Runge-Lenz method for the energy level of hydrogen atom Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: January 06, 2015)
Runge-Lentz (or Laplace-Runge-Lentz) vector In classical mechanics, the Runge–Lenz vector (or simply the RL vector) is a vector used chiefly to describe the shape and orientation of the orbit of one astronomical body around another, such as a planet revolving around a star. For two bodies interacting by Newtonian gravity, the LRL vector is a constant of motion, meaning that it is the same no matter where it is calculated on the orbit; equivalently, the LRL vector is said to be conserved. More generally, the RL vector is conserved in all problems in which two bodies interact by a central force that varies as the inverse square of the distance between them; such problems are called Kepler problems. The hydrogen atom is a Kepler problem, since it comprises two charged particles interacting by Coulomb's law of electrostatics, another inverse square central force. The RL vector was essential in the first quantum mechanical derivation of the spectrum of the hydrogen atom, before the development of the Schrödinger equation. However, this approach is rarely used today. http://en.wikipedia.org/wiki/Laplace%E2%80%93Runge%E2%80%93Lenz_vector
Wolfgang Pauli in 1926 used the matrix mechanics of Heisenberg to give the first derivation of the energy levels of hydrogen and their degeneracies. Pauli's derivation is based on the Runge- Lenz vector multiplied by the particle mass. [W. Pauli, Z. Physik 36, 336 (1926).]
1. Kepler's law of planetary motion The Kepler's laws (I, II, and III) describe the motion of planets around the Sun, (I) The orbit of a planet is an ellipse with the Sun at one of the two foci. (II) A line segment joining a planet and the Sun sweeps out equal areas during equal interval of time. (III) The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.
The Sun is at the one focus of the ellipse (the planet orbit). The ellipse orbit is described by
x2 y 2 1. a2 b2
where a is the semi-major axis, b is the semi-minor axis, and e is the eccentricity (0 1 K0 aba p0 ae ae aaF1 ae,0 O F2 ae,0 r1 r2 (i) The eccentricity e From the definition of ellipsoid, we have F1K0 F2K0 2a When F1K0 F2K0 , we have F1K0 F2K0 a. We apply the Pythagorean theorem to the triangle F1OK0 , a2 b2 a2e2 , Then we have the expression for the eccentricity b2 1 a2 Note that 2 b a 1 e2 . (ii) The perihelion and aphelion The focus is at (ae,0) and (-ae,0). For simplicity, we assume that Sun is located at focus (- ae,0). The perihelion (r1) the point nearest the Sun r1 a(1 e) , The aphelion ( r2) the point farthest the Sun r2 a(1 e) (iii) Area of the ellipsoi The area of the ellipse orbit is given by A ab a2 1 e2 . (iv) The mathematical formula the ellipsoid: Here we show that mathematically, an ellipse can be represented by the formula p r 0 , 1 ecos where p0 is the semi-latus rectum, and e is the essentricity of the ellipse, r is the distance from the Sun to the planet, and is the angle to the planet's current position K0 from its closest point P (perihelion), as seen from the Sun at F1. We use p0 instead of p since p is typically a linear momentum in physics. We also use instead of , for convenience. 