1 2X2 Unitary Matrix, S&N, P. 256, Problem 3

P6574 HW #3 - solutions Due March 1, 2013

1 2X2 unitary matrix, S&N, p. 256, Problem 3

Consider the 2X2 matrix deﬁned by a + iσ a U = 0 · , a iσ a 0 − · where a0 is a real number and a is a three-dimensional vector with real components.

1. Prove that U is unitary and unimodular. [Solution: Deﬁne A a + iσ a. Then ≡ 0 · a + ia a + ia A = 0 3 2 1 a + ia a ia − 2 1 0 − 3 so det A = a + ia 2 + a + ia 2 = a2 + a 2 = det A where we use the well know properties | 0 3| | 2 1| 0 | | † σ = σ and (σ ˆn)2 = 1 for any unit vector ˆn. Thus, A 1 = 1 A . We ﬁnd † · − det A † A AA AA U = = = A† A†A det A Thus 1 2 2 1 UU † = A (A†) = AA† = 1 = U †U, (det A)2 det A 1 2 so U is unitary. Furthermore, det U = (det A)2 det A = 1, so U is unimodular. ] 2. In general, a 2X2 matrix represents a rotation in three dimensions. Find the axis and angle

of rotation appropriate for U in terms of a0, a1, a2, and a3. [Solution: We now write

1 2 Generators

1. Show that in any representation where Jx and Jz are real matrices (therefore symmetrical), Jy is a pure imaginary matrix (therefore antisymmetrical). [Solution: Recall the angular momentum algebra:

[Jx,Jy] = i~Jz, [Jy,Jz] = i~Jx, [Jz,Jx] = i~Jy

Given a basis i , the matrix representations for the J is [J ] i J j , and likewise for | i x x ij ≡ h | x | i Jy and Jz. Thus, since Jx and Jz have real matrix elements, we ﬁnd:

i J j = j J i , i J j = j J i h | x | i h | x | i h | z | i h | z | i

where we use the Hermiticty of Jx and Jz. Taking the matrix elements of the third commutator given above, we ﬁnd: i [Jx,Jz] j = i~ i Jy j h | | i − h | | i where

i [J ,J ] j = ( i J k k J j i J k k J j ) h | x z | i h | x | i h | z | i − h | z | i h | x | i Xk = ( j J k k J i j J k k J i ) h | z | i h | x | i − h | x | i h | z | i Xk = j [J ,J ] i − h | x z | i Therefore, i J j = j J i h | y | i − h | y | i so, since Jy is Hermitian, Jy has imaginary matrix elements.]

2. Show that if any operator commutes with two components of an angular momentum vector, it commutes with the third. [Solution: Suppose that [ ,J ] = [ ,J ] = 0 for some operator . Therefore, we ﬁnd: O x O y O [ , [J ,J ]] = [J , [ ,J ]][J , [J , ]] = 0 O x y y O x x y O where we apply the Jacobi identity, [A, [B,C]] + [B, [C,A]] + [C, [A, B]] = 0. Applying the

above result to the angular momentum algebra [Jx,Jy] = iJz, we conclude that

[ ,J ] = 0 O z so the result is proven. ]

3. Let u, v, w be three unit vectors forming a right-handed Cartesian system. Show that the inﬁnitesimal rotation 1 1 Rˆ R− ()R− ()R ()R () ≡ v u v u

2 diﬀers from R ( 2) only by terms of higher order than 2. w − [Solution: We have iθn J/~ R (θ) e− · n ≡ where n is a unit vector and Rn(θ) represents a (right-handed) rotation by an angle θ about the axis deﬁned by n. We wish to compute

ˆ 1 1 iv J/~ iu J/~ iv J/~ iu J/~ R = Rv− ()Ru− ()Rv()Ru() = e · e · e− · e− ·

where u, v, and w form a right-handed coordinate system (u v = w, etc), and is an × inﬁnitesimal angle. To solve this problem, we will use the identity

A B A+B+ 1 2[A,B]+ (3) e e = e 2 O

B Multiplying on the left by e− and applying the original identity to the RHS, we ﬁnd:

B A B B A+B+ 1 2[A,B]+ (3) A+2[A,B]+ (3) e− e e = e− e 2 O = e O

A Multiplying on the left by e− and applying the original identity once more, we ﬁnd:

A B A B 2[A,B]+ (3) e− e− e e = e O

This can also be written as:

A B B A 2[A,B]+ (3) e e = e e e O

where the higher-order terms are not present if [A, B] commutes with A and B. For the problem at hand, we put in A = iv J/~ and B = iu J/~, so that − · − · 1 i i [A, B] = viuj[Ji,Jj] = ijkujvjJk = w J −~2 ~ ~ · where the last step follows from u v = w, or u v = w , and we use the commutator × ijk i j k [Ji,Jj] = iijkJk. Therefore, applying the identity we derived above, we ﬁnd i Rˆ = exp 2w J + (3) = R ( 2) + (3) · O w − O ~ which is the desired result.]

3 Electron-Positron system, S&N, p. 256, Problem 4

The spin dependent Hamiltonian of an electron-positron system in the presence of a uniform mag- netic ﬁeld in the z-direction can be written as

− + eB − + H = AS(e ) S(e ) + (S(e ) S(e )). · mc z − z (e−) (e+) Suppose the spin function of the system is given by χ+ χ . −

3 1. Is this an eigenfunction of H in the limit A 0, eB/mc = 0? If it is, what is the energy → 6 eigenvalue? If it is not, what is the expectation value of H? [Solution: In the limit A 0, → eB H = (S1 S2) mc z − z and

1 2 eB ~ 1 2 eB 1 2 Hχ+χ = (1 ( 1))χ+χ = ~χ+χ ] − mc 2 − − − mc −

2. Same problem when eB/mc 0,A = 0. → 6 [Solution: Write

1 2 1 1 2 2 12 22 1 2 Hχ+χ = A (S + S ) S S χ+χ − 2 − − − 1 2 3 3 1 2 = A~ (2 or 0) χ+χ 2 − 4 − 4 − According to a table of CG coeﬃcients

1 2 1 1 χ+χ = 11 + 00 − r2| i r2| i Therefore

2 1 2 2 1 1 ST χ+χ = ST 11 + 00 − r2| i r2| i! ~2 = 2 11 √2 | i 2 = ~ √2 11 | i So

1 1 2 2 2 1 H = ( 11 + 00 ) ST S1 S2 ( 11 + 00 ) h i 2 *r2 h | h | | − − | r2 | i | i + 1 = ( 11 + 00 ) S2 S2 S2 ( 11 + 00 ) 4 h | h | | T − 1 − 2 | | i | i ~2 3 3 3 3 = (2 + 0 ) 4 − 4 − 4 − 4 − 4 ~2 = − 4

4 4 Rigid body, S&N, p. 256, Problem 6

Let the Hamiltonian of a rigid body be

1 K2 K2 K2 H = 1 + 2 + 3 2 I I I 1 2 3 where K is the angular momentum in the body frame. From this expression obtain the Heisenberg equation of motion for K and then ﬁnd Euler’s equation of motion in the correspondence limit. [Solution:

∂K i = − [K, H] ∂t ~ 2 ∂K1 i 1 Kj = [K1x,ˆ ] ∂t −~ 2 Ij ! 1 (K K + K K ) = j k k j ˆi 2 1jk I j 1 (K K + K K ) (K K + K K ) = 2 3 3 2 3 2 2 3 2 I − I 2 3 1 (2K K + [K ,K ]) ([K ,K ] + 2K K ) = 2 3 3 2 2 3 3 2 2 I − I 2 3 1 (2K2K3 i~K1) (i~K1 + 2K3K2) = − 2 I − I 2 3 1 1 = (ω K)1 + i~K1( ) × I − I 3 2 = (ω K) →cassical limit × 1 ∂K = (ω K)] ∂t ×

5 Euler rotations, S&N, p. 256, problem 9

Consider a sequence of Euler rotations represented by

iσ α iσ β iσ γ (1/2)(α, β, γ) = exp − 3 exp − 2 exp − 3 D 2 2 2 e i(α+γ)/2 cos β e i(α γ)/2 sin β − 2 − − − 2 =  i(α γ)/2 β i(α+γ)/2 β  e − sin 2 e cos 2   Because of the group properties of rotations, we expect that this sequence of operations is equivalent to a single rotation abut some axis by an angle θ. Find θ.

