3 Angular Momentum and Spin

Transcription

1 In this chapter we review the notions surrounding the different forms of angular momenta in quantum mechanics, including the spin angular momentum, which is entirely quantum mechanical in nature. Some of the material presented in this chapter is taken from Auletta, Fortunato and Parisi, Chap. 6 and Cohen-Tannoudji, Diu and Laloë, Vol. II, Chaps. IX and X. 3. Orbital Angular Momentum Orbital angular momentum is as fundamental in quantum mechanics as it is in classical mechanics. In quantum mechanics, when applied to the realms of atoms and molecules, it can come in many different forms and flavours. For example, in molecular quantum mechanics the electronic angular momentum for the motion of electrons about the atomic nuclei and the rotational angular momentum for the motions of nuclei about the molecular centre of mass arise in the analysis of such systems. We define the quantum mechanical orbital angular momentum in the same manner as its classical counterpart ˆL ˆr ˆp, 3. with ˆr and ˆp the position and linear momentum observables, respectively. It follows that in quantum mechanics, the orbital angular momentum is also an observable. If we introduce the components ˆx j and ˆp j for the position and linear momentum, where j, 2, 3 i.e., in Cartesian coordinates ˆx ˆx, ˆx 2 ŷ and ˆx 3 ẑ, and similarly for ˆp j, such that ˆx j ˆp j e j ˆx j 3.2 e j ˆp j, 3.3 with e j the different unit basis vectors, then ˆL j,k e j e k ˆx j ˆp k j,k i e i ε ijk ˆx j ˆp k 6

2 i,j,k e i ε ijk ˆx j ˆp k. 3.4 In equation 3.4 we have introduced the elements of the very important Levi-Civita tensor ε ijk, which are defined as follows +, even permutation of ijk 23, 23, 32 ε ijk, odd permutation of ijk 23, 32, , if i j, j k or j k. This tensor is particularly useful for expressing the elements of a vectorial cross-product i.e., a b i j,k ε ijka j b k, and it is also related to the Kronecker delta and thus the scalar product through ε ijk ε imn δ jm δ kn δ jn δ km. 3.6 i It is straightforward through equation 3.6 to verify that [a b c j a c b j a b c j. Coming back to equation 3.4, since it is also the case that ˆL i e i ˆLi, 3.7 it then follows that ˆL i j,k ε ijk ˆx j ˆp k. 3.8 There are several important and useful commutation relations involving the orbital angular momentum, we derive some here. First, the commutations with the position and linear momentum components [ˆx j, ˆL k m,n ε kmn [ˆx j, ˆx m ˆp n m,n ε kmn ˆx m [ˆx j, ˆp n + [ˆx j, ˆx m ˆp n i m,n ε kmnˆx m δ jn i m ε jkmˆx m, 3.9 and 62

3 [ˆp j, ˆL k m,n ε kmn [ˆp j, ˆx m ˆp n m,n ε kmn ˆx m [ˆp j, ˆp n + [ˆp j, ˆx m ˆp n i m,n ε kmn ˆp n δ jm i n ε jkn ˆp n 3. where [ˆx j, ˆp k i δ jk was used for both derivations. components of the angular momentum For the commutation between [ˆLj, ˆL k [ε jmnˆx m ˆp n, ˆL k m,n m,n [ ε jmn ˆx m ˆp n, ˆL k + [ˆx m, ˆL k ˆp n i m,n,r ε jmn ˆx m ε nkr ˆp r + ε mkr ˆx r ˆp n i ε njm ε nrk ˆx m ˆp r + ε mnj ε mkr ˆx r ˆp n m,n,r [ i δ jr δ mk δ jk δ mr ˆx m ˆp r + δ nk δ jr δ nr δ jk ˆx r ˆp n m,r n,r i δ jr δ mk ˆx m ˆp r + δ nk δ jr ˆx r ˆp n m,r n,r i n,r δ jr δ kn δ jn δ kr ˆx r ˆp n i n,r i i ε ijk ε irnˆx r ˆp n ε ijk ˆLi, 3. where equations 3.6 and were used. Considering the square of the angular momentum [ˆL2, ˆL j [ˆLk ˆLk, ˆL j k k ˆLk [ˆLk, ˆL j + [ˆLk, ˆL j ˆLk 63

