Quantum Mechanics, Spin, Lorentz Group. Amer Iqbal LUMS SCHOOL OF SCIENCE AND ENGINEERING

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1 Quantum Mechanics, Spin, Lorentz Group Amer Iqbal LUMS SCHOOL OF SCIENCE AND ENGINEERING 1

2 LINEAR ALGEBRA FOR QUANTUM MECHANICS Vector Space A vector spacev overcis a set which: 1. is a commutative group under addition. closed under multiplication by complex numbers such that z( v + w = z v +z w, (z 1 +z v = z 1 v +z v, (z 1 z v = z 1 (z v 1 v = v for all z,z 1, C and v, w V. If{ 1,,, n } is a basis ofv then every v V can be expressed as v = v 1 1 +v + +v n n (1 This expression for v is unique because of linear independence of the the basis vectors. If the basis is understood we can also write the vector in terms of its coordinates: v v = v 1 1 +v + +v n n.. v n v 1 ( Linear Transformation T : V W is a linear transformation if T(α v +β u = αt( v +βt( u (3 for every v, u V and α,β C. Dual Space The dual vector spacev to a vector spacev is the set of all linear transformations L : V C Let { 1,,, n } be a basis ofv. For v = v 1 1 +v + +v n n L( v = v 1 L( 1 +v L( + +v n L( n (4

3 Thus the action oflon an arbitrary vector is completely determined by the action oflon the basis vectors. Thusncomplex numbersα i = L( i determine the linear transformation completely. We can label the linear transformationlby these numbers so that L α1,α,,α n ( i = α i, i = 1,,,n. (5 It is easy to check that L α1,α,,α n +L β1,β,,β n = L α1 +β 1,α +β,,α n+β n γl α1,α,,α n = L γα1,γα,,γα n L α1,α,,α n = α 1 L 1,0,0,0 +α L 0,1,0,,0 + +α n L 0,0,0,,1 ThusV = {L : V C L is linear} is a vector space with basis{l 1,0,0,0,L 0,1,0,,0,,L 0,0,0,,1 } and hence is of dimension n. Let us denote L i = L 0,0,,10,,0 (1 is at the i-th place. Then {L 1,L,,L n } is a basis dual to the basis{ 1,,, n } ofv since L i ( j = δ ij (6 Since V and V have the same dimension they are isomorphic. There is no natural isomorphism. One way of defining an isomorphism is as follows: v = v 1 1 +v + +v n n L v = v 1 L 1 +v L + +v n L n (7 This isomorphism gives us a Hermitian inner product onv: (, : V V C ( v, w = L v ( w = v 1 w 1 +v w + +v n w n (8 This inner product satisfies the following properties: ( v, w = ( w, v ( v,a w +b u = a( v, w +b( v, u (a v +b u, w = a ( v, w +b ( v, u ( v, v 0, ( v, v = 0 v = 0 A vector space with a Hermitian inner product with respect to which it is a complete metric space is called a Hilbert space 1. C n with the Hermitian inner product given above is a complete metric space, hence a Hilbert space, and this is the prototype example of a finite dimensional Hilbert space. Dirac Notation We will denote the linear transformationl v as v. Thus ( v, w = L v ( w = v w (9 v = v 1 1 +v + +v n n 1 A metric space is complete if every Cauchy sequence, in the metric space, converges to a point of the metric space. 3

4 Adjoint If : V V is a linear operator then the adjoint ofâdenoted byâ is a linear operator  : V V defined by the relation ( =  (Â+ B =  + B (λâ = λ Â, λ C ( B = B  ( v,â w = ( v, w (10  is called Hermitian if  =   is called Unitary if   =  = 1 Matrix Representation of Operators Let H be a Hilbert space with an orthonormal basis{ 1,,, n } i j = δ ij { 1,,, n } being the basis of the dual space: i : H C A linear operatorâ : H H is completely determined by its action on the basis ofh  j = n A ij i, A ij C. i=1 The numbersa ij form an n n matrix which is the matrix representation of the operatorâin the basis{ 1,,, n }. The numbersa ij are also called the matrix elements of the operatorâ: ( i  j = i A kj k = i A kj,k k k = A kj i k = A kj δ ik k k (11 = A ij (1 4

