Chapter 13. Angular Momentum: General Theory

Transcription

1 13. Angular Momentum Theory, February 10, Chapter 13. Angular Momentum: General Theory Section Definition and Basic Properties of the Angular Momentum 1 Introduction The definition of the angular momentum operator 3 ˆL commutes with ˆL x, ˆL y, and ˆL z 4 The eigenstates of ˆL and ˆL z 5 ˆL and ˆL z commute with other operators 6 A more convenient notation Section 13.. The Eigenvalues of ˆL and ˆL z 7 A summary of our task 8 The ladder operators 9 Find expressions for ˆL + L z ; j, m and ˆL L z ; j, m 10 The constants C + (j, m) and C (j, m) 11 The matrix elements of ˆL +, ˆL, ˆL x, ˆL y in the basis set j, m 1 Determining j and m 13 For a given j, m has a highest and a lowest value 14 The values of m and j 15 Summary Section Eigenvalue Problems for Angular Momentum Operators 16 Introduction 17 The matrix-and-vector representation for angular momentum problems 18 The case j = 1 19 The eigenvalues and eigenvectors of ˆL x for j = 1 0 The energy of the electron in the hydrogen atom exposed to a magnetic field

2 13. Angular Momentum Theory, February 10, Physical consequences Using a dumb choice of axis Appendix A Compact Form of the Commutation Relations Appendix 13.. Commutator Algebra c 014 Horia Metiu Section Definition and Basic Properties of the Angular Momentum 1 Introduction. There are many reasons why you should study angular momentum in quantum mechanics. It describes rotational motion, which is essential for understanding microwave and infrared spectra of molecules. Moreover, the spin of a particle is an angular momentum. A particle whose spin differs from zero has a magnetic moment which means that we can change its energy by acting on it with an external magnetic field. NMR and ESR spectroscopies are based on this property. You cannot understand these techniques without a through understanding of the properties of angular momentum. Finally, the total angular momentum is conserved during a collision between two molecules. This conservation places limitations on the reaction probability and the energies of the products after reaction. In this chapter we define angular momentum through the commutation relations between the operators representing its projections on the coordinate axes. These commutation relations allow us to determine the eigenstates of the angular momentum operator and to derive all matrix elements needed in calculations. The present chapter derives the general theory of angular mo-

3 13. Angular Momentum Theory, February 10, mentum. The subsequent ones apply this knowledge to derive the properties of spin and of the orbital angular momentum. The definition of the angular momentum operator. In quantum mechanics, we encounter two kinds of angular momenta. The orbital angular momentum is similar to the angular momentum in classical mechanics, which is defined by L = r p (1) where r is the position of a particle and p is its momentum. To define the operator ˆ L that represents the orbital angular momentum in quantum mechanics, we replace, in Eq. 1, r and p with the appropriate operators ˆ r and ˆ p. This means that ˆ L = ˆ r ˆ p () In addition to the orbital angular momentum, the particles studied in quantum mechanics have an angular momentum called spin. This is not caused by the motion of the particle (e.g. by its rotation); it is instead an intrinsic property of the particle similar to charge or mass. This angular momentum cannot be defined by Eq.. Born, Heisenberg, and Jordan 1 noted that the angular momentum defined by Eq. satisfies the commutation relations [ˆL x, ˆL y ] = i hˆl z (3) 1 M. Born, W. Heisenberg, and P. Jordan, Zur Quantenmechanik II, Zeitschrift für Physik 35 (8 9), (published August 196; received November 16, 195). English translation in: B. L. van der Waerden, editor, Sources of Quantum Mechanics (Dover Publications, 1968, ISBN ).

4 13. Angular Momentum Theory, February 10, [ˆL y, ˆL z ] = i hˆl x (4) [ˆL z, ˆL x ] = i hˆl y (5) The angular momentum is a vector and the operators ˆL x, ˆLy and ˆL z are the components of this vector on a Cartesian coordinate system. They recommended that Eqs. 3 5 be used as a definition: a vector operator whose components satisfy Eqs. 3 5 represents an angular momentum. This definition is valid for both the orbital angular momentum and the spin. A compact way of writing these expressions, using the Levi-Civita tensor, is explained in Appendix That notation is helpful when programming angularmomentum calculations in Mathematica. It also allows a more compact, but harder to decipher, notation. In this chapter we will use often a variety of expressions involving commutators; Appendix 13. reviews several useful relationships involving them. 3 ˆL commutes with ˆL x, ˆL y, and ˆL z. We know from classical mechanics that the rotational energy is proportional to the square of the angular momentum vector. Since this is an important quantity, we are naturally led to examine the operator ˆL = ˆL x + ˆL y + ˆL z (6) I show below that this operator commutes with ˆL x, ˆL y, and ˆL z. We already know that the components of the angular momentum do not commute with each other and therefore they cannot have joint eigenvalues. Physically this means that there is no state in which two of the projections of the angular momentum have well defined values. However, there are states that are eigenstates of the angular momentum squared and one of its projections (e.g. ˆL z ).

5 13. Angular Momentum Theory, February 10, This means that one can create in the laboratory states in which a measurement of either the square of the angular momentum or of its projection on a (Cartesian) coordinate axis will result in unique, well defined values (in other words the probability that these values are determined in a measurement is equal to 1). Showing that ˆL commutes with ˆL x, and ˆL y and ˆL z, is a matter of patiently using the commutator properties. We start with (use Eq. 6) [ˆL z, ˆL ] = [ˆL z, ˆL x ] + [ˆL z, ˆL y ] + [ˆL z, ˆL z ] (7) Any operator commutes with a power of itself, so [ˆL z, ˆL z ] = 0 (8) Next I use the relationship, given in Appendix 13. (Eq. 16), [Â, ˆB ] = [Â, ˆB] ˆB + ˆB[Â, ˆB], which gives [ˆL z, ˆL x ] = [ˆL z, ˆL x ]ˆL x + ˆL x [ˆL z, ˆL x ] (9) Using Eq. 5 in Eq. 9, to obtain [ˆL z, ˆL x] = i hˆl y ˆLx + i hˆl xˆly = i h (ˆLy ˆLx + ˆL x ˆLy ) (10) You can show similarly that [ˆL z, ˆL y] = i h (ˆLy ˆLx + ˆL x ˆLy ) (11) Using Eqs. 8, 10, and 11 in Eq. 7 gives [ˆL, ˆL z ] = 0 (1)

6 13. Angular Momentum Theory, February 10, In the same way, you can show that [ˆL, ˆL x ] = 0 (13) and [ˆL, ˆL y ] = 0 (14) Common sense tells us that if Eq. 1 is true, then Eqs. 13 and 14 must also be true, because there is no physical agent that differentiates between the OZ, and OX, and OY directions, unless one acts with a vectorial external field on the system. 4 The eigenstates of ˆL and ˆL z. Since ˆL commutes with ˆL z, there exists a set of eigenfunctions of ˆL that are also eigenfunctions of ˆL z. Because ˆL z does not commute with ˆL x or with ˆL y, a joint eigenstate of ˆL and ˆL z cannot be an eigenstate of ˆL x or of ˆL y. Let us examine in more detail what this means. Since ˆL and ˆL z commute, there exists a set of kets L z ; α, β that satisfy the eigenvalue equations for those two operators: ˆL L z ; α, β = α L z ; α, β (15) ˆL z L z ; α, β = β L z ; α, β (16) The symbol L z inside the ket L z ; α, β reminds me that this is an eigenket of ˆL z, not of ˆL x or ˆL y. If the system is in the state L z ; α, β, we are guaranteed to get the value α if we measure L and the value β if we measure L z. α is related to the rotational energy. β is the value that the projection of the angular momentum on the Z-axis can take, which is related to the orientation of the vector L.