3 K0 rr1 P a,0 O F1 ae,0 F2 ae,0 In the triangle F1F2K0 (see the above figure), we have r r1 2a , (1) from the definition of the ellipse. Using the cosine law, we have 2 2 2 2 2 2 2 r1 r 4a e 4aer cos( ) r 4a e 4aer cos . (2) where is the angle between the vector F1P and F1K0 . From Eqs.(1) and (2), we get (2a r)2 r 2 4a2e2 4aer cos or 4a2 4ar r 2 r 2 4a2e2 4aer cos or 2 r(1 ecos ) a(1 e ) p0 4 or p r 0 , 1 ecos with 2 p0 a(1 e ) , where (r, ) are heliocentric polar coordinates for the planet, p0 is the semi latus rectum, and e is the eccentricity, which is less than one. For = 0 the planet is at the perihelion at minimum distance: p a(1 e2 ) r 0 a(1 e) . (perihelion) 1 1 e 1 e For , 2 r p0 . (semi latus rectum) For , the planet is at the aphelion at maximum distance: p r 0 a(1 e) . (aphelion) 2 1 e Note that 1 1 1 e 1 e 2 . r1 r2 p0 p0 p0 The semi-major axis a is the arithmetic mean between r1 and r2, r r a 1 2 . 2 The semi-minor axis b is the geometric mean between r1 and r2, 5 p0 b r1r2 1 e2 ((Note)) The meaning of semi latus rectum The chord through a focus parallel to the conic section directrix of a conic section is called the latus rectum, and half this length is called the semilatus rectum (Coxeter 1969). "Semilatus rectum" is a compound of the Latin semi-, meaning half, latus, meaning 'side,' and rectum, meaning 'straight.' 2. Kepler problem in classical mechanics (hydrogen atom) The classical Hamiltonian for the Kepler problem is 1 H p2 , 2m r where m is a reduced mass and is a positive quantity. For the case of the hydrogen-like atom, we can identify Ze 2 . This system is invariant under the rotation. So the angular momentum is conserved. The classical orbit of the particle is elliptical. The Runge-Lentz vector is defined as 1 M r p L , r m or A mM mer p L , where L (= r p ) is the angular momentum and p is the linear momentum. 6 7 Fig. A (=mM) (denoted by the arrows with purple) on the ellipse orbits. The vector A points in the direction of the perihelion. The magnitude is constant. The angular momentum L is always perpendicular to the orbit. The perihelion, point of the orbit the nearest to the focal point F). The aphelion, the point of orbit from the focal point. It is obvious from the definition that L M 0 , (M lies in the plane of motion). p M p r , r and 8 1 r M r ( r p L) r m 1 r r ( p L) m . 1 r L2 m 1 r l 2 m or r A r mM mr l 2 , since r ( p L) L (r p) L2 , A mM . Since the angular momentum (along the z axis) is conserved, we can calculate L l r1 p1 , where p1 is the linear momentum at the perihelion. M is a conserved quantity since d d 1 1 d M r ( p L) dt dt r m dt p r(e p) 1 [ r ] [F L p (r F)] m r r 2 m p r(e p) [ r ] [e (r p) p (r e )] m r r 2 mr 2 r r p r(e p) [ r ] [(e p)r rp ( p e )r ( p r)e ] m r r 2 mr 2 r r r p r(e p) p ( p e )r [ r ] [ r ] m r r 2 m r r 2 0 where d 1 1 p r(e p) r [ r ] , dt r m r r 2 9 d p F e dt r 2 r and d τ L r F . (torque). dt ((Mathematica)) Proof of dM/dt = 0. Clear "Global`" ;r t_ : x t , y t , z t ; R t_ : r t .r t ; p t_ : m D r t , t ;L t_ : Cross r t ,pt ; Ft_ : r t ; R t 3 eq1 r t 1 D ,t Cross F t ,L t R t m 1 Cross p t , Cross r t ,F t m FullSimplify 0, 0, 