Physics 342 Lecture 23. Angular Momentum. Lecture 23. Physics 342 Quantum Mechanics I

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1 Physics 342 Lecture 23 Angular Momentum Lecture 23 Physics 342 Quantum Mechanics I Monday, March 31st, 2008 We know how to obtain the energy of Hydrogen using the Hamiltonian operator but given a particular E n, there is degeneracy many ψ nlm r, θ, φ) have the same energy. What we would like is a set of operators that allow us to determine l and m. The angular momentum operator ˆL, and in particular the combination L 2 and L z provide precisely the additional Hermitian observables we need Classical Description Going back to our Hamiltonian for a central potential, we have H = p p 2 m + Ur). 23.1) It is clear from the dependence of U on the radial distance only, that angular momentum should be conserved in this setting. Remember the definition: L = r p, 23.2) and we can write this in indexed notation which is slightly easier to work with using the Levi-Civita symbol, defined as in three dimensions) 1 for ijk) an even permutation of 123). ɛ ijk = 1 for an odd permutation 23.3) 0 otherwise Then we can write 3 3 L i = ɛ ijk r j p k = ɛ ijk r j p k 23.4) j=1 k=1 1 of 9

2 23.2. QUANTUM COMMUTATORS Lecture 23 where the sum over j and k is implied in the second equality this is Einstein summation notation). There are three numbers here, L i for i = 1, 2, 3, we associate with L x, L y, and L z, the individual components of angular momentum about the three spatial axes. To see that this is a conserved vector of quantities, we calculate the classical Poisson bracket: [H, L i ] = H j L i p j H p j L i j and p j p k ɛ ijk = 0 1. Noting that j = du dr j ɛ ikj r k p j 2 m ɛ ijk p k 23.5) = r j r, we have the first term in the bracket above proportional to ɛ ijk r j r k, and this is zero as well. So we learn that along the dynamical trajectory, dl i dt = 0. The vanishing of the classical Poisson bracket suggests that we take a look at the quantum commutator of the operator form ˆL and the Hamiltonian operator Ĥ. If these did commute, then we know that the eigenvectors of ˆL can be shared by Ĥ Quantum Commutators Consider the quantum operator: ˆL = r ) i which, again, we can write in indexed form: We have seen that the replacement [, ] P B [, ] i 23.7) ˆL i = ɛ ijk r j i k. 23.8) leads from the Poisson bracket to the commutator, and so it is reasonable to ask how ˆL i commutes with Ĥ we expect, from the classical consideration, that [Ĥ, ˆL i ] = 0 for i = 1, 2, 3. To see that this is indeed the case, let s use a test function fr n ) shorthand for fx, y, z)), and consider the commutator directly: [Ĥ, ˆL i ] fr n ) = Ĥ ˆL i fr n ) ˆL i Ĥ fr n ). 23.9) 1 the object p j p k is unchanged under j k, while ɛ ijk flips sign ɛ ijk p j p k = ɛ ijk p k p j = ɛ ikj p k p j = ɛ ijk p k p j = ɛ ijk p j p k, 23.6) and then we must have ɛ ijk p j p k = 0 since it is equal to its own negative. 2 of 9

3 23.2. QUANTUM COMMUTATORS Lecture 23 The first term, written out, is ) i Ĥ ˆL i fr n ) = 2 2 m q q + Ur) ɛ ijk r j k fr n )) = 2 2 m q ɛ ijk δ jq k fr n ) + ɛ ijk r j q k fr n )) + Ur) ɛ ijk r j k fr n )) = 2 2 m ɛ iqk q k fr n ) + ɛ ijk δ jq q k fr n )) + ɛ ijk r j q q k fr n )) + Ur) ɛ ijk r j k fr n )) 23.10) and note that the first two terms involving second derivatives are automatically zero. Then, in vector notation, we can write: Ĥ ˆL fr n ) = ] [ 2 i 2 m r 2 f) + Ur) r f ) Now consider the other direction, ) i ˆL i Ĥ fr n ) = ɛ ijk r j k 2 2 m q q fr n ) + Ur) fr n ) ) = 2 U 2 m ɛ ijk r j k q q fr n ) + ɛ ijk r j fr n ) + Ur) k fr n ) k 23.12) where, once again, k r k so the term involving the derivative of U will vanish, leaving us with ˆL Ĥ fr n) = ] [ 2 i 2 m r 2 f) + Ur) r f ) So it is, indeed, the case that [Ĥ, ˆL i ] = ˆLi Commutators Before exploring the implications of the shared eigenfunctions of Ĥ and ˆL i, let s calculate the commutators of the components of ˆL i with themselves. Consider, for example 1 2 ˆL i ˆLj fr n ) = ɛ ipq r p q ɛ juv r u v fr n )) = ɛ ipq r p ɛ jqv v fr n ) + ɛ ipq r p ɛ juv r u q v fr n ) 23.14) 3 of 9