7 13. Angular Momentum Theory, February 10, But ˆL also commutes with ˆL x, so there must exist a set of kets L x ; α, β such that ˆL L x ; α, β = α L x ; α, β (17) ˆL x L x ; α, β = β L x ; α, β (18) As long as no electric or magnetic field acts on the system there is no physical difference between the OX and the OZ axes. Therefore, the values that β can take when the system is in the state L x ; α, β are the same as the values it can take when the system is in L z ; α, β. In what follows, we concentrate on the joint eigenstates of ˆL and ˆL z. At this point there is no reason to prefer ˆL z over ˆL x or ˆL y, but we must pick one. Suppose that we have performed an experiment that places the system in the state L z ; α, β. We can measure L and get the value α, or we can measure L z and get the value β. However, if we measure L x, when the system is in state L z ; α, β, the probability that L x has the value β and L has the value α is L x ; α, β L z ; α, β When the system is in the state L z ; α, β, we know L and L z with certainty but we only know the probability that L x takes a given value. For someone familiar with quantum mechanics, this is not surprising. ˆL z and ˆL x do not commute so the eigenstate L z ; α, β cannot be an eigenstate of ˆL x. In addition, because ˆL z and ˆL x do not commute, the uncertainties in their values are connected by the Robinson inequality. These results are incomprehensible within classical physics, where there

8 13. Angular Momentum Theory, February 10, is no physical law to prevent us from knowing with certainty all three projections of any vector once we know the state of the system (i.e. the values of the position and of the momentum). 5 ˆL and ˆL z commute with other operators. One difficulty in grasping the physics of angular momentum comes from the fact that in many concrete problems, ˆL x, ˆL y, and ˆL z are not the only operators describing the system. For example, we might examine two particles interacting through a central force. This means that the potential energy V (r) depends only on the distance r between the particles and is independent of their orientation in space. You know two examples of such systems: the hydrogen atom and the diatomic molecule. The eigenvalue problem for the Hamiltonian of the hydrogen atom in spherical coordinates is (see Metiu, Quantum Mechanics, p. 06) h µ ( ) 1 r ψ r r r + ˆL µrψ + V (r)ψ = Eψ (19) We see that besides the angular momentum squared ˆL, the equation contains the potential energy V (r) and the radial kinetic energy (the first term in Eq. 19). You have also learned that the Hamiltonian Ĥ of the hydrogen atom commutes with ˆL and with ˆL z. This means that Ĥ, ˆL, and ˆL z have common eigenstates, denoted by n, j, m. In the state n, j, m, the energy of the atom is (see Metiu, p. 300) the square of the angular momentum is E n = µe z 1 (4πɛ 0 ) h n, (0) h j(j + 1), (1)

9 13. Angular Momentum Theory, February 10, and the projection of the angular momentum on the z-axis is hm () In other words, we have Ĥ n, j, m = µe z 1 (4πɛ 0 ) h n, j, m n (3) ˆL n, j, m = h j(j + 1) n, j, m (4) ˆL z n, j, m = hm n, j, m (5) Here j indicates the eigenvalue of ˆL, and m indicates the eigenvalue of ˆL z. We learn from this example that the eigenstates of ˆL and ˆL z may also be eigenstates of other observables that commute with ˆL and ˆL z (in this case, the energy). These eigenstates are labeled by additional quantum numbers (n, in this case); they are n, j, m, not just j, m. A similar situation appears in the case of the diatomic molecules. If we make the rigid-rotor approximation and the harmonic approximation, the eigenvalue equation for the energy, Eq. 19, becomes (see Metiu, Chapter 16) h µ ( ) 1 r ψ + ˆL ψ + 1 r r r µr0 k(r r 0) ψ = Eψ (6) The eigenstates are ψ = v, j, m, in which the molecule has the vibrational energy ( E v = hω v + 1 ) (7) and the values of ˆL and ˆL z given by Eqs. 1 and. Eqs. 4 and 5 hold for this example also. In general, when ˆL and ˆL z commute with other operators, their eigenstates must be eigenstates of those operators. When we need to make this

10 13. Angular Momentum Theory, February 10, explicit, these eigenstates will be labeled a, j, m, where a tells us the values that the quantities represented by these other operators take, when the system is in a pure state a, j, m and a measurement of those other quantities is made. 6 A more convenient notation. I now change notation by writing α h j(j + 1) (8) β hm (9) L z ; α, β L z ; j, m (30) At this point pretend that you have never heard of angular momentum and that these equations are a change of variable to be justified because they simplify the algebra that follows. In this state of blissful ignorance you do not know the values of α and β and therefore you do not know the values of j and m. In what follows, I will derive the possible values of j and m (i.e. the eigenvalues of ˆL and ˆL z ). In the new notation, the eigenvalue equations Eqs. 15 and 16 become ˆL L z ; j, m = h j(j + 1) L z ; j, m (31) and ˆL z L z ; j, m = hm L z ; j, m (3) Setting α = h j(j + 1) needs further examination. We must have α 0 (33) because α is a value that L can take. Our change of notation connects j to α through the equation α = h j(j + 1) which has two solutions for j:

11 13. Angular Momentum Theory, February 10, and j 1 = 1 j = 1 ( ) α/ h ( ) α/ h (34) (35) Because α/ h 0, both roots j 1 and j are real, j 1 is negative, and j is nonnegative. Remember that α is a physical observable while j is an arbitrary mathematical quantity introduced for convenience. You can verify that the negative values of j give the same values of α as the non-negative ones. We can choose either for indexing the kets, and we choose j 0 (36) Note also that m must be a real number ( hm is a value of L z ). Because L z is equally likely to be positive or negative, we expect m to be able to take both positive and negative values. Moreover, we guess that if m is an allowed value, then m is also allowed since the projection of the angular momentum vector on the OZ axis can be oriented along OZ (m > 0) or opposite to it (m < 0), and in a field-free region of space these two states are equally probable. Section 13.. The Eigenvalues of ˆL and ˆL z 7 A summary of our task. To set up the theory of angular momentum, we need to do the following. 1. Since most calculations in quantum mechanics are done by representing operators by matrices, we need to use the kets L z ; j, m as a basis set

