5.73 Lecture 22: 3-D Central Force Problems II, Levi-Civita

5.73 Lecture #22 22 - 1 3D-Central Force Problems II. Levi-Civita: �ijk. Last time: * [x,p] = i! → use to obtain vector commutation rules: generalize from 1-D to 3-D * we have conjugate position and momentum components in Cartesian coordinates Correspondence Principle Recipe Cartesian coordinates and vector analysis Symmetrize (make it Hermitian) classical mechanics in ! → 0 limit Derived key results: ∂f dr ∂f x ∂f ⎡⎣ f ( x ) ,px ⎤⎦ = i! = i! ⇒ ⎡⎣ f (r),p x ⎤⎦ = i! ∂r dx ∂r r ∂x 1/2 ∂f ∂f ⎛ ∂r ∂r ∂r ⎞ 2 2 2 ⎤ ⎡ ⎡⎣ f ( r ) ,q ⋅ p ⎤⎦ = i! r based on + + + y + z and r = x ⎦ ⎣ ∂r ∂r ⎜⎝ ∂x ∂y ∂z ⎟⎠ *p r = r −1 ( q ⋅ p − i!) (came from symmetrization in Cartesian coordinates) *p 2 = p 2r + r −2 L2 operator algebra gave simple separation of variables *L = q × p not necessary (or possible) to symmetrize separated p ! from p ⊥ p 2r ⎡ L2 ⎤ *H = +⎢ + V (r)⎥ 2 2µ ⎣ 2µr ⎦ need to show we can ignore the order of L2 and r2 Vℓ (r) radial effective potential We do not yet know anything about the eigenstates and the eigenvalues and eigenstates of L2 and Li. TODAY [purpose is mostly to practice commutation rule [,], and angular momentum algebras] * Obtain angular Momentum Commutation Rules → Block diagonalize H * εijk Levi-Civita Antisymmetric Tensor useful in derivations, vector commutators, and remembering stuff. Next Lecture: Begin derivation of all angular momentum matrix elements starting from the Commutation Rule definitions of angular momentum components. 1 5.73 Lecture #22 22 - 2 GOALS 1. ⎡⎣Li , f (r)⎤⎦ = 0 any scalar function of scalar r. difficult - need to use ε ijk ! 2. ⎡⎣Li ,pr ⎤⎦ = 0 3. ⎡⎣Li ,p2r ⎤⎦ = 0 4. ⎡⎣Li ,L2 ⎤⎦ = 0 (but ⎡⎣L ,L ⎤⎦ ≠ 0!) 5. C.S.C.O. H,L2 ,Li → enable block diagonalization of H i 2 j L2 and Li block-diagonalize H according to different eigenvalues of L2 and Li. Items 1-4 are chosen to show that all terms in H commute with L2 and Li Li: choose Lz for example 1. [ Lz ,f(r)] = [xpy − ypx ,f(r)] = x[ py ,f ] + [ x, f ]py − y[px , f ] − [y, f ]px [ x, f ] = 0, ! [ y, f ] = 0 because [q,f(r)] = 0 î + 0 ĵ + 0k̂ ∂f ∂r ∂f x recall ⎡⎣ f (r ),p x ⎤⎦ = i ! = i! ∂r ∂x ∂r r ∂f ⎡ y x⎤ ⎡⎣L z , f (r )⎤⎦ = − i ! ⎢ x − y ⎥ = 0 r⎦ ∂r ⎣ r 2. –1 ⎡⎣L z ,pr ⎤⎦ = ⎡⎣L z ,r −1 ( q ⋅p − i! ) ⎤⎦ = ⎡⎣L z ,r -1q ⋅p ⎤⎦ (already know that [Lz,r i!] = 0) 0 1 is f (r) and we just showed this = ⎡⎣L z ,r −1 ⎤⎦ q ⋅p + r −1 ⎡⎣L z ,q ⋅p⎤⎦ r commutation rule = 0 " " ⎡⎣L z ,q ⋅p⎤⎦ = q ⋅ ⎡⎣L z ,p⎤⎦ + ⎡⎣L z ,q⎤⎦ ⋅p two vector commutators on RHS " Note that vector q is a not scalar f(r)! need to define special notational