0 ______We note that p L p (r p) p2r ( p r) p Then the vector M (A mM ) always points in the direction of the perihelion from the focal point. At the perihelion, p r 0 We have 10 A mM 2 mer p r1er l 2 (m )er r1 l 2 (m )(ex ) r1 A0 (ex ) where er ex and l 2 A0 m . r1 In general case, we have the orbital equation as 2 r A A0r cos mr l , where is the angle between r and the perihelion direction. From this equation we get 2 r(A0 cos m) l , or A l 2 r( 0 cos 1) m m l 2 p r m 0 , A 1 0 cos 1 ecos m Then we have p0 is given by l 2 p . (semi latus rectum) 0 m and 11 A A0 me . l 2 Since A0 m , we have r1 l 2 m(1 e) r1 3. Derivation of the Runge-Lentz vector The equation of motion of a particle of mass m in the attractive potential is dp F e . dt r 2 r We take the cross product of both sides with the angular momentum L. dp L F L r . dt r3 Since L is constant in time, the left-hand side can be written as dp d L (L p) . dt dt From the explicit form dr L r p mr , dt we get m dr L r r L r (r ) r 3 r 3 r 3 dt m dr dr [(r )r r 2 ] r 3 dt dt d r m dt r Then we have 12 d d r (L p) m 0 , dt dt r or d r (L p m ) 0 . dt r The Runge-Lentz vector is obtained as A mM mer L p mer p L . which is a constant of motion. 4. Energy of the system We note that 2H M 2 L2 2 . m The proof is given by using the Mathematica. ((Mathematica)) Clear "Global`" ;r x, y, z ;p px, py, pz ; L Cross r, p ;R r.r ; 1 M1 r Cross p, L ; R m 1 H1 p.p ; 2 m R 2 eq1 M1.M1 H1 L.L FullSimplify m 2 The energy can be derived as follows. 13 A2 m2 M 2 2Hml 2 m2 2 m2 2e2 , or m 2 H E (1 e2 ) (<0), 2l 2 2a with A me , L2 l 2 , and H E . 4. Construction of the diagram by Mathematica K3 K2 K4 A p L H2 P 2 m K5 K1 K0 P O F H1 Fig. Simplified version of the diagram. is the angle between A and r. The direction of A is from the focal point to the perihelion (P). We consider a point (K0) of the orbit of the ellipsoid which is expressed by 14 x2 x2 1, a2 b2 where the essentricity is given by b2 e 1 . a2 where a>b. The co-ordinate of the focal point F is (ae,0) . We also consider a point (H2) on the circle which is expressed by x2 y2 a2 . We assume that the angle H2OH1 is . The co-ordinates of the points K0 and H2 OH 2 (acos,asin ) , OK0 (acos,bsin) . The slope of the tangential line (K0K4) at the point K0 on the ellipsoid is given by dy bcos b slope= d cot tan , dx asin a d where the angle K4K0K1 is ; b arctan( cot ) . a The unit vector along the vector K0 K2 , K K (acos ae,bsin ) u 0 2 , 2 2 2 K0 K2 (acos ae) (bsin ) The unit vector along the vector K0 K2 , 15 K0K3 u3 (sin,cos) . K0K3 We note that K0K3 p L s3u3 K0K2 K2K3 mu2 Aex , where A em , ex (1,0 ). Then we have s3 m(u2 u3 ex u3 ) . The co-ordinate of the point is given by OK3 OK0 K0K3 (acos,bsin) s3u3 ((Mathematica)) 16 17 18 19 K3 K2 K4 A p L H2 P 2 m K5 K1 K0 O F H1 5. Runge-Lenz vector in quantum mechanics Here we introduce a Runge-Lenz vector which in quantum mechanics is defined by Ze2 1 M r ( p L L p) . r 2m This operator is Hermitian since .( p L) (L p ) Note that r [=(x, y, z)] is the position vector, L is the orbital angular momentum, L r p . M commutes with the Hamiltonian H, [H,M i ] 0 , 20 where H is the Coulomb Hamiltonian 1 Ze2 H p2 . 