12 13. Angular Momentum Theory, February 10, and find out how to calculate the matrix elements L z ; j, m Ô L z; j, m for any operator Ô that is a function of ˆL x and/or ˆL y.. We also need to find the values of j and m allowed by the eigenvalue equations Eqs To reach these objectives, we only have the commutation relations Eqs. 3 5 and our wits. The strategy is simple but tedious. We know how ˆL and ˆL z act on L z ; j, m. Therefore we must use the commutation relations to express ˆL x and ˆL y in terms of ˆL and ˆL z. Determining the allowed values of j and m is more subtle and cannot be reduced to a few sentences. 8 The ladder operators. In vector calculus, one often describes a vector {v x, v y, v z } through its spherical components {v x +iv y, v x iv y, v z }. It might therefore be useful to take a look at the operators ˆL + ˆL x + iˆl y (37) and ˆL ˆL x iˆl y (38) along with ˆL z and ˆL. If we find out how ˆL + and ˆL act on L z ; j, m, then it is easy to use and ˆL x = ˆL + + ˆL ˆL y = ˆL + ˆL i (39) (40) to find out how ˆL x and ˆL y act on L z ; j, m. Working with ˆL + and ˆL will prove fruitful.

13 13. Angular Momentum Theory, February 10, Our calculations rely on the commutation relations and I derive here a number of relations involving commutators that we need. The commutators of ˆL + and ˆL with ˆL and ˆL z are easily derived from the definitions, Eqs. 37 and 38, and the original commutation relations, Eqs They are [ˆL z, ˆL + ] = hˆl + (41) and We also have [ˆL z, ˆL ] = hˆl (4) [ˆL, ˆL + ] = [ˆL, ˆL ] = 0 (43) because ˆL commutes with both ˆL x and ˆL y, hence with both ˆL + and ˆL. To carry out our program of determining what happens when ˆL + and ˆL act on L z ; j, m, we need to find expressions that contain ˆL + and/or ˆL in the left-hand side and only ˆL z and ˆL in the right-hand side. I derive below several such expressions. One of them is ˆL ˆL+ = (ˆL x iˆl y )(ˆL x + iˆl y ) (used Eqs. 37 and 38) = ˆL x + ˆL y + i(ˆl xˆl y ˆL y ˆL x ) = ˆL x + ˆL y + i[ˆl x, ˆL y ] = ˆL x + ˆL y + i hˆl z (used Eq. 3) = ˆL ˆL z hˆl z (used ˆL = ˆL x + ˆL y + ˆL z) (44) This is the kind of equation we seek because it expresses ˆL + and ˆL in terms of ˆL and ˆL z. This is good because I know how the latter two operators act on ˆL z ; j, m. Similarly ˆL + ˆL = ˆL ˆL z + hˆl z (45)

14 13. Angular Momentum Theory, February 10, We can add Eqs. 44 and 45 to obtain ˆL +ˆL + ˆL ˆL+ = ˆL ˆL z (46) Using the fact that the operators ˆL x, ˆL y, and ˆL z represent observables and are therefore Hermitian (so that ˆL x = ˆL x and (iˆl x ) = iˆl x, etc.), we have ˆL + = ˆL (47) and ˆL = ˆL + (48) I can rewrite Eq. 46, with the help of Eqs. 47 and 48, as ˆL ˆL + ˆL + ˆL + = ˆL ˆL z (49) These manipulations were made to ensure that the right-hand side contains operators whose action on L z ; j, m is known (see Eqs. 31 and 3). 9 Find expressions for ˆL + L z ; j, m and ˆL L z ; j, m. For convenience, I will drop the marker L z from the eigenkets L z ; j, m and reinstate it only when there is some possibility of confusion. The fact that ˆL and ˆL + commute leads to ˆL ˆL+ j, m = ˆL ˆL + j, m = h j(j + 1)ˆL + j, m (50) This equation tells us that if j, m is an eigenstate of ˆL with the eigenvalue h j(j +1) then ˆL + j, m is also an eigenstate of ˆL, with the same eigenvalue h j(j + 1). Since [ˆL z, ˆL + ] = hˆl + (Eq. 41), we have (ˆLz ˆL+ ˆL ) +ˆLz j, m = hˆl+ j, m (51)

15 13. Angular Momentum Theory, February 10, Use ˆL z j, m = hm j, m in the left-hand side and rearrange the terms to obtain: ˆL z (ˆL+ j, m ) = h(m + 1) (ˆL+ j, m ) (5) Eq. 5 tells us that ˆL + j, m is either zero, which is uninteresting, or it is an eigenstate of ˆL z with the eigenvalue h(m + 1). This means that ˆL + j, m = C + (j, m) j, m + 1 (53) where C + (j, m) is a number that may or may not depend on j and m. It is prudent to assume that it does. Note that in Eq. 53, ˆL + acts on j, m and does not affect the value of j but changes m. This is in agreement with Eq. 50, which says that ˆL + j, m is an eigenfunction of ˆL with the eigenvalue h j(j + 1). A similar argument starts from [ˆL z, ˆL ] = hˆl (Eq. 4) and concludes that ˆL j, m = C (j, m) j, m 1 (54) where C (j, m) is another, unknown constant. Also, ˆL j, m is an eigenfunction of ˆL corresponding to the eigenvalue h j(j + 1). The operators ˆL + and ˆL are called ladder operators; ˆL + is called a raising operator and ˆL, a lowering operator. When ˆL + acts on j, m, it increases m by 1 and multiplies the result by C + (j, m); when ˆL acts on j, m, it decreases m by 1 and multiplies the result by C (j, m). The operators ˆL + and ˆL do not affect j in j, m, because both ˆL + and ˆL commute with ˆL and therefore they have common eigenstates with ˆL. In other words, they must convert a state in which ˆL has the value h j(j + 1) into a state in which ˆL has the same value h j(j + 1).

