11. Perturbation Theory 11.1 Time-independent perturbation theory 11.1.1 Non-degenerate case 11.1.2 Degenerate case 11.1.3 The Stark effect 11.2 Time-dependent perturbation theory 11.2.1 Review of interaction picture 11.2.2 Dyson series 11.2.3 Fermi’s Golden Rule 11.1 Time-independent perturbation theory Because of the complexity of many physical problems, very few can be solved exactly (unless they involve only small Hilbert spaces). In particular, to analyze the interaction of radiation with matter we will need to develop approximation methods36 . 11.1.1 Non-degenerate case We have an Hamiltonian = + ǫV H H0 where we know the eigenvalue of the unperturbed Hamiltonian and we want to solve for the perturbed case H0 = 0 + ǫV , in terms of an expansion in ǫ (with ǫ varying between 0 and 1). The solution for ǫ 1 is the desired Hsolution. H → We assume that we know exactly the energy eigenkets and eigenvalues of : H0 k = E(0) k H0 | ) k | ) As is hermitian, its eigenkets form a complete basis k k = 11. We assume at first that the energy spectrum H0 k | )( | is not degenerate (that is, all the E(0) are different, in the next section we will study the degenerate case). The k L eigensystem for the total hamiltonian is then ( + ǫV ) ϕ = E (ǫ) ϕ H0 | k)ǫ k | k)ǫ where ǫ = 1 is the case we are interested in, but we will solve for a general ǫ as a perturbation in this parameter: (0) (1) 2 (2) (0) (1) 2 (2) ϕ = ϕ + ǫ ϕ + ǫ ϕ + ..., E = E + ǫE + ǫ E + ... | k) k k k k k k k ) ) ) 36 A very good treatment of perturbation theory isin Sakurai’s book –J.J. Sakurai “Modern Quantum Mechanics”, Addison­ Wesley (1994), which we follow here. 107 (0) (0) where of course ϕ = k . When ǫ is small, we can in fact approximate the total energy E by E . The energy k | ) k k ) (0) shift due to the perturbation is then only ∆k = Ek Ek and we can write: − (0) (0) ( + ǫV ) ϕ = (E + ∆ ) ϕ (E ) ϕ = (ǫV ∆ ) ϕ H0 | k) k k | k)ǫ → k − H0 | k) − k | k) Then, we project onto k : ( | k (E(0) ) ϕ = k (ǫV ∆ ) ϕ ( | k − H0 | k) ( | − k | k) The LHS is zero since k ϕ = k E(0) ϕ , and from the RHS k (ǫV ∆ ) ϕ = 0 we obtain: ( | H0 | k) ( | k | k) ( | − k | k) k V ϕk ∆ = ǫ ( | | ) ∆ = ǫ k V ϕ k k ϕ → k ( | | k) ( | k) where we set k ϕk = 1 (a non-canonical normalization, although, as we will see, it is approximately valid). ( | ) 1 2 2 Using the expansion above, we can replace ∆ by ǫE + ǫ E + ... and ϕ by its expansion: k k k | k) ǫE1 + ǫ2E2 + = ǫ k V ( k + ǫ ϕ(1) + ǫ2 ϕ(2) + ... ) k k · · · ( | | ) k k ) ) and equating terms of the same order in ǫ we obtain: n (n 1) E = k V ϕ − k ( | k ) This is a recipe to find the energy at all orders based only on the knowledge of the eigenstates of lower orders. (n 1) However, the question still remains: how do we find ϕk − ? We could think of solving the equation: ) (E(0) ) ϕ = (ǫV ∆ ) ϕ ( ) k − H0 | k) − k | k) ∗ for ϕ , by inverting the operator (E(0) ) and again doing an expansion of ϕ to equate terms of the same | k) k − H0 | k) order: (1) (0) 1 (1) k + ǫ ϕ + = (E )− (ǫV ∆ )( k + ǫ ϕ + ... ) | ) k · · · k − H0 − k | ) k ) ) Unfortunately this promising approach is not correct, since the operator (E(0) ) 1 is not always well defined. k 0 − (0) 1 − H (0) 1 Specifically, there is a singularity for (E )− k . What we need is to make sure that (E )− is never k − H0 | ) k − H0 applied to eigenstates of the unperturbed Hamiltonian, that is, we need ψ = (ǫV ∆ ) ϕ = k for any ϕ . | k) − k | k) 6 | ) | k) We thus define the projector Pk = 11 k k = h=k h h . Then we can ensure that ψ the projected state − | )( | 6 | )( | ∀ | ) ψ ′ = P ψ is such that k ψ ′ = 0 since this is equal to | ) k| ) ( | ) L k P ψ = k ψ k k k ψ = 0 ( | k| ) ( | ) − ( | )( | ) (0) 1 Now, using the projector, (Ek 0)− Pk ψ is well defined. We then take the equation ( ) and multiply it by Pk from the left: − H | ) ∗ P (E(0) ) ϕ = P (ǫV ∆ ) ϕ . k k − H0 | k) k − k | k) (0) (0) Since Pk commutes with 0 (as k is an eigenstate of 0) we have Pk(Ek 0) ϕk = (Ek 0)Pk ϕk and