The first change comes in how we associate operators with classical observables. In one dimension, we had. p p ˆ

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1 VI. Angular momentum Up to this point, we have been dealing primaril with one dimensional sstems. In practice, of course, most of the sstems we deal with live in three dimensions and 1D quantum mechanics is at best a useful model. In this section, we will focus in particular on the quantum mechanics of 3D sstems. Man of the elements we discovered for one dimensional problems will carr over directl to higher dimensions; however, we will encounter certain effects that are qualitativel new, and we will spend most of our time eploring these new phenomena. The first change comes in how we associate operators with classical observables. In one dimension, we had q q ˆ p p ˆ = i q In three dimensions the position and momentum are vectors and so we must substitute vector calculus for the single variable results: r rˆ ( i ˆ j ˆ k ˆ ) p pˆ Δ ip ˆ jp ˆ k p ˆ «Where r is the position vector and vector quantities will alwas be indicated in bold face. Note that the operators that correspond to different aes (i.e p ˆ and ˆ ) commute with one another, while the position and momentum along a given ais (i.e p ˆ and ˆ ) obe the normal commutation relation. We can summarie this in a few equations: [p ˆ i, p ˆ j ]= 0 [ˆ, r i r ˆ j ]= 0 [ˆ, ri p ˆ j ]= i Zδ ij where i and j can take the value 1, or 3 to indicate the ˆ, ˆ and ˆ components of each vector. a. Rotations The first difference between 3D and 1D is the possibilit of performing a rotation of our sstem about one of the three aes. Let us denote a rotation of an angle θ about a unit vector n b R n (θ ). Clearl, R n (θ )

2 is a matri (it transforms vectors to vectors). Further, it is clear that R (θ )R (θ ) R (θ )R (θ ) (rotations about different aes do not n m m n commute). Note that this has nothing to do with quantum mechanics and everthing to do with geometr! It is eas to verif that the rotation operators associated with the three Cartesian aes are: R (θ )= Δ0 cos θ sin θ Δ 0 sin θ cos θ «cos θ 0 sin θ R (θ ) = Δ Δ 0 cos θ «sin θ cos θ sin θ 0 R (θ )= Δ sin θ cos θ 0 «0 0 1 Note that the rotation matrices for and can be obtained from the matri b the cclic permutation,,. This must alwas be the case, because our labeling of the, and aes is totall arbitrar! The onl thing we must be careful of is that the triple product ( ) is alwas 1. This defines the handedness, or chiralit, of our coordinates. Cclic permutations preserve the handedness while a simple interchange of two aes (i.e. ) will flip the sign of the triple product, reverse the handedness of our coordinates and give us the wrong answer (tr it and see). This cclic invariance is ver important because it reduces the work we need to do b a factor of 3, but we must be careful to appl it correctl. In the future, we can therefore state the result for the ais and then infer the results for and b cclic permutations. These rotations are unitar (i.e. R T R = 1) and like man unitar i J / Z transformations the can be written in the form e θ where J is called a generator. For eample, the generator of rotations about the ais is

3 0 i 0 J = ZΔ i 0 0 «0 0 0 θ This can be verified b actuall computing e i J / Z and checking that it gives the rotation operator discussed earlier. b. Commutation Relations We now wish to compute the commutator between J and J : i J J = ZΔ0 0 i Δ = Z Δ «0 i 0 «i 0 0 « i J J = Z Δ Δ0 0 i = Z Δ0 0 0 «i 0 0 «0 i 0 «0 0 0 Ω [J,J ]= i ZJ As discussed previousl, all other commutators between the elementar generators of rotations can be deduced from the above relation b cclic permutations of the indices. These are the fundamental commutation relations for angular momentum. In fact, the are so fundamental that we will use them to define angular momentum: an three transformations that obe these commutation relations will be associated with some form of angular momentum. It is also useful to define the vector J J i J j J k and the scalar J = J J = J J J. It is eas to show that while the elementar generators do not commute with J the do commute with J : [J ] = [J ] [J ] = [ ] J [ ] [ ( i ZJ ) J J ( i ZJ ), J, J, J J, J J J, J J, J J J, J = ( i ZJ ) J J = 0 ]J [ ] ( i ZJ )