2m r The commutation relations between the angular momentum L and linear momentum p are given by [Li , p j ] i ijk pk , which leads to the expression L p p L 2ip . Then we have Ze2 1 i M r ( p L) p . r m m We note that L M M L 0. We also note that 2 2 r ( p L) L , ( p L) r L 2ip r , ( p L)2 p2 L2 . We find that 2 M 2 Z 2e4 H (L2 2 ) . m 2 The term arises from the non-commutativity of quantum operators. These equations can be easily proved using the Mathematica (see below in detail). We know that the angular momentum operators satisfy the commutation relations, 21 [.Li , Lj ] iijk Lk We also get [,M i , Lj ] iijk M k and 2i [M ,M ] HL . i j m ijk k The Li’s generate rotations and define a closed algebra. But Li.s and Mi’s do not form a closed algebra since the last relation involves the Hamiltonian. However, we consider the case of specific bound states. In this case, the vector space is truncated only those that are eigenstates of H, with eigenvalue E<0. In this case we replace H with –E, and the algebra is closed. We introduce a vector N such that m N ( )1/ 2 M . 2E In this case we have the closed algebra, [Li , Lj ] iijk Lk , [,Ni , L j ] i ijk Nk and [,Ni , N j ] iijk Lk 2 [M i , L ] 0 . 6. Dynamic Symmetry operation We consider the symmetry operation generated by the operators L and N, which corresponds to the rotation in four spatial dimensions. We introduce (x1, x2 , x3, x4 ) and ( p1, p2 , p3, p4 ) . Note that x4 and p4 are fictitious and cannot be identified with dynamical variables. L1 Lx L23 x2 p3 x3 p2 , 22 L2 Ly L31 x3 p1 x1 p3 , L3 Lz L12 x1 p2 x2 p1 , N1 L14 x1 p4 x4 p1 , N2 L24 x2 p4 x4 p2 , N3 L34 x3 p4 x4 p3 . This algebra is isomorphic (equivalent) to that of rotations in four dimensions [or the O(4) group]. 7. Energy level of the hydrogen-like atom We define two sets of new generators as 1 1 I (L N ) , K (L N ) . i 2 i i i 2 i i and Li Ii Ki . Then we get [,Ii , I j ] i ijk Ik [,Ki , K j ] iijk Kk [Ii , K j ] 0 . Furthermore we have [Ii , H ] [Ki , H ] 0 . 23 So that these operators are also conserved. In this basis the algebra becomes equivalent to that of two decoupled algebras of angular momenta. The eigenvalues of the operators I2 and K2 will have the eigenvalues, 2 2 2 2 I ; i(i 1) , K ; k(k 1) . where i and k are either integer or half integers. We note that 1 C I 2 K 2 (L2 N 2 ) , 2 C' I 2 K 2 L.N 0 . The second relation implies that i(i 1) k(k 1) , or i = k. Correspondingly, the allowed values of C are 2 C: 2 k(k 1) , 2E 2 M 2 N 2 Z 2e4 E(L2 2 ) , m m or mZ 2e4 N 2 (L2 2 ) , 2E or 1 mZ 2e4 2 (L2 N 2 ) 2 2k(k 1) , 2 4E 2 or 24 mZ 2e4 2 2 [4k(k 1) 1] (2k 1)2 , 4E 2 2 mZ 2e4 mZ 2e4 E 2 2 2 2 . (energy level of hydrogen atom). 2 (2k 1) 2 n where n 2k 1. Since L I K with i= k, l = i + k, i + k - 1,…….., |l - k|, or l = 2k, 2k - 1,…….., 0. or l = n - 1, n – 2,…….., 0. ______((Mathematica)) 25 Clear "Global`" ; ux 1, 0, 0 ;uy 0, 1, 0 ;uz 0, 0, 1 ;r x, y, z ; R x2 y2 z2 ; L: — Cross r, Grad , x, y, z & Simplify; P: — Grad , x, y, z &; Lx : ux.L &; Ly : uy.L &; Lz : uz.L &; Px : ux.P &; Py : uy.P &; Pz : uz.P &; PSQ : Px Px Py Py Pz Pz &; LSQ : Lx Lx Ly Ly Lz Lz &; 1 — Z1 e12 Mx : Py Lz Pz Ly Px x &; m m R 1 — Z1 e12 My : PzLx PxLz Py y &; m m R 1 — Z1 