5 [Solution: In terms of a rotation by θ about an axis ˆn,

= exp ( iˆn Jθ/~) = I cos θ/2 iˆn σ sin(θ/2) D − · − · cos κ sin κe iφ = I cos θ/2 i sin(θ/2) − − sin κeiφ cos κ − where ˆn = ˆnx sin κ cos φ + ˆny sin κ sin φ + ˆnz cos κ. Then cos(θ/2) i sin(θ/2) cos κ i sin(θ/2) sin κe iφ = − − − D i sin(θ/2) sin κeiφ cos(θ/2) + i sin(θ/2) cos κ − Compare the trace of the two forms. We ﬁnd

cos(β/2) cos(α + γ)/2 = cos(θ/2)]

6 Angular momentum

A simultaneous eigenstate of J 2 and J is denoted j, m . z | i

1. Show that the expectation values of the operators Jx and Jy for this state are zero. [Solution: There are at least two ways to approach this problem:

1 1 Using the ladder operators, Jx = (J+ + J ) and Jy = (J+ J ). When Jx (resp. • 2 − 2i − − J ) acts on j, m , it produces a linear combination of two ket states j, m + 1 and y | i | i j, m 1 , both of which are orthogonal to j, m . So if we put the bra j, m with the | − i | i h | ket J j, m (resp. J j, m ) together, the bracket must vanish: that is, the expectation x| i y| i value of J (resp. J ) in the state j, m is 0. x y | i It is also possible to exploit the commutation relations of the J alone without invoking • i the ladder operators. Since [Jy,Jz] = i~Jx, and Jz j, m = m~ j, m , we can compute | i | i the expectation value of J in j, m as follows: x | i 1 j, m Jx j, m = j, m [Jy,Jz] j, m h | | i i~ h | | i 1 = ( j, m JyJz j, m j, m JzJy j, m ) i~ h | | i − h | | i 1 = (m~ j, m Jy j, m m~ j, m Jy j, m ) i~ h | | i − h | | i = 0.

Essentially the same arguments go to show that j, m J j, m = 0.] h | y | i 2. Show that if any operator commutes with 2 components of an angular momentum operator, it must commute with the third component. [Solution: For this part we use:

The commutation relation [Ji,Jj] = i~ijkJk (i, j, k = 1, 2.3). •

6 The Jacobi identity of commutators (Lie brackets): [A, [B,C]] = ([B, [C,A]]+[C, [A, B]]). • − Thus if an operator commutes with both J and J (i = j), then it must commute with the O i j 6 third component: ijk ijk [ ,Jk] = [ , [Ji,Jj]] = [Ji,[Jj, ]] + [Jj,[ ,Ji]] = 0.] O i~ O − i~ { O O } 7 Spin precession

We’re given a spin-1/2 particle ψ in a B ﬁeld | i

B(t) = B cos(ωt)î + B sin(ωt)ˆj + B0kˆ which consists of a static z-component and an oscillating component in the xy-plane. In this lab frame, ψ evolves according to the Schrödingerequation | i d i~ ψ(t) = H(t) ψ(t) = γ(S B(t)) ψ(t) . dt| i | i − · | i Due to the time-dependence of B (or H), it is more complicated to write down the solution ψ(t) | i in the lab frame. So instead we shifts to a frame which co-rotates with the oscillating field, and introduce ψ (t) = e iωSzt/~ ψ(t) , which should see a time-independent B field. What is the | r i − | i effective Hamiltonian Hr in the rotating frame? A direct calculation shows that d d iωSzt/~ i~ ψr(t) = i~ e− ψ(t) dt| i dt | i d iωSzt/~ iωSzt/~ = i~ ( iωSz/~) e− ψ(t) + i~e− ψ(t) − | i dt| i iωSzt/~ = ωS ψ (t) + e− H(t) ψ(t) z| r i | i iωSzt/~ iωSzt/~ = ωS + e− H(t)e ψ (t) . z | r i h i =Hr

iωSzt/~ iωSzt/~ To go on we must unravel the operator| e− {z H(t)e .} This can be done by considering its matrix representation in the eigenbasis of S , + , : Clearly z {| i | −i} e iωt/2 0 iωSzt/~ − iωSzt/~ iωSzt/~ † e− = iωt/2 and e = e− . 0 e ! Meanwhile,

3 ~γ H(t) = γ(S B(t)) = σ B − · − 2 j j Xj=1 ~γ B0 B[cos(ωt) + i sin(ωt)] = − 2 B[cos(ωt) i sin(ωt)] B − − 0 ! iωt ~γ B0 Be = . − 2 Be iωt B − − 0 !