4 i i,k ε ikj ˆLk ˆLi + ˆL i ˆLk ε ikj ˆLk ˆLi + ε kij ˆLk ˆLi i i,k i i,k ε ikj ˆLk ˆLi ˆL k ˆLi ˆ. 3.2 and [ˆL2, ˆx j [ˆLk ˆLk, ˆx j k k ˆLk [ˆLk, ˆx j + [ˆLk, ˆx j ˆLk ˆLk ε ijk ˆx i + ε ijk ˆx i ˆLk i i,k i i,k ε ijk 2ˆx i ˆLk [ˆx i, ˆL k i,k 2i ε ijk ˆx k ˆLi ε ijk 2 ε ikmˆx m m 2 i ε jkiˆx k ˆLi + 2ˆx j 3.3 i,k since k,i ε ijkε ikm k δ jkδ km δ jm δ kk 2δ jm, while similarly [ˆL2, ˆp j 2 i ε jki ˆp k ˆLi + 2 ˆp j, 3.4 i,k A little more work will establish that see the Second Problem List [ˆL2, ˆx j ˆx k [ˆL2, ˆp j ˆp k [ 2 i ˆx j ε kmn + ˆx k ε jmn ˆx m ˆLn 2 ˆx 2 δ jk 3ˆx j ˆx k 3.5 m,n [ 2 i ˆp j ε kmn + ˆp k ε jmn ˆp m ˆLn 2 ˆp 2 δ jk 3ˆp j ˆp k, 3.6 m,n which also imply that 3. [ˆL2, ˆx 2 [ˆL2, ˆp 2. Finally, we find using equations

5 [ˆLj, ˆx [ˆLj, ˆp Eigenvalues of the Angular Momentum The fact that the three components of the angular momentum ˆL x, ˆLy, ˆLz commute with its square ˆL 2, from equation 3.2, implies that we can find a common set of eigenvectors { ψ } for ˆL 2 and one component of ˆL the three components cannot share the same eigenvectors since they do not commute with each other; see equation 3.. This can be verified as follows. If a is an eigenvector of ˆL 2 and ψ one of its eigenvector, then ˆL 2 ˆLj ψ ˆL ˆL2 j ψ aˆl j ψ 3.9 implies that the ket ˆL j ψ is also a eigenvector of ˆL 2. If this eigenvalue is non-degenerate, then ˆL j ψ b ψ, with b a constant, and ψ is also an eigenvector of ˆL j and b its corresponding eigenvalue. Similarly, we can assert that ˆL 2 ψ is an eigenvector of ˆLj by swapping the order of the two operators in the first line of equation 3.9; they, therefore, share the same set of eigenvectors. It is common to choose ˆL j ˆL z and we denote by { l 2, m } for the set of eigenvectors, with l 2 and m the quantum numbers associated to ˆL 2 and ˆL z, respectively. As a matter of fact, we set l 2 2 as an eigenvalue of ˆL 2 and m one for ˆL z although both l 2 and m still need to be determined. We now introduce the raising and lowering operators ˆL ± ˆL x ± iˆl y 3.2 for which the following commutation relations can easily be established [ˆLz, ˆL ± [ˆL2, ˆL ± [ˆLz, ˆL x ± i [ˆLz, ˆL y i ˆLy iˆl x ± ˆL ± 3.2, 3.22 from equations 3. and 3.3. Using equation 3.2 we can calculate that ˆL z ˆL± l 2, m ˆL± ˆLz ± ˆL ± l 2, m m ± ˆL ± l 2, m

