integer i, j. This object is zero ij unless i = j, when it is 1. 2 Discrete here means that we can label the basis vectors by a finite or infinite set of integers. It contrasts to continuous bases, as discussed in the next section.8 CHAPTER2. LINEARVECTORSPACES For such an orthonormal basis the completeness relation can be written as  (e ) (e ) = d , (2.9) i a i å b ab i where(e ) denotes the ath component of ith basis vector e . i a i 2.2.2 Problems 2. Use the definition of independence to show that Eq. (2.1) holds for any set of independent functions. a 3. Show that M , defined by Eq. (2.7), is a symmetric matrix. 1/2 4. Harder: Show that the eigenvalues of the overlap matrix N are positive, and deduce that N is Hermitian; the matrix is defined up to a choice of sign for the square root. 2.3 Function spaces 2.3.1 Continuous basis functions: Fourier Transforms 3 For a vector space of complex valued functions f :R7C one can choose basis functions 1 ikx e = f (x)=p e , ¥ k¥, k k 2p and expand in these functions, Z ¥ ˜ f(x)= dkf (x) f(k). (2.10) k ¥ The expansion coefficient is nothing but the Fourier transform, Z ¥  ˜ f(k)= dxf (x) f(x). (2.11) k ¥ ikx In much of physics, one traditionally does not normalise thef , but usesf (x)= e . In that k k case an explicit factor 2p enters in the Fourier transforms, but not in the inverse one, Z ¥ dk ikx ˜ f(x)= e f(k), 2p ¥ Z ¥ ikx ˜ f(k)= dx e f(x). ¥ In order to figure out the orthogonality relations, we substitute (2.11) into (2.10), which gives the relation Z Z ¥ ¥ 0  0 0 f(x)= dkf (x) dx f (x ) f(x ), (2.12) k k ¥ ¥ which must hold for any f(x). We now swap the order of integration, and find   Z Z ¥ ¥ 0  0 0 f(x)= dx dkf (x)f (x ) f(x ). (2.13) k k ¥ ¥ 3 If we work in a real function space we should use the real and imaginary parts as a basis for real functions, but it is often easier to deal even with real functions as if they are complex. Think about the Fourier transform of a real function, see below.2.3. FUNCTIONSPACES 9 0 We call the object between square brackets the “Dirac delta function”d(x x ), whered(y) can be defined using the explicit definition of the functionsf , as k Z ¥ 1 iyz d(y)= dz e . (2.14) 2p ¥ (See the appendix to this chapter, 2.4, for additional properties.) From the definition (2.14) of the delta function, we can show that the basis states satisfy the orthonor- mality relation Z ¥  0 (f ,f 0)= dxf (x)f 0(x)= d(k k ) k k k k ¥ and the completeness relation Z ¥  0 0 dkf (x)f (x )= d(x x ). k k ¥ 2.3.2 General orthogonality and completeness in function spaces 4 We start with a space of functions, where the scalar product is assumed to be defined by Z  (f,y)= dxf (x)y(x), where the space from whichf andy are chosen is such that the integral is finite for any pair. In general: Any vector space of functions with a scalar product, where all functions have finite norm, is called a Hilbert space. Suppose in this space we have a discrete (i.e., labelled by integers), but let us assume infinite, set of basis functionsf , chosen to be orthonormal (similar to Eq. (2.8)) n Z  (f ,f )= dxf (x)f (x)= d , m n n nm m using orthogonalisation if necessary. Then an arbitraryy(x) can be decomposed as y(x)= c f (x). (2.15) å n n n The coefficients c can be determined from the orthogonality relation, multiplying (2.15) from the left n  withf , integrating with respect to x, exchanging the order of the summation and the integration on the m right-hand side, we find that (f ,y)= c (f ,f )= c d , m å n m n å n mn n n from which we conclude that c =(f ,y). (2.16) m m Substituting Eq. (2.16) into Eq. (2.15) we find Z 0 0  0 y(x)= dx f (x ) y(x )f (x) n n å n " Z 0 0  = dx f (x ) f (x) y(x) , å n n n 4 More general definitions are possible, but apart from some minor changes