U but the eigenvalue is sta2onary created odd it must

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Unformatted text preview: The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again. u༇  Example: Find the eigenstates of Sx. u༇  If we prepare an Sx eigenstate, spin wll precess around a B field in the z direc2on. 35 The Spin Gyromagne2c Ra2o The electron is a point par2cle so the magne2c moment from spin cannot be calculated classically in any way. u༇  From the rela2vis2c version of the Schrodinger equa2on (Dirac equa2on), we see that g=2. u༇  o  spin ½ electron magne2c moment is one Bohr Magneton despite half the angular momentum. o  there are (radia2ve) correc2ons from QED which we can calculate and are much bigger than any as yet unobserved structure of the electron. u༇  This gyromagne2c ra2o makes the rela2onship between magne2c moment and angular momentum more complicated. 36 Hydrogen Fine Structure u༇ The n=2 states are split due to rela2vity and the spin- orbit interac2on. u༇ The correc2on is down by α2 with respect to Hydrogen binding energy. u༇ But all the different m states are degenerate o  un2l a magne2c field is applied o  which splits the states with different z component of angular momentum. 37 The Spin Orbit Interac2on u༇ The B- field generated by the orbital angular momentum will interact with the spin magne2c moment (or vice versa) and generate an energy difference between spin parallel to L and spin an2- parallel to L. u༇ One can calculate this classically to within about a factor of 2 of the correct formula. u༇ Thomas calculated it correctly in 1926. Ze 2 gs ΔH = L⋅S 223 16πε 0 m ec r 38 Total Angular Momentum J J = L+S J = L + S + 2L ⋅ S J 2 − L2 − S 2 ( j ( j + 1) − ( + 1) − s(s + 1)) 2 L⋅S = = 2 2 ( j ( j + 1) − ( + 1) − 3 ) 2 4 L⋅S = 2 1 j = ± 2 2 2 2 u༇  We can write the spin- orbit interac2on in terms of quantum numbers. u༇  Since the spin vector is short, it can only change L by ½ o  count the states 2(l+½ )+1 + 2(l- ½ )+1 = 2(2l+1) o  this can all be done rigorously 39 We have a New Coordinate u༇  Sz is a new coordinate o  with far fewer eigenvalues than x u༇  With the new coordinate, we will need a new quantum number to label our energy eigenstates o  a simple answer is ms u༇  But we will find that the energy eigenstates, in the presence o...
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