4 23.2. QUANTUM COMMUTATORS Lecture 23 and the other direction, obtained by interchanging i j: 1 ˆL 2 j ˆLi fr n ) = ɛ jpq r p ɛ iqv v fr n ) + ɛ jpq r p ɛ iuv r u q v fr n ), 23.15) then the commutator is 1 2 [ˆL i, ˆL j ] fr n ) = ɛ ipq r p ɛ jqv v fr n ) ɛ jpq r p ɛ iqv v fr n ) ) Using the identity ɛ qip ɛ qjv = δ ij δ pv δ iv δ pj, we can rearrange 1 2 [ˆL i, ˆL j ] fr n ) =δ iv δ pj δ jv δ pi )] r p v fr n ) = δ ip δ jv δ iv δ jp ) v fr n ) = ɛ lij ɛ lpv r p v fr n ) 23.17) or, finally = i ɛ lij ˆL l, [ˆL i, ˆL j ] = i ɛ lij ˆLl ) We learn that, for example, [ˆL x, ˆL y ] = i L z. This tells us that it is impossible to find eigenfunctions of L x that are simultaneously eigenfunctions of L y and/or L z. So returning to the issue of [Ĥ, ˆL i ] = 0, we can, evidently, choose any one of the angular momentum operators, and have shared eigenfunctions of Ĥ and ˆL i, but we cannot also have these eigenfunctions for ˆL j. That may seem strange after all, if a vector is an eigenvector of Ĥ and ˆL x, and we can also make an eigenvector that is shared between Ĥ and ˆL z, then surely there is an eigenvector shared by all three? The above says that this is not true, and we shall see some explicit examples next time. For now, let s look for combinations of angular momentum operators that simultaneously commute with Ĥ and ˆL z, for example we are preferentially treating ˆL z as the operator that shares its eigenstates with Ĥ, but this is purely a matter of choice). The most obvious potential operator is ˆL 2 : [ˆL 2, ˆL z ] = [L 2 x + L 2 y + L 2 z), L z ] = L x [L x, L z ] i L y L x + L y [L y, L z ] + i L x L y + 0 = L x i L y ) + L y i L x ) i L y L x + i L x L y = ) and the same is true for [ˆL 2, ˆL x ] and [ˆL 2, ˆL y ]. In addition, it is the case that [ˆL 2, Ĥ] = 0, clearly, since the individual operators do. 4 of 9

5 23.3. RAISING AND LOWERING OPERATORS Lecture Raising and Lowering Operators We can make a particular algebraic combination of ˆL x and ˆL y that plays a role similar to the a ± operators from the harmonic oscillator example. Take then the commutator of these with L z is L ± L x ± i L y, 23.20) [L ±, L z ] = [L x, L z ] ± i [L y, L z ] = i L y L x = L x ± i L y ) = L ± ) For an eigenfunction of L z, call it f such that L z f = α f, the operator L + acting on f is also an eigenfunction, but with eigenvalue: L z L ± f) = L ± L z ± L ± ) f = α ± ) L ± f) 23.22) so acting on f with L ± increases or decreases) the eigenvalue by. Similarly, if we take the function f to be simultaneously an eigenfunction of L 2 which we know can be done), the L ± f is still an eigenfunction, with 2 L 2 L ± f) = L ± L 2 f = β L ± f 23.23) for eigenvalue β in other words, L ± f is an eigenfunction of L 2 with the same eigenvalue as f itself. The last observation we need is that there must be a minimal and maximal state for the L ± operators since the state L ± f has the same L 2 eigenvalue as f does, while its eigenvalue w.r.t. L z is going up or down, there will be a point at which the eigenvalue result of measurement) of L z is greater than the total magnitude of the eigenvalue w.r.t. L 2 the result of a measurement of total angular momentum) so we demand the existence of a top L + f t = 0 and, for the same reasons, a bottom L f b = 0 So what? So remember that for the harmonic oscillator, we wrote H in terms of the products a ± a, and that allowed us to build a lowest energy state from which we could define 0 as a wavefunction. The same type of game applies here, we will write L 2 in terms of L ± L : L ± L = L 2 x i L x L y ± i L y L x +L 2 }{{} y) = L 2 x + L 2 y ± L z 23.24) = i [L x,l y] 2 I m going to do a commutator down here too many in the main portion gets boring. Since [L 2, L i] = 0, we have [L 2, L x ± il y] = 0. 5 of 9