5 v v = v 1 1 +v + +v n n.., v i = i v C v n A 11 A 1.. A 1n v 1 w 1 A 1... A n v  v = w = w.. A n1 A n.. A nn v n w n The linearity property of the linear operator allows us to express an arbitrary linear operator as linear combination of operators that map a basis state to a basis state. The operator that maps a basis state j to a basis state i is given by i j : v 1 i j : H H ( i j v = i v j. There are n such operators and in a given basis an operator  is determined by n numbers A ij. Thus we can write Thus  = n α ij i j, α ij C (13 i,j=1 A ij = i  j = i ( n = k,l=1 IfA ij = δ ij we get the identity operator thus α kl k l j = n α kl i k l j n α kl δ ik δ lj = α ij (14 k,l k,l  = n i,j A ij i j (15 1 = i i i (16 IfÂand B are two linear operators then ( (AB ij = i  B j = i Â1 B j = i  k k B j = A ik B kj (17 5 k k

6 Thus the matrix representation of product of two operators is the product of matrix representations of the two operators. Let us see what the adjoint of i j is: ( v,( i j w = v ( i j w = v i j w = vi w j = vi ( = ( j i v, w ( ( j, w = v i j, w (18 Thus it follows that From Eq(15 and Eq(19 we get ( i j = j i (19  = i,j A ij j i = i,j A ji i j (0 Thus the matrix of operator  is the adjoint (complex conjugate, transpose of the matrix of operatorâ. Eigenvalues and Eigenvectors of a Hermitian operator Let Ĥ be a Hermitian operator with eigenvector v and corresponding eigenvalueλ: Ĥ v = λ v = v Ĥ = λ v (1 SinceĤ = Ĥ. Thus the matrix element v Ĥ v can be evaluated in two different ways v Ĥ v = { λ v v λ v v = λ = λ ( Thus a Hermitian operator has real eigenvalues Let v and w be two eigenvectors of Ĥ with eigenvalues λ and µ respectively then the matrix element w Ĥ v can be evaluated in two different ways w Ĥ v = { λ w v µ w v = (λ µ w v = 0 (3 Thus the eigenvectors of a Hermitian operator with distinct eigenvalues are orthogonal Eigenvalues and Eigenvectors of a Unitary operator 6

7 Let Û be a unitary operator with eigenvector v and corresponding eigenvalueλ: Û v = λ v = v Û = λ v (4 SinceÛ Û = 1. Thus the matrix element v Ĥ v can be evaluated in two different ways v v = v Û Û v (5 = v λ λ v = λ v v = λ = 1 Thus a unitary operator has eigenvalues which are of the typee iθ Tensor Product Given two vector spaces V and W we can form a new vector space called the tensor product ofv and W denoted by V W. If { v 1,, v n } is a basis of V and { w 1,, w m } is a basis of W then a basis ofv W is given by thus { v 1 w 1, v 1 w v n w m 1, v n w m } (6 dim(v W = dim(vdim(w If L : V V and K : W W are linear transformations then L K : V W V W is a linear transformation defined by L K( v w = L v K w In the above basis L is a n nmatrix and K is a m m matrix whose entries we will denote by L ij and K ab respectively (i,j = 1,, n&a,b = 1,, m. The linear transformation L K (in the basis 6 is a nm nm matrix given by L 11 K L 1 K L 1n K L 1 K L K L n K.... L n1 K L n K L nn K where L ijk := x = n i=1 xi v i and y = m a=1 ya w a then L ij K 11 L ij K 1 L ij K 1m L ij K 1 L ij K L ij K m.... L ij K m1 L ij K m L ij K mm (7 x y = i,a x i y a v i w a (8 n = dimv, m = dimw 7

8 This implies that x x 1 x. x n x y, y x 1 y 1 x 1 y. x n y m 1 x n y m y 1 y. y m then (9 Exercise 1: Let H be a three dimensional Hilbert space with an orthonormal basis { 1,, 3 } (Orthonormal means that the basis vectors are unit vectors and orthogonal to each other i.e., i j = δ ij. a. Find the matrix representation of the following operators: i b. Which of the above operators are Hermitian. c. Let  = i 1 i Show that  is Hermitian and find the eigenvalues and eigenstates ofâ. d. Show that the eigenstates ofâare orthogonal. e. Find the matrix representation of the operators given in part a in the basis of normalized eigenstates ofâ. 8

