(2) Orbital angular momentum

Transcription

1 (2) Orbital angular momentum Consider SS = 0 and LL = rr pp, where pp is the canonical momentum Note: SS and LL are generators for different parts of the wave function. Note: from AA BB ii = εε iiiiii AA jj BB kk we get LL xx = yypp zz zzpp yy, L y = zp x xp z, L z = xp y yp x (A) Prove that LL ii, LL jj = iiεε iiiiii LL kk For instance, LL xx, LL yy = iill zz by applying the commutation relations xx ii, pp jj = iiδδ iiii This vector satisfies the required angular momentum commutation relations (B) In addition, it can be shown directly that this combination is the generator for the rotation group in the real space. Proof: iiiiφφll zz xx, yy, zz = iiiiii xx pp yy + iiiiiiyy pp xx xx, yy, zz = iiiiii xxpp yy + iiiiiiyypp xx xx, yy, zz = xx yyyyyy, yy + xxxxxx, zz Applying the definition of a translation TT yy xx, yy, zz = ee iiiiiipp xx xx, yy, zz = xx + dddd, yy, zz, etc., where pp is the canonical (not kinematic) momentum From (A) and (B), we can conclude that this vector is the angular momentum xx, yy, zz may be written as rr, θθ, φφ Because these operators are generators of rotations, only θθ, φφ are relevant variables (drop rr) When writing xx, yy, zz = rr, θθ, φφ, we will write down explicitly the θθ, φφ part only

2 (2) The direction kets basis set xx is a good basis set It may be made of xx, yy, zz or rr, θθ, φφ = rr θθ, φφ = rr nn, where n nn = nn ( direction eigenket, standing in for θθ, φφ) [it appeared in σσ nn already] The direction nn kets form a complete ddω nn nn nn = and orthogonal nn ii nn jj = δδ iiii basis set In practice θθ, φφ are measurement variables. Here, will learn how to represent operators in this practical basis set. nn = θθ, φφ The representation of LL in the llll basis set is the same as for SS = (they both transform with the SO(3) rotation group). It is another way of saying that LL is a vector. Here we introduced a new basis set and are interested in the representation LL in this new basis set. Because this is a double infinite-dimensional basis set, we will get functions of 4 variables θθθθ LL xx θθ φφ

3 (2) Representation of LL 2, LL zz, LL ± in the nn basis set The task is made easier because (uncharacteristically) we know what the rotation operators (which are usually the more complicated ones, compared to the generators) do to states of this basis set. Slightly different derivation from the textbook θθ, φφ iiiiiill zz αα = θθ, φφ dddd αα θθ, φφ LL zz αα = ii θθ,φφ αα [minus sign because it acts on a bravector] θθ, φφ LL zz θθ φφ = iiii δδ θθ θθ δδ(φφ φφ ) θθ, φφ iiiiφφ xx LL xx αα = θθ + sin φφφφφφ xx, φφ + cot θθ cos φφ ddφφ xx αα θθ, φφ LL xx αα = ii sin φφ cot θθ cos φφ θθ, φφ αα θθ, φφ LL xx θθ φφ = ii sin φφ cot θθ cos φφ δδ θθ θθ δδ(φφ φφ ) [looks a lot like the representation for momentum pp, which is expected as LL has pp in it] θθ, φφ iiiiφφ yy LL yy /ħ αα = θθ cos φφ δδφφ yy, φφ + cot θθ sin φφ δδφφ yy αα θθ, φφ LL yy αα = ii coooo φφ cot θθ sin φφ θθ, φφ LL ± αα = by applying LL ± = LL xx ± iiii yy θθθθ αα We see again that the representation of LL zz is much simpler than that of LL xx,yy because, just as for the jjjj basis set, our angles are measured from the zz axis Applying LL 2 = LL zz LL LL + + LL + LL can get θθ, φφ LL 2 αα = ħ 2 sin θθ φφ + sin θθ sin θθ θθ, φφ αα

4 (3) Transformation function between nn and llll basis sets Call nn llll θθ, φφ llll YY mm ll (θθ, φφ), a function of 4 variables [analog of xx pp = 3 2 ee iiiiii ħ, which is a 2πππ function of 2, 4 or 6 variables in,2,3d] We will show immediately that this function is the spherical harmonic, justifying the notation Note: YY mm ll nn = nn llll are the eigenstates llll of the LL 2 and LL zz operators, represented in the nn basis set Note: YY mm ll nn is the amplitude to be found along nn if you are in the llll state [from the rules of probability PP = θθ, φφ llll 2 ] llll = ddω nn YY mm ll (θθ, φφ) nn where ddω = dd cos θθ dddd Applying the QM relations between llll and Dirac rules, it can be shown that the properties of the function θθ, φφ llll are compatible with those of the spherical harmonics YY mm ll (θθ, φφ). For instance, mm () YY ll θθ, φφ YYll mm θθ, φφ dd cos θθθθθθ = dd cos θθθθθθ llll θθθθ θθθθ ll mm = llll ll mm = δδ mmmmm δδ lllll (2) LL zz llll = mm llll θθ, φφ LL zz llll = mmyy mm ll iiħ YY ll mm θθ,φφ = mmyy mm ll θθ, φφ YY mm ll (θθ, φφ) ee iiiiii (3) LL 2 llll = ll(ll + ) llll nn LL 2 llll = ll(ll + )YY mm ll the differential equation for YY mm ll (θθ, φφ) in the θθ variable Solution: a mathematical function YY ll mm (θθ, φφ) = and mm 0 ll 2 ll ll! 2ll+ 4ππ ll+mm! ll mm! eeiiiiii sin θθ mm dd ll mm dd cos θθ ll mm sin θθ 2ll with ll an integer only (unlike spin) Examples: YY 0 0 = θθθθ 00 = 4ππ, YY 0 = θθθθ 0 = 3 4ππ cos θθ, YY ± = θθθθ ± = 3 8ππ sin θθee±iiii, etc. Note: define YY ll mm θθ, φφ = mm YY ll mm θθ, φφ or θθθθ ll, mm = mm llll θθθθ Conclusion: can avoid solving complex differential equations for rotation in QM by using the notation θθ, φφ llll and applying the economical Dirac notation and rules (just as we did for instance for the SHO, where we found the spectrum and wavefunctions without applying Hermite polynomials at all)