7 Thus

γ e iωt/2 0 B Beiωt eiωt/2 0 iωSzt/~ iωSzt/~ ~ − 0 e− H(t)e = − 2 0 eiωt/2 Be iωt B 0 e iωt/2 ! − − 0 ! − ! iωt/2 iωt/2 iωt/2 ~γ e− 0 B0e Be = − 2 0 eiωt/2 Be iωt/2 B e iωt/2 ! − − 0 − ! ~γ B0 B = − 2 B B − 0 ! = γ(B S + BS ). − 0 z x Putting it together, we get

d ˆ ω ˆ i~ ψr(t) = γ(S Br) ψr(t) where Br = Bir + B0 k. dt| i − · | i − γ Now that the eﬀective Hamiltonian H = γ(S B ) is time-independent, we may easily write down r − · r the time evolution of any state in the rotating frame as

iHrt/~ iγ(S Br)t/~ ψ (t) = e− ψ (0) = e · ψ (0) . | r i | r i | r i

iγ(S Br)t/~ To explicit compute the action of the unitary evolution operator U(t) = e · on an arbitrary state, it helps to exploit the following identities: Deﬁne

θ θ nˆ; + = cos + + eiφ sin | i 2 | i 2 | −i θ θ nˆ; = sin + + eiφ cos , | −i − 2 | i 2 | −i where nˆ is the unit vector in R3 with azimuthal angle θ (from +kˆ) and polar angle φ. Then

Therefore it has eigenkets nˆ; with eigenvalues Br ~/2. By the functional calculus of operators | ±i ±| | (see #1, PS1), the operator eiγ(S Br)t/~ has the same eigenkets nˆ; with eigenvalues e iγ Br t/2. · | ±i ± | | This means that

iγ Br t/2 iγ Br t/2 ψ (t) = e | | nˆ; + ψ (0) nˆ; + + e− | | nˆ; ψ (0) nˆ; . | r i h | r i| i h − | r i| −i

8 Now suppose the initial ket is ψ (0) = ψ(0) = + , per the problem. Then in the rotating | r i | i | i frame its time evolution is given by

iγ Br t/2 iγ Br t/2 ψ (t) = e | | nˆ; + + nˆ; + + e− | | nˆ; + nˆ; | r i h | i| i h − | i| −i iγ Br t/2 θ iγ Br t/2 θ = e | | cos nˆ; + e− | | sin nˆ; 2 | i − 2 | −i iγ Br t/2 θ θ θ = e | | cos cos + + sin 2 2 | i 2 | −i iγ Br t/2 θ θ θ e− | | sin sin + + cos − 2 − 2 | i 2 | −i γ B t γ B t γ B t = cos | r| + i sin | r| cos θ + + i sin | r| sin θ 2 2 | i 2 | −i ω t ω ω ω t γB ω t = cos r + i 0 − sin r + + i sin r , 2 ω 2 | i ω 2 | −i r r where we have used the shorthands ω = B /γ and ω = B /γ. 0 0 r | r| Back in the lab frame, the state would read

ψ(t) = eiωSzt/~ ψ (t) | i | r i ω t ω ω ω t γB ω t = cos r + i 0 − sin r eiωSzt/~ + + i sin r eiωSzt/~ 2 ω 2 | i ω 2 | −i r r ωrt ω0 ω ωrt iωt/2 γB ωrt iωt/2 = cos + i − sin e + + i sin e− . 2 ω 2 | i ω 2 | −i r r Note that ψ(0) = + , as required. It follows that Sz(0) = ψ(0) Sz ψ(0) = ~/2. | i | i h i h | | i If ω = ω0 (”on resonance”), the effective field Br in the rotating frame consists of the transverse component Bîr only, so the spin state precesses about the îr axis at angular frequency at ωr = ω0 = γB. From the perspective of the lab frame, the state evolves as