to the algebra, the final results hold for all scalar products10 CHAPTER2. LINEARVECTORSPACES where we have interchanged the summation and integration (which mathematicians will tell you may be incorrect). From this we conclude that 0  0 f (x ) f (x)= d(x x ), n n å n which is the form of the completeness relation for a basis labelled by a discrete variable. If the basis is labelled by a continuous label, as for the Fourier transformation, we get the completeness relations discussed for that case, Eqs. (2.12,2.13). This leads to (f ,f )= d(0). k k In this case we do not speak of a Hilbert space, since the basis is not normalisable, the norm of the basis states is necessarily infinite. We might instead speak of a “pre-Hilbert” space, a vector space of functions endowed with a scalar product, but without the assumption of a finite norm. 2.3.3 Example from Quantum Mechanics Much of what we have stated in this section can be illustrated for quantum mechanical problems. Like all linear wave problems, QM relies heavily on the principle of superposition: For any physical system, ify (x, t) andy (x, t) are possible wave functions, then so is 1 2 y(x, t)= ly (x, t)+my (x, t), 1 2 a wherel andm are arbitrary complex numbers. a This superposition principle is not a QM property, but one that is common to all (wave) solutions to linear wave equations. This implies that the space of all y is a linear vector space (over x, since t is a parameter that describes the time evolution). A very important statement can now be found in the fact that: Theorem 2.3. The eigenfunctions y (x) of any physical operator form a complete set. (Can be proven for specific n cases only.) This implies that y(x, t)= c (t)y (x). å n n n A very clear example are the eigenstates of the harmonic oscillator Hamiltonian, 2 2 h ¯ d 1 2 2 ˆ H = + mw x . 2 2m dx 2 ¨ The solutions of the Schrodinger equation ˆ Hy(x, t)= ih ¯¶ y(x, t) t 2 2 are y (x) = exp(x /(2b ))H (x/b), with H a Hermite polynomial. In this case the time-dependence n n n is determined by the eigenenergies, and we conclude that i(n+1/2)wt y(x, t)= a e y (x). å n n n2.4. APPENDIX 11 1 2 Figure 2.1: A sketch of sin(Lx) for a few values of L. This function converges to a Dirac delta function 2p x Figure 2.2: A sketch of a piecewise constant function, that in the limit a 0 gives the Dirac delta function 2.4 The Dirac delta function The Dirac delta functiond(x) is defined by the “reproducing” property, i.e., Z 0 0 0 dx d(x x ) f(x )= f(x) 5 for any function f(x). This is equivalent to the following explicit definition (further forms are discussed in the Mathematica examples), see also Fig. 2.1. Z Z ¥ L 1 1 1 2 izx izx d(x)= dz e  lim dz e = lim sin(Lx) . 2p 2p L¥ 2p L¥ x ¥ L It is often useful to think of the d function as the limit of a simple function, and one example is an infinitely narrow spike, as in Fig. 2.2 for a 0. Important properties 0 0 Since integration with thed function “samples” f(x ) at the single point x = x, we must conclude that 0 0 d(x x )=0 for x6= x . 0 The area under the d function is 1, as can be seen from taking f(x ) = 1. Combining the last two results leads to an interesting expression for the area under the curve, Z x+e 0 0 dx d(x x )=1 for anye 0. xe 5 Thed function is strictly speaking not a function, it is only defined inside an integral A “function” of this kind is usually called a distribution, but we won’t labour the distinction in this course.12 CHAPTER2. LINEARVECTORSPACES A very useful relation is obtained when we scale the variables in the delta function, writing y = ax. Now, for a 0, x¥ corresponds to y¥, so that Z Z ¥ ¥ 1 0 0 0 0 0 0 dx d(a(x x )) f(x )= sign(a) dy d(y y ) f(y /a) , a ¥ ¥ where sign(a) is the so-called signum function, sign(a)=+1 for a 0 