Angular momentum in spherical coordinates

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1 Anguar momentum in spherica coordinates Peter Haggstrom December 6, Introduction Anguar momentum is a deep property and in courses on quantum mechanics a ot of time is devoted to commutator reationships and spherica harmonics. However, many basic things are actuay set for proof outside ectures as probems. For instance, one of the standard quantum physics textbooks [1, pp ] deas with the issue this way: Appying the cassica technique of changing variabes, we obtain, from formuas D-1 and D-2, the foowing expressions the cacuations are rather time-consuming but pose no great probems: L x i sin φ θ + cos φ L y i cos φ θ + sin φ L z i 1 which yied: L θ θ + 1 sin 2 θ L + e iφ θ + i cot θ

2 L e iφ θ + i cot θ 4 The starting point for 1 are the Cartesian expressions for the anguar momentum components: L x y i z z y L y z i x x z L z x i y y x 5 The spherica coordinate transformation is as foows: x r cos φ y r sin φ z r cos θ 6 with: r 0 0 θ π 0 φ < 2π 7 2 The derivations The fundamenta formua is this: x i r r + x i θ θ + x i x i 8 where is a pacehoder. In principe there is nothing particuary difficut about performing the reevant cacuations but it is very easy to make sma mistakes. The Cartesian coordinates are reated to the spherica coordinates as foows: 2

3 r x 2 + y 2 + z 2 z cos θ x 2 + y 2 + z 2 9 tan φ y x r x 1 2 x2 + y 2 + z 2 1 x 2 2x cos φ 10 r r y 1 2 x2 + y 2 + z 2 1 y 2 2y r r z z r sin φ 11 cos θ 12 θ x cos 1 z x x 2 + y 2 + z z z x2 + y 2 + z x x 2 +y 2 +z 2 zxr x 2 + y 2 1 r 3 r cos θ r cos φ r 2 r cos θ cos φ r 13 Note that x 2 + y 2 r cos φ 2 + r sin φ 2 r 2 sin 2 θcos 2 φ + sin 2 φ r. Using 13 we can go straight to: θ y zy r 2 r r cos θ r sin φ r 2 r cos θ sin φ r 14 3

4 θ z cos 1 z 1 z 1 2 r r 1 r2 sin 2 θ r 3 r z x 2 + y 2 + z 2 x 2 +y 2 +z 2 1 r z2 r 3 r 2 z 2 r 3 1 r + z 1 2 x2 + y 2 + z z x tan 1 y x x y2 x 2 y y 1 x 2 x 2 + y 2 r sin φ r 2 sin 2 θ sin φ r y tan 1 y y x y2 x x 2 x x 2 + y 2 r cos φ r 2 sin 2 θ cos φ r z

5 Thus the Cartesian operators have the foowing form using 5: x cos φ r + 1 r cos φ cos θ θ 1 sin φ r y sin φ r + 1 r sin φ cos θ θ + 1 cos φ r z cos θ r 1 r θ 19 It is now ony a matter of abouriousy making the reevant substitutions in 2. L x y i z z y [ r sin φ cos θ i r 1 r θ sin 2 θ sin φ cos 2 θ sin φ i θ i i sin φ θ cos φ sin φ θ + cos φ ] [ r cos θ sin φ r + 1 r sin φ cos θ θ + 1 r cos θ cos φ 20 cos φ ] L y z i x x z [ r cos θ cos φ i r + 1 r cos φ cos θ θ 1 sin φ r cos φ cos 2 θ + cos φ sin 2 θ i θ sin φ cos φ i θ sin φ i cos φ θ + sin φ ] [ r cos φ cos θ r 1 r ] θ 21 5

6 L z x i y y x [ r cos φ sin φ i r + 1 r sin φ cos θ θ + 1 cos φ r [ r sin φ cos φ r + 1 r cos φ cos θ θ 1 r cos 2 φ + sin 2 φ i i sin φ ] ] 22 The foowing symbos are used in what foows to cut down keystrokes: r r θ θ φ 23 Now we have that L is the orbita momentum of a spiness partice [see [1], p. 660] and the operator L 2 is defined to be: L 2 L 2 x + L 2 y + L 2 z 24 Now a we have to do is make the reevant substitutions in 24. This is straightforward, but error prone, so each component wi be done separatey. 6

7 L 2 x 2 sin φ θ + cos φ φ sin φ θ + cos φ φ 2 2 sin 2 φ 2 θ + sin φ θ cot θ cos φ φ + cot θ cos φ [ ] sin 2 φ θ 2 + sin φ cos φ φ csc 2 θ + cot θ cos φ θ φ + ] cot θ cos φ [ θ cos φ + sin φ φ θ + cot 2 θ cos 2 φ 2 sin φ θ + cot 2 θ cos 2 φ 2 2 sin 2 φ θ 2 sin φ cos φ csc 2 θ }{{} φ + sin φ cot θ cos φ θ φ + cot θ cos 2 φ }{{}}{{}}{{} θ cot θ cos φ sin φ φ θ + cot 2 θ cos 2 φ 2 }{{}}{{} L 2 y 2 cos φ θ + cot θ sin φ φ cos φ θ + cot θ sin φ φ 2 cos 2 φ θ 2 cos φ cot θ sin φ φ cot θ sin φ cos φ θ + cot 2 θ sin 2 φ 2 θ 2 [ ] [ ] cos 2 φ θ 2 cos φ sin φ φ csc 2 θ + cot θ sin φ θ φ cot θ sin φ θ sin φ + cos φ φ θ + cot 2 θ sin 2 φ 2 φ 2 cos 2 φ θ 2 + cos φ sin φ csc 2 θ }{{} φ cos φ cot θ sin φ θ φ + cot θ sin 2 φ θ cot θ sin φ cos φ }{{}}{{}}{{} φ θ }{{} cot 2 θ cos 2 φ 2 φ }{{} 6 26 L 2 z i i 2 2 }{{} 6 27 Now pairing up the terms 1-6 we have: 7