ω0t iω0t/2 ω0t iω0t/2 ψ(t) = cos e + + i sin e− | i 2 | i 2 | −i iω0t/2 ω0t iπ/2 ω0t iω0t = e cos + + e sin e− 2 | i 2 | −i = eiω0t/2 nˆ(t); + , | i where nˆ(t) is the unit vector with associated angles θ(t) = ω t and φ(t) = (π/2) ω t. The 0 − 0 orientations of the spin state form a ﬁgure-8 on the Bloch sphere (Fig. 1). Note that the up state + ﬂips into the down state in a duration T = π/ω , and vice versa. | i | −i 0 For general ω, the z-magnetization of the state ψ(t) at time t is | i S (t) h z i = ψ(t) S ψ(t) h | z | i ωrt ω0 ω ωrt iωt/2 γB ωrt iωt/2 = cos i − sin e− + i sin e 2 − ω 2 h | − ω 2 h− | r r

9 PHYS 6572 - Fall 2010 PS7 Solutions

Back in the lab frame, the state would read

ψ(t) = eiωSzt/! ψ (t) | ! | r ! ω t ω ω ω t γB ω t = cos r + i 0 − sin r eiωSzt/! + + i sin r eiωSzt/! 2 ω 2 | ! ω 2 | −! ! " # " r # " #$ " r # " # ωrt ω0 ω ωrt iωt/2 γB ωrt iωt/2 = cos + i − sin e + + i sin e− . 2 ω 2 | ! ω 2 | −! ! " # " r # " #$ " r # " #

Note that ψ(0) = + , as required. It follows that Sz(0) = ψ(0) Sz ψ(0) = !/2. | ! | ! # ! # | | ! If ω = ω0 (”on resonance”), the effective field Br in the rotating frame consists of the transverse component Bîr only, so the spin state precesses about the îr axis at angular frequency at ωr = ω0 = γB. From the perspective of the lab frame, the state evolves as

ω0t iω0t/2 ω0t iω0t/2 ψ(t) = cos e + + i sin e− | ! 2 | ! 2 | −! " # " # iω0t/2 ω0t iπ/2 ω0t iω0t = e cos + + e sin e− 2 | ! 2 | −! ! " # " # $ = eiω0t/2 nˆ(t); + , | ! where nˆ(t) is the unit vector with associated angles θ(t)=ω t and φ(t)=(π/2) ω t.The 0 − 0 orientations of the spin state form a ﬁgure-8 on the Bloch sphere (Fig. 1). Note that the up state + ﬂips into the down state in a duration T = π/ω , and vice versa. | ! | −! 0

ˆ ˆ ˆ FigureFigure 1: Time 1: Time evolution evolution of aof spin-1 a spin-1/2/2 state state in in a field fieldBB(t()t)= = B Bcos(cos(ωt)ωi +t)îB+sin(Bωtsin()j +ωtB)ˆj0k+whenB0kˆ when on resonanceon resonance (ω = (ωγ=B γB).0 Initial). Initial state state is is the the ”up” ”up” state state ++==zˆ; +zˆ;+. Dots. Dots on the on Bloch the Bloch sphere sphere 0 | | i ! | | i ! indicateindicate the successive the successive spin spin orientations orientations in in the the lab lab frame. frame.

For general ω,thez-magnetizationωrt ω0 ω of the stateωrt ψiωt/(t2) at timeγBt is ωrt iωt/2 Sz cos + i − sin |e !+ + i sin e− × 2 ω 2 | i ω 2 | −i r r Sz(t) 2 2 ~ ωrt ω0 ω ωrt γB ωrt # = ! cos + i − sin sin2 = ψ(t) 2 S ψ(t2) ω 2 − ω 2 # | " z | ! r r # 2 2 ~ ωrt (ω0 ω) (γB) ωrt = cos 2 + − − sin2 2 2 ω2 2 r 4 2 2 ~ 1 + cos(ωrt) (ω0 ω) (γB) 1 cos(ωrt) = + − − − 2 2 ω2 2 r 2 2 ~ 2(ω0 ω) 2(γB) = − + cos(ω t) 2 2ω2 2ω2 r r r (ω ω)2 (γB)2 = S (0) 0 − + cos(ω t) . h z i (ω ω)2 + (γB)2 (ω ω)2 + (γB)2 r 0 − 0 − From this we deduce that the up state reverses orientation in a duration π π T = = . ωr (ω ω)2 + (γB)2 0 − p

10 8 Angular momentum of an unknown particle, S&N p.258, problem 17)

1. It helps to rewrite ψ(x) in spherical coordinates:

ψ(x) = rf(r)(sin θ cos φ + sin θ sin φ + 3 cos θ).