5 (3) From the representation of LL zz in the llll basis set to the one in the nn basis set In the llll basis set, we have llll LL zz ll mm = mmδδ mmmmm δδ lllll Let s find the representation of LL zz in the θθ φφ basis set with the similarity transformation θθθθ LL zz θθ φφ = llll mmmmm θθθθ llll llll LL zz llllll ll mm θθ φφ = YY mm ll θθ, φφ mmδδ mmmm δδ llll YY mm llll mmmmm ll θθ, φφ = YY mm mm llll ll θθ, φφ mmyy ll (θθ φφ ) We have YY mm ll θθ, φφ = ee iiiiii FF(θθ). Therefore, mmyy mm ll θθ, φφ = YY ll mm (θθ,φφ) iiiiii We also have YY mm mm llll ll θθ, φφ YY ll θθ φφ = llll θθ, φφ llll llll θθ, φφ = θθ, φφ θθ, φφ = δδ(cos θθ cos θθθ)δδ(φφ φφφ) Applying this, and bringing the derivative outside of the sum, we get θθθθ LL zz θθ φφ ii = YY φφ ll mm mm llll θθ, φφ YY ll θθ φφ ii = δδ cos θθ cos θθθ δδ φφ φφ φφ = ii δδ cos θθ cos θθ δδ φφ φφ = ii δδ θθ θθ δδ φφ φφ Therefore, ddω θθθθ LL zz θθ φφ θθ φφ αα = ii θθθθ αα Or θθθθ LL zz αα = ii θθθθ αα [the textbook expression]

6 (3) Representation of DD zz RR in the θθθθ basis set This is not difficult for rotation about the zz axis θθθθ DD zz αα θθ φφ = θθθθ ee iill zzαα θθ φφ = θθθθ θθ, φφ + αα = δδ θθ θθ δδ(φφ φφ αα) Note: algebra is more difficult for generators LL xx,yy and rotation DD xx,yy RR about the xx and yy axes, but relatively straightforward. We have the representations in the llll basis set (use the angular momentum from previous lectures) and this is just a change of basis from llll to θθθθ

7 (3) Example Earlier for ss =, we applied the finite rotation operator to + to find SS 2 xx, +, SS yy, +, etc. The same exercise can be done for LL, but now with a rotation in real space. Question: take LL =. What is the wave function ψψ θθ, φφ and probability distribution P = ψψ θθ, φφ 2 for a state with mm = 0 along an axis zz making an angle ββ with axis zz? Note: this may be a bound electron to an atom or an unbound γγ ray, or an unbound neutron, etc. The result will be very general. Answer If mm = 0 were about the zz axis, we would know that ψψ θθ, φφ YY mm=0 ll= θθ, φφ cos θθ. The probability then is PP(θθ, φφ) cos θθ 2. This function is plotted below. The problem however asks for a state with mm = 0 about a different, zz axis. This is just as for ss = spin case where, given state SS 2 zz = +, we 2 were asking for the state SS xx = +, for instance. 2 The plan is to use Dirac notation as much as possible. We first orient the xx, yy axes so that the ββ rotation can be done about the yy axis. We start with ll =, mm = 0, rotate by an angle ββ with RR yy ββ to ll =, mm = 0, then calculate θθθθ ll =, mm = 0 and take to absolute value square to find PP. zz axis xx zz zz ββ yy

8 It remains to do the algebra. We use the llll basis set, made of eigenkets of LL 2 and LL zz (about the zz axis). ll =, mm = and ll =, mm = 0 RR yy ββ 0 0 = 2 sin ββ cos ββ sin ββ we found before, or ll =, mm = 0 = sin ββ + cos ββ 0 + sin ββ 2 2 Then θθθθ ll =, mm = 0 = sin ββ YY 2 θθθθ + cos ββ YY 0 θθθθ + sin ββ YY 2 θθθθ θθθθ ll =, mm = 0 = 3 4ππ cos θθ cos ββ + sin θθ sin ββ cos φφ 2 with the expression for RR yy (ββ) The plot of PP = θθθθ ll =, mm = 0 2 for ββ = ππ is shown below. 8 Not surprisingly, it is the same shape as the one on previous page, rotated by ββ. We could have predicted this, but not as easily written down the algebraic expression for PP above. The machinery of QM rotations we developed works very well when tested. With a little practice, you will find these once intimidating problems almost straightforward. zz axis zzz axis