trick to evaluate these DIFFICULT COMMUTATORS Either A' = A or HAA' = 0. If A = A', it is still possible to find linear combination of different eigenstates of A (with same-A eigenvalues of A) that diagonalizes the associated block of H. 2 5.73 Lecture #22 22 - 3 ⎡⎣ H, A⎤⎦ = 0 0 = A ⎡⎣ H, A⎤⎦ A′ = A ⎡⎣ −AH + HA⎤⎦ A′ = ( − A + A′ ) H AA′ so either A = A′ or H AA′ = 0 Now for something very special and useful ε ijk Levi-Civita Symbol cyclic order ε xyz = ε yzx = ε zxy = +1 adjacent interchange ε yxz = ε zyx = ε xzy = −1 2 repeated indices ε xxy = etc. = 0 [ ] (anti-cyclic order ) I claim L i ,p j = i! ∑ ε ijkpk . This will become the definition of a “vector operator” k with respect to L! [[vector = 1st rank tensor] Nonlecture: Verify claim for 1 of 3 × 3 = 9 possible cases let i = x, j = y ⎡⎣L x ,p y ⎤⎦ = ⎡⎣ yp z − zp y ,p y ⎤⎦ = y ⎡⎣p z ,p y ⎤⎦ + ⎡⎣ y,p y ⎤⎦ p z − z ⎡⎣p y ,p y ⎤⎦ − ⎡⎣ z,p y ⎤⎦ p y = 0 + i!p z − 0 − 0 Now check this using εijk [L ,p ] = i!∑ ε x y xyk [ 0 0 pk = i! ε xyx px + ε xyy py + ε xyzpz k = i!pz . ] OK All other 8 cases go similarly. Feel the power of �ijk! Other important Commutation Rules [ Li ,p j ] = i! ∑k ε ijkpk ⎫⎪ ⎬ L ,q = i! ε q ∑ [ i j ] k ijk k ⎪⎭ [ Li ,L j ] = i!∑ ε ijkLk general definition of a “vector” operator ! ! q and p are examples of vector operators. Classify as vectors with respect to L! general definition of an “angular momentum.” Works even for spin where a q × p definition cannot exist. This is the MOST IMPORTANT STEP All angular momentum matrix elements will be derived next lecture from these commutation rules. k FOR THE READER: VERIFY ONE COMPONENT OF EACH OF THE THREE ABOVE COMMUTATORS 3 5.73 Lecture #22 22 - 4 [ Li ,L j ] = i! ∑k ε ijk Lk is identical to L × L = i!L (expect 0! because vector cross product ! ! A × B =|A||B|sin θ AB note reversal of x and ⎛ î ĵ k̂ ⎞ ⎛ ⎞ ⎛z terms ⎛ ⎞ ⎞ ⎜ ⎟ L × L = L x L y Lz = î ⎜ Ly Lz − L z Ly ⎟ + ĵ⎜ L z Lx − Lx Lz ⎟ + k̂ ⎜ Lx Ly − L yLx ⎟ ⎟ ⎜ ⎟ ⎟ ⎜ ⎟ ⎜ ⎜ ⎝ ⎠ ⎠ L ⎠ ⎝ ⎝ L L ⎝ x y z⎠ [ ] = i! îLx + ĵLy + k̂Lz = i!L This vector cross product definition of L is more general than q × p because there is no way to define spin in q × p form but S × S = i!S is quite meaningful. Can one generalize that, if L × L = i!L (instead of 0), and the [Li, Lj] and [Li, pj] commutation rules have similar forms, that L × p = i!p? NO! Check for yourself! 2. Continued. use p r = r −1 ( q ⋅p − i! ) already know that this commutes with Li ! ! ! −1 ! −1 L ,p = r ,p + r L , q ⋅p q⋅ L [ [ ] [ ] z r] z z evaluate the first term ! [L i ,p ] = i !∑ iˆε ixk + jˆε iyk + kˆε izk p k sum of 3 terms k ! q⋅ [L i ,p ] = i !