Appendix: SU(2) spin angular momentum and single spin dynamics

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1 Phys 7 Topics in Particles & Fields Spring 03 Lecture v0 Appendix: SU spin angular momentum and single spin dynamics Jeffrey Yepez Department of Physics and Astronomy University of Hawai i at Manoa Watanabe Hall, 505 Correa Road Honolulu, Hawai i yepez@hawaiiedu wwwphyshawaiiedu/ yepez Dated: January 9, 03 Contents I Introduction II Operators III Unitary Rotations IV Eigenvalues of J and J i V Angular Momentum Eigenvectors VI Spin- Operators 3 VII General Spin- Operator 4 VIII Two-Level Energy System 4 IX Interaction Hamiltonian in the Presence of a Magnetic Field 5 X Spin-/ Dynamics 5 XI Larmor precession 6 I INTRODUCTION Presented is a review of angular momentum and angular momentum ladder raising and lowering operators From the matrix representations for the spatial components of the angular momentum operators, one finds irreducible blocks of the rotation group, each block providing its own unique representation of the group We consider the simplest block of the rotation group: the fundamental spin one-half representation Two ideas that are essential to understanding the behavior of qubits are reviewed for the spin one-half group First, we review the well-known isomorphism between a two-level energy quantum system and the spin one-half group in quantum mechanics Second, we review spin one-half dynamics in an external magnetic field II OPERATORS Let us denote the angular momentum operator by J The commutation relation may be written J J i J where J J x, J y, J z The more conventional way of expressing the commutation relation uses the Levi-Cevita symbol [J i, J j ] i ε ijk J k The square the of the angular momentum operator is also important It is simply J J x +J y +J z and as a direct consequence of, J commutes with J i [J, J i ] 0 3 We introduce the raising and lowering operators J + and J in analogy with the simple harmonic oscillator problem J ± J x ± ij y 4 Clearly, the raising and lowering operators satisfy the following commutation relations [J z, J ± ] ± J ± 5 [J, J ± ] 0 6 To find the angular momentum operators eigenvalues, we will make use of the important identity III UNITARY ROTATIONS J J ± J J z J z 7 Let us construst the infinitesimal rotation as follows U[Rδθ]ϕr ϕr δθ r 8 Now if we expand ϕ to first order we find U[Rδθ]ϕr ϕr δθ r ϕ 9 Then, interchanging the cross and dot operations of the triple product and using J i r, we obtain U[Rδθ]ϕr i δθ J ϕr 0

2 V ANGULAR MOMENTUM EIGENVECTORS To affect a finite rotation, we undergo an infinite number of infinitesimal rotations θ Nδ θ The result is simply U[Rθ] lim i θ J N e iθ J/ N N With the expression for an infinitesimal rotation, U[Rδθ] iδθ J/, we may directly compute the commutators of J i and J with the Hamiltonian operator If the Hamiltonian is invariant under arbitrary rotations, U [R] H U[R] H, then the angular momentum operators commute with the Hamiltonian IV EIGENVALUES OF J AND J i [H, J i ] 0 [H, J ] 0 3 We now wish to find the eigenvalues of J and J i, which we denote by α and m, respectively: J αm α αm 4 J z αm m αm 5 We may do a quick check to test the effect of the raising and lowering operators J z J ± αm J ± J z ± J ± αm m ± J ± αm 6 Since J z α, m ± m ± α, m ±, we immediately identify J ± αm C ± α, m ± 7 Therefore, the raising and lowering operators act as intended Now we return to the problem of finding the eigenvalues, α and m We employ the following construction αm J J z αm αm J x + J y αm 8 α m 0 9 Since m is bounded by α, there must exist states αm + and αm that cannot be raised or lowered, respectively J ± αm ± 0 0 Operating with J and using J J ± J J z J z, we get J J z J z αm ± 0 α m ± m ± αm ± 0 α m ± m ± ± 3 From this we conclude that m + m Since we can raise αm to αm + in β steps, it follows that or m + m m + β 4 m + β, 5 where β 0,,, We may then write α as α β β +, 6 where β/ specifies the angular momentum quantum number To conform with convention, we redefine our quantum number as j β/, so j takes on integer and half-integer values The angular momentum eigenvalue equations may then be written as J jm jj+ jm, j 0,,, 3, 7 J z jm m jm, m j, j,, j +, j 8 We may now find the eigenvalues of the raising and lowing operators We begin with the previous result J ± jm C ± j, m ± 9 Multiplying this equation by its congugate, we determine the coefficient C ± jm J J ± jm 30 C ± jm J J z J z jm 3 C ± jj + mm ± j mj ± m + 3 Therefore, we have the important result J ± jm [j mj ± m + ] j, m ± 33 V ANGULAR MOMENTUM EIGENVECTORS In this section we find the matrix representations of the angular momentum eigenvectors The trick used is the same as in the case of the simple harmonic oscillator: express J i and J in terms of J + and J, and use 33 to calculate the matrix components Doing this we find the matrix elements of J x to be j m J x jm j m J + + J jm 34

