Time Independent Perturbation Theory

apr_0-may_5.nb: 5/4/04::9:56:8 Time Independent Perturbation Theory Note: In producing a "final" vrsion of these notes I decided to change my notation from that used in class and by Sakurai. In class, I used the notation that unperturbed eigenstates are labeled by the subscript 0 as in Sn 0, and the eigenstates of the full H are given simply as Sn. Here I will alter this, and use Sn for the unperturbed states and Sn'for the perturbed eigenstates. In some cases, the subscript 0 is still used, in which case it explicitly denotes the 0 th -order term in a perturbative expansion, and in this case one usually has Sn 0 =Sn. Similarly, E n' = E n +de n. There are two reasons for this decision. ) make the notation consistent with that used for time dependent perturbation theory, where Sn denotes unperturbed states. ) The new notation reduces the need for multiple subscripts. ü Introduction We may think of two typical problems in perturbation theory. In both cases we imagine that the full problem is defined by a Hamiltonian H which may be divided into two parts H = H 0 + H, where H 0 describes a solved problem and H is, in some sense, small. When one says that H 0 is a solved problem, one means that the eigenstates Sn and eigenvalues E n are known; i.e. H 0 Sn = E n Sn. Similarly, the term "small" implies that the full solutions can be approximated by the known solutions of the exact problem. There are two main cases, time independent and time dependent perturbation theory. In time independent perturabtion theory, H is independent of time. In this case, one is mainly interested in finding more exact solutions to the spectrum of eigenstates; i.e. one wants to find the states Sn' and the eigenvalues E n' such that H Sn' = E n' Sn', or H 0 + H Sn + Sdn = E n + de n Sn + Sdn. where the second expression explicitly gives the result in the form of corrections to the known eigenstates for H 0. The corrections depend on H. If H is small the corrections should also be small, i.e. they should be a perturbation to the simpler system. Time independent perturbation theory amounts to finding approximate solutions for de n and Sdnin the case where H is small. The notation suggests that if one could adjust the strength of H, that in the limit where the perturbation vanishes the perturbed state and energy labeled by n' become synonomous with the unperturbed quantities labeled by n. In time dependent perturbation theory the goals are somewhat different. H is typically a harmonic or transient phenomenon. In either case, one is interested in transitions from some initial state Si, often an eigenstate of H 0, to a final state S f, also typically another (different) eigenstate of H 0, i.e. one is interested in matrix elements of the form n f, t f H 0 + H t n i, t i Here, the definition of small is either that the transtion rate to go from state Si Ø S f is small, or the full probability to make the transition is small.

apr_0-may_5.nb: 5/4/04::9:56:8 ü Two state problem as an illustration of the main results Before turning to perturbation theory, it is useful to look at the two state system with the Hamiltonian H = E0 + D d d E 0 - D Although this problem can be solved exactly, one can acquire some insight into perturbation theory by examining the solutions when d and/or D are "small". For example, if d + D, then this becomes an example of non-degenerate perturbation theory with H 0 = E 0 + D 0 and H = 0 0 E 0 - D d d 0 or, if D is small, the problem can be treated as an example of degenerate perturbation theory, with H 0 = E 0 0 and H 0 E = +D 0 d d -D An example of such a system would be a spin- ÅÅÅÅ system in a magnetic field with D, d B z, B x In either event, the full eigenvalues are E = E 0 H D + d If d is small, then the energies may be approximated by E = E 0 D ÅÅÅÅÅÅÅ d D This is a typical result for non-degenerate perturbation theory. The first order shift is given by the diagonal components to H i.e. n H n = D. The energy perturbation due to off-diagonal components of H are nd order in the perturbation and inversely proportional to the energy separation of the states. Another comment is that the perturbation causes the energy "eigenstates" to repel each other, i.e. the splitting between the states is increased by H. On the other hand, if D = 0, then one finds an example of degenerate perturbation theory. The energy shift is st order in the off-diagonal perturbation, E = E 0 d The eigenstates for this two state system can be written in terms of a mixing angle q, + = cosq + sinq - = -sinq + cosq Defining D = H D + d, the mixing coefficients are given by cosq = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ D + DI H D sinq = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ D - DI H D In this case, if d ÅÅÅÅ D is small, then q is a perturbative quantity -