and1 for a 0: the sign-change that appears for a 0 comes from reversing the limits of integration and the factor of 1/a arises from the change of variables. From this we find Z ¥ f(y/a) f(x) 0 0 0 dx d(a(x x )) f(x )= = . jaj jaj ¥ We can interpret the result as the contribution from the slope of the argument of the delta function, which appears inversely in front of the function at the point where the argument of the d function is zero. Since the d function is even, the answer only depends on the absolute value of a. Also note that we only need to integrate from below to above the singularity; it is not necessary to integrate over the whole infinite interval. This result can now be generalised to a d-function with a function g(x) as argument: we need to sum over all zeroes of g inside the integration interval, and the quantity a above becomes the slope dg/dx at each of the zeroes,   Z b f(x) dx f(x)d(g(x))= å jdg/dxj a x=x i i where the sum extends over all points x such that g(x )= 0 and a x b. i i i Example 2.6: Calculate the integral Z ¥ 2 2 2 dx f(x)d(x c t ). ¥ Solution: 2 2 2 2 2 2 Let us first calculate the zeroes of x c t , x =ct. The derivative of x c t at these points is2ct, and thus Z ¥ 1 2 2 2 dx f(x)d(x c t )= ( f(ct)+ f(ct)) . 2cjtj ¥ Integrals such as these occur in electromagnetic wave propagation.Chapter 3 Operators, Eigenvectors and Eigenvalues 3.1 Linear operators A linear operator L acts on vectors a, b . . . in a linear vector space V to give new vectors La, Lb, . . . such 1 that L(la+mb)= lLa+mLb Example 3.1: 1. Matrix multiplication of a column vector by a fixed matrix is a linear operation, e.g.   1 2 3 Lx= x. 4 8 1 2. Differentiation is a linear operation, e.g., d L f(x)= f(x). dx 3. Integration is linear as well, Z x 0 0 (L f)(x)= f(x )dx , 1 0 Z 1 0 0 0 (L f)(x)= G(x, x ) f(x )dx , 2 0 are both linear (see example sheet). 3.1.1 Domain, Codomain and Range If the operators L maps the vector f on the vector g, Lf = g, the vector space of f’s (the domain) can be different from the vector space of g’s (the codomain or target). L is an operator which maps the domain onto the codomain, and even though it is defined for every element of the domain, the image of the domain (called the “range of L” or the “image of L”) is in general only a subset of the codomain, see Fig. 3.1, even though in many physical applications the range and codomain do in fact coincide. 1 This changes in the most general case where multiplication is not commutative 1314 CHAPTER3. OPERATORS,EIGENVECTORSANDEIGENVALUES Figure 3.1: The definition of domain, codomain and range Example 3.2: n 1. The operator L:C C defined by La=(b, a) with b a fixed vector, is a linear operator.   3 2 1 3 2 2. The matrix maps from the spaceR of 3-vectors to the codomainR of 2- 6 4 2   1 vectors. The range is the 1D subset of vectorsl ,l2R. 2 3. The (3D) gradient operatorr maps from the space of scalar fields ( f(x) is a real function of 3 variables) to the space of vector fields (r f(x) is a real 3-component vector function of 3 variables). 3.1.2 Matrix representations of linear operators Let L be a linear operator, and y = lx. Let e , e , . . . and u , u , . . . be chosen sets of basis vectors in the 1 2 1 2 domain and codomain, respectively, so that x= e x , y= u y . å i i å i i i i Then the components are related by the matrix relation y = L x , j å ji i i where the matrix L is defined by ji   T Le = u L = L u . (3.1) i å j ji å j ij j j Notice that the transformation relating the components x and y is the transpose of the matrix that connects the basis. This difference is related to what is sometimes called the active or passive view of transformations: in the active view, the components change, and the basis remains the same. In the passive