3 VI SPIN- OPERATORS j m J x jm {δ jj δ m m+[j mj + m + ] + δjj δ m m [j + mj m + ] } 35 Similarly, we find the matrix elements of J y to be j m J y jm j m J + J jm 36 i j m J y jm i {δ jj δ m m+[j mj + m + ] δjj δ m m [j + mj m + ] } 37 Finally, we may immediately write down the matrix elements of J z and J by using 7 and 8 j m J z jm m δ jj δ m m 38 j m J jm jj + δ jj δ m m 39 Using these relations, we find the matrix representations of the angular momentum operators J x i i J y i i 0 i i J z J VI SPIN- OPERATORS In the previous section we found the matrix representations of the angular momentum operators Although J x and J y are not diagonal, they are both block diagonal Then, each block of the angular matrices forms an irreducible representation of the rotation group In this section, we deal with the simplest block representing j rotations This is an important special case of the rotation operator, and we will use a convention where the symbol S in lieu of J The spin- operators in matrix form are then simply S x S y S z i i Since j and m,, the spin- state space is spanned by the two kets:, and, Conventionally, these kets are denoted by + and and are called the spin up and spin down kets, respectively As a special case of 7 and 8 for j, these kets satisfy the following eigenvalue equations S ± 3 4 ±, j S z ± ± ±, m, Actually, since these spin up and down kets are eigenkets of S z, we will attach a subscript to them to avoid confusion with the eigenkets of S x and S y, which we will find later So, we write the S z eigenkets as + z 49 0 z

4 VIII TWO-LEVEL ENERGY SYSTEM VII GENERAL SPIN- OPERATOR Let us find the matrix representation of a spin operator in an arbitrary direction specified by the spherical angles θ and ϕ The general unit vector pointing along this direction is simply: u sin θ cos ϕ, sin θ sin ϕ, cos θ The spin operator S u along u is then u S u S x u x + S y u y + S z u z z u x iu y u x + iu y u z 5 S u cos θ sin θe iϕ sin θe iϕ 5 cos θ Knowing the matrix representation of S u, we may now find its eigenkets satisfying the equation S u ± u ± ± u by using the half-angle relations The results are tan θ cos θ sin θ 53 cot θ + cos θ 54 sin θ + u cos θ e i ϕ + z + sin θ ei ϕ z 55a 55 allow us to directly write down the eigenkets ± x and ± y Letting θ π and ϕ 0, we find the x- directed eigenkets ± x ± 56 Letting θ π and ϕ π, we similarly find the y-directed eigenkets ± y e i π 4 ± i VIII TWO-LEVEL ENERGY SYSTEM 57 An alternate method of deriving 55 is to work in the energy representation We consider a two-level system described by the eigenvalue equation H ± E ± ± 58 We write the Hamiltonian generally as H H H 59 H H u sin θ e i ϕ + z + cos θ ei ϕ z 55b The trick is to split the hamiltonian in to two parts H H + H 0 0 H + H H H + H H H H 60 so the separated Hamiltonian may thus be written H H + H + H H K 6 where K is defined as K H H H H H H 6 Now since the Hamiltonian is Hermitian, we require H H So the off-diagonal components of the Hamiltonian matrix differ only in phase We introduce the spherical azimuthal angle as the phase of H ; that is, H H e iϕ Furthermore, we introduce the spherical polar angle by constructing a right triangle with adjacent leg H H and opposite leg H In this way tan θ H /H H Then in terms of the spherical angles K becomes tan θe K iϕ tan θe iϕ cos θ sin θe iϕ k sin θe iϕ cos θ 63 where we have defined k / cos θ We have discovered that K ks u / Now, from the definition of K, we find its eigenvalues, which based on our triangle construction are simply ±k, and may be written as k ± H H [H H + 4 H ] 64 Therefore, we find or K ± k S u ± ±k ± 65 S u ± ± ±, 66 4