6 which implies that whenever ˆL ± acts on a eigenvector it transforms it to another eigenvector of eigenvalue m ± for ˆL z. It is therefore apparent that ˆL ± are ladders operators that will allow us to span the whole set of eigenvectors associated to a given eigenvalue l 2 2 i.e., since ˆL ± commutes with ˆL 2 it shares the same eigenvectors. Moreover, we deduce that m must have lower and upper bounds, since the following relation must be obeyed m l If we denote the upper and lower bounds of m as m + and m, respectively, then we can write Furthermore, we have ˆL ± l 2, m ± ˆL ˆL± l 2, m ± ˆL2 x + ˆL 2 y ± i [ˆLx, ˆL y l 2, m ± ˆL2 ˆL 2 z ˆL z l 2, m ± l 2 m 2 ± m ± 2 l 2, m ± From this result we obtain the two equations from which we can take the difference and transform it to l 2 m 2 + m 3.27 l 2 m 2 + m m + + m m + m The term in the second set of parentheses must be greater than zero because m + m, which implies that the term within the first set of parentheses must cancel and m + m 3.3 m + m 2l, 3.3 where 2l has to be some positive integer number, since m + m corresponds to the number of times i.e., 2l times ˆL ± is applied to l 2, m to span the whole set of kets. It is now clear from equation 3.3 that m + l and from equation 3.28 l 2 l l

7 Furthermore, equation 3.23 and tell us that m is quantized and can either be an integer with or a half-integer such that m l, l +,...,,,,..., l, l 3.33 m l, l +,..., 2,,..., l, l Given these results, we will write from now on the set of eigenvectors for ˆL 2 and ˆL z as { l, m } with ˆL 2 l, m l l + 2 l, m 3.35 ˆL z l, m m l, m We know from our previous discussion focusing on equation 3.23 that ˆL ± l, m l, m ±. Then considering the following, we have using the projector l k l l, k l, k ˆP l l, m ˆL ˆL± l, m l l, m ˆL l, k l, k ˆL ± l, m k l l, m ˆL l, m ± l, m ± ˆL ± l, m l, m ± ˆL 2 ± l, m, 3.37 since ˆL± l, m l, m ˆL. On the other hand we also have see equation 3.26 l, m ˆL ˆL± l, m l, m ˆL 2 ˆL 2 z ˆL z l, m [l l + m m ± 2, 3.38 which with equation 3.37 implies that It also follows that ˆL ± l, m l l + m m ± l, m ± ˆL ˆL± l, m l l + m ± m ± l l + m m ± 2 l, m [l l + m m ± 2 l, m [l m l ± m + 2 l, m

8 Alternatively, we can write the matrix elements associated to the different angular momentum observables l, m ˆL 2 l, m l, m ˆL z l, m l, m ± ˆL ± l, m l, m ˆL ˆL± l, m l l m 3.42 l l + m m ± 3.43 [l m l ± m All other matrix elements are null. Exercise 3.. Write the matrix representations for ˆL z, ˆL ± and ˆL 2 for l, /2 and. Solution. The dimension of the subspace or the matrices associated to a given l is determined by the number of values the quantum number m can take. Referring to equations we find that m takes 2l + values as it goes from l to l with unit steps. i For l the dimension is 2l + and the only matrix elements are L z L ± L 2 and. ii For l /2 the dimension is 2l + 2. Using the following basis representation for the eigenvectors equations , we have 2, 2, 2, 2 ˆL z 2 ˆL + ˆL ˆL We can also readily verify from these matrices that ˆL 2 2 iii For l the dimension is 2l + 3 and with,,,,, ˆL+ ˆL + ˆL ˆL+ + ˆL 2 z,

9 we have ˆL z ˆL ˆL ˆL Eigenfunctions of the Angular Momentum To derive the eigenfunctions we start by writing down the action of the different angular momentum operators on a wave function ψ r with see equation.84 r ˆL x ψ r ˆL y ψ r ˆL z ψ r ŷˆp z ẑ ˆp y ψ i y z z ψ r y 3.55 r ẑ ˆp x ˆxˆp z ψ i z x x ψ r y 3.56 r ˆxˆp y ŷˆp x ψ i x y y ψ r. x 3.57 Within the present context, where the three components of the angular momentum have the same standing i.e., no particular orientation is linked to a symmetry, we would do well to use spherical coordinates, which are related to the Cartesian coordinates through x r sin θ cos φ 3.58 y r sin θ sin φ 3.59 z r cos θ