16 13. Angular Momentum Theory, February 10, The constants C + (j, m) and C (j, m). To determine C + (j, m) or C (j, m), we need to derive first a few helpful results. Consider λ =  x (55) where x is an arbitrary ket and  is an arbitrary operator. We have λ λ = Âx Âx = x   x (56) I used here the property Âx z = x  z of adjoint operators ( x and z are arbitrary kets). In addition, if  x = a y (57) where a is a number, then λ λ = ay ay = a a y y (58) Let us use Eqs. 56 and 58 to determine C +. To do this I apply Eq. 56 to λ ˆL + j, m (59) and obtain λ λ = j, m ˆL + ˆL + j, m (60) We also have (see Eq. 53) λ = ˆL + j, m = C + j, m + 1 (61) Use this to calculate λ λ = C +C + j, m + 1 j, m + 1 = C +C + (6)

17 13. Angular Momentum Theory, February 10, I used the fact that j, m + 1 j, m + 1 = 1. The two expressions for λ λ must be equal and therefore j, m ˆL ˆL + + j, m = C+C + (63) I am very pleased with this equation, for two reasons. First, I obtained a formula for C +. Second, this equation contains ˆL ˆL + +, which, according to Eq. 44 is (use ˆL + = ˆL ) ˆL ˆL+ = ˆL ˆL z hˆl z This expresses ˆL ˆL + in terms of ˆL and ˆL z. I know how ˆL and ˆL z act on j, m and therefore I can readily calculate the matrix element in Eq. 63. This gives j, m ˆL ˆL+ j, m = h j(j + 1) h m hm (64) Eq. 63 determines the absolute value of C + but not its phase. Since all physical properties of a ket are independent of a phase factor in front of it, I can take in C + any phase I want. I choose it so that C + (j, m) = Now use Eq. 44 in Eq. 65: C + (j, m) = j, m ˆL + ˆL + j, m (65) j, m (ˆL ˆL z hˆl z ) j, m (66) Using the eigenvalue equations for ˆL and ˆL z (Eqs. 31 3), we rewrite Eq. 66 as (we used j, m j, m = 1). C + (j, m) = h j(j + 1) m(m + 1) (67)

18 13. Angular Momentum Theory, February 10, Finally, putting C + (j, m) given by Eq. 67 into Eq. 53 gives ˆL + j, m = h j(j + 1) m(m + 1) j, m + 1 (68) Note that ˆL + j, m is an eigenstate of ˆL corresponding to the eigenvalue hj(j + 1) and an eigenstate of ˆL z corresponding to the eigenvalue (m + 1) h. However, j, m is not an eigenstate of ˆL +. It should not be, because ˆL + does not commute with ˆL or with ˆL z. To calculate C (j, m), we follow the reasoning used to find C + (j, m). This leads to C (j, m) = h j(j + 1) m(m 1) (69) and ˆL j, m = h j(j + 1) m(m 1) j, m 1 (70) We can now calculate (use Eqs. 67, 70, 39) ˆL x j, m = 1 (ˆL + + ˆL ) j, m = 1 [C +(j, m) j, m C (j, m) j, m 1 ] (71) Similarly (use Eqs. 67, 70, 40) ˆL y j, m = 1 i [C +(j, m) j, m + 1 C (j, m) j, m 1 ] (7) 11 The matrix elements of ˆL +, ˆL, ˆL x, ˆL y in the basis set j, m. Now that we know how ˆL + and ˆL act on j, m (Eqs. 68 and 70), we can calculate the matrix elements j, m ˆL + j, m = h j(j + 1) m(m + 1) j, m j, m + 1 = h j(j + 1) m(m + 1) δ j jδ m,m+1 (73)

19 13. Angular Momentum Theory, February 10, The only non-zero matrix elements are those for which m = m+1 and j = j. Therefore only j, m + 1 ˆL + j, m = h j(j + 1) m(m + 1) (74) differs from zero. The matrix elements of ˆL are j, m ˆL j, m = h j(j + 1) m(m 1) δ j jδ m,m 1 (75) and only j, m 1 ˆL j, m = h j(j + 1) m(m 1) (76) differs from zero. Since (see Eq. 39) we have ˆL x = ˆL + + ˆL j, m ˆL x j, m = 1 (C +(j, m)δ j jδ m,m+1 + C (j, m)δ j jδ m,m 1) (77) Only the matrix elements j, m + 1 ˆL x j, m = 1 C +(j, m) (78) and j, m 1 ˆL x j, m = 1 C (j, m) (79) differ from zero. Using (see Eq. 40) ˆL y = ˆL + ˆL i

20 13. Angular Momentum Theory, February 10, with Eqs. 73 and 75 leads to j, m ˆL y j, m = 1 i (C +(j, m)δ j jδ m,m+1 C (j, m)δ j jδ m,m 1) (80) Only the matrix elements j, m + 1 ˆL y j, m = 1 i C +(j, m) (81) and j, m 1 ˆL y j, m = 1 i C (j, m) (8) differ from zero. This covers all the matrix elements used in applications. However, we still don t know what values j and m can take, since we do not know the eigenvalues of ˆL and ˆL z. Next, we start working to find them. 1 Determining j and m. We have decided to use the eigenstates j, m of ˆL and ˆL z as a basis set and have managed to calculate the matrix elements of ˆL x, ˆL y, ˆL +, and ˆL. The remaining task is to determine the allowed values of j and m. We will find that j can only be integer or half-integer and that m in ˆL x ; j, m can only take integer or half-integer values between j and j. For example, if j = 1 then m can only be one of 1, 0, 1; if j = 1 then m can only be one of 1, For a given j, m has a highest and a lowest value. A classical interpretation of angular momentum tells us that if ˆL is fixed then ˆL z must have an upper and a lower bound. Indeed the projection of a vector on an axis can at most be equal to the length l of the vector: the projection has the upper bound l and the lower bound l. Since quantum mechanics is not

21 13. Angular Momentum Theory, February 10, classical mechanics, it will be reassuring to prove that, for a given j, m has an upper and a lower bound. and Recall that (Eqs. 46 and 49) ˆL +ˆL + ˆL ˆL+ ˆL ˆL + ˆL + ˆL + = ˆL ˆL z (83) = ˆL ˆL z (84) From properties of the inner product, we know that for any ket x and any operator Â, we have x   x 0 (with equality only when  x = 0). Taking the matrix element of the operator in Eq. 84 with the non-zero ket j, m gives j, m ˆL ˆL j, m + j, m ˆL + ˆL + j, m = h (j(j + 1) m ) (85) Since the left-hand side cannot be negative, we conclude that j(j + 1) m 0 (86) This means that if j is fixed then m has both a lower bound and an upper bound. We don t know what the bounds are but we know that they exist, so let us denote them by m l and m u. This gives us two new equations to play with: ˆL + j, m u = 0 (87) ˆL j, m l = 0 (88) To derive them, observe that, according to Eq. 53, ˆL + j, m u = C + (j, m u ) j, m u + 1. Since no value of m can be larger than m u, the state j, m u + 1 cannot exist and the ket is zero (which proves Eq. 87). One arrives at Eq. 88 similarly.