5 X SPIN-/ DYNAMICS as expected Therefore, the general spin- kets ± are seen to be the energy eigenkets of the two-level system Finally, as a matter of completeness, we write down the energy eigenvalues E ± H + H ± k H H 67 IX INTERACTION HAMILTONIAN IN THE PRESENCE OF A MAGNETIC FIELD Here we calculate the time evolution of the quantum state describing the spin-space dynamics of an electron in an external magnetic field We begin with the Hamiltonian of an electron in an electromagnetic field H p e c A m p m + e A e p A+A p 68 Now the commutator of the momentum operator and vector potential vanishes, ie [p, A] i A 0, in the Coulomb gauge Since in this gauge p and A commute, we find that the interaction Hamiltonian is H int e p A 69 We wish to write H int directly in terms of the magnetic field To do this, let us suppose our background magnetic is uniform Then we may write the vector potential as A B r 70 Substituting this into the interation Hamiltonian, we obtain H int e p B r 7 If we permute the vectors of the triple product and substitute J r p, we obtain H int e B J 7 There are a couple issues with 7 The first issue is that we have shown 7 to be valid only in an uniform magnetic field What is the interaction Hamiltonian in a non-uniform field magnetic field? The second issue is much more serious For the spin-/ case, if we replace J with S, the interaction Hamiltonian is incorrect by a factor of one half The origin of this famous problem is the anomalous from a classical standpoint gryomagnetic ratio of the electron To obtain the correct expression for H int we follow another route Since the torque on a magnetic moment is [p, A] p A A p and p i µ B, we find the interaction Hamiltonian by integrating over the spin angle H int dθ T θ dθ µb sin θ µb cos θ 73 Therefore, with µ es/, we obtain the correct expression X SPIN-/ DYNAMICS H int µ B e B S 74 To describe the dynamics of a single spin in a uniform background magnetic field, we start with the Schroedinger equation, where we use the interaction Hamiltonian is specified by 74 i t ψt e B S ψt 75 The unitary evolution operator is et ψt e i B S ψ e i ebut Su ψ 76 Here we have taken the magnetic field to point in a general direction specified by the unit vector u The spin operator S u is given in matrix form by 5 Let us define ω eb u / and write S u σ u Then 76 becomes ωt i ψt e σu ψ 77 That is, the Schroedinger equation governing single spin- / dynamics is i t ψt ω σ u ψt 78 We must find the matrix representing the exponential operator in 77 We may accomplish this by first diagonalizing σ u The eigenvalues of σ u are just ±, and the associated eigenvectors are simply the spin kets + u and u Therefore, we can diagonalize σ u by a similarity transformation using the matrix Λ u and its conjugate, where Λ u is formed from the spin eigenkets That is, using 55, we form Λ u + u u cos θ e i ϕ sin θ e i ϕ sin θ ei ϕ cos θ ei ϕ Λ + u cos θ u ei ϕ sin θ e i ϕ u sin θ ei ϕ cos θ e i ϕ One may readily verify that Λ u Λ u Λ u Λ u and that Λ u σ u Λ u σ z With this understanding, we can cast 77 in the form ωt i ψt Λ u e σz Λ u ψ 8 5

6 XI LARMOR PRECESSION Now the exponent is diagonal, so the exponential operator reduces to a diagonal matrix itself e i ωt 0 ψt Λ u 0 e i ωt Λ u ψ 8 We may perform the indicated matrix multiplication to calculate the time-evolution operator matrix cos ωt ωt U u t i cos θ sin i sin θe iϕ sin ωt i sin θe iϕ sin ωt cos ωt ωt + i cos θ sin 83 defined such that ψt U u t ψ As a special case of 83, θ 0 and ϕ 0, we have U z t e i ωt σz cos ωt + iσ z sin ωt 84 where we have explicitly written the Hamiltonian describing the magnetic interaction 85 Now, expresses the unitary finite rotation operator also in terms of the angular momentum U[Rθ] e iθ J/ 88 On physical grounds time evolution of a spin is equivalent to rotation in spin-space, we have Ut U[Rθ] 89 Then, equating exponents of and 87, the vector angle θ can be related directly to the magnetic field by XI LARMOR PRECESSION The time-dependent interaction Hamiltonian is H int t γbt J, 85 where γ is the gyromagnetic ratio For a time dependent Hamiltonian, the quantum state ψt must be written ψt e i dt Ht ψ 86 So the evolution operator, Ut, defined by ψt Ut ψ is simply Ut e i γj t dt Bt, 87 θt γ t dt Bt 90 For the case of the electron, γ e/, so we have dθt dt e Bt 9 This is Larmor precession with angular frequency of magnitude ω dθ/dt eb/ The orientation of a spin- / particle precesses about a magnetic field 6