9 QUANTUM MECHANICS OF TWO STATE SYSTEMS Let H be a two dimensional Hilbert space with orthonormal basis { 1, }. Let Ĥ be the Hamiltonian of the system. In the basis{ 1, } the Hamiltonian is a Hermitian matrix: ( Ĥ { 1, } H11 H H = 1, H H 1 H 1 = H1, H 11,H are real. (30 In general this is not a diagonal matrix and therefore the basis states 1 and are not the eigenstates of the Hamiltonian. We denote the eigenstates of the Hamiltonian by + and. These are the stationary states of the system. When a measurement of the energy of the system is made it will be found in one the stationary states with energy equal to the corresponding eigenvalue of the stationary state. Thus possible energies of the system are the eigenvalues given by E ± = H 11 +H ± (H 11 H +4H 1 H 1 (31 The Hermitian matrixh has four real numbers in it. We can write this matrix as a linear combination (with real coefficients of four Hermitian matrices (which form a basis of space of Hermitian matrices. These four matrices are 1,σ 1,σ,σ 3. The matrices σ 1,,3 are called Pauli matrices and are given by σ 1 = ( 0 1, σ 1 0 = ( ( 0 i 1 0, σ i 0 3 = 0 1 The Pauli matrices satisfy some very interesting properties which are very useful in calculations: (3 σ a = 1 σ 1 σ σ 3 = i1 (33 {σ a,σ b } = δ ab Tr(σ a = 0 σ a σ b = iσ c, where a,b,c is a cyclic permutation of 1,,3 [σ a,σ b ] = iɛ abc σ c (summation overc σ a σ b = δ ab +iɛ abc σ c (summation overc Define σ = (σ 1,σ,σ 3 then ( n σ = ( n n 1 ( n σ( m σ = ( n m1+i σ ( n m ( exp iθˆn σ = cos(θ+iˆn σ sin(θ, ˆn is a unit vector 9

10 The Hamiltonian can be written as a linear combination of the identity matrix and the three Pauli matrices: ( ( ( ( ( H11 H i 1 0 = n H 1 H 0 +n n 1 0 +n i 0 3 ( H = n 0 1+n 1 σ 1 +n σ +n 3 σ 3 = n 0 1+ n σ n 0 = 1 TrH = H 11 +H, n = 1 Tr(H σ = 1 (H 1 +H 1,i(H 1 H 1,H 11 H Since ( n σ = ( n n 1 therefore the eigenvalues of n σ are ± n. This implies that the eigenvalues of the HamiltonianH are: E ± = n 0 ± n (35 The eigenvectors ofh will also be eigenvectors of n σ since every vector is an eigenvector of 1: ( n3 n n σ = 1 in (36 n 1 +in n 3 ( ( ( n3 n 1 in α± α± = (E n 1 +in n ± n 0 3 β ± Solving the above linear equations we get the following two eigenvectors ofh: ( ( α± ±(n1 in =, with eigenvalue E n n ± = n 0 ± n. (37 3 β ± These eigenvectors are not normalized so are not unit vectors. Since n is a set of three real numbers we can express these numbers in terms of spherical polar coordinates n = (n 1,n,n 3 = n (sinθ cosφ, sinθ sinφ, cosθ (38 n = H (H 11 H 11 H +H 1 H 1, cosθ =, cosφ = H 1 +H 1 (H 11 H +H 1 H 1 H 1 In terms of the spherical polar coordinates(θ,φ the normalized eigenvectors ofh are given by cos( θ, with eigenvalue E + = n 0 + n, (39 e iφ sin( θ e iφ sin( θ, with eigenvalue E = n 0 n, cos( θ β ± 10