22 13. Angular Momentum Theory, February 10, 014 Now let us act with ˆL on Eq. 87 and with ˆL + on Eq. 88, to obtain ˆL ˆL+ j, m u = 0 (89) ˆL + ˆL j, m l = 0 (90) Why do I do that? Because I know that (Eq. 44) ˆL ˆL+ = ˆL ˆL z hˆl z ; the right-hand side of this equation contains only operators that act on j, m according to known rules, which means that I can calculate 0 = ˆL ˆL+ j, m u = (ˆL ˆL z hˆl z ) j, m u = ( h j(j + 1) h m u h m u ) j, m u (91) This implies that j(j + 1) m u m u = 0 (9) Similarly, using (see Eq. 45) ˆL + ˆL = ˆL ˆL z + hˆl z in Eq. 90 gives j(j + 1) m l + m l = 0 (93) Solving Eq. 93 for j(j + 1) and using the result in Eq. 9 gives m u (m u + 1) = m l (m l 1) (94) This equation has two solutions m u = m l (95)

23 13. Angular Momentum Theory, February 10, and m u = m l 1 (96) The second one is excluded by the fact that m u m l by definition. We see therefore that m u = m l. Symmetry requires the possible projections of ˆL along OZ to be equal to and of opposite sign to the possible projections in the direction opposite to OZ. Eq. 95 agrees with this requirement. 14 The values of m and j. We have learned a lot but we still don t know which values j and m can take. Here we settle this question. Let us consider a ket j, m 0 where m 0 is an allowed value of m for the given j. Acting on that ket with ˆL + repeatedly, we obtain ˆL + j, m 0 = C + (j, m 0 ) j, m 0 + 1, ˆL + j, m 0 = C + (j, m 0 )C + (j, m 0 + 1) j, m 0 +, etc. After n u steps, this procedure must generate a ket proportional to j, m u. If it did not, then we could apply ˆL + indefinitely and generate kets in which m > m u ; this contradicts the fact that m has a largest value, namely m u. Therefore we must have m u = m 0 + n u (97) where n u is a non-negative integer. Similarly, applying ˆL to j, m 0 some n l times produces a ket proportional to j, m l, so that m l = m 0 n l (98) with n l a non-negative integer. From Eqs. 97 and 98, it follows that m u m l = (m 0 + n u ) (m 0 n l ) = n u + n l n (99)

24 13. Angular Momentum Theory, February 10, where n is a non-negative integer. Since we already know that m u = m l (Eq. 95), we conclude that m u = n (100) We also know that (see Eq. 9) j(j + 1) = m u + m u = m u (m u + 1) (101) This has two solutions and j = m u (10) j = m u 1 (103) Eq. 103 is not compatible with the facts that m u 0 and j 0, so we can eliminate it as irrelevant. We also know that m u = m l, which used in Eq. 10 gives m l = j (104) Now combining m u = n/ (Eq. 100) with Eq. 10 gives j = n (105) We have obtained a startling result: j can be either a non-negative integer (if n is even) or a half-integer (if n is odd). In addition, the largest value of m is j (Eq. 10) and the smallest value of m is j (Eq. 104). Therefore m can take the j + 1 values j, j + 1,..., j 1, j. The theory of orbital angular momentum (see Metiu, Chapter 14, p. 0ff), based on the equation L = r p, tells us that half-integer values of j are not allowed for orbital angular momentum. However, by using the commutation

25 13. Angular Momentum Theory, February 10, relations to define the angular momentum, we have found that j can be either integer or half-integer. We know of no additional condition that we can use to conclude that half-integers are not allowed in general. Therefore the postulate that the have a half-integer intrinsic angular momentum (the spin) is in harmony with conclusions derived from the commutation relations. 15 Summary. This lengthy derivation led us to the following results. ˆL and ˆL z have common eigenstates ˆL z ; j, m for which ˆL ˆL z ; j, m = h j(j + 1) ˆL z ; j, m (106) and ˆL z ˆL z ; j, m = hm ˆL z ; j, m. (107) j can be either a non-negative integer or a positive half-integer (that is, of the form integer/). m can take j + 1 discrete values m = j, j + 1, j +,..., j, j 1, j (108) For example, if j = 3 then m can take = 4 values, namely m = 3, 3 + 1, 3 1, 3 = 3, 1, 1, 3 If j = 3 then m takes the = 7 values 3,, 1, 0, 1,, 3. We have also found that ˆL x ˆL z ; j, m = 1 [C +(j, m) ˆL z ; j, m C (j, m) ˆL z ; j, m 1 ] (109) and ˆL y ˆL z ; j, m = 1 i [C +(j, m) ˆL z ; j, m + 1 C (j, m) ˆL z ; j, m 1 ] (110)

26 13. Angular Momentum Theory, February 10, with and C + (j, m) = h j(j + 1) m(m + 1) (111) C (j, m) = h j(j + 1) m(m 1). (11) Note that j, m = 0 if m [ j, j]. The matrix elements needed in computations are ˆL z ; j, m ˆL z ˆL z ; j, m = δ jj δ mm hm (113) ˆL z ; j, m ˆL ˆL z ; j, m = δ jj δ mm h j(j + 1) (114) [ ] ˆL z ; j, m C+ (j, m)δ m ˆL x ˆL z ; j, m = δ,m+1 + C (j, m)δ m,m 1 jj (115) [ ] ˆL z ; j, m C+ (j, m)δ m ˆL y ˆL z ; j, m = δ,m+1 C (j, m)δ m,m 1 jj (116) i Note that all these matrix elements are zero if j j. This means that in the basis set ˆL z ; j, m, the matrices of ˆLx and ˆL y consist of smaller matrices strung along the diagonal; there are no non-zero matrix elements j, m Ôj, m in which j j if Ô = ˆL x, ˆL y, ˆL +, or ˆL or any function of them. Section Eigenvalue Problems for Angular Momentum Operators 16 Introduction. We have seen several times that kets are very useful for theoretical analysis, but practical calculations often use the representation of kets by vectors and of operators by matrices. In this section I review briefly the general theory and then apply it to the case of j = 3 as an example.

27 13. Angular Momentum Theory, February 10, Let us assume that on physical grounds we have decided that a certain physical system can be described by a ket space generated by the orthonormal basis set η 1, η,..., η N. An arbitrary ket ψ is then represented as N N ψ = η i η i ψ η i c i (117) i=1 i=1 and an arbitrary operator Ô by N N N N Ô = η i η i Ô η k η k η i O ik η k (118) i=1 k=1 i=1 k=1 Since we are assumed to know η i, i = 1,,..., N, knowing the vector ψ = {c 1, c,..., c N } (119) is equivalent to knowing ψ (because of Eq. 117). Eq. 118 tells us that we know how to operate with Ô if we know the matrix elements O ik η i Ô η k (10) The basis set kets η i are represented by the vectors η 1 = {1, 0,..., 0}. η N = {0, 0,..., 1} (11) and therefore ψ = N c i η i (1) i=1 The eigenvalue problem Ô ψ = o ψ is represented by the eigenvalue problem for the matrix O O ik c k = oc i, k

