The top and bottom lines on the extreme right show the energies of the states 1

# The top and bottom lines on the extreme right show

• Harvard University
• PHYS 143
• Test Prep
• migabrael2
• 277
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left side of the diagram quantifies the Zeeman effect (weak field). The top and bottom lines on the extreme right show the energies of the states | 1 , 1 )| + ) and | 1 , 1 )|−) , which are eigenstates of the full Hamiltonian for all values of B/A . The levels on the extreme right of Figure 9.2 show the energies described by this formula in the case that l = 1 and s = 1 2 . The fact that in a strong magnetic field an atom’s energies depend on m l and m s in this way is known as the Paschen–Back effect . Zeeman effect In a sufficiently weak magnetic field, H SO affects the atom more strongly than H Bs . Then spin-orbit coupling assigns different energies to states that differ in j . Consequently, when we use perturbation theory to calculate the smaller effect of an imposed magnetic field, the degenerate eigenspace in which we have to work is that spanned by the states that have given values of j,l and s but differ in their eigenvalues m of J z . Fortunately, H Bs is already diagonal within this space because [ J z ,S z ] = 0. So the shift in the energy of each state is simply E b = ( j,m,l,s | H Bs | j,m,l,s ) = μ B ( m + ( j,m,l,s | S z | j,m,l,s ) ) . (9 . 27) As we saw in § 7.5, our basis states do not have well-defined values of S z – in general they are linear combinations of eigenstates of L z and S z : | j,m,l,s ) = s summationdisplay m = s c m | l,m m )| s,m ) , (9 . 28) where the coefficients c m are Clebsch–Gordan coefficients (eq. 7.146). In any concrete case it is straightforward to calculate the required expectation value of S z from this expansion. However, a different approach yields a general formula that was important historically. In the classical picture, spin-orbit coupling causes the vector S to precess around the invariant vector J . Hence, in this picture the expectation value of S is equal to the projection of S onto J . 4 The classical vector triple product formula enables us to express S in terms of this projection: J × ( S × J ) = J 2 S ( S · J ) J so J 2 S = ( S · J ) J + J × ( S × L ) . (9 . 29) In the classical picture, the expectation value of the vector triple product on the right side vanishes. If its quantum expectation value were to vanish, 4 This heuristic argument is often referred to as the vector model .
9.2 Variational principle 189 the expectation value of the z component of the equation would relate ( S z ) , which we require, to the expectation values of operators that have the states | j,m,l,s ) as eigenstates, so our problem would be solved. Motivated by these classical considerations, let’s investigate the operator G J × ( S × L ) so G i summationdisplay jklm ǫ ijk J j ǫ klm S l L m . (9 . 30) It is straightforward to check that its components commute with the angular- momentum operators J i in the way we expect the components of a vector to do: [ J i ,G j ] = i summationdisplay k ǫ ijk G k . (9 . 31) From equation (9.30) it is also evident that J · G = 0. In Problem 7.24 identical conditions on the operators L and x suffice to prove that ( x ) = 0 in any state that is an eigenket of L 2 . So the steps of that proof can be retraced with L replaced by J and x replaced by G to show that for the states of interest ( G ) = 0.

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