11 ( 1 Since1 = 0 and == ( 0 1 therefore the eigenstates ofĥ are: ( θ ( θ + = cos 1 + e iφ sin (40 ( θ ( θ = e iφ sin 1 cos Since i j = δ ij,i,j = 1, therefore it follows that: + + = = 1 and + = + = 0 and ( θ ( θ 1 = cos + + e iφ sin (41 ( θ ( θ = e iφ sin + e iφ cos Exercise : A spin- 1 is also a two state system. The spin can be either parallel to the z-axis or anti-parallel to it. We denote these twos states by + and. The magnetic moment for a spin- 1 particle is given by µ = 1 γ σ. γ is the gyromagnetic ratio. If such a particle is placed in a constant magnetic field B 0 the Hamiltonian of the system is given by Ĥ0 = µ B. Suppose that the magnetic field is in the z-direction so that the matrix representation of the Hamiltonian (in the basis +, is given by H 0 = γ B 0 σ 3 = γ B ( If we now perturb the system by introducing a time varying magnetic fieldb 1 (t = B 1 (cos(ωt i sin(ωt j. The matrix representation of the new Hamiltonian in the basis +, is given by H = H 0 µ B 1 (t = ( ω0 ω 1 e iωt ω 1 e ωt (4 ω 0 whereω 1 = γb 1 is called the Rabi frequency. We will calculate the probability of transition from state to the state +. a. Express the HamiltonianĤ as linear combination of operators + +, +, +,. a. Arbitrary state at timetis given by ψ(t = c + (t + +c (t where c + (t + c (t = 1. Using the Schrödinger equation find the coupled differential equations satisfied by the coefficients c ± (t. ( b. Using the ansatzc ± (t = γ ± (te iλ ±t/ obtain the differential equation satisfied byγ ± (t. λ ± are the eigenvalues ofĥ0. In the absence ofĥ1 γ ± (t would be independent of timet. 11

12 c. Convert the set of coupled first order differential equations satisfied by γ ± (t into a set of uncoupled second order differential equations. d. Solve the second order differential equations satisfied by γ ± (t subject to the initial condition γ + (0 = 0,γ (0 = 1. e. Show that the probability of transition from state at time t = 0 to the state + at time t is given by ω ( 1 ω P(t = 1 +(ω ω 0 t ω1 +(ω ω 0 sin 1

13 ANGULAR MOMENTUM & SPIN In classical mechanics the angular momentum is given by L = r p (43 In quantum mechanics angular momentum is an observable hence there is a corresponding operator L = ( L 1, L, L 3 = ( x p 3 x 3 p, x 3 p 1 x 1 p 3, x 1 p x p 1 (44 The position and momentum operators satisfy non trivial commutation relations [ x a, x b ] = 0, [ p a, p b ] = 0, [ x a, p b ] = i δ ab (45 therefore it follows that 3 [ L a, L b ] = i ɛ abc Lc (46 c=1 The only non-zero values ofɛ abc are: ɛ 13 = ɛ 31 = ɛ 31 = 1 ɛ 13 = ɛ 31 = ɛ 13 = 1 The eigenvalues of these operators L a have dimensions same as that of so we can defined operatorsĵa whose eigenvalues are dimensionless by Ĵ a = L a [Ĵa,Ĵb] = i 3 ɛ abc Ĵ c c=1 (47 From the commutation relation it follows that the operator Ĵ = Ĵ 1 + Ĵ + Ĵ 3 commutes with Ĵ a, a = 1,,3, [Ĵ,Ĵa] = 0, a = 1,,3. (48 Thus the we can pick a basis of the Hilbert spacehin which the operatorĵ is diagonal. SinceĴ3 also commutes with Ĵ therefore we can pick the vectors in the basis of H to be eigenvectors of both Ĵ and Ĵ3. We label the basis vectors with the eigenvalues of Ĵ and Ĵ3 (which are real since these are Hermitian operators and take these vectors to be orthonormal: Ĵ λ,m = λ λ,m (49 Ĵ 3 λ,m = m λ,m λ,m λ,m = δ λ,λ δ m,m. 13