28 13. Angular Momentum Theory, February 10, which in vector-and-matrix notation is written as Oψ = oψ (13) or O 11 O 1N.. O N1 O NN c 1. c N = o c 1. c N (14) The solution of Eq. 14 provides N eigenvectors ψ(α) {c 1 (α),..., c N (α)}, α = 1,..., N (15) and N eigenvalues o(α), α = 1,..., N. From these we can calculate the N eigenkets ψ(1),..., ψ(n) by using N ψ(α) = c k (α) η k, α = 1,,..., N (16) k=1 17 The matrix-and-vector representation for angular momentum problems. It is natural to use for angular momentum calculations the basis set j, m. In this basis set, the matrix elements of ˆL z, ˆL, ˆLx, and ˆL y are given by Eqs When looking at these equations, we notice that all matrix elements in which j j are zero. The matrices of the operators ˆL and ˆL z

29 13. Angular Momentum Theory, February 10, are diagonal and those of ˆL x and ˆL y are block diagonal : All matrix elements other than the dots are also zero. We have arranged the rows and the columns in the following order: j = 0, m = 0 j = 1, m = 1 j = 1, m = 0 j = 1, m = +1 etc. The state with j = 0 generates a 1 1 block, the states with j = 1 generate a 3 3 block,... ; in general, the states having an arbitrary j generate a (j + 1) (j + 1) block. This break-up into blocks makes this basis set particularly simple to use in calculations. Physically it signifies that states

30 13. Angular Momentum Theory, February 10, with different values of j are not coupled by any of the operators ˆL, ˆL z, ˆL x, and ˆL y. Because of this block property, we can treat physical problems involving a specific j within a reduced space generated by the basis set j, j, j, j + 1,..., j, j 1, j, j (17) 18 The case j = 1. Let us solve some problems for j = 1. The basis set is 1, 1, 1, 0, 1, 1 (18) Since we know that j = 1, the first index is superfluous and we use instead the notation 1, 0, 1. The matrix representing an operator Ô in this subspace is 1 Ô 1 1 Ô 0 1 Ô 1 O = 0 Ô 1 0 Ô 0 0 Ô 1 (19) 1 Ô 1 1 Ô 0 1 Ô 1 The look of a matrix depends on the order of the kets in the basis set; the physical conclusions derived by using the matrix do not. However, vectors and matrices must be derived with the same basis set in order for us to use ordinary matrix and vector algebra. Eqs. 113 and 114 tell us that the matrix elements 1, m ˆL 1, m and 1, m ˆL z 1, m are zero if m m. Therefore the corresponding matrices are diagonal: L = h (1)(1 + 1) (130) 0 0 1

31 13. Angular Momentum Theory, February 10, and L z = h h (131) No surprise here. The basis set consists of eigenstates j, m (with j = 1) of ˆL and ˆL z and therefore the matrices of these operators, in that basis set, are diagonal and the diagonal elements are the eigenvalues of the operator. The matrices corresponding to ˆL x and ˆL y are more interesting. Let us look at ˆL x, and use Eq. 115 together with Eqs. 111 and 11. I calculate, as an example, 1 ˆL x 1. We have m = 1 ˆL x m = 1 = 0 (13) because (see Eq. 115) δ m,m+1 = δ 1, 1+1 = δ 1,0 = 0 and Another example is (use Eq. 115) with δ m,m 1 = δ 1, 1 1 = δ 1, = 0 m = 1 ˆL x m = 0 = C +(1, 0)δ 1,0+1 + C (1, 0)δ 1,0 1 = C (1, 0) (133) C (1, 0) = h 1(1 + 1) 0(0 1) = h (134) Together these give 1 ˆL x 0 = h (135)

32 13. Angular Momentum Theory, February 10, Patiently calculating all the matrix elements leads to L x = 0 h 0 h 0 h 0 h 0 (136) These matrix elements were calculated 3 in WorkBook13.Angular Momentum. The matrix must be Hermitian because ˆL x represents an observable, and therefore we need only calculate the matrix elements in the lower-left triangle; the ones in the upper-right triangle are then obtained from m Ô m = m Ô m. In addition, the Kronecker deltas in Eq. 115 tell us that the diagonal elements (that is, those with m = m) are zero and that in the lower-left triangle, only 0 ˆL x 1 and 1 ˆL x 0 are nonzero. Exercise 1 Show that for j = 1 L y = h 0 i 0 i 0 i 0 i 0 (137) Is this matrix Hermitian? The top row is m = 1, the middle row is m = 0, and the bottom row is m = 1; the left column is m = 1, the middle column is m = 0, and the right column is m = 1. In all entries j = 1. 3 Before Mathematica was invented these calculations were done by hand and were so tedious that many a student developed a sturdy dislike for quantum theory of angular momentum. I have friends who avoided any research on subjects in which angular momentum was important. Nowadays the proofs are still tedious but at least the calculations are easier.

33 13. Angular Momentum Theory, February 10, Exercise Verify that the matrices representing the operators satisfy the commutation relations L x L y L y L x = i hl z L y L z L z L y = i hl x L z L x L x L z = i hl y and that L = L x + L y + L z 19 The eigenvalues and eigenvectors of ˆL x for j = 1. Now that we have a matrix representation of ˆL x, we can calculate the eigenvalues and eigenvectors of this operator. We need to know them because they appear in some problems in spectroscopy. I solved the eigenvalue problem in Cell 6 of the Mathematica file WorkBook13.Angular momentum. The results are (Cell 6 of WorkBook 13): eigenvalue eigenvector h { 1, 1, 0 { 1, 0, 1 } 1 } h { 1, 1, 1 } The eigenvectors are normalized. I remind you that in this basis set the eigenvectors of L z are {1, 0, 0}, {0, 1, 0}, and {0, 0, 1}. We are pleased to see that L x has the same eigenvalues as L z. Since there is no force on the system that breaks spherical symmetry, there is no

34 13. Angular Momentum Theory, February 10, preferred direction in space. The possible values of the projection of the vector L on the x-axis must be the same as those of the projection on the z-axis. Exercise 3 Show that L y has the same eigenvalues as L z. 6) The eigenstates of ˆL x in the basis set L z ; 1, m are (see WorkBook13, Cell eigenvalue eigenket h L x ; 1, 1 = 1 L z; 1, 1 1 L z ; 1, L z; 1, 1 (138) 0 L x ; 1, 0 = 1 L z ; 1, L z ; 1, 1 (139) h L x ; 1, 1 = 1 L z; 1, L z ; 1, L z; 1, 1 (140) I have had to augment the notation with a new index: L z ; j, m are eigenkets of ˆL z, and L x ; j, m are those of ˆL x. If we manage to place the system in the state L x ; 1, 1, a measurement of L x is guaranteed to give the result h. However, if we measure L z, when the system is in the state L x ; 1, 1, we obtain the value h with the probability P 1 L z ; 1, 1 L x ; 1, 1 [ 1 = L z; 1, 1 L z ; 1, 1 1 L z ; 1, 1 L z ; 1, L z; 1, 1 L z ; 1, 1 ] = 1 4 (141)

