3D-Central Force Problems II

Transcription

1 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 Catesian and vecto analysis Symmetize (mae it Hemitian) classical in h 0 limit Deived ey esults: [ f( x), px]= ih x [ f() q p]= i, h based on d f d x 1 * p = ( q p ih) * p = p + opeato * = q p * H p = + + V () µ l µ and = [ x + y + z ] 1 / (came fom symmetization in Catesian coodinates) algeba gave simple sepaation of vaiables not necessay (o possible) to symmetize V adial effective potential We do not yet now anything about nd i. TODAY [pupose is mostly to pactice [,] and angula momentum algeba] Obtain angula Momentum Commutation Rules Bloc diagonalize H ε i evi-civita Antisymmetic Tenso useful in deivations, vecto commutatos, and emembeing stuff. Next ectue: Begin deivation of all angula momentum matix elements stating fom Commutation Rule definitions of angula momentum. evised 1 Octobe, 10:19 AM

2 5.73 ectue # GOAS 1. [ i, f ]= 0. [ i, p]= 0 3. [ i, p]= 0 [ i ]= [ i ] 4., 0 but, 0! 5. CSCO.... H,, any scala function of scala. difficult - need ε i i bloc diagonalize H - These 1-4 ae chosen to show that all tems in H commute with and i 1. [ z,f()]= [ xp y yp x,f()]= xp [ y,f]+[ x,f]p y yp [ x,f] [ y,f]p x [ x,f]=0, [ y,f]=0 because [ q,f() ]=0î +0ĵ +0ˆ x ecall [ f(), px] x [, () ]= = z f ih x y y x [ z, p]= [ z, ( q p ih) ]=[ z, q p] = [ z, ] qp + [ z, qp ] [ z, q p]= q [ z, p]+ [ z, q] p two vecto commutatos on RHS Note that qis not f! need to define special notational tic to evaluate these DIFFICUT COMMUTATORS evised 1 Octobe, 10:19 AM

3 5.73 ectue # - 3 evi-civita Symbol cyclic ode adacent intechange epeated indices [ ]= I claim, i p p. ih εi ε i ε xyz = ε yzx = ε zxy =+1 ε yxz = ε zyx = ε xzy = 1 ε xxy = etc.= 0 This will become the definition of a vecto opeato with espect to. Nonlectue: Veify claim fo 1 of 3 3 = 9 possible cases let i = x, = y [ x,p y ]= [ yp z zp y,p y ]= [ yp z,p y ]+0 0 = [ y,p y ]p z + yp z,p y p z [ ] Now chec this using ε i 0 0 [ x, py]= i h εxyp ε p + ε p + ε p p. All othe 8 cases go similaly Othe impotant Commutation Rules [ xyx x xyy y xyz z] z OK [ i,p ] ε i p [ i,q ] ε i q [ i, ] ε i geneal definition of a vecto opeato geneal definition of an angula momentum. Wos even fo spin whee q p definition is inapplicable All angula momentum matix elements will be deived fom these commutation ules. FOR THE READER: VERIFY ONE COMPONENT OF EACH OF THE THREE ABOVE COMMUTATORS evised 1 Octobe, 10:19 AM

4 5.73 ectue # [ i, ] ε i is identical to =ih - 4 expect 0! because vecto coss poduct A B = A B sinθeˆ AB î ĵ ˆ = x y z x y = î y z z y + ĵ z x x z + ˆ x y y x z [ î x + ĵ y + ˆ z ] This vecto coss poduct definition of is moe geneal than q p because thee is no way to define spin in q p fom but S S S is quite meaningful. Can one genealize that, if (instead of 0), and the [ i, ] and [ i, p ] commutation ules have simila foms, that p p? NO! Chec fo youself!. Continued. 1 1 z, p q z, p z, q p [ i, p] ˆiε + ˆε + ˆε p [ ]= [ ]+ [ ] sum of 3 tems ix iy iz q [, p] xε + yε + zε p i, ε i qp ix iy iz (1) vecto commutatos only one of these tems is nonzeo evised 1 Octobe, 10:19 AM

5 5.73 ectue # - 5 and the othe tem [ i, q ] p [ i, q] ˆi ˆ ix iy ˆ ε + ε + εiz q [ i, q] p ε q p + ε q p + ε q p [ ] [ ix x iy y iz z] ε qp= ih i,, putting Eqs. (1) and () togethe = ih ε, i ε qp qp i ( labels pemuted) () switch ode of and [ ] =0! q [ i,p]+ [ i,q] p ε i q p ε i q p, Elegance and powe of ε i notation! We have shown that: * [ i,p ] = 0 fo all i * easy now to show [ i,p ] = 0 Finally i, [ ] = [ i, ] = [ i, ]+ [ i, ] ( ] = ih Σ ε i + ih Σ ε i same tic: pemute indices in second tem ε = =0 i ε i But be caeful: [ i, ] = [ i, ]+ i, ih Σ ε i [ ] Σ ε i +Σε i because this is a sum only ove, so can t combine and cancel tems. evised 1 Octobe, 10:19 AM

6 5.73 ectue # - 6 fo i = x, = y x, y [ ] 0! [ ] y z + z y so we have shown [, i ] =0 [,f()] =0 [ i,f()]=0 [,p ] =0 [ i,p ]=0, i, H all commute Complete Set of Mutually Commuting Opeatos eigenfunction of with eigenvalue h + 1 So what does this tell us about H =? BOCK DIAGONAIZATION OF H! Basis functions ψ = χ(), z = nm Next time I will show, stating fom adial special angula univesal eigenfunctions of z eigenfunctions of which adial eigenfunction? * * [ i, ] Σε i, that = nm = h + 1 nm 01,, z nm = hm nm M =, + 1, + also deive all and matix elements in nm basis set. x y evised 1 Octobe, 10:19 AM