14 The eigenvalue ofĵ is positive (and therefore we have written it as λ since λ,m Ĵ λ,m = Ĵ1 λ,m + Ĵ λ,m + Ĵ3 λ,m (50 We will determine these eigenvaluesλandm. Let us define Ĵ± = Ĵ1 ± iĵ then [Ĵ,Ĵ±] = 0, [Ĵ3,Ĵ±] = ±Ĵ± (51 Ĵ + Ĵ = (Ĵ1 +iĵ(ĵ1 iĵ = Ĵ 1 +Ĵ i[ĵ1,ĵ] = Ĵ Ĵ 3 +Ĵ3 Ĵ Ĵ + = Ĵ Ĵ 3 Ĵ3 Ĵ ± have been defined since there action on λ,m is very interesting: Ĵ 3 Ĵ ± λ,m = (Ĵ±Ĵ3 ± Ĵ± λ,m = (m±1ĵ± λ,m (5 Ĵ ± λ,m λ,m±1. From the definition of these new operators it is clear thatĵ+ = Ĵ. Therefore Ĵ+ λ,m = λ,m Ĵ Ĵ+ λ,m = λ,m (Ĵ Ĵ 3 Ĵ3 λ,m (53 = λ,m Ĵ λ,m λ,m Ĵ 3 λ,m λ,m Ĵ3 λ,m = λ m +m 0 and similarly Ĵ λ,m = λ,m Ĵ+Ĵ λ,m = λ,m (Ĵ Ĵ 3 +Ĵ3 λ,m (54 = λ,m Ĵ λ,m λ,m Ĵ 3 λ,m + λ,m Ĵ3 λ,m = λ m m 0 From the above two inequalities we see that λ m = λ m +λ (55 Thus for a given eigenvalues ofĵ the eigenvalues ofĵ3 are bounded. Hence for a given eigenvalue λ ofĵ we have a maximum and minimum eigenvalues ofĵ3 which we will denote bym max and m min respectively. Ĵ + λ,m max = 0 (56 Ĵ λ,m min = 0 14

15 From Eq(53 and Eq(54 it follows that { m λ = max +m max m = m min m max = m min (57 max SinceĴ+ takes us from m min to m max in steps of one therefore Let us call m max = j then m max m min = m max Z 0, (58 λ = j(j +1, j m +j (59 Ĵ + j,m = C + j,m j,m+1, Ĵ j,m = C j,m j,m 1 (60 C + j,m = Ĵ+ j,m = j,m Ĵ Ĵ+ j,m = j(j +1 m m C j,m = Ĵ j,m = j,m Ĵ+Ĵ j,m = j(j +1 m +m Thus the Hilbert spaces H is a direct sum H 0 : H = H 0 H1 H 1 H3 (61 H j is spanned by { j, j, j, j +1, j, j +,, j,j 1, j,j } dim C H j = j +1 It is one dimensional with basis vector 0,0. Thus the operators are all numbers (1 1 matrices: ( ( Ĵ { 0,0 { 0,0 0, Ĵ a 0, a = 1,,3. (6 H1 It is two dimensional with basis vectors{ 1, 1, 1, 1 }. The operators are now matrices: ( Ĵ { 1,1, 1, 1 } 1 ( (1 +1 (63 ( { 1 Ĵ,1, 1, 1 } { 1 Ĵ,1, 1, 1 } ( = σ 3 = σ 1, Ĵ { 1,1, 1, 1 } 15 ( 0 i = σ i 0

16 In the problem set 1 we saw that the Pauli matrices satisfy the following commutation relations [ σ a, σ b ] = iɛ σ c abc H 1 It is three dimensional with basis vectors { 1,1, 1,0, 1, 1 }. The operators are now 3 3 matrices: 1( Ĵ { 1,1, 1,0, 1, 1 } 0 1(1+1 0 ( ( { 1,1, 1,0, 1, 1 } Ĵ Ĵ 1 { 1,1, 1,0, 1, 1 } { 1,1, 1,0, 1, 1 } 1 0 1, Ĵ i 0 1 i 0 i 0 i 0 H3 It is four dimensional with basis vectors{ 3, 3, 3, 1, 3, 1, 3, 3 }. The operators are now4 4 matrices Ĵ { 3,3, 3,1, 3, 1, 3,3 } { 3 Ĵ,3, 3,1, 3, 1, 3,3 } 3 { 3 Ĵ,3, 3,1, 3, 1, 3,3 } 1 { 3 Ĵ,3, 3,1, 3, 1, 3,3 } 3 ( ( ( , i i 3 0 i 0 0 i 0 i i (3 +1