10 We must express the Cartesian partial derivatives in equations in terms of spherical coordinates. There are several ways to accomplish this, but we will proceed as follows. We can use the chain rule for the total differential of a Cartesian coordinate dx i as a functions of the spherical coordinates α j r, θ, φ dx i j x i α j dα j 3.6 and then write it down in matrix form dx dy dz x r y r z r x θ y θ z θ x φ y φ z φ dr dθ dφ sin θ cos φ r cos θ cos φ r sin θ sin φ sin θ sin φ r cos θ sin φ r sin θ cos φ cos θ r sin θ dr dθ dφ On the other hand, the partial derivatives can be written as x i j j α j x i α j xi, 3.63 α j α j where x i / α j is the inverse of the matrix of equation 3.62 since i α j x i x i α k α j α k δ jk We should note that since dx i is an element of a column vector, then / x i is an element of a row vector or a covector. Therefore, calculating the inverse of matrix of equation 3.62 we find x y z r θ φ sin θ cos φ sin θ sin φ cos θ r cos θ cos φ r cos θ sin φ r sin θ r sinφ sinθ r cosφ sinθ, 3.65 The inverse of a matrix is the transpose of its adjoint, divided by its determinant the adjoint is the matrix made of the cofactors. 7

11 which implies that x sin θ cos φ r + r cos θ cos φ θ sin φ r sin θ φ 3.66 sin θ sin φ y r + r cos θ sin φ θ + cos φ r sin θ φ 3.67 z cos θ r r sin θ θ Inserting equations and into equations we find as well as ˆL x i ˆL y i + cos φ cot θ φ sin φ θ cos φ + sin φ cot θ θ φ ˆL z i φ, 3.7 ˆL ± e ±iφ ± + i cot θ θ φ [ ˆL [ 2 sin θ θ sin θ θ + θ 2 + cot θ θ + sin 2 θ φ 2 2 sin 2 θ φ The eigenvalue problem for ˆL 2 and ˆL z then lead to the following differential equations [ 2 θ 2 + cot θ θ + sin 2 θ 2 φ 2 ψ r, θ, φ l l + ψ r, θ, φ 3.75 i ψ r, θ, φ mψ r, θ, φ φ But since these equations are purely angular in nature and do not depend on r, we can express the wave function as a product of radial and angular functions with the required normalization property ψ r, θ, φ R r Y lm θ, φ,

12 ˆ 2π ˆ R r 2 r 2 dr 3.78 ˆ π dφ Y lm θ, φ 2 sin θ dθ The separation of variable expressed in equation 3.77 transform the set of differential equations to be solved to [ 2 θ 2 + cot θ θ + sin 2 θ 2 φ 2 Y lm θ, φ l l + Y lm θ, φ 3.8 i φ Y lm θ, φ my lm θ, φ. 3.8 The solution of this system of equations is rendered easier upon the realization that equation 3.8 allows for a further separation of the two angular variables. More precisely, we can set Y lm θ, φ F lm θ e imφ The functions Y lm θ, φ are called spherical harmonics and form a basis in the corresponding two-dimensional angular space, which implies that they must possess a closure relation l l m l Y lm θ, φ Ylm θ, φ δ [ cos θ cos θ δ φ φ beyond their orthonormality condition ˆ 2π sin θ δ θ θ δ φ φ 3.83 ˆ π dφ Yl m θ, φ Y lm θ, φ sin θ dθ δ l lδ m m Any function f θ, φ can therefore be expanded with a series of spherical harmonics f θ, φ l l m l with the coefficients of the expansion given by c lm ˆ 2π c lm Y lm θ, φ, 3.85 ˆ π dφ Ylm θ, φ f θ, φ sin θ dθ

13 Although we will not demonstrate this here, the solution of equation 3.75, given equation 3.82, can be obtained by setting, for example, m l and applying equation 3.72 for ˆL + to show that Y ll θ, φ c l sin l θ, 3.87 and then successively act with ˆL to recover the functions Y lm θ, φ. Here are some examples of spherical harmonics for the three lowest values of l Y, θ, φ 4π Y, θ, φ cos θ π 3 Y,± θ, φ 8π sin θ e±iφ [ Y 2, θ, φ 3 cos 2 θ 3.9 6π 5 Y 2,± θ, φ 8π sin θ cos θ e±iφ Y 2,±2 θ, φ 32π sin2 θ e ±i2φ Exercise 3.2. The diatomic molecule. Let us consider two atoms of masses m and m 2 that are bound together into a diatomic molecule. Express the Hamiltonian as a function of the angular momentum of the system. We assume that the distance r between the two nuclei is constant. Solution. Let us start with the classical version of the problem. We set the origin at the centre of mass of the system, which implies that the positions of the nuclei positive, by definition are given by r r 2 m 2 r 3.94 m + m 2 m r m + m 2 The kinetic energy of the molecule is entirely in the form of rotational motion relative to the centre of mass and can be expressed with T 2 Iω2, 3.96 where ω is angular velocity about an axis perpendicular to the axis of symmetry and I the moment of inertia about the same axis 73

