Time Independent Perturbation Theory Contd.

Transcription

1 Time Independent Perturbation Theory Contd. A summary of the machinery for the Perturbation theory: H = H o + H p ; H 0 n >= E n n >; H Ψ n >= E n Ψ n > E n = E n + E n ; E n = < n H p n > + < m H p n > }{{} E n E m m n }{{} nd order Ψ n >= n > + δψ >; δψ >= m n < m H p n > E n E m n > Degenerate Cases Where n = m the Hamiltonian is degenerate. An example is the Hydrogen Atom that can be described with an eigenstate nlm >, the energy depends only on n, i.e E n = Ry 1 n and only its ground state is non-degenerate. For the degenerate states, we need to make sure that the overlap integral < m H p n > is zero whenever E n = E m. To do that, we take the Hilbert space and split it into sub-spaces, each spanned by a complete set of eigenstates to H o with the same degenerate energy E n : V = V E1 V E V E3... basis set for the Hilbert space with V E1 = 1, α >; V E =, α >... So that for the Hydrogen atom we can now write the basis for V E1 : 100 >, the basis for V E : 00 >, 11 >, 10 >, 1 1 >, and similarly for the additional subspaces (higher energies). Each of the subspaces is a finite dimensional subspace, and any operator can be written as an n n matrix if restricted within a finite dimension. Given H p, the expression < E, α H p E, α > 1

2 gives all elements of the n n matrix for H p in that subspace. For each subspace for a degenerate eigenvalue of H 0 we diagonalize this matrix representing H p. A transformation can always be made from the old basis set to the new basis set following the steps: find the matrix elements of each subspace linearly combine the previous basis vectors to have H p diagonal (unitary transformation yielding a new orthonormal basis of the subspace that diagonalizes both H 0 and H p.) Example: We have a perturbation due to an electric field: H p = eez. For the ground state 100 > (no degeneracy, no first-order change in energy) δψ > = ee n>1 E 1 = E1 +e E }{{} n>1 0 < n10 z 100 > Ry(1 1 n ) n10 > < n10 z 100 > Ry(1 1 n ) For the state 00 > We have 4 orthogonal sub states so we have to check the matrix elements of H p and see if we can diagonalize. The selection rules for the process are that 1- z can only couple states with m = m and - the element matrix must have odd parity. Since z has odd parity if it s taken between states with the same parity we get a zero. The elements to calculate are thus < 10 z 00 >= 3a o 00 > 10 > 11 > 1 1 > < a o ee 0 0 < 10 3a o ee < <

3 From the table values, H p is not diagonal, so we have to combine the elements in such a way as to make it so. Eigenvalues of the matrix : 3a o ee Eigenstates, with corresponding eigenvectors: 1 ( 00 > ± 10 >) Rewriting the unperturbed state: a >= 1 ( 00 > + 10 >) b >= 1 ( 00 > 10 >) with similar combinations for the other states, H p in the new states is now diagonalized and we can evaluate the equation 3a o ee H p = 0 +3a o ee E =< a H p a >= 3a o ee E =< b H p b >= +3a o ee There is no dipole moment in the ground state for the perturbation. To the nd order, there is an induced dipole moment which interacts linearly with the electric field and causes the nd order perturbation. However, the 1st excited state (n = ) apparently does have a non-zero permanent dipole moment, so the energy changes in first order. More specifically: combining states creates a dipole moment if there s no electric field, the Hydrogen atom is always symmetric (standard. wave functions) in even the slightest electric field, degeneracy causes the combination of states (since it costs no energy ), creating the dipoles that interact with the field. 3

4 Intrinsic Perturbations of the H atom The eigenstates to the unperturbed hydrogen atom hamiltonian can be written as : nlm l m (e) which completely specifies the state of the Hydrogen atom (often, the proton spin and even the electron spin are suppressed when writing down these states, but they are of course always there implicitly). nlm l m (e) s ignoring the spin of the proton, we have >= nlm l > m (e) nlm l m (e) which gives an additional terms to the Hamiltonian Relativistic corrections The kinetic energy T k P m, instead T k = m c 4 + p c mc ( ) = mc 1 + P m c 1 = mc (1 + P m c 1 P 4 8 m 4 c... 1) 4 P m P 4 8m 3 c }{{} H p The H p is very small at normal velocities. LS Coupling The electron has a magnetic moment µ = γs that interacts with the external magnetic field the electron sees in its own rest frame where the proton is moving (and, as any moving charge, will produce a magnetic field). This gives a second term due to interaction the angular momentum and spin of the electron as: e H p = m c r S 3 e L e The scalar product can be expressed as S e L e = (S + L) L S 4 = J L S

5 If we rewrite the eigenstates in terms of eigenstates to J, L, S, we can automatically diagonalize the perturbing Hamiltonian (and evaluate its matrix elements). Total angular momentum: J = (L + S) For the electron S = 1 4 ; S = 3 4 h We rewrite the state in terms of J as { l(l + 1) eigenvalue to l nljm j > with j, m j eigenstate to J Rules for J: So that J = { l + 1 l 1 if l 1 n, l, j, m j >= m l,m s < l, m l, 1 m s j, m j > n, l, m l, m (e) Sum over all possible states that can contribute with only the ones for which the C.G.C are not zero with m j = m l + m s 5