∑ k ( ( xε vector commutators ) ixk ) + yε iyk + zε izk p k call this an index-j sum = i !∑ ε ijk q j p k (1) j ,k 4 only one of these terms is nonzero (but use simpler form) ) 5.73 Lecture #22 22 - 5 ! ! and evaluate the second term [L i , q ] ⋅p ! L , q [ i ] = i!∑ ⎡⎣iˆε ixk + ĵε iyk + k̂ε izk ⎤⎦ q k k ! [Li, q ] ⋅ p = i!∑ ⎡⎣ε ixk q k p x + ε iyk q k p y + ε izk q k p z ⎤⎦ (a 2nd –index sum) k sum is over j and k, so can permute the k↔j labels = i!∑ ε ijk q k p j = i!∑ ε ikjq jp k j,k k, j = −i!∑ ε ijk q j p k k, j putting Eqs. (1) and (2) together (2) switch order of j and k ! ! ! ! q⋅ ⎣⎡Li ,p ⎤⎦ + ⎡⎣Li ,q ⎤⎦ ⋅p = i"∑ ⎡⎣ ε ijk q j pk − ε ijk q j pk ⎤⎦ = 0! The 2 terms from the [L, p·q] are combined here. j ,k Elegance and power of εijk notation! We have shown, for pr = r–1(q· p–i!), that: * [Li ,pr] = 0 for all i 2 * easy now to show [Li,pr ] = 0 ( Finally ⎡⎣L i ,L2 ⎤⎦ = ∑ ⎡⎣L i ,L2j ⎤⎦ = ∑ L j ⎡⎣L i ,L j ⎤⎦ + ⎡⎣L i ,L j ⎤⎦ L j ⎤ ⎦ j j ( )( sum is over j and k, so can permute the j & k indices ) − ( i! Σ ε L ) L = ∑ ⎡⎢L j i! Σ ε ijk L k + i! Σ ε ijk L k L j ⎥⎤ k k ⎣ ⎦ j same trick: permute j ↔ k indices in second term Thus ε ijk = −ε ikj k ijk j k ⎡⎣L i ,L2 ⎤⎦ = 0 But be careful: Li and Lj2 do not commute even though Li and L2 do commute ( ( ) ) ⎡L i ,L2j ⎤ = L j ⎡L i ,L j ⎤ + ⎡L i ,L j ⎤ L j = i! L j Σ ε ijk L k + Σ ε ijk L k L j ≠ 0 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ k k because this is a sum only over k, can’t combine and cancel terms. See detail on next page. 5 5.73 Lecture #22 22 - 6 for i=x, j=y 2 ⎤ ⎡ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎣L x ,L y ⎦ = L y ⎣ L x ,L y ⎦ + ⎣ L x ,L y ⎦ L y = i! ⎣L y L z + L z L y ⎦ ≠ 0! so we have shown [ L2 ,Li ] = 0 [ L2 ,f(r)] = 0 [ Li ,f(r)] = 0 [ L2 ,pr ] = 0 [ Li ,pr ] = 0 ∴ L2, Li, H all commute — Complete Set of Mutually Commuting Operators eigenfunction of L2 with eigenvalue ! 2 L ( L +1) So what does this tell us about L H L′ = ? also M L H M ' L Lz LM L = !M L LM L BLOCK DIAGONALIZATION OF H! Basis functions Both H LL′ = 0 and H M ′ L ,ML =0 ψ = χ (r ) L2 , Lz = nLM L radial special eigenfunctions of L2 which radial eigenfunction? Next time I will show, starting from ⎡⎣ Li , L j ⎤⎦ = i! Σ ε ijk L k k eigenfunctions of Lz angular universal , that * L2 nLM L = ! 