we can rewrite the equation Has | ) H − H | ) − H | ) (0) 1 P ϕ = (E )− P (ǫV ∆ ) ϕ k | k) k − H0 k − k | k) We can further simplify this expression, noting that Pk ϕk = ϕk k k ϕk = ϕk k (since we adopted the normalization k ϕ = 1). Finally we obtain: | ) | ) − | )( | ) | ) − | ) ( | k) (0) 1 ϕ = k + (E )− P (ǫV ∆ ) ϕ ( ) | k) | ) k − H0 k − k | k) ∗∗ 108 This equation is now ready to be solved by using the perturbation expansion. To simplify the expression, we define the operator Rk (0) 1 h h R = (E )− P = | )( | k k − H0 k E0 E0 h=k k h L6 − Now using the expansion k + ǫ ϕ(1) + = k + R ǫ(V E1 ǫE2 ... )( k + ǫ ϕ(1) + ... ) | ) | k ) · · · | ) k − k − k − | ) | k ) we can solve term by term to obtain: 1st order: ϕ(1) = R (V E1) k = R (V k V k ) k = R V k | k ) k − k | ) k − ( | | ) | ) k | ) (where we used the expression for the first order energy and the fact that Rk k = 0 by definition). We can now calculate the second order energy, since we know the first order eigenstate:| ) h h E2 = k V ϕ(1) = k VR V k = k V | )( | V k k ( | | k ) ( | k | ) ( | E0 E0 | ) h=k k h L 6 − or explicitly V 2 E2 = | kh| k E0 E0 h=k k h L6 − Then the second order eigenstate is 2nd order: ϕ2 = R VR V k k k k | ) A. Formal Solution We can also find a more formal expression that can yield the solution to all orders. We rewrite Eq. (**) using Rk and obtain ϕ = k + R (ǫV ∆ ) ϕ = R H ϕ | k) | ) k − k | k) k 1 | k) where we defined H = (ǫV ∆ ). Then by iteration we can write: 1 − k ϕ = k + R H ( k + R H ϕ ) = k + R H k + R H R H ϕ | k) | ) k 1 | ) k 1 | k) | ) k 1 | ) k 1 k 1 | k) and in general: ϕ = k + R H k + R H R H k + + (R H )n k + ... | k) | ) k 1 | ) k 1 k 1 | ) · · · k 1 | ) This is just a geometric series, with formal solution: 1 ϕ = (11 R H )− k | k) − k 1 | ) B. Normalization In deriving the TIPT we introduced a non-canonical normalization k ϕ = 1, which implies that the perturbed ( | k) state ϕ is not normalized. We can then define a properly normalized state as | k) ϕk ψk = | ) | ) ϕ ϕ ( k| k) so that k ψ = 1/ ϕ ϕ . We can calculate perturbativelyv the normalization factor ϕ ϕ : ( | k) ( k| k) ( k| k) ✟ 2 v 1 1 ✟1 2 1 1 2 Vkh ϕk ϕk = k + ǫϕk + ... k + ǫϕ + ... = 1 + ǫ✟k ϕk + .. + ǫ ϕk ϕk + .. = 1 + ǫ | | ( | ) ( | ) ( | ) ( | ) (E0 E0)2 h=k h k L 6 − Notice that the state is correctly normalized up to the second order in ǫ. 109 C. Anti-crossing 0 0 Consider two levels, h and k with energies Ekand E h and assume that we apply a perturbation V which connects only these two states (that is, V is such that l V j = 0 and it is different than zero only for the transition from h to k: h V k = 0.) ( | | ) If the (perturbation| | ) 6 is small, we can ask what are the perturbed state energies. The first order is zero by the choice of V , then we can calculate the second order: 2 2 V V E(2) = | kj | = | kh| k E0 E0 E0 E0 j=k k j k h L6 − − and similarly 2 2 V V E(2) = | hj | =| kh| = E.(2) h E0 E0 E0 E0 − k j=h h j h k L6 − − 0 This opposite energy shift will be more important (more noticeable) when the energies of the two levels Ek and 0 Eh are close to each other. Indeed, in the absence of the perturbation, the two energy levels would “cross” when 0 0 Ek= E h. If we add the perturbation, however, the two levels are repelled with opposite energy shifts. We describe what is happening as an “anti-crossing” of the levels: even as the levels become connected by an interaction, the levels never meet (never have the same energy) since each level gets shifted by the same amount in opposite directions. D. Example: TLS energy splitting from perturbation 0 Consider the Hamiltonian = ωσz + ǫΩσx. For ǫ = 0 the eigenstates are k = 0 , 1 and eigenvalues Ek = ω. We also know how to solveH exactly this simple problem by diagonalizing the| )entire{| )matrix:| )} ± E = ω2 + ǫ2Ω2 , 1,2 ± ϕ = cos(ϑ/2) 0 + sin(ϑ/2) 1 , ϕ = cos(vϑ/2) 1 sin(ϑ/2) 0 with ϑ = arctan(ǫΩ/ω) | 1) | ) | ) | 2) | ) − | ) For ǫ 1 we can expand in series these results to find: ≪ ǫ2Ω2 E (ω + + ..