14 I m r 2 + m 2 r 2 2 m r r 2, 3.97 with the reduced mass m r m m 2 / m + m 2. As is well known, for such a system there is correspondence between linear and rotational motions, where the mass, velocity and linear momentum make way to the moment of inertia, angular velocity and angular momentum It follows that the classical Hamiltonian can be written as L Iω H L2 2I, 3.99 which is similar in form to the Hamiltonian of a free particle. The transition to the quantum mechanical version of the problem is straightforward and yields the corresponding Hamiltonian Ĥ ˆL 2 2I. 3. The eigenvectors of this Hamiltonian consist of the set { l, m } previously derived, with the corresponding eigenvalues E l 2 l l +, 3. 2I which are 2l + times degenerate since the energy is independent of m. We find that the energy difference between two adjacent energy levels is given by E l+,l 2 [l + l + 2 l l + 2I 2 I l Transitions between these states will be accompanied by the emission or absorption of a photon of frequency when the two atoms forming the molecules are different, these are the only allowed molecular transitions for an electric dipole interaction with the rotational constant B h/ 8π 2 I. ν l+,l E l+,l h 2B l +,

15 3.4 The Spin Intrinsic Angular Momentum If we consider a classical electron of charge q and mass m e exhibiting some orbital motion in an atom of molecule, then we can calculate it magnetic moment µ by the product of the associated electric current and the area contained within the circuit traced by the charge. Assuming a circular orbit we have µ q ω 2π πr2 q m e vr 2m e q L, 3.4 2m e where we used v ωr for the orbital speed. We therefore see that the magnetic moment of the electron is proportional to its angular momentum. When the atom or molecule is subjected to an external magnetic induction field B the Hamiltonian of the system can be written H H µ B, 3.5 where H is the Hamiltonian of the free atom or molecule i.e., when unperturbed by the external field and we have generalized the field-matter interaction using the total vectorial magnetic moment µ, which potentially contains a contribution from all the components i.e., particles of the systems. If we now define the magnetic field as being oriented along the z-axis, i.e., B Be z, then for the simple case where only one electron contributes to the total magnetic moment with the Bohr magneton H H µ B L zb, 3.6 We now transition to the quantum world and write µ B q 2m e. 3.7 Ĥ Ĥ ˆµ B Ĥ µ B ˆL z B, 3.8 where we still treat the magnetic field as a classical entity for that reason analyses such as this one area often called semi-classical. Assuming that Ĥ has the set { l, m } for its eigenvector with the eigenvalue, say, E l,m, we find that the interaction with the magnetic field changes this energy by an amount E l,m µ B mb

16 We therefore find that the energy of the system is altered by a quantity that is proportional to the magnetic quantum number m and the external magnetic field. This is a very important example as it clearly shows that the 2l + times degenerate l, m state is lifted by the presence of the magnetic field. For an atom or molecule, this implies that a spectral line associated with the principal quantum number l should split into 2l + separate fine structure lines. This is the so-called normal Zeeman effect. 2 Although this effect is observed experimentally, it does not take account for all possible results. For example, for systems where l and therefore m it is found e.g., for the hydrogen atom that there can still exist a spectral splitting, contrary to what would be expected through equation 3.9. For this reason, this is called the anomalous Zeeman effect. It was found that this effect could be explained if one postulates that the electron and other particles possesses an intrinsic angular momentum or spin Ŝ, to which an intrinsic magnetic moment is associated ˆµ s γ µ B Ŝ, 3. where γ 2 for the electron. Comparison with equations 3.4 and 3.7 reveals that the electron spin has a gyromagnetic ratio that is approximately twice that of its orbital angular momentum. Although the existence of the electron spin can be demonstrated from the relativistic Dirac equation, we can incorporate it in our formalism for non-relativistic quantum mechanics through the addition of a new postulates to the ones introduced in Chapter Seventh Postulate The spin operator is an intrinsic angular momentum with commutation relations similar to those obtained for the orbital angular momentum, i.e., [Ŝj, Ŝk [Ŝ2, Ŝj i i ε ijk Ŝ i 3. ˆ. 3.2 The spin possesses its own space, along with its eigenvectors s, m s and associated eigenvalues such that Ŝ 2 s, m s s s + 2 s, m s 3.3 Ŝ z s, m s m s s, m s Although we have considered a one-electron atom at least, as far as the external interaction is concerned, this formalism can be extended to any quantum mechanical system possessing a magnetic moment. For example, the case of the diatomic molecule considered previously is an example where such considerations can come into play. 76