To check whether ψ is an eigenfunction of L2, we may carry out a direct computation. First note that the operator L2 can be explicitly written in spherical coordinates [e.g. Sakurai Eq. (3.6.15)]: 2 2 1 1 L ψ(x) = ~ ∂φφ + ∂θ[(sin θ)∂θ] ψ(x) − sin2 θ sin θ ∂ ψ(x) = rf(r)(sin θ cos φ + sin θ sin φ) φφ − ∂ ψ(x) = rf(r)(cos θ cos φ + cos θ sin φ 3 sin θ) θ − ∂ [(sin θ)∂ ] ψ(x) = rf(r)∂ sin θ cos θ(cos φ + sin φ) 3 sin2 θ θ θ θ − = rf(r)[cos(2 θ)(cos φ + sin φ) 3 sin(2θ)] − So

2 2 1 1 L ψ(x) = ~ [ sin θ(cos φ + sin φ)] + [cos(2θ)(cos φ + sin φ) 3 sin(2θ)] rf(r) − sin2 θ − sin θ − 2 = 2~ rf(r)[ sin θ(cos φ + sin φ) 3 cos θ] − − − 2 = 2~ ψ(x).

Thus ψ(x) is an eigenfunction of L2 with eigenvalue l(l + 1)~2 = 2~2, or l = 1. 2. Alternatively, we can re-express ψ(x) as a linear combination of spherical harmonics. Using the normalized spherical harmonics

0 3 1 3 iφ Y1 = cos θ , Y1± = sin θe± , r4π ∓r8π we ﬁnd eiφ + e iφ eiφ e iφ ψ(x) = rf(r) sin θ − + − − + 3 cos θ 2 2i 1 iφ 1 iφ = rf(r) (1 i) sin θe + (1 + i) sin θe− + 3 cos θ 2 − 2 2π 2π 1 1 √ 0 = rf(r) (1 i)Y1 + (1 + i)Y1− + 2 3πY1 . "−r 3 − r 3 # In one fell swoop, we’ve shown that ψ(x) is an eigenfunction of L2 with eigenvalue 1(1 + 1)~2 = 2~2, and expanded ψ(x) in the eigenbasis of the j = 1 Hilbert space, i.e,. ψ(x) = 1 m rf(r) m= 1 cmY1 where − P 2π 2π c1 = (1 i) , c0 = 2√3π , c 1 = (1 + i). −r 3 − − r 3

11 It ought to be clear that the probability of ψ being found in the state 1, m is given by | i c 2 P (m) = | m| . 1 2 m= 1 cm − | | 2 2 2 P Since c0 = 9 c1 = 9 c 1 , we have | | | | | − | 1 9 1 P (1) = ,P (0) = ,P ( 1) = . 11 11 − 11

3. Recall that the Laplacian in R3 can be written as

2 1 2 1 1 1 1 2 L ∆ = ∂r r ∂r + ∂φφ + ∂θ ((sin θ)∂θ) = ∂r r ∂r . r2 r2 sin2 θ sin θ r2 − 2 ~ So the time-independent Schr¨odingerequation in R3 takes the form

~2 L2 ∂ r2∂ + + V (r) Ψ(x) = EΨ(x). (1) −2mr2 r r 2mr2 Now suppose the energy eigenstate Ψ(x) is the known wavefunction ψ(x) = rf(r)ζ(Ω), where ζ(Ω) = sin θ cos φ+sin θ sin φ+3 cos θ. From Part (a) we already saw that L2ψ(x) = 2~2ψ(x). Meanwhile,

∂rψ(x) = [f(r) + rf 0(r)]ζ(Ω) 2 2 ∂r(r ∂rψ(x)) = 2r[f(r) + rf 0(r)]ζ(Ω) + r [2f 0(r) + rf 00(r)]ζ(Ω) 2 = r[2f(r) + 4rf 0(r) + r f 00(r)]ζ(Ω).