17 Notice that in each of the above case the matrices satisfy the same commutation relation and that not more than one matrix is diagonal (since otherwise commutation relation will not be satisfied. The components of the orbital angular momentum are the generators of rotations around the coordinate axis. Hence a rotation of π around any axis should be represented by identity matrix. However, notice that Therefore exp (iθj 3 = e iθj e iθ(j e iθj (65 { 1, j = 0,1,,3, exp (iπj 3 = 1, j = 1.3, 5, (66 This implies that orbital angular momentum correspond to integer j values only. Thus half-integer values represent angular momentum which is not orbital and it is called the spin angular momentum. Exercise 3: The operator Ŝ n of the spin projection on an arbitrary direction n is defined as S n, WhereS = (Ŝx,Ŝy,Ŝz is the spin angular momentum operator and n = (sinθcosφ,sinθsinφ,cosθ is a unit vector. (a. Find the eigenkets 1,m of the operatorŝn = S n for a spin 1 particle. (b. A particle of spin 1 is prepared in an eigenstate of S n 0 with the eigenvalue + ħ, where n 0 = (sinθ,0,cosθ. We denote this state by 1,+1 n 0. Supposes x, the eigenvalue ofŝx, is measured. What is the probability of obtaining+ ħ, i.e. finding the particle in the state 1,+1 î? (c. Calculate (Ŝx Ŝx for the state of part (b. Exercise 4: A beam of spin 1 atoms goes through a series of Stern-Gerlach-type magnets with the following setup. The first magnet is oriented in the+z direction and the ħ component of the beam is directed into a beam dump (i.e., stopped. The second magnet is aligned at an angleθ with respect to the z-axis and the ħ component is dumped (i.e., stopped. The third magnet is oriented in the+z direction and the+ ħ component is dumped (i.e., stopped. (a. What is the intensity of the final s z = ħ beam with respect to the initial beam intensity? (b. How must the second magnet be oriented in order to maximize the intensity of the final s z = ħ beam? 17

18 ADDITION OF ANGULAR MOMENTUM Consider two non-interacting systems with angular momentum quantum numbers j 1 and j. The Hilbert space of the two systems is denoted byh j1 andh j (we only consider the angular momentum quantum numbers and ignore possible others. The Hilbert space of the total system, denoted by H, is the tensor product ofh j1 and H j. The total angular momentum is given by: J = J J (67 Components of J act on H = H j1 H j. We will usually not use the tensor product notation and just write the above as J = J 1 + J with the understanding that operators labeled by 1 only act on states in H j1 and operators labeled by only act on states in H j. The components of the total angular momentum also satisfy the angular momentum commutation relation: Now we have eight operators acting onh: [J a,j b ] = iɛ abc J c (68 {J 1,x,J 1,y,J 1,z,J 1,J,x,J,y,J,z,J } (69 Of these eight operators we can find four which are mutaully commuting and give us the largest set of commuting operators: Another combination of four operators which commute are: {J 1,J 1,z,J,J,z } (70 {J,J z,j 1,J } (71 we will label the corresponding eigenstates by the eigenvalues of the operators: {J 1,J 1,z,J,J,z } j 1,m 1,j,m = j 1,m 1 j,m (7 {J,J z,j 1,J } j,m,j 1,j Sincej 1,j are fixed we will simplify the above notation for the eigenstates and simply write j 1,m 1,j,m = m 1,m (73 j,m,j 1,j = j,m (74 It is clear that dim C H = (j 1 +1(j +1 and 73 gives two different basis ofh: j,m = C j,m m 1 m m 1 m (75 18