35 13. Angular Momentum Theory, February 10, To get this, I used the orthonormality condition L z ; 1, m L z ; 1, m = δ mm. Similarly, the probability P 0 that L z is zero is given by P 0 L z ; 1, 0 L x ; 1, 1 = 1 (14) and the probability P 1 that L z is 1 by P 1 L z ; 1, 1 L x ; 1, 1 = 1 4 (143) P 1 + P 0 + P 1 = 1, which is a must. This result is consistent with the fact that ˆL z does not commute with ˆL x and therefore there is no state in which the value of both is known with certainty. We can also calculate the average value of L z, L z L x ; 1, 1 ˆL z L x ; 1, 1, when the system is in the state L x ; 1, 1. Using Eq. 138 in the right-hand side (for L x ; 1, 1 ) gives L x ; 1, 1 ˆL z L x ; 1, 1 = L x ; 1, 1 ˆL ( ) 1 z L z ; 1, 1 ( ) L x ; 1, 1 ˆL 1 z L z ; 1, 0 + L x ; 1, 1 ˆL ( ) 1 z L z ; 1, 1 Using Eq. 138 again, as well as L z ; j, m ˆL z L z ; j, m = δ mm hm, leads to ( ( ( ) ( ( L z = ( h) (0) + ( h) = 0 (144) ) ) ) ) This is the same as L z = 1 m= 1 hmp m = ( h) (0)1 + ( h)1 = 0, (145) 4

36 13. Angular Momentum Theory, February 10, as it should be. This result is physically reasonable. Since no forces break spherical symmetry, there is no bias to prefer positive values of L z over negative values. The average is zero because the value h is as probable as h. 0 The energy of the electron in the hydrogen atom exposed to a magnetic field. When we discuss the hydrogen atom, I will show that when the atom is exposed to a magnetic field B, the energy of its electron changes. The Hamiltonian is Ĥ = Ĥ0 e m e ˆ L B (146) where m e is the mass of the electron, e is the proton charge, and Ĥ 0 is the Hamiltonian of the atom in the absence of the field. I am ignoring other terms in the Hamiltonian since I only want to illustrate how we handle this kind of problem. The question is: what happens to the energy of the electron when it is exposed to the magnetic field? I answer first the question for the case when I was smart enough to pick the z-axis along B. In this case B = {0, 0, B} and ˆ L B = ˆLz. The Hamiltonian is Ĥ = Ĥ0 e m e ˆLz B, where B is the magnitude of the field B. Since ˆL z and Ĥ0 commute, they have joint eigenstates, denoted by n, j, m, where n gives the energy E n in the absence of the field. I want to know what happens to the energy of the states, j, m when the field is on. To answer this question, I will use these states as a basis set. Therefore I have Ĥ n, j, m = Ĥ 0 n, j, m eb m e ˆL z n, j, m

37 13. Angular Momentum Theory, February 10, = ( E n eb ) m n, j, m (147) m e We see that n, j, m are eigenstates of the Hamiltonian defined by Eq The energy spectrum is given by E n eb m e m (148) As noted above, E n is the energy of the atom in the absence of the field: E n = Ry n, (149) where Ry is the Rydberg constant. For the case n =, in the absence of the field, the atom has four degenerate states:, 0, 0,, 1, 1,, 1, 0, and, 1, 1. If the field is on, the energy of these states changes to that given by Eq. 148 and depends on the quantum number m. The states are state n, j, m energy change, 0, 0 Ry/4 unchanged, 1, 1 Ry/4 + (eb/m e ) h increased, 1, 0 Ry/4 unchanged, 1, 1 Ry/4 (eb/m e ) h decreased The energy-level diagram changes, when the field is turned on, from having four states of the same energy E to having three distinct energy levels, E + eb h/m e, E, and E eb h/m e. The level having the energy E is doubly degenerate: the states, 0, 0 and, 1, 0 have energy E. 1 Physical consequences. In the absence of the field, an atom excited in any one of the levels of energy E will emit a photon of frequency (E E 1 )/ h = [ Ry 4 ( Ry 1 )] / h = Ry h ( ) (150)

38 13. Angular Momentum Theory, February 10, When the atom is in a magnetic field, the E levels corresponding to j = 1 splits into three levels as shown in Fig. 1. In the absence of the magnetic field, the atom excited in a state with n = emits a photon of frequency Ω. When the field is present, three emission frequencies, Ω 1, Ω, and Ω 3 are possible. Not all of them will be observed because of selection rules that render the rate of some transitions equal to zero. 4 E,1,1,1 E,0,1,0 and,,0 E,-1,1,-1 W 1 W W 3 E 1,0 1,0,0 Figure 1: The energy levels of a hydrogen atom in a magnetic field for n = 1 and n = 4 The spectrum of the hydrogen atom in a magnetic field will be discussed in detail in a future chapter. For a complete discussion we need to take into account electron spin, proton spin, and some relativistic effects, which add terms to the Hamiltonian in addition to the ones used here.

39 13. Angular Momentum Theory, February 10, Using a dumb choice of axis. We decided to use as a basis set the eigenstates L z ; n, j, m of ˆL z. Then we wisely decided that the z-axis coincides with the vector B, so that the interaction energy between the electron and the field became (e/m e )BˆL z. The Hamiltonian Ĥ = Ĥ0 (e/m e )BˆL z commutes with ˆL z and with ˆL and has L z ; n, j, m as eigenstates. This makes the matrix of Ĥ, in the chosen basis, diagonal. The eigenvalue problem is then trivial. What would happen if we were not so clever and decided to take the x-axis along B? We have B = {B, 0, 0} and the Hamiltonian is Ĥ = Ĥ 0 (e/m e )BˆL x (151) The basis set is still L z ; n, j, m. Ĥ no longer commutes with ˆL z and the basis-set kets are no longer eigenstates of Ĥ. The matrix of Ĥ, in the basis set L z ; n, j, m, is no longer diagonal. The matrix elements of the Hamiltonian are given by L z ; n, j, m Ĥ L z ; n, j, m = Ry n eb m e L z ; n, j, m ˆL x L z ; n, j, m (15) For n =, we have j = 0 and m = 0, or j = 1 and m = 1, 0, 1. We have listed the equations giving the matrix elements L z ;, j, m ˆL x L z ;, j, m in 15 (see Eqs. 115, 111, 11). The matrix H corresponding to the Hamiltonian in Eq. 151 is (see WorkBook 13)