4 Note that these J matrices are not quantum operators the are simpl transformations of 3D space. However, we can use this as a definition of angular momentum in the quantum case. Specificall, we assume that the quantum operators (which act in Hilbert space) obe the same commutation rules as the classical transformations (which act in real space). Hence, quantum angular momentum operators obe [Jˆ, ]= i Z and cclic permutations thereof. This seems strange at first, but momentaril we will show that this rule for associating operators with classical variables is consistent with our definitions of r ˆ and p ˆ, which strongl supports the new quantiation rule. Further, we will later see that the same commutation rules appl to a particle s intrinsic spin angular momentum, which cannot be described as some function of r ˆ and p ˆ. Hence the commutation relation above actuall generalies the standard quantiation rules. Classicall, angular momentum is given b L ˆ = r ˆ p ˆ. Using our standard prescription, this means the corresponding quantum operator should be L ˆ = r ˆ p ˆ. We proceed to verif that the components of L ˆ obe the epected commutation relations. Lˆ L ˆ = ( ˆ p ˆ ˆpˆ ˆ ˆ ˆ p ˆ ) = ˆp ˆ ˆ pˆ ˆ p ˆ ˆ pˆ ˆ p ˆ ˆ p ˆ ˆ pˆ )( p p Lˆ L ˆ = ( ˆ pˆ ˆp ˆ )( ˆ p ˆ ˆ p ˆ ) = ˆp ˆ ˆ p ˆ ˆ pˆ ˆ p ˆ ˆ p ˆ ˆ p ˆ ˆ p ˆ Ω [L ˆ, Lˆ ]= ˆ p ˆ [p ˆ, ˆ ] ˆ p ˆ [ ˆ, p ˆ ] ˆ ˆ ˆ p ˆ = i Z ˆ p ˆ i Z ˆ p ˆ = i ZL ˆ c. Eigenstates Since Ĵ and commute, the share common eigenstates. We will denote the eigenvalues of Ĵ and b α and β, respectivel so that: α, β = α α, β α, β = β α, β

5 It is convenient to define the raising and lowering operators (note the similarit to the Harmonic oscillator!): i Jˆ Which satisf the commutation relations: [ ˆ ˆ ˆ ˆ ˆ ˆ ˆ, J ]= ZJ [J, J ]= ZJ [J, J ]= 0 The raising and lowering operators have a peculiar effect on the eigenvalue of : ( α, β ) = ([J ˆ, ] Jˆ ) α, β = (β Z)( α, β ) Thus, ( ) raises (lowers) the eigenvalue of b Z, hence the names. Since the raising and lowering operators commute with Ĵ the do not change the value of α and so we can write α, β α, β Z and so the eigenvalues of are evenl spaced. What are the limits on this ladder of eigenvalues? Recall that for the harmonic oscillator, we found that there was a minimum eigenvalue and the eigenstates could be created b successive applications of the raising operator to the lowest state. There is also a minimum eigenvalue in this case. To see this, note that: ˆ 1 (α β =) α, β J α, β = α, β ( ) α, β 1 = α β (, ) α, β 1 1 = α, β α, β 0 Hence β α and therefore α β α. Which means that there are both maimum and minimum values that β can take on for a given α. If we denote these values b β ma and β min, respectivel, then it is clear that α, β ma = 0 J α, β m i n = 0. We can then use this knowledge and a trick to determine the relationship between α and β ma (or β min ): Ω, ˆ α β ma = 0 J α, β m i n = 0 ˆ

6 ˆ ˆ Ω ( Ĵ i ( J J )) α, β = 0 ( Ĵ i ( ma )) α, β min = 0 Ω ( Z ) α, β = 0 ( ZĴ ) α β = 0 ma, min Ω (α β ma Zβ ma )= 0 (α β min Zβ min )= 0 Ω α = β (β Z) = β (β Z) ma ma min min Ω β ma = β min Zj So we have that Zj β Zj. Further, since we can get from the lowest to the highest eigenvalue in increments of Z b successive applications of the raising operator, it is clear that the difference between the highest and lowest values [ Zj ( Zj) = Zj ] must be an integer multiple of Z. Thus, j itself must either be an integer or a half-integer. Putting all these facts together, we conclude (Define m b /Z): ˆ 1 3 J j, m = Z j( j 1 ) j, m j = 0,, 1,,... and j, m = mz j, m m = j, j 1... j 1, j where in the first equation, we have noted that 0 J ˆ = Z j( j 1) implies j 0. These are the fundamental eigenvalue equations for all forms of angular momentum. The other matri elements we might be interested in are those of the raising and lowering operators. As we saw before Ĵ j, m j, m 1 and so j ', m ' j, m = C j,m δ j, j δ m, m 1 and one onl needs to determine the value of C j,m. To this end, ˆ ˆ ˆ j, m = j, m J J j, m = j, m J j, m = j, m Ĵ _ZĴ j, m Ĵ _ = Z [ j( j 1) m m]= C j,m the phase of C j,m is undetermined, because the phase of the eigenstate j, m is arbitrar. We will choose the phase of j, m so that C j,m is real and positive which leads to:

7 j ', m ' j, m = Z j ( j 1 ) m ( m 1 ) δ j, j δ, m m 1.