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Lecture 12 Particle on a sphere (c) So Hirata, Department of Chemistry, University of Illinois at Urbana-Champaign. This material has been developed and made available online by work supported jointly by University of Illinois, the National Science Foundation under Grant CHE-1118616 (CAREER), and the Camille & Henry Dreyfus Foundation, Inc. through the Camille Dreyfus Teacher-Scholar program. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the sponsoring agencies. The particle on a sphere Main points: the particle on a sphere leads to a two-dimensional Schrödinger equation and we must use the separation of variables. One of the resulting onedimensional equations is the particle on a ring. For the other, we seek mathematicians’ help: we introduce associated Legendre polynomials. The product of these and the particle on a ring eigenfunctions are spherical harmonics. The particle on a sphere The particle on a sphere The potential is zero – only kinetic energy: 2 2 ˆ H E 2m In the Cartesian (xyz) coordinates, the ‘del squared’ is 2 2 2 2 2 2 2 x y z What is it in spherical coordinates? The particle on a sphere The answer is: 2 2 2 1 1 1 2 2 2 2 sin 2 r r r r sin sin The derivation is analogous to that for cylindrical coordinates. You are invited to derive! The particle on a sphere The value of r is held fixed. The derivatives with respect to r vanish. 2 2 1 1 2 1 2 2 2 sin 2 r r r r sin sin 2 The Schrödinger equation is 2 1 2 1 2 E sin 2 2 2mr sin sin Two variables θ and φ. The particle on a sphere Two variables – let us try the separation of variables technique. ( , ) ( )( ) Substituting 2 1 2 1 2 E sin 2 2 2mr sin sin For the separation to take place, we must be able to cleanly separate the equation into two parts, each depending on just one variable. The particle on a sphere The differentiation with respect to θ for example acts on Θ alone. Therefore 2 2 2 E sin 2 2 2mr sin sin Dividing the both sides by ΘΦ æ 1 ¶ 2F 1 ¶ ¶Q ö + sin q =E ÷ 2ç 2 2 Qsin q ¶q ¶q ø 2mr è Fsin q ¶f 2 The particle on a sphere Multiplying both by sin2θ 2 1 2 sin 2 sin E sin 2 2 2mr Function of φ Function of θ Function of θ Subtracting the RHS from both sides æ 1 ¶ 2F sin q ¶ ¶Q ö 2 + sin q E sin q =0 ÷ 2ç 2 Q ¶q ¶q ø 2mr è F ¶f 2 Function of φ Function of θ Constant The particle on a sphere Two independent one-dimensional equations! These two parts of the equation must be constant. 2 1 2 E 2 2 2mr 2 sin 2 sin E sin E 2 2mr We have already solved the first equation. 2 2 ml2 2 iml ; e 2mr 2 2 2mr 2 The particle on a sphere 2 1 2 1 2 E sin 2 2 2mr sin sin Separation of variables 1 2 sin 2 sin E sin E E 2 2 2 2mr 2mr 2 2 2D rotation, ml is introduced 2 2 m iml l ; e 2mr 2 2 2mr 2 2 2 Custom-made Spherical harmonics Custom-made solutions Associated Legendre polynomials l and ml are quantum numbers ( , ) ( )( ) The particle on a sphere To summarize: the Schrödinger equation is 2 1 2 1 2 E sin 2 2 2mr sin sin The eigenfunctions are spherical harmonics specified by two quantum numbers l (= 0, 1, 2, …) and ml (= –l, … l), having the form ( , ) N lml Ylml ( , ) N lml lml ( ) ml ( ) Normalization Spherical harmonics Associated Legendre eim φ l The particle on a sphere Some low-rank spherical hamonics are given on the right. Spherical harmonics are orthogonal functions. They as fundamental to spherical coordinates as sin and cos to Cartesian coordinates. Spherical harmonics Spherical harmonics are the standing waves of a sphere surface (e.g., soap bubble, earthquake). Imagine a floating bubble. It vibrates – the amplitudes of the vibration is a linear combination of spherical harmonics. GNU Image from Wikipedia The particle on a sphere The total energy is determined by the quantum number l only: l (l 1) 2 El ; l 0,1,2, 2 2mr Out of this, the energy arising from the φ 2 2 rotation is ml Eml 2mr 2 The latter cannot exceed the former. ml = l, l -1,… , 0,… , -l +1, -l The particle on a sphere Parameter l is called the orbital angular momentum quantum number. Parameter ml is the magnetic quantum number. Energy is independent of ml. Therefore, a rotational state with l is (2l +1)-fold degenerate because there are (2l +1) permitted integers ml can take. The particle on a sphere Let us verify that the associated Legendre polynomial is indeed the solution for l = 1 and ml = 1. 2 2 m 2 sin 2 l sin E sin 2mr 2 2mr 2 2 sin ( sin ) 2 2 sin sin 2 2 2mr sin mr 2 2 2 ( sin cos ) sin 2 2 2mr mr 2 2 2 2 2 2 (sin cos ) sin 2 2 2mr 2mr 2 2 m 2 2 2 l 2 ( sin cos ) 2 2 2mr 2mr E l (l 1) 2 2mr Spherical harmonics This is a breathing mode of a bubble This is a a-candy-inmouth mode of a bubble This is an accordion mode of a bubble Summary The spherical harmonics are the most fundamental functions in a spherical coordinates. We have encountered a differential equation whose solution involves associated Legendre polynomials. The eigenfunctions of the particle on a sphere are spherical harmonics and characterized by two quantum numbers l and ml. The energy is determined by l only and is proportional to l(l + 1).