40 13. Angular Momentum Theory, February 10, , 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1 Ry/ Ry/4 eb m e h/ 0 0 eb m e h/ Ry/4 eb m e h/ 0 0 eb m e h/ Ry/4 The eigenvalues of this matrix were calculated in Cell 7 of WorkBook 13. They are Ry/4, Ry/4, Ry/4 eb m e h, and Ry/4 + eb m e h. They are exactly the same as the values obtained when we took the z-axis along B. They must be, because the coordinate system and its axes are only in our heads. The atom does not know about them. The energies are measurable quantities and if they depend on our choice of axes, the theory would be erroneous. Note however that the eigenvectors of H with B taken along OX differ from the eigenvectors of H with B taken along OZ. This is all right. The eigenvectors are not measurable quantities. However, any observable calculated with these eigenvectors is the same whether B is oriented along OX or along OZ. You see here a conspicuous feature of quantum mechanics. The mathematics depends on the choice of basis set and coordinate axes, but the physics does not. Appendix A Compact Form of the Commutation Relations It is sometimes useful to write the commutation relations Eqs. 3 5 in a more compact form. To do this we write the components of the vector ˆ L as

41 13. Angular Momentum Theory, February 10, ˆL 1, ˆL, and ˆL 3, with ˆL 1 ˆL x, ˆL ˆL y, and ˆL 3 ˆL z. Then we can write 3 [ˆL α, ˆL j ] = i h ε αjkˆl k, α = 1,, 3 (153) k=1 Here ε αjk is the Levi-Civita tensor. Since this symbol appears often in physics, it is worth explaining it in detail. To do so, we need to introduce the concept of permutation. We start with an ordered list of symbols, such as {1,, 3}, or {A, B, C} or {m, α, 3}. A permutation is an operation that changes the order of the symbols in the list. For example, suppose the permutation Π 1 acts on {1,, 3} to produce {1, 3, }. We can write this as Π 1 {1,, 3} = {1, 3, }. Π 1 has nothing to do with the numbers and 3; it simply exchanges the second and third objects in the list. That is, it takes the object in position and places it in position 3, and it takes the object in position 3 and places it in position. Because of this, Π 1 {A, B, C} = {A, C, B} and Π 1 {m, α, 3} = {m, 3, α} (154) A transposition is a permutation that interchanges the position of two elements in the list. Π 1 is a transposition, but Π {1,, 3} = {, 3, 1} (155) is not. Any permutation can be written as a succession of transpositions. For example, we can reach {, 3, 1} from {1,, 3} by two transpositions: {1,, 3} {, 1, 3} {, 3, 1} (156) This decomposition of a permutation into successive transpositions is not unique. The permutation Π in Eq. 155 is also equivalent to the following

42 13. Angular Momentum Theory, February 10, sequence of transpositions: {1,, 3} {1, 3, } {3, 1, } {3,, 1} {, 3, 1} (157) This takes four transpositions, while Eq. 156 took only two, but the outcome is the same Π. One can prove the following theorem: the number of transpositions through which a given permutation is expressed is either odd or even. In other words, Π might be expressed using two or four or six transpositions, but never using one, or three, or five. A permutation that can be decomposed into an even number of transpositions has the signature +1 and is called even ; one that can be decomposed into an odd number of transpositions has the signature 1 and is called odd. Mathematica has a function Permutations[list] that generates all permutations of the objects in the list. For example, the lists generated by Permutations[{1,,3}] are Permutations Signature {1,,3} +1 {1,3,} 1 {,1,3} 1 {,3,1} +1 {3,1,} +1 {3,,1} 1 The Mathematica function Signature[a] gives the signature of the permutation that creates the list a from the ordered version of the list a.

43 13. Angular Momentum Theory, February 10, We can now return to the Levi-Civita symbol. It is defined by signature[{i, j, k}] if i j and i k and j k ε ijk = 0 otherwise (158) Thus ε 11 = 0, ε 13 = 1, ε 13 = 1, etc. In many books, Eq. 153 is written as [ˆL α, ˆL j ] = i hε αjk ˆLk with the understanding that repeated indices are summed over. This is called Einstein s convention. Exercise 4 Show that the cross-product w = v u of vector algebra can be written as 3 3 w i = ε ijk v j u k, i = 1,, 3 j=1 k=1 Appendix 13.. Commutator Algebra This chapter relies heavily on commutators so I collect here some information about them. These relationships are useful in many areas of quantum mechanics. From the definition of the commutator [Â, ˆB] = Â ˆB ˆBÂ, it is obvious that [Â, ˆB] = [ ˆB, Â] (159) and [Â, ˆB + Ĉ] = [Â, ˆB] + [Â, Ĉ] (160)

44 13. Angular Momentum Theory, February 10, It is also easy to prove that [f(â), Â] = 0 (161) if  is the operator of an observable and f is an arbitrary function. Not so obvious, but still easy to prove, is [Â, ˆBĈ] = [Â, ˆB]Ĉ + ˆB[Â, Ĉ] (16) Indeed, we have (from the definition of a commutator) [Â, ˆBĈ] =  ˆBĈ ˆBĈ (163) Since (from the definition)  ˆB = [Â, ˆB] + ˆBÂ, we can rewrite Eq. 163 as from which it follows that This is what we wanted to prove. [Â, ˆBĈ] = [Â, ˆB]Ĉ + ˆBÂĈ ˆBĈÂ, [Â, ˆBĈ] = [Â, ˆB]Ĉ + ˆB(ÂĈ ĈÂ) = [Â, ˆB]Ĉ + ˆB[Â, Ĉ] If we take Ĉ = ˆB in Eq. 16, we have Then taking Ĉ = ˆB in Eq. 16: [Â, ˆB ] = [Â, ˆB] ˆB + ˆB[Â, ˆB] (164) [Â, ˆB 3 ] = [Â, ˆB ˆB ] = [Â, ˆB] ˆB + ˆB[Â, ˆB ] (use Eq. 16) = [Â, ˆB] ˆB + ˆB([Â, ˆB] ˆB + ˆB[Â, ˆB]) (use Eq. 16) = [Â, ˆB] ˆB + ˆB[Â, ˆB] ˆB + ˆB [Â, ˆB]

45 13. Angular Momentum Theory, February 10, Repeating this procedure, we obtain [Â, ˆB n ] = n 1 s=0 ˆB s [Â, ˆB] ˆB n s 1 (165) For fun, let s apply this to  = ˆx and ˆB = ˆp, for which we know that where Î is the unit operator. We have [ˆx, ˆp n ] = n 1 s=0 [ˆx, ˆp] = i hî (166) ˆp s [ˆx, ˆp]ˆp n s 1 = n 1 s=0 ˆp s i hî ˆpn s 1 = i hˆp n 1 + ˆp(i h)ˆp n + + ˆp n 1 (i h) = i h(nˆp n 1 ) = i h ˆpn ˆp Exercise 5 Show that if f(ˆp) = f nˆp n n=0 where f n, n 0, are numbers, then [ˆx, f(ˆp)] = i h f(ˆp) ˆp Exercise 6 Show that [Â, [ ˆB, Ĉ]] + [ ˆB, [Ĉ, Â]] + [Ĉ, [Â, ˆB]] = 0