e12 Mz : Px Ly Py Lx Pz z &; m m R M1 ux Mx uy My uz Mz &; Proof Mx Lx + My Ly + Mz Lz = 0 Mx Lx x, y, z My Ly x, y, z Mz Lz x, y, z FullSimplify 0 Proof Lx Mx + Ly My + Lz Mz = 0 Lx Mx x, y, z Ly My x, y, z Lz Mz x, y, z FullSimplify 0 Proof 2 [Mi ,L ]∫0 26 Mx LSQ x, y, z LSQ Mx x, y, z Simplify 1 e12 mxZ1 x, y, z e12 mxzZ1 0,0,1 x, y, z 2 —2 m 2 2 2 2 2 2 x y z x y z e12 mxyZ1 0,1,0 x, y, z x —2 0,0,2 x, y, z 2 2 2 x y z e12 my2 Z1 1,0,0 x, y, z x —2 0,2,0 x, y, z 2 2 2 x y z e12 mz2 Z1 1,0,0 x, y, z —2 1,0,0 x, y, z 2 2 2 x y z 3z—2 1,0,1 x, y, z x2—2 1,0,2 x, y, z y2 —2 1,0,2 x, y, z 3y—2 1,1,0 x, y, z 2yz—2 1,1,1 x, y, z x2 —2 1,2,0 x, y, z z2 —2 1,2,0 x, y, z 2x—2 2,0,0 x, y, z 2xz—2 2,0,1 x, y, z 2xy—2 2,1,0 x, y, z y2 —2 3,0,0 x, y, z z2 —2 3,0,0 x, y, z Proof x P μ L + y P μ L +z P μ L = L2 x y z 27 x Py Lz x, y, z Pz Ly x, y, z y Pz Lx x, y, z Px Lz x, y, z z Px Ly x, y, z Py Lx x, y, z Lx Lx x, y, z Ly Ly x, y, z Lz Lz x,y, z FullSimplify 0 Proof 2 P μ L x x + P μ L y y + P μ L z z = L + 2 i Ñ (Px x+Py y +Pz z) Py Lz x x, y, z Pz Ly x x, y, z Pz Lx y x, y, z Px Lz y x, y, z Px Ly z x, y, z Py Lx z x, y, z Lx Lx x, y, z Ly Ly x, y, z Lz Lz x, y, z 2 — Px x x, y, z 2 — Py y x, y, z 2 — Pz z x, y, z FullSimplify 0 Proof Px P μ L x + Py P μ L y + Pz P μ L z =0 P. (P x L)=0 28 Px Py Lz x, y, z Px Pz Ly x, y, z Py Pz Lx x, y, z Py Px Lz x, y, z Pz Px Ly x, y, z Pz Py Lx x, y, z FullSimplify 0 Proof 2 2 2 P μ L x Px + P μ L y Py + P μ L z Pz = 2 i Ñ (Px + Py + Pz ) (P x L). P. = 2 i Ñ P2 Py Lz Px x, y, z Pz Ly Px x, y, z Pz Lx Py x, y, z Px Lz Py x, y, z Px Ly Pz x, y, z Py Lx Pz x, y, z 2 — Px Px x,y, z Py Py x, y, z PzPzx, y, z FullSimplify 0 29 Proof P L 2 P2 L2 K1 Py Lz Py Lz x, y, z Py Lz Pz Ly x, y, z Pz Ly Py Lz x, y, z Pz Ly Pz Ly x, y, z Pz Lx Pz Lx x, y, z Pz Lx Px Lz x, y, z Px Lz PzLxx, y, z PxLzPx Lzx, y, z PxLy Px Ly x, y, z Px Ly Py Lx x, y, z Py Lx Px Ly x, y, z Py Lx Py Lx x, y, z FullSimplify; K2 PSQ LSQ x,y, z Simplify; K1 K2 Simplify 0 Definition of Hamiltonian 2 2 H = 1 (Px2 + Py2 +Pz2) - Z e 2 m r 30 1 Z1 e12 H1 : PSQ &; 2 m R Proof [H, Mx]=0, [H, My]=0, [H, My]=0, H1 Mx x, y, z Mx H1 x, y, z FullSimplify 0 H1 My x, y, z My H1 x, y, z FullSimplify 0 Proof Mx2 + My2 + Mz2 - Z2 e4 = 2 H (L2 + Ñ2) m MSQ1 Mx Mx x, y, z My My x, y, z Mz Mz x, y, z Z12 e14 x, y, z Simplify; 2 eq1 H1 LSQ x, y, z —2 H1 x, y, z ; m MSQ1 eq1 FullSimplify 0 31 Proof [Lx, Ly] = i Ñ Lz, [Mx, Ly] = i Ñ Mz, [My, Lz] = i Ñ Mx, Lx Ly x, y, z Ly Lx x, y, z — Lz x, y, z Simplify 0 Mx Ly x, y, z Ly Mx x, y, z — Mz x, y, z Simplify 0 My Lz x, y, z Lz My x, y, z — Mx x, y, z Simplify 0 Proof [Mx, My] =- i 2 Ñ H Lz, m [My, Mz] =- i 2 Ñ H Lx, m Mx My x, y, z My Mx x, y, z 2 — H1 Lz x, y, z Simplify m 0 My Mz x, y, z Mz My x, y, z 2 — H1 Lx x, y, z Simplify m 0 32 ______REFERENCES L.I. Schiff, Quantum Mechanics (McGraw-Hill Book Company, Inc, New York, 1955). E. Merzbacher, Quantum Mechanics, third edition (John Wiley & Sons, New York, 1998). Ramamurti Shankar, Principles of Quantum Mechanics, second edition (Springer, New York, 1994). A. Das, Lectures on Quantum Mechanics, 2nd edition (Hindustan Book Agency, 2012). S. Weinberg, Lectures on Quantum Mechanics (Cambridge