Plugging the various terms into (1) yields

2 2 ~ ¨ 2 ~ [2f¨(r) + 4rf 0(r) + r f 00(r)]ζ(Ω) + [rf(r)ζ(Ω)] + V (r)[rf(r)ζ(Ω)] = E[rf(r)ζ(Ω)]. −2mr ¨ mr2 Thus

2 2 2 1 ~ 2 ~ 4rf 0(r) + r f 00(r) V (r) = E + 4rf 0(r) + r f 00(r) = E + . rf(r) 2mr 2mr2 f(r) 9 Rotated angular momentum, S&N, p. 258, problem 20

A state ψ rotated by an angle β about the y-axis becomes e iJyβ/~ ψ . So the probability for | i − | i the new state to be in 2, m (m = 0, 1, 2) is given by the modulus squared of the projection | 0i 0 ± ± of e iJyβ/~ l = 2, m = 0 onto the subspace l = 2, m , i.e., − | i | 0i 2 2 (2) iJyβ/~ 0 (α = 0, β, γ = 0) = 2, m0 e− 2, 0 , Dm 0 | | D E

12 where α, β, and γ are the Euler angles. At this stage we may invoke Sakurai Eq. (3.6.52):1

(l) 4π m m0(α, β, γ = 0) = Yl ∗(β, α). D r2l + 1 m Using the expressions Y2 (θ, φ) in Appendix A, we ﬁnd

(2) 4π 2 4π 15 2 3 2 2,0(α = 0, β, γ = 0) = Y2 ∗(β, 0) = (sin β) = sin β, D r 5 r 5 r32π r8 (2) 4π 1 4π 15 3 1,0(α = 0, β, γ = 0) = Y2 ∗(β, 0) = (sin β cos β) = (sin β cos β), D r 5 r 5 "−r8π # −r2

(2) 4π 0 4π 5 2 1 2 0,0(α = 0, β, γ = 0) = Y2 ∗(β, 0) = (3 cos β 1) = (3 cos β 1), D r 5 r 5 r16π − 2 − (2) (1) 1,0(α = 0, β, γ = 0) = 1,0(α = 0, β, γ = 0), D− −D (2) (1) 2,0(α = 0, β, γ = 0) = 2,0(α = 0, β, γ = 0). D− D Thus 2 (2) 3 4 2,0(α = 0, β, γ = 0) = sin β, D± 8 2 (2) 3 2 2 1,0(α = 0, β, γ = 0) = sin β cos β, D± 2 2 1 (2)(α = 0, β, γ = 0) = (3 cos2 β 1)2. D0,0 4 − 2 2 (2) It is straightforward to check that m= 2 m0( α = 0, β, γ = 0) = 1. − D

P 10 Rotation matrix for j = 1 states, S&N p. 260, problem 26

1 1. Since Jy = (J+ J ), it is clear that the matrix element j, m0 Jy j, m vanishes for any 2i − − h | | i m, m where m m = 1. Also j, m J j, m = ( j, m J j, m ) by hermiticity of J . 0 | − 0| 6 h | y | 0i h 0 | y | i ∗ y So for j = 1, it is enough to compute the matrix elements 1, 0 J 1, 1 and 1, 0 J 1, 1 . h | y | i h | y | − i From 1 1 i~ Jy 1, 1 = (J+1, 1 J 1, 1 ) = (√2~) 1, 0 = 1, 0 ; | i 2i | i − −| i −2i | i √2| i 1 1 i~ Jy 1, 1 = (J+ 1, 1 J 1, 1 ) = (√2~) 1, 0 = 1, 0 , | − i 2i | − i − −| − i 2i | i −√2| i

we deduce that 1, 0 Jy 1, 1 = (i~)/√2 and 1, 0 Jy 1, 1 = (i~)/√2. So the matrix h | | i h | | − i − representation of J in the 1, 1 , 1, 0 , 1, 1 basis reads y {| i | i | − i} 0 i~ 0 √2 0 √2i 0 − ~ − J (j=1) =  i~ 0 i~  = √2i 0 √2i . y √2 − √2 2  −  i~ 0 √2i 0  0 √ 0   2        1Please read Sakurai Eqs. (3.6.46) through (3.6.51) and the accompanying text for the derivation.