6 23.4. EIGENSTATES OF L 2 Lecture 23 so that we can write L 2 = L ± L + L 2 z L z ) Our first goal is to relate the spectrum of L 2 to that of L z consider the top state f t with eigenvalue L z f t = l f t the is in there just to make the eigenvalue under lowering a bit nicer), and L 2 f t = α f t. Now from the above factorization of L ), we have L 2 f t = L L + + L 2 z + L z ) f = l l) f = α f 23.26) so the maximum eigenvalue for L z is related to the eigenvalue α via α = 2 l l + 1). Similarly, if we take the minimum eigenvalue for L z corresponding to the state f b : L z f b = l f b, then we use the other factorization and get α = l l 1), which tells us that l = l. So L + takes us from f b with eigenvalue l to f t with eigenvalue l it does so in integer steps, since L z L + f b ) = l + 1) L + f b, so l itself must either be an integer, or, potentially, a half-integer. We have described the eigenstates of L 2 and L z, then, in terms of two numbers: l = 0, 1 2, 1, 3 2, 2,... and m which, for each l takes integer values m [ l, l]. The states can be labelled fl m, and, if we are lucky, these will be the simultaneous eigenstates of H Eigenstates of L 2 To find the eigenstates of L 2, let s first work out, carefully, the classical value then we will attempt to make scalar operators which will be easy to write down in spherical coordinates: L 2 = ɛ ijk r j p k ) ɛ imn r m p n ) = δ jm δ kn δ jn δ km ) r j p k r m p n = r j p k r j p k r j p k r k p j ) There are some obvious simplifications here, but remember that while it is true, classically, that r j p k = r k p j, this is not true when we move to operators, so let s keep the ordering as above, and input p k i k etc. We will introduce the test function f to keep track: L 2 f = 2 [r j k ) r j k f) r j k ) r k j f)] = 2 [r j δ jk k f + r j r j k k f r j k ) j r k f) δ jk f)] = 2 [ r f + r 2 2 f r j j k r k f) r k k f) ] = 2 [ r f + r 2 2 f r 3 f + r f) r f) ], 6 of )

7 23.5. EIGENSTATES OF L Z Lecture 23 and combining terms, the final scalar form can be written: L 2 f = 2 [ r 2 2 f r f r ) r f) ] ) Now, think of the operator r f in spherical coordinates, that s just r f and we know the 2 operator, so in spherical coordinates, we can write the above trivially: L 2 f = 2 [ = 2 [ 1 sin θ r 2 f θ ) + 1 sin θ sin θ f θ sin θ f ) + 1 θ θ ) + 1 sin 2 θ 2 f φ 2 ]. sin 2 θ 2 f φ 2 r f r r f )] 23.30) But this is precisely the angular portion of the Laplacian itself and we know the solutions to this that have separation constant l l + 1), they are precisely the spherical harmonics, conveniently indexed appropriately. We have L 2 Y m l = 2 l l + 1) Y m l ) 23.5 Eigenstates of L z The fastest route to the operator expression for L z comes from the observation that classically, if we had a constant angular momentum pointing along the ẑ axis, then we have counter-clockwise rotation in the x y plane. If we use cylindrical coordinates with x = s cos φ, y = s sin φ), then the only motion is in the ˆφ direction. We might mimic this on the operator side by considering a test function that depends on φ only. Then L z fφ) =r x p y r y p x ) fφ) = i and from φ = tan 1 y/x), we have φ x = sin φ r [ r cos φ f φ f r sin φ φ x φ and φ y = cos φ r ] φ y, then 23.32) L z fφ) = i cos 2 φ + sin 2 φ ) f φ 23.33) or, to be blunt, L z = i φ ) 7 of 9

8 23.5. EIGENSTATES OF L Z Lecture 23 The eigenstates of L z are defined by L z f = m f for integer m now m is an integer aside from any periodicity concerns, it comes to us as an integer from the algebraic approach). The solution is f = e i m φ, and of course, this is the φ-dependent portion of the spherical harmonics, L z Y m l = i A m l φ P l m cos θ) ei m φ) = m Yl m 23.35) where we have written the ugly normalization as A m l. In the end, the Yl m θ, φ) are eigenstates of both the L 2 and L z operators. In addition, to the extent that they form the angular part of the full Hamiltonian, ψ nlm R n r) Yl m θ, φ) is an eigenstate of H: Hψ nlm = E n ψ nlm, and of course, the angular momentum operators do not see the radial function at all there being no radial derivatives in L 2 or L z ), so L 2 ψ nlm = 2 ll + 1) ψ L z ψ nlm = m ψ nlm, 23.36) and these separated wavefunctions spherical infinite well, Hydrogen, etc.) are eigenstates of all three H, L 2 and L z. Homework Reading: Griffiths, pp Problem 23.1 Using only the commutator for position and momentum: [r j, p k ] = i δ jk and the definition of angular momentum in terms of the Levi-Civita symbol: L i = ɛ ijk r j p k : a. Show that [L i, L j ] = i ɛ kij L k 23.37) b. Show that the components of L commute with L 2 : for i = 1, 2, 3. [L i, L 2 ] = ) 8 of 9

9 23.5. EIGENSTATES OF L Z Lecture 23 Problem 23.2 Griffiths Here you will explore the classical correspondence of torque and angular momentum. 9 of 9