Lecture 13 Highlights - Quantum Physics II | PHYS 402, Study notes of Quantum Physics

Download Lecture 13 Highlights - Quantum Physics II | PHYS 402 and more Study notes Quantum Physics in PDF only on Docsity! Lecture 13 Highlights The eigenfunctions of can be expressed as linear combinations of states with different values of and using the world-famous Clebsch-Gordan coefficients ( ) as: 2J lm sm js mmm js C l l s mmm js mmmj msmCmj js js∑ =+ = l l l l l Where the ket ll m represents the spherical harmonics . The C-G coefficient values are given in Table 4.8 on page 188 of Griffiths. Remember that the all of the coefficients should appear under a square root, with the minus sign (if any) out front. Also note that we have dropped the radial part of the wavefunction ( ) because it plays no role in combining angular momenta. Don’t forget to put it back later. l l mY lnR Where do these coefficients come from? Consider starting with a product wavefunction at the top of the ladder (it is a product of the wavefunctions with maximum values of and ). Now apply the lowering operator, and construct orthonormal states on lower rungs of the ladder. The coefficients on the terms of those wavefunctions are the C-G coefficients. jm lm sm −J We did a specific example of a hydrogen atom with 1=l and spin . In this case the angular momentum vector and spin vector can either be “parallel” or “anti- parallel.” Consider the two cases: 2/1=s 1) “Parallel” L r and S r : The maximum value of is +1, while the value of is +1/2 for the “parallel” case. This means that lm sm 2/3=+= sj mmm l . This is the state at the top of the ladder. There must also be states with 2/3,2/1,2/1 −−+=jm . This is a set of 4 states on the ladder of . Thus the eigenvalues of for this ladder must be 2/3=j 2J 22 4 15)1 2 3( 2 3 hh =+ . Note that 0>• SL rr is this case, giving a positive spin-orbit Hamiltonian perturbation. 2) “Anti-Parallel” L r and S r : The maximum value of is +1, while the value of is -1/2 for the “anti-parallel” case. This means that lm sm 2/1=+= sj mmm l . This is the state at the top of the ladder. There must also be a state with 2/1−=jm . This is a set of 2 states on the ladder of . Thus the eigenvalues of for this ladder must be 2/1=j 2J 22 4 3)1 2 1( 2 1 hh =+ . Note that 0<• SL rr is this case, giving a negative spin-orbit Hamiltonian perturbation. There are a total of 6 states possible by simply combining the orbital angular momentum with and spin angular momentum with1=l 2/1=s ! Just imagine what happens when you combine 3 or more angular momentum vectors! Now for an example of how to construct states that are simultaneous eigenfunctions of , , and . Take the case again of hydrogen with and spin . How do we find the state with 2L 2S 2J zJ 1=l 2/1=s 2/3=j and 2/1−=jm in terms of the and spinors? Look at the ll mY 2 11× Table on page 188. We are led to this table because we are combining an angular momentum vector with 1=l and spin vector with . Now look under the column labeled “ “. It says: 2/1=s 2/1 2/3 − s mm mm mmC s s 2 11 2 1 2 3 2/1 2/32/11 2/1∑ −=+ −=− l l l 2 1 2 111 3 1 2 1 2 101 3 2 2 1 2 3 −+−=− This can be written in a more familiar way in terms of spherical harmonics and spinors as: + − − +=− χχ 1 1 0 1 3 1 3 2 2 1 2 3 YY One can move back and forth between the coupled and un-coupled representations using the Clebsch-Gordan table on page 188. Here is the schematic layout for the CG table for combining two spins (called 21, SS rr ) to form a total spin 21 SSS rrr += ( has eigenvalue ): 2S 2)1( h+ss 21 SS × 21 ss mm #CG General Schematic of the C-G Table s sm 21 SSS rrr += Coupled Representation Un-Coupled Representation