8 L 2 L 2 x + L 2 y + L 2 z 2 2 θ + cot θ θ + cot 2 θ φ 2 θ θ + cos2 θ + sin 2 θ sin 2 θ 2 θ θ + 1 sin 2 θ 2 φ 2 φ 28 Thus the textbook answer is indeed obtained. Equations 3 and 4 are easiy derived once it is known that [ [1], p.647]: and L + L x + il y 29 Thus: L L x il y 30 L + L x + il y i sin φ θ + cot θ cos φ φ cos φ θ + cot θ sin φ φ cos φ θ + i cot θ cos φ φ + i sin φ θ cot θ sin φ φ cos φ + i sin φ θ + i cot θ φ e iφ θ + i cot θ φ 31 Simiary: L L x + il y i sin φ θ + cot θ cos φ φ + cos φ θ + cot θ sin φ φ cos φ θ + i cot θ cos φ φ + i sin φ θ + cot θ sin φ φ cos φ i sin φ θ + i cot θ φ e iφ θ + i cot θ φ 32 8

9 2.1 Soving the partia differentia equations There is a substantia preiminary overhead in estabishing that the eigenvaues of L 2 are nd those of L z are m see [1], pages Using 2 we have: 2 θ θ + 1 sin 2 θ 2 φ ψr, θ, φ ψr, θ, φ θ θ + 1 sin 2 θ 2 φ ψr, θ, φ + 1 ψr, θ, φ 33 Using 22 we have: i ψr, θ, φ m ψr, θ, φ i ψr, θ, φ m ψr, θ, φ 34 Because r does not appear as a differentia operator in either 33 or 34 we assume an eigenfunction which depends ony on the anguar variabes θ and φ. Note that if we assumed a soution of the form ψr, θ, φ Y m θ, φ fr see [2], p.314 the term fr simpy cances because of the ack of derivatives in r or viewed as a constant of integration. However, once 33 and 34 have been soved for Y m θ, φ the eigenfunctions wi be of the form ψ,m r, θ, φ Y m θ, φ fr. Denoting the common eigenfunction of L 2 and L z by Y m θ, φ we have that: θ θ + 1 sin 2 θ 2 φ Integration of 36 eads straightaway to: Y m θ, φ Y m θ, φ 35 i Y m θ, φ m Y m θ, φ 36 Y m θ, φ F m θ e imφ 37 Noting here that F m θ is ike a constant given the ack of θ dependence in 36. A wave function must be continuous throughout space if the differentia deveopment of the operators is to make any sense. This is the case because differentiabiity impies continuity, thus if the wave function were not continuous it coud not be differentiabe and hence none of the above deveopment woud make any sense. So et s take some boundary vaues and expoit the continuity: 9

10 Y m θ, φ 0 Y m θ, φ 2π 38 thus we have that: Y m θ, φ 0 F m θ Y m θ, φ 2π F m θ e 2imφ 39 Thus we must have: e 2imφ 1 40 Because m is integra or haf-integra see [1],pages , 40 shows that orbita anguar momentum must be integra and because m and must be either both integra or haf-integra, it foows that must be integra too. On the basis of genera theory see [1], pages the Y m θ, φ must satisfy: Using 31 and 37 we have: L + Y θ, φ 0 41 L + Y θ, φ 0 e iφ θ + i cot θ φ F θ eiφ 0 e iφ e iφ d dθ F θ + i cot θ F θ ieiφ 0 e i+1φ d dθ cot θ F θ 0 d dθ cot θ F θ 0 42 To sove 42 note that: cot θ dθ d 43 Thus we have: 10

11 df θ cot θ F θ dθ F θ θdsin df θ F θ n F d θ n + c F θ c and so Y θ, φ c e iφ 44 This is just touching the surface of the detaied treatment of eigenfunctions represented as spherica harmonics. Courant and Hibert give proofs, for instance, of how one can expand a function in terms of spherica harmonics see [2], page 513. Reference [1] covers the ground we with many detaied cacuations but the authors often eave out specific justifications eg for expansions in terms of spherica harmonics. 3 References 1. Caude Cohen-Tannoudji, Bernard Liu, Franck Laoë, Quantum Mechanics, Voume 1, John Wiey and Sons, R Courant and D Hibert, Methods of Mathematica Physics, Voume 1, John Wiey and Sons, History Created 06/12/