view, the components remain the same but the basis changes. Both views represent the same transformation If the two basis setsfeg andfug are both orthonormal, we can find the matrix elements of L as an inner i j product, L =(u , Le ). (3.2) ji j i3.1. LINEAROPERATORS 15 Example 3.3: d Find a matrix representation of the differential operator in the space of functions on the dx interval(p,p). Solution: Since domain and codomain coincide, the bases in both spaces are identical; the easiest and ¥ most natural choice is the discrete Fourier basis 1,fcos nx, sin nxg . With this choice, using n=1 0 0 (cos nx) =n sin nx and(sin nx) = n cos nx, we find 0 1 0 1 0 1 1 0 1 B C B C B C cos x sin x cos x B C B C B C B C B C B C sin x cos x sin x d B C B C B C T = = M . B C B C B C cos 2x 2 sin 2x cos 2x dxB C B C B C B C B C B C sin 2x 2 cos 2x sin 2x A A A . . . . . . . . . We can immediately see that the matrix representation ”M” takes the form 0 1 0 0 0 0 0 . . . B C 0 0 1 0 0 B C B C 0 1 0 0 0 B C T M = . B C 0 0 0 0 2 B C B C 0 0 0 2 0 A . . . . . . Matrix representation of the time-independent Schrodinger ¨ equation Another common example is the matrix representation of the Schrodinger ¨ equation. Suppose we are ¥ ˆ given an orthonormal basisffg for the Hilbert space in which the operator H acts. By decomposing an i i=1 eigenstatey of the Schrodinger ¨ equation, ˆ Hy(x)= Ey(x) in the basisf (x) asy= c f , we get the matrix form å j j j j H c = Ec , (3.3) å ij j i j with Z  ˆ H =(f , Hf )= dxf(x) Hf (x) . ij i j i j This is clearly a form of Eq. (3.2). The result in Eq. (3.3) is obviously an infinite-dimensional matrix problem, and no easier to solve than the original problem. Suppose, however, that we truncate both the sum over j and the set of coefficients c to contain only N terms. This can then be used to find an approximation to the eigenvalues and eigenvec- tors. See the Mathematica notebookheisenberg.nb for an example how to apply this to real problems. 3.1.3 Adjoint operator and Hermitian operators You should be familiar with the Hermitian conjugate (also called adjoint) of a matrix, the generalisation of transpose: The Hermitian conjugate of a matrix is the complex conjugate of its transpose, †  † T  (M ) =(M ) , or M =(M ) . ij ji16 CHAPTER3. OPERATORS,EIGENVECTORSANDEIGENVALUES Figure 3.2: The definition of a matrix and its Hermitian conjugate Thus         † † 0 i 0 i 0 i 0 i = , = . i 0 i 0 i 0 i 0 We can also define the Hermitian conjugate of a column vector, if 0 1 v 1 B . C †   . v= , v =(v , . . . , v ). A n . 1 v n This allows us to write the inner product as a matrix product, † (w, v)= w v. The most useful definition of Hermitian conjugate, which will be generalised below, is through the inner product: † The Hermitian conjugate M of a matrix M has the property that for any two vectors a and b in the codomain and domain of M, respectively, † (a, Mb)=(M a, b). Thus, with a little algebra,   †  †  † (a, Mb)= a M b = a (M ) b = (M a ) b =(M a, b), (3.4) å ij j å j å i j i i ji ji ij ij ij † see Fig. 3.2. From the examples above, and the definition, we conclude that if M is an m n matrix, M is an n m matrix. We now use our result (3.4) above for an operator, and define † †  8a2 codomain, b2 domain : (a, Lb)=(L a, b)=(b, L a) where the last two terms are identical, as follows from the basic properties of the scalar product, Eq. (2.2). † A linear operator L maps the domain onto the codomain; its adjoint L maps the codomain back on to the domain. As can be gleaned from Fig. 3.3, we can also use a basis in both the domain and codomain to use the matrix representation of linear operators (3.1,3.2), and find that the matrix representation of an operator satisfies the same relations as that for a finite matrix,