2 L ( L +1) nLM L * L z nLM L = !M L nLM L L = 0,1,… M L = −L, −L +1,…+ L also derive all L x and L y matrix elements in nLM L basis set. 6 5.73 Lecture #22 22 - 7 Translation and Rotation Operators We are interested in QM operators that cause translation or rotation of an initially localized state: |x0,y0,z0〉 or |α0, β0, γ0〉 (where α, β, γ are Euler angles that relate the body-fixed axis system to the laboratory-fixed axis system. Translations are related to p̂x ,p̂ y ,p̂z operators and rotations are related to L̂x ,L̂ y ,L̂z operators. How do we demonstrate these relationships? Begin by asking what does e |x0,y0, z0〉. −ip̂x δ ! do to an initially localized state An initially localized state is an eigenfunction of x̂ , ŷ,ẑ operators xˆ x 0 , y0 , z0 = x 0 x 0 , y0 ,z0 similarly for yˆ and zˆ −ip̂x δ ! do to |x0,y0,z0〉? We want to know to what eigenvalue(s) of x̂ −ip̂ δ ! does e x 0 , y0 ,z0 belong? We ask for x̂ ⎡⎣e x x 0 , y0 ,z0 ⎤⎦ and we use the df commutation rule ⎡⎣ x̂ , f p̂x ⎤⎦ = i! dpx What does e −ip̂x δ ! ( ) x̂ ⎣⎡e −ip̂x δ ! ( ) −ip̂ δ ! −ip̂ δ ! x 0 , y0 ,z0 ⎤⎦ = e x x̂ + ⎡⎣ x̂ ,e x ⎤⎦ x 0 , y0 ,z0 ( ) −ip̂ δ/! f p̂x = e x dfˆ −ip̂ δ/! = ( −iδ / ! )e x dpx ( ) ⎡ x̂ , f p̂ ⎤ = (i!)( −iδ / ! )e −ip̂x δ/! = δe −ip̂x δ/! x ⎦ ⎣ Put it all together: x̂ ⎡⎣e −ip̂x δ/! x̂ ⎡⎣e −ip̂x δ/! −ip̂ δ/! x 0 , y0 ,z0 ⎤⎦ = e x [x 0 + δ] x 0 , y0 ,z0 −ip̂ δ/! x 0 , y0 ,z0 ⎤⎦ = (x 0 + δ) ⎣⎡e x x 0 , y0 ,z0 ⎦⎤ 7 5.73 Lecture #22 This means that e −ip̂x δ/! 22 - 8 x 0 , y0 ,z0 belongs to the (x0 + δ) eigenvalue of x̂ ! e −ip̂x δ/! x 0 , y0 ,z0 = x0 + δ, y0 ,z0 So we know how to build an operator that causes translations of a localized state in the x, y, or z directions: T̂x ,T̂y ,T̂z . But we know that !"̂! , "̂ "% = 0 for all components of linear momentum. This means that for all linear translations, !()! , ()" % = 0. The sequence of the linear translations does not matter! What about rotations of the initially localized state| ,#, -#, .#⟩? ' What does 0 $!%&!/ℏ do to | ,#, -#, .#⟩? Consider α̂ ⎡⎣e −iφL̂z /! α 0 ,β0 ,γ 0 ⎤⎦ . Following an argument similar to that for the translational operators α̂e α̂e −iφL̂z /! −iφL̂z /! ( ) α 0 ,β0 ,γ 0 = α 0 + φ e −iφL̂z /! α 0 ,β0 ,γ 0 α 0 ,β0 ,γ 0 belongs to the α0 + φ eigenvalue of α̂ . Now show something beautiful: that infinitesimal rotations about different axes do not commute! e −iφL̂z /! −iθL̂ y /! e ( α 0 ,β0 ,γ 0 = e −iθL̂ y /! −iφL̂ /! z e 8 + ⎡⎣e −iφL̂z /! ,e −iθL̂ y /! ) ⎤⎦ α ,β ,γ 0 0 0 5.73 Lecture #22 22 - 9 Expand the exponentials for infinitesimal θ,φ: 1st two terms in the power series expansion of e–iα: 1 – iα. ⎡⎣e −iθL̂ y /! ,e −iφL̂z /! ⎤⎦ = [1,1] − ⎡1,−iφL̂ / ! ⎤ + ⎡ −iθL̂ / !,1⎤ − ⎛⎜ − i ⎞⎟ ⎡φL̂ ,θL̂ ⎤ ⎣ ⎦ ⎣ z y ⎦ ⎝ !⎠ ⎣ z y ⎦ ⎛ i⎞ = 0+ 0+ 0+ ⎜ − ⎟ ( φθ )i!L̂x ⎝ !⎠ Reversing the order of the rotations about the y and z axes results in a non-zero rotation by �� about the x-axis!& 9