13 2. A direct computation shows that

2 0 √2i 0 0 √2i 0 2 2 0 2 ~ − − ~ − [J (j=1)]2 = √2i 0 √2i √2i 0 √2i = 0 4 0 y 2  −   −  2   0 √2i 0 0 √2i 0 2 0 2      −        and

3 0 √2i 0 2 0 2 0 √2i 0 ~ − − ~3 − [J (j=1)]3 = √2i 0 √2i 0 4 0 = √2i 0 √2i . y 2  −    2  −  0 √2i 0 2 0 2 0 √2i 0    −          (j=1) 3 (j=1) In other words, Jy /~ = Jy /~, which means that for positive integers n, h i (j=1) n Jy / , n odd (j=1) ~ Jy /~ = 2 .  hJ (j=1)/ i , n even h i  y ~ h i Therefore 

2n+1 2n 2n+1 (1) 2n (1) (1) ∞ ( iβ) J ∞ ( iβ) J iJy β/~ y y e− = 1 + − + − (2n + 1)! (2n)! n=0 ~ ! n=1 ~ ! X X 2 n 2n+1 (1) n 2n (1) ∞ ( 1) β Jy ∞ ( 1) β Jy = 1 i − + − − (2n + 1)! (2n)! n=0 ~ ! n=1 ~ ! X 2 X J (1) J (1) = 1 i y sin β + y (cos β 1). − ~ ! ~ ! −

(1) (1) iJy β/~ 3. The matrix representation d (β) of e− reads

0 √2i 0 2 0 2 i sin β − 1 2 − d(1)(β) = 1 √2i 0 √2i + (cos β 1) 0 4 0 − 2  −  − 2   0 √2i 0 2 0 2    −  1 (1 + cosβ) 1 sin β 1 (1 cos β)   2 − √2 2 − =  1 sin β cos β 1 sin β  . √2 − √2 1 1 1  2 (1 cos β) √ sin β 2 (1 + cos β)   − 2    11 Rotations and Clebsch Gordon coeﬃcients

Prove

j j1 j2 mm0 (R) = j, m, j1, j2 j1, m1, j2, m2 Dm m0 (R) m ,m0 (R) j1, m10 , j2, m20 j, m0, j1, j2 D h | i 1 1 D 2 2 | m ,m0 ,m ,m0 1 X1 2 2

14 1 Use this along with the known values of the rotation matrices d for j = 2 and j = 1 to compute

3 3 3 2 2 2 d 3 3 (θ), d 3 1 (θ), d 1 1 (θ) 2 , 2 2 , 2 2 , 2 [Solution: Write

j, m, j , j = j , m , j , m j, m, j , j j , m , j , m (2) | 1 2i | 1 1 2 2ih 1 2 | 1 1 2 2i m ,m X1 2 Rotate by R

(R) j, m, j , j = (R) j , m D(R) j , m j, m, j , j j , m , j , m D | 1 2i D | 1 1i | 2 2ih 1 2 | 1 1 2 2i m ,m X1 2 Multiply from the left by the dual of 2. j, m0, j , j (R) j, m, j , j = j , m0 (R) j , m j , m0 D(R) j , m h 1 2 |D | 1 2i h 1 1 |D | 1 1ih 2 2 | | 2 2i m0 m0 m1,m2 X1 2 X j , m0 , j , m0 j, m0 j, m, j , j j , m , j , m × 1 1 2 2 | h 1 2 | 1 1 2 2i j j1 j2 m0,m = m0 m m0 m j1, m10 , j2, m20 j, m0 j, m, j1, j2 j1, m1, j2, m2 D D 1 1 D 2 2 | h | i m0 m0 m1m2 X1 2 X

Then

3 1 2 1 2 1 3 3 1 3 3 d 3 3 = 0 0 1, m0 , , m0 , 1, m1, , m2 , , m m1 m m2 1 2 2 2 D 1 D 2 2 | 2 2 2 | 2 2 X 1 1 2 1 1 3 3 1 1 3 3 = 11 1 1 1, 1, , , 1, 1, , , D D 2 2 2 2 | 2 2 2 2 | 2 2 i(1+ 1 )α 1 i(1+ 1 )γ = e− 2 (1 + cos β) cos(β/2)e− 2 2