19 The coefficients C j,m m 1 m are called the Clebsch-Gordon coefficients. Example of two spin- 1 particles: Before discussing the general case lets consider the case of two spin 1 particles. In this case the Hilbert space H is 4 dimensional. The basis m 1m states are: 1 ( ( = ( ( = ( ( = ( ( = The above are not the eigenstates of J and J z. A linear combination of the above are the eigenstates ofj andj z : 1,1 = + 1,0 = 1, 1 = 0,0 = Thus the total angular momentum quantum number takes value j = 0 and j = 1 each with multiplicity 1: H1 H1 = H 0 H 1 Two particles with spin j 1 and j : Let us consider two particles with spin j 1 and j and corresponding Hilbert spaces H j1 and H j. In this case we would like to determine the possible values 19

20 of the total angular momentum quantum number. Suppose that in the total angular momentum takes the valuej with multiplicityn j i.e., H j1 H j = N j H j (76 j=0, 1,1, Since N j = # of timesh j occurs in the product J z = J z J z (77 therefore the state m 1 m is an eigenstate ofj z with eigenvaluem 1 +m. let n(m = # of states withj z eigenvalue equal to m (78 Since for a angular momentum quantum numberj the values ofmgo from j to+j in steps on1 therefore n(0 = N 0 +N 1 +N +N 3 + n( 1 = N1 +N3 + n(1 = N 1 +N +N 3 + n( 3 = N3 +N5 +. = N j = n(j n(j +1 We can construct a generating function forn j (S = {0, 1,1, 3,, } j S N j q j = j S(n(j n(j +1q j = n(0q +n( 1 q 1 +(1 q j S n(jq j.(79 Notice that m S Tr H q Jz = n(mq m m {,, 3, 1,0,1,1,3,, } = n(mq m + n(mq m m S m {, 3, 1 ] } n(mq m = [Tr H q Jz + (80 0

21 where[ ] + indicates that only non-negative powers ofq are kept. From Eq(79 it follows that N j q j = n(0q +n( 1 q 1 +(1 q j S j S = n(0q +n( 1 ] q 1 +(1 q [Tr H q Jz + = n(0q +n( 1 ] q 1 +q (q 1 [Tr H q Jz [ ] = q (q 1 Tr H q Jz + n(jq j (81 + (q 1Tr H q Jz = (q 1Tr Hj1 q J 1,z Tr Hj q J,z (8 = (q 1(q j 1 + +q +j 1 (q j + +q +j = (q 1q j 1 j 1 q 4j 1+1 q 4j+ 1 q 1 q = q j 1 j q 4j q (1 q 4j + It is easy to find the non-negative powers in the above equation assumingj 1 j : [ ] ( q (q 1 Tr H q Jz = q q (j 1 j + +q (j 1 j q (j 1+j + + = q (j 1 j +q (j 1 j + + +q (j 1+j (83 Threfore j S N jq j = q (j 1 j +q (j 1 j + + +q (j 1+j Thus in the total angular momentum quantum number j takes values from j 1 j to (j 1 + j with multiplicity1. 1

22 Exercise 5: Consider the 4 state system consisting of two spin- 1 particles. The Hilbert space of the system is spanned by the4orthonormal states: where the arrows refer to the direction of the spin along the z-axis and the subscript 1 and refer to the particle. Suppose that the Hamiltonian of this system is given by H = γ (S 1,z + S,z + γ S 1 S. a. Write the above Hamiltonian in terms S 1,±,S 1,z,S,±,S,z. b. Using the form of the Hamiltonian found in part (a find the matrix of H in the basis given above. c. Write the Hamiltonian in terms of the S total where S total = S 1 + S. d. Find the energies and stationary states of the Hamiltonian. e. If the system is in the state at time t = 0 what is the probability of finding the system in the singlet state at timet. Exercise 6: Consider a system of three particles. Particle 1 has spin 1, particle has spin 1 and particle 3 has spin1. This system has 1 states: a. What are the possible eigenvalues ofj total? b. For each of the eigenvalues found in part (a what are possible eigenvalues ofj total,z. c. Determine(J 1,x +J,x 1 and J 1,x J 3,y 1. d. Write down the normalized state with total angular momentum eigenvalue 0 in terms of the individual spin states given above.