University Press, 2013). J.J. Sakurai and J. Napolitano, Modern Quantum Mechanics, second edition (Addison-Wesley, New York, 2011). J.M.Finn, Classical Mechanics Infinity Science Press LLC (Hingham, Ma 2008). E.D. Commins, Quantum Mechanics: An Experimentalist’s Approach (Cambridge, 2014). APPENDIX Kepler’s law For convenience, we use the following diagram for the ellipse orbit. P x,y r O F A.1 The angular momentum In general case 33 2 ar r r 1 d 2 a r 2r (r ) r dt vr r v r Since the gravitational force is directed toward the origin (so called central field), Mm ma G r r 2 r 2 ma 0 where mMG . In other words, 1 d 2 a r 2r (r ) 0 r dt v r or v l mr 2 mr 2 mrv = constant. r The angular momentum is defined as L r p r (,mv) m(rer ) (vrer v e ) mrv ez Lzez or Lz mv r , dL d d (r p) (r mv) r F 0 , dt dt dt 34 since F is a central force (r//F), Lz is a constant of motion. 2 l mrv mr . The velocity v is given by l l v r r . mr 2 mr A.2 Physical meaning What is the physical meaning of the constant angular momentum? We now consider the dA/dt, where dA is the partial area of the ellipse. 1 dA r 2d 2 dA 1 d 1 l r 2 r 2 const dt 2 dt 2 2m since l mr 2 const . The period T is given by 2m 2m 2m T dt dA ab a2 1 e2 l l l 2m since dt dA. l A.3. The effective potential The total energy is a sum of the kinetic energy and the potential energy 35 1 1 E mv2 m(v 2 v 2 ) 2 r 2 r r or 2 2 1 2 l 1 2 l E m(r 2 2 ) mr 2 , (1) 2 m r r 2 2mr r The energy is dependent only on r (actually one dimensional problem). l 2 U . (effective potential) eff r 2mr 2 The effective potential energy Ueff has a local minimum m 2 U min , eff 2l 2 at l 2 p r 0 . min m 2 Since E = constant, we have an equation of motion dE l 2 mrr 2 r 3 r dt r mr l 2 (mr 2 3 )r 0 r mr l 2 mr 2 3 0 . (equivalent 1D problem) r mr ______Plot of the effective potential as a function of r 36 1 0 0.2 0.4 0.6 0.8 1.0 -1 -2 -3 -4 -5 Fig. The effective potential vs r with l = 0.1 – 0.5 A.4 Perihelion and aphelion When r 0 for the perihelion (nearest from the Sun) and the aphelion (farthest from Sun) r1 and r2 are the roots of Eq.(1). l 2 r 2 r 0 . E 2mE There are the relations between r1 and r2. r r 2a 1 2 E E l 2 l 2 r r 1 2 2mE 2m E where the total energy is given by E . (E<0, bound state) 2a and 37 r a(1 e) 1 r2 a(1 e) From this we have r r l 2 l 2 2a l 2 1 e2 1 2 a2 a2 2m E 2ma2 ma or p0 is the semi latus rectum, l 2 p a(1 e2 ) 0 m A.5 Kepler’s Third Law 2m 2m 2 2ma3 / 2 T a2 1 e2 a2 , l l ma m or 4 2m2a3 4 2ma3 T 2 , m or a3 GmM GM sun sun , T 2 4 2m 4 2m 4 2 or 4 2 T 2 a3 GM sun or [T(year)]2 [a(AU )]3 . 38 A.6 Derivation of the Kepler's First Law We start with 2 mr mr 2 r mr 2 l cons tant Here we have ldt mr 2d . Note that r depends only on . d d l d dt dt mr 2 d d d l d l d ( ) ( ) dt dt mr 2 d mr 2 d or l d l dr l 2 ( ) . mr 2 d mr 2 d m2r3 mr 2 1 We define u as u , r 1 dr d 1 du ( ) . r 2 d d r d Then we have l 2 d du l 2 ( ) , m2r 2 d d m2r 3 mr 2 or d 2u m u . d 2 l 2 39 The solution of this equation is given by 1 m u (1 C cos ) , r l 2 1 where A is constant, r1 a(1 e) for = 0, and r2 a(1 e) for = 1 m 1 m 2 (1 C1) , 2 (1 C1) , r1 l r2 l or l 2 C e m e . 1 a(1 e2 ) Then we have p r 0 , 1 ecos with l 2 p a(1 e2 ) . 0 m 40