17 Furthermore, the complete state of a particle is the direct product of a ket ϕ t, which is a function of its orbital characteristics, etc., and a ket or a linear combinations of kets s, m s associated to its spin. Finally, any spin observable commutes with all orbital observables. For example, the electron has a spin one-half, i.e, s /2. We define the basis ± 2, ± 2 with + and + +, and have that Ŝ 2 ± ± 3.5 Ŝ z ± ± ± The most general ket for the spin state of an electron is therefore ψ c c, 3.7 with c ± complex numbers c c 2. We can also define the raising and lowering operators which when acting on the basis yield see equation 3.43 Ŝ ± Ŝx ± iŝy, 3.8 If we associate the following matrices with the spin basis Ŝ Ŝ Ŝ Ŝ the we find the following Pauli matrices ˆσ j, +, 3.23 ˆσ + ˆσ ˆσ z and 77

18 ˆσ x ˆσ y 3.27 i, 3.28 i when with ˆσ e xˆσ x + e y ˆσ y + e z ˆσ z Also, the matrix for Ŝ2 is given by Ŝ ˆσ Ŝ If, for example, the basis for the electron s position is { r }, then its compound basis is given by the set of vectors χ r ε r, ε, 3.3 with ε ±. And from our earlier spin postulate [ˆx j, ŝ k, 3.32 as well while ˆx j r, ε x j r, ε 3.33 Ŝ 2 r, ε r, ε 3.34 Ŝ z r, ε ε r, ε, r, ε r, ε δ ε εδ r r The closure relation for the orbital-spin compound system is given by ˆ d 3 x r, ε r, ε ˆ ε An arbitrary state vector ψ for a compound system can be expanded using the { r, ε } basis in the usual manner 78

19 ψ ε ε ˆ ˆ d 3 x r, ε r, ε ψ d 3 xψ ε r r, ε, 3.38 with ψ ε r r, ε ψ. 3.5 Composition of Angular Momenta Given a classical system composed of two particles possessing the angular momenta L and L 2, respectively, we know from Newtonian mechanics that the total angular momentum L L +L 2 will be a conserved quantity if the system is not subjected to a net external torque. But the lack of an external torque does not imply the absence of internal torques. For example, the two particles could be interacting and affecting each other s angular momentum. That is, it is possible that dl /dt and dl 2 /dt, however it will be the case in the absence of a net external torque that dl /dt dl 2 /dt, such that dl/dt. In quantum mechanics we can verify the conservation of the total angular momentum in the following manner. In the case where the two particles do not interact, the total Hamiltonian of the system can be written as the sum of the individual Hamiltonians of the two particles with j, 2 Ĥ Ĥ + Ĥ Ĥ j ˆp2 j 2m j + ˆV ˆr j. 3.4 For simplicity we assume that the potential energies ˆV j are only a function of the distance of the corresponding particle to the origin of the system i.e., ˆr j ˆx 2 j + ŷ2 j + ẑ2 j. Because of this dependency it is possible to expand the potential about the origin with a Taylor series in ˆx 2 j, ŷ2 j and ẑ2 j, i.e., ˆV ˆV ˆx 2 j, ŷ2 j, ẑ2 j. With the commutation relations given in equations , it is straightforward to verify that [ˆL2 Ĥ, [ˆL2 Ĥ2, [ˆL Ĥ, [ˆL Ĥ2, 3.4, 3.42 and similarly for ˆL 2 2 and ˆL 2. It follows that the two individual angular momenta and their squares commute with the total Hamiltonian of the system Ĥ, and are therefore conserved quantities see equations.8 and