23 LORENTZ GROUP The set of4 4 matrices which preserve the quadratic form c t x y z form a group known as the Lorentz groupo(1,3. Since c t x y z = ( ct x y z ct x y z, therefore g O(1, 3 is such that g T ηg = η This implies that det(g = ±1 and therefore O(1,3 is not connected 3. The identity of the group is the4 4 identity matrix and belongs to the component with determinant+1, this component is denoted withso(1,3. There are two important transformations which are not in SO(1,3: Time Reversal T : T = det(t = 1 (84 Parity P : P = det(p = 1 (85 DIMENSION: To determine the dimension of the group notice that g T ηg is a symmetric matrix therefore has 10 independent components which are quadratic functions of 16 components of g. The condition g T ηg = η therefore imposes 10 constraints on 16 components of g leaving 6 independent components. Therefore the Lorentz group has dimension 6 and SO(1, 3 is generated by 6 generators. Among these 6 generators 3 generate rotation aroundx,y and z axis. The remaining3 3 A set is connected if two points of the set can be connected by a continuous curve. In this case the group has two connected components one with determinant equal to +1 and the other with determinant equal to 1. 3

24 generate boosts along the x, y and z axis. The component with g O(1, 3 such that det(g = 1 is given by including the discrete transformationst,p. Generators: The generators ofso(1,3 are J 1 = , J = K 1 = ,K = These generators satisfy the following commutation relations, J 3 =, K 3 = (86 [J a,j b ] = ɛ abc J c (87 [K a,k b ] = ɛ abc J c [K a,j b ] = ɛ abc K c If we define new generators A a = Ka+iJa and B a = Ka+iJa then [A a,a b ] = iɛ abc A c, [B a,b b ] = iɛ abc B c, [A a,b b ] = 0 (88 Thus the new generatorsa a and B b each satisfy the angular momentum commutation relation and commute with each other. Thus we can use the result of the angular momentum commutation relation derived earlier and label the states with two angular momentum quantum numbers, one corresponding toa,a 3 and other corresponding tob,b 3,j 1 andj. Thus representations of the Lorentz group are labeled by two quantum numbers(j 1,j withj 1, {0,, 1,1, 3, }. (j 1,j = (0,0 is the Lorentz scalar (j 1,j = ( 1,0 is the chiral -component spinor (j 1,j = (0, 1 is also chiral -component spinor (j 1,j = ( 1, 1 is the 4-vector (j 1,j = (1,0 is the self-dual -form,f µν + (j 1,j = (0,1 is the antiself-dual -form Fµν Spinor representation: Chiral, Dirac and Majorana Chiral -component spinor ( 1,0 transform in an irreducible representation of the Lorentz group. Acting on this -component spinor A a = σa, Ba = 0 J a = i σ a, K a = σ a 4 We denote the rotation generators byj 1,,3 and boost generators byk 1,,3. 4

25 ψ L = ψ L = ( ψ1 ψ ( ψ1 ψ rotation e iθ ˆn σ boost e β ˆn σ ( ψ1 ψ ( ψ1 ψ (89 Chiral -component spinor (0, 1 also transform in an irreducible representation of the Lorentz group. Acting on this -component spinor A a = 0, B a = σa J a = i σ a, K a = σ a ( ( ψ1 ψ R = ψ rotation e iθ ˆn σ ψ1 ψ ( ( ψ1 ψ R = boost e β ˆn σ ψ1 ψ The Dirac spinor ψ D transforms in ( 1,0 (0, 1 representation of the Lorentz group which is a reducible representation: ( ( ( ψl e ψ D = iθ ˆn σ 0 ψl ψ R rotation (91 0 e iθ ˆn σ ψ R ( ( ( ψl e ψ D = β ˆn σ 0 ψl ψ R boost 0 e β ˆn σ ψ R Parity: ψ (90 ψ(r,t W ψ( r,t (9 where W is a 4 4 matrix representing action of parity on ψ. Since P = 1 therefore W = 1. Also P commutes with the generators of rotation but anti-commutes with the generators of boost. This implies that It is easy to see that Thus under parity W S rot ( nw 1 = S rot ( n W S boost ( nw 1 = S boost ( n W = ( = γ 0 ψ(r,t γ 0 ψ( r,t (93 5