3D-Central Force Problems I

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Valerie Thompson
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1 5.73 Lectue #1 1-1 Roadmap 1. define adial momentum 3D-Cental Foce Poblems I Read: C-TDL, pages fo next lectue. All -Body, 3-D poblems can be educed to * a -D angula pat that is exactly and univesally soluble * a 1-D adial pat that is system-specific and soluble by 1-D techniques in which you ae now expet Next 3 lectues: Coespondence Pinciple Commutation Rules all matix elements p = -1 ( q p-ih) L = q p. define obital angula momentum also L L = ihl and [ L i,l j ]= ih ε ijk L k k geneal definition of angula momentum and of vecto opeatos 3. sepaate p into adial and angula tems: p = p + L 4. find Complete Set of Commuting Obsevables (CSCO) useful fo blockdiagonalizing H H,L = L,L i 5. = H,L i =0 H,L,L i CSCO L,M L univesal basis set sepaate adial pat of H: 1-D Schöd Eq. H l () = p µ + V() + h ll+1 ( ) µ V l () effective adial potential 6. ALL Matix Elements of Angula Momentum Components Deived fom Commutation Rules. 7. Spheical Tenso Classification of all opeatos. 8. Wigne-Eckat Theoem all angula matix elements of all opeatos. I hate diffeential opeatos. Replace them using exclusively simple Commutation Rule based Opeato Algeba. evised Octobe 1, 10:15 AM

2 5.73 Lectue #1 1 - Lots of deivations based on classical VECTOR ANALYSIS much will be set aside as NONLECTURE q 1 q Cental Foce Poblems: bodies whee inteaction foce is along the vecto 1 q 1 1 q cm CM q1 q = q 1 + q 1 q 1 = q q 1 = îx ( x 1)+ ĵy ( y 1)+ ˆk( z z 1 ) oigin q q 1 = x x 1 [( ) + ( y y ) 1 + ( z z ) ] 1/ 1 also C.M. Coodinate system 1 = q 1 q cm 1 = m M v = q q cm [ = m 1 M] H = H tanslation + H cente of mass LAB BODY fee tanslation of C of M of system of mass M = m 1 + m H motion of fictitious paticle of mass mm 1 µ= m + m 1 in coodinate system with oigin at C of M (CTDL page 713) Ptans tanslation = + constant ( m m ) + V 1 1 HCM.. = P + () µ cm V { fee otation (no θφ, dependence) GOAL IS TO SIMPLIFY P cm because that is only place whee θ,φ degees of feedom appea. fee tanslation of system with espect to lab (not inteesting) motion of paticle of mass µ with espect to oigin at c. of m. evised Octobe 1, 10:15 AM

3 5.73 Lectue # Define Radial Component of P cm Coespondence Pinciple * classical mechanics * Catesian Coodinates * symmetize to avoid failue to satisfy Commutation Rules ** veify that all thee deived opeatos, p, p and L ae Hemitian satisfy [q,p]=ih Pupose of this lectue is to walk you though the standad vecto analysis and Quantum Mechanics Coespondence Pinciple pocedues q îx + ĵy + ˆkz p îp x + ĵp y + ˆkp z 1/ = q q 1/ = q x + y + z find adial (i.e. along q) pat of p poject p onto q q p= qp cosθ q θ p cos( q,p {) = q p q p θ adial component of p is obtained by pojecting p onto q p = p cosθ = p q p q p = q p so fom standad vecto analysis we get p = 1 q p evised Octobe 1, 10:15 AM

4 5.73 Lectue #1 This is a tial fom fo p, but it is necessay, accoding to Coespondence Pinciple, to symmetize it. p = ( q p + p q)+ ( q p + p q) 1 aange tems in all possible odes! NONLECTURE (except fo Eq. (4)) SIMPLIFY ABOVE Definition to p = 1 q p ih ( ) q, p + z,p z p q = q p q, p is a vecto commutato be caeful [ q, p ] = [ x,p x ]+ y,p y =3ih ( is not a vecto) 1-4 p = ( q p [ q, p ])+( q p [ q, p ]) 1 = q p 1 q p q p 1 6ih 1 add and subtact 1 q p = 1 q p 3 ih [ q p, 1 ] (1) () (3) LEMMA: need moe geneal Commutation Rule fo which is a special case 1st simplify: [ f(),q p]= q [ f(), p ]+ [ f(), q ] p 0 [ q p, 1 ] evised Octobe 1, 10:15 AM