20 Let us now assume that the two particles interact through an Hamiltonian Ĥint ˆr ˆr 2, which is a function of the distance between them as is usually the case. We now have for the Hamiltonian of the system Ĥ Ĥ + Ĥint ˆr ˆr Because we have ˆr ˆr 2 ˆx ˆx ŷ ŷ ẑ ẑ 2 2, 3.44 [ it follows that, for example note that ˆx, Ĥint ˆ, etc., [ˆLz, Ĥ [ˆLz Ĥint, [ ˆx [ˆp y, Ĥint ŷ ˆp x, Ĥint, 3.45 which when acting on an arbitrary ket ψ and projected on the { r } basis yields r [ˆLz, Ĥ ψ { [ Hint ψ i x i i r y H int x y H int r H int ψ H int y ψ y H int x [ } Hint ψ ψ y H int x x [x y y 2 y x x 2 ψ 3.46 i x y 2 + y x r r We therefore find that an individual angular momentum does not commute with the Hamiltonian and its eigenvectors can therefore not be advantageously used to diagonalize the Ĥ. It is not a conserved quantity. On the other hand, because of the form of equation 3.44 we find that r [ˆL2z, Ĥ ψ H int i x 2 y 2 H int H int y 2 ψ x 2 i x 2 y + y 2 x r r r [ˆLz, Ĥ ψ, 3.49 which implies that r [ˆLz + ˆL 2z, Ĥ ψ

21 Since the same kind of calculations can be done for the x and y components of the individual angular momenta, we conclude that the total angular momentum ˆL ˆL + ˆL 2 commutes with the Hamiltonian, is a conserved quantity and its eigenvectors are shared with the Hamiltonian. Furthermore, the square of the total momentum can be written as ˆL 2 ˆL 2 + 2ˆL ˆL 2 + ˆL 2 2, such that [ˆL2, Ĥ [ˆL2, Ĥ + [ˆL2 2, Ĥ 2 ˆL [ˆL2, Ĥ + 2 [ˆL ˆL 2, Ĥ + [ˆL, Ĥ ˆL 2 ˆ, 3.5 on the account of equations It is therefore clear that we should concentrate on the total angular momentum, and not its individual components, when dealing with a quantum mechanical system The Compound System Before we consider the total angular momentum it will be instructive to combined the two particles and angular momenta through the direct product formalism. If the two respective bases for the individual particles and their Hamiltonians as in the no-interaction case of equations 3.39 are { l, m } and { l 2, m 2 }, then a natural compound base is given by their direct product l, m l 2, m 2 l, l 2 ; m, m This compound basis contains 2l + 2l 2 + eigenvectors of the angular momenta ˆL ˆL ˆ ˆL 2 ˆL 2 ˆ ˆL 2 ˆ ˆL ˆL 2 2 ˆ ˆL From now on we will drop this notation for the extension of operators from a single to the compound space and use ˆL for ˆL, etc., as there will be no ambiguity in what follows. The total angular momentum ˆL ˆL + ˆL 2 leads to the commutation relation [ˆL2, ˆL z, 3.57 as is the case for a single particle, but also [ˆL2, ˆL 2 [ˆL2, ˆL

22 [ˆLz, ˆL 2 [ˆLz, ˆL Note, however, that [ˆLz, ˆL 2 and [ˆL2z, ˆL 2. Because of the vanishing commutators of equations we have that ˆL 2, ˆL 2 2, ˆL 2 and ˆL z share the same eigenvectors also with the Hamiltonian of equation 3.43, which we will denote with { l, m } we omit the the quantum numbers l and l 2 for reasons that will become clearer later on. The action of ˆL 2 and ˆL z on these eigenvectors is defined to be ˆL 2 l, m l l + 2 l, m 3.62 ˆL z l, m m l, m, 3.63 with m changing by unit steps on the range m l. But we also have, with ˆL z ˆL z + ˆL 2z, which implies that ˆL z l, l 2 ; m, m 2 m + m 2 l, l 2 ; m, m 2, 3.64 m m + m It is important to realize that different values for m, m 2 or m imply a different orientation for the corresponding angular momenta vectors. We therefore expect that there will be several possibilities for both the orientation of the total angular momentum ˆL as well as its magnitude l l + 2. To get a better sense of this let us consider the case where l, m,,, and l 2 /2, m 2 /2, /2. Equation 3.65 tells us that the following values for m are realized m 3 2,, }{{} 2 2,, }{{ 2} 2, 3. }{{ 2} m m m m 2 ± 2 m 2 ± 2 m 2 ± These values for the magnetic total quantum number can be grouped as follows to find the realized values for l according to equations l 3 2, m 3 2, 2, 2, 3 2 l 2, m 2, This result can be generalized to any pair of angular momenta of any kind with l l 2 l l + l m l,

23 where successive values for l differ by. Now that we have the determined the eigenvalues of ˆL 2 and ˆL z, we need to express the corresponding eigenvectors l, m as a function the set of l, l 2 ; m, m 2. The recipe for accomplishing this task is, in principle, simple to apply. One can start with the eigenvector l l + l 2, m l + l 2 l + l 2, l + l 2 and then act upon it with the lowering operator to obtain see equation 3.39 or l + l 2, l + l 2 l, l 2 ; l, l ˆL ˆL + ˆL ˆL l + l 2, l + l 2 2 l + l 2 l + l 2, l + l ˆL + ˆL 2 l, l 2 ; l, l 2 2 l + l 2 2l l, l 2 ; l, l 2 + 2l 2 l, l 2 ; l, l 2 2 l + l 2 l l + l 2 l, l 2 ; l, l 2 + l2 l + l 2 l, l 2 ; l, l Repeating this process will yield the remaining eigenvectors l + l 2, m belonging to l l + l 2. We would then proceed to the subspace l l +l 2, in which the vector l + l 2, l + l 2 is unique non-degenerate and can only correspond to l + l 2, l + l 2 a l, l 2 ; l, l 2 + b l, l 2 ; l, l 2, 3.74 with a 2 + b 2. However, this eigenvector much be orthogonal to l + l 2, l + l 2 and from equation 3.73 we have l + l 2, l + l 2 l + l 2, l + l 2 a + b l + l 2 l + l 2, 3.75 which admits a l / l + l 2 and b l 2 / l + l 2. We thus find that l l2 l + l 2, l + l 2 l, l 2 ; l, l 2 l, l 2 ; l, l 2. l + l 2 l + l 2 l l

24 We can then repeated apply ˆL to find the remaining eigenvectors for the l l + l 2 subspace, and proceed similarly for l l + l 2 2, etc. More generally, we can concisely relate the two basis with the expansions l, m l, l 2 ; m, m 2 l m l l +l2 l 2 l l l 2 m l m 2 l 2 l, l 2 ; m, m 2 l, l 2 ; m, m 2 l, m 3.77 l l, m l, m l, l 2 ; m, m 2, 3.78 where the scalar products l, l 2 ; m, m 2 l, m l, m l, l 2 ; m, m 2 are the so-called Clebsch-Gordon coefficients, for which extensive tables can be found in the literature. Exercise 3.3. Let us consider the case of the composition of two one-half spins s s 2 /2. Express the eigenvectors s, m s corresponding to Ŝ Ŝ + Ŝ2 as a function of the basis s, s 2 ; m, m 2 associated to Ŝ and Ŝ2. Solution. Since s s 2 we will omit these quantum numbers in the basis { s, s 2 ; m, m 2 } and write the corresponding eigenvectors as ±, ±, where ± stands for m,2 ±/2. We start the calculation with s m s and write Using the lowering operator we have, +, or Ŝ, 2, Ŝ Ŝ2 + +, + Applying the lowering operator once more we have, + + +, 3.8, 2, + + +,. 3.8, 2 Ŝ, Ŝ + Ŝ2 2, + + +, 2 2, +,,

25 The three eigenvectors {,,,,, } form a triplet. Finally, we set, a +, + b, +, 3.83 as it is the only possible linear combination consistent with m. This ket is obviously orthogonal to, +, + and,,, but in order to have,, it must be that a b / 2. The singlet state is therefore given by, 2 +,,