5 5.73 Lectue #1 but, fom 1-D, we could have shown h h f( x), p φ f( x) φ ( f( x) φ ) i x i x h = [ i f ( x) φ f φ f φ ]= i h f ( x) φ = f [ f( x), p]= ih x fo 1 - D 1-5 Thus, in 3-D, the chain ule gives RESUME HERE f f(), p i f = h i j k f + + x y z evaluate these fist = [ + + ] = = + + x x x x y z 1 / x x y z 1 / / etc. fo & y z f Thus f(), i x i y p j k z = h i + + = f q h f x + y + z [ f(), q p]= q [ f(), p]= ih i = f h [ f(),q p]= ih f But we wanted to evaluate the commutation ule fo f() = 1 [ 1, q p]= ih i = h 1 1 plug this esult into (3) p = 1 q p 3 ih ( ) ih 1 p = 1 ( q p ih) this is a scala, not a vecto, equation This is the compact but non-symmetic esult we got stating with a caefully symmetized stating point as equied by Coespondence Pinciple. (4) (5) (6) evised Octobe 1, 10:15 AM

6 5.73 Lectue #1 * This esult is identical to esult obtained fom standad vecto analysis IN THE LIMIT OF h Still must do things: * [ ( )] 1 [ p, ]=, q p ih show [,p ] = ih show p is Hemitian =, q p, ih, q p =, q p [ q, p]= ih [, p ]= ih Use Eq. (4) 0 0 ( ih) * we do not need to confim that p is Hemitian because it was constucted fom a symmetized fom which is guaanteed to be Hemitian. Coespondence Pinciple!. Veify that Classical Definition of Angula Momentum is Appopiate fo QM. L = q p = î ĵ ˆk x y z p x p y p z (7) We will see that this definition of an angula momentum is acceptable as fa as the coespondence pinciple is concened, but it is not sufficiently geneal. NONLECTURE What about symmetizing L? L x = yp z zp y = p z y p y z p q = L = ( p q ) x PRODUCTS OF NON-CONJUGATE QUANTITIES evised Octobe 1, 10:15 AM

7 5.73 Lectue #1 q p + p q =0 q p p q =L But is L Hemitian as defined? symmetization is impossible! antisymmetization is unnecessay! 1-7 BE CAREFUL: ( q p) p q! go back to definition of vecto coss poduct L x = yp z zp y L x = pz y py z = pz y p y z = yp z zp y = L x (p,q ae Hemitian) RESUME L is Hemitian as defined. 3A. Sepaate p into adial and angula tems. GOAL: p = p + L vecto analysis p= p + p (8) ( and with espect to q) p p Classically L p = 678 qq ( p) q (q p) component component in q,p to q q p plane which is to q (is the sign coect?) (9) is needed in both tems to emain dimensionally coect. evised Octobe 1, 10:15 AM

8 5.73 Lectue #1 1-8 talk though this vecto identity 1st tem ( p ): q p = q p cosθ q / q = unit vecto along q nd tem ( p ): q p points up out of pape thumb finge palm thumb } q Is it necessay to symmetize Eq. (9)? NONLECTURE q { p is in plane of pape in opposite diection of p, finge hence minus sign. Examine Eq. (9) fo QM consistency x component p x = [ xxp ( x + yp y + zp z ) ( yl z zl y )] but yl z zl y = yxp ( y yp x )+ zxp ( z zp x ) p x = ( x + y + z 0 0 )p x + ( xy yx)p y + ( xz zx)p z = p x similaly fo p y, p z Symmetize? No, because pats of p ae aleady shown to be Hemitian. RESUME evised Octobe 1, 10:15 AM

9 5.73 Lectue #1 3B. Evaluate p p [ ( )] p = p qq ( p) q q p goal is p = p + L 1-9 (10) commute p though - to be able to take advantage of classical vecto tiple poduct NONLECTURE [ p, ] = ih î x + ĵ y + ˆk z =ih 4 q Recall [ f(x),p x ]= ih f x because x = 3 x = 3 1 x = x4 ( ) ( ) qq p thus p = p +ih q p = p +ih q [ ( ) q ( q p) ] (11) (1) get 4 tems p = ( p q) q p ( ) p [ q ( q p) ]+ ( ih) ( q q) ( q p) ( ih) q q q p I II III IV [ ( )] p + ih I = ( q p 3ih) ( q p) 1 ( q p ih) ( q p)= p ( q p) III = ( ih) ( q p) II = ( p q) ( q p)= ( ± L )= L 4 0 IV = ( ih) ( q q) ( q p) p = p ( + )+ = [ ] + + = + p ih L p ih p pih L p L p -[,p ] (13) evised Octobe 1, 10:15 AM

10 5.73 Lectue # RESUME This p = p + L equation is a vey useful and simple fom fo p sepaated into additive adial and angula tems! If H can be sepaated into additive tems, then the eigenvectos can be factoed. SUMMARY p = 1 ( q p ih) adial momentum p = p + L sepaation of adial and angula tems H = p µ + L µ + V() eventually V l () = h l(l +1) µ + V() Next: popeties of L i, L CSCO evised Octobe 1, 10:15 AM

3D-Central Force Problems II

3D-Central Force Problems II 5.73 ectue # - 1 3D-Cental Foce Poblems II ast time: [x,p] vecto commutation ules: genealize fom 1-D to 3-D conugate position and momentum components in Catesian coodinates Coespondence Pinciple Recipe