apr_0-may_5.nb: 5/4/04::9:56:8 q = d ÅÅÅÅÅÅÅ D For a non-degenerate perturbation problem, the mixing is first order in the perturbation. Note, however, that this is the amplitude for mixing, and that the probability of overlap between the perturbed and original basis states scales is proportional to the square of the perturbation strength. In the degenerate case, as D Ø 0, the mixing becomes large, and both amplitude and probability are of order, even if d + E 0, i.e. it is the size of perturbation with respect to the energy splitting that matters, not with respect to the absolute energy scale. ü Comment: "just" a transformation of basis The two state problem illustrates an important point. The effect of making a change to the Hamiltonian does not change the degrees of freedom for the system, but it does change which linear combination of basis states will be eigenstates of the Hamiltonian. Thus, if the eigenstates of the unperturbed system H 0 form a set of basis functions Snfor the vector space describing the system, then the eigenstates for the full H are just another set of basis functions Sn' which, with some caveates, span the same underlying vector space. Thus, it is generally possible to find a transformation Sn' = k Sk k n' = k c nk Sk In the two state problem it is possible to find the c nk explicitly, but in general there may be a large number of degrees of freedom mixed by the perturbation and an exact solution is not practical. Even in this case, however, one can make progress if the perturbation is "small". For example, suppose the strength of the perturbation is characterized by a parameter l, then one expects that Lim c nk l = d nk, or Sn' Ø Sn, and Lim E n Ø E 0 n, i.e. the perturbed states and lø0 lø0 energies should return to their unperturbed values as the strength of the perturbation goes to zero. Viewed from this perspective, perturbation theory is a technique for finding approximate eigenvectors and eigenvalues of matrices with a large (perhaps infinite) number of degrees of freedom. Time-independent (TI) Perturbation Theory ü Statement of the problem for TI perturbation theory The general problem is to solve the eigensystem problem for the Hamiltonian H, i.e. find both E n and n in the equation. H Sn' = E n' Sn' where H is of the form H = H 0 + H and H 0 Sn = E n Sn is a solved problem. The eigenstates and energies can then be expressed as differences Sdn and de n from the unperturbed solution.

apr_0-may_5.nb: 5/4/04::9:56:8 4 H 0 + H Sn + Sdn = E n + de n Sn + Sdn. As noted above, Sn' can be expressed as a linear combination of the basis states, Sn' = k c nk Sk or Sn' = Sn + S c nk Sk, where in the second relation, the Sn state has been separated from the sum, so that k n Sdn = S c nk Sk. With this definition, the perturbation is orthogonal to the unperturbed state n dn = 0, but the k n perturbed state is not normalized to unity n' n' = nw + S c* nl lw Sn + S c nk Sk l n k n = n n + S S c* nl c nk l k l n k n = n n + S S c* nl c nk d lk l n k n = + S k n Sc nk W The question of normalization is discussed further below. ü Order by order, iteratively To emphasize that H is small, it is customary to write H = H 0 + l H where l can be adjusted from 0 < l <. The perturbed solutions can then be expressed as a Taylor series in l, Sn' = Sn 0 + l Sn + l Sn + E n' = E n 0 + ld n, + l D n, + where by Sn i one means the i th -order perturbation to the state Sn. It should be understood that the 0 th -order terms are identicle to the unperturbed terms for the same n, i.e. Sn 0 = Sn. Given that n dn = 0, in the perturbation series it must be that n n i = 0 for each order l i, i in the perturbation series. It follows that for i, each term in the perturbation series for the state can be rewritten as a sum over the orthogonal unperturbed states, i.e. Sn i = S i Sk, with expansion coefficients that depend on i. For l = 0, c 0 nk = d nk. With this preamble, the eigenvalue problem becomes H 0 + l H Sn 0 + l Sn + l Sn + = E n 0 + ld n, + l D n, + Sn 0 + l Sn + l Sn + Since the value of l is arbitrary, the equation must be satisfied order by order in l. The first three equalities are k n c nk l 0 : l : l : H 0 Sn 0 = E n 0 Sn 0 H 0 Sn + H Sn 0 = E n 0 Sn + D n, Sn 0 H 0 Sn + H Sn = E n 0 Sn + D n, Sn + D n, Sn 0 The l 0 equation is just the eigenvalue equation for the unperturbed system, H 0 Sn = E n Sn. ü First order energy shift The first order energy shift can now be determined by operating on the l equation by nw from the left.

apr_0-may_5.nb: 5/4/04::9:56:8 5 n H 0 n + n H n = n E n n + n D n, n Given the orthogonality n n = 0 for each order in the perturbation series, D n, = n H n ü General result for D n,i Since the orthogonality n n i = 0 holds order by order in l, the more general relation for the i th order energy shift is found by considering nw H 0 + l H Sn 0 + l Sn + l Sn + = nw E n + ld n, + l D n, + Sn 0 + l Sn + l Sn + On the left nw H 0 + l H Sn 0 + l Sn + l Sn + = nw E n + l H Sn 0 + l Sn + l Sn + = E n + l nw H Sn 0 + l nw H Sn + l nw H Sn + while on the right nw picks out just the Sn 0 term nw E n + ld n, + l D n, + Sn 0 + l Sn + l Sn + = E n + ld n, + l D n, + Equating terms order by order in l i, D n, i = n H n i- To find the i th -order energy shift one needs to know the i - th order correction to the state. ü First order correction to the state The first order expansion coefficients can be determined by operating on the l equation with the state kw. Following the steps k H 0 n + k H n = k E n n + k D n, n E k k n + k H n = E n k n E n - E k k n = k H n or = k n = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ kh n c nk E n -E k The expansion coefficient is proportional to the matrix element and inversely proportional to the splitting between the two states. This is consistent with the expression q = ÅÅÅÅÅÅÅ d D for the two state system. ü nd order energy shift Given the first order correction to the state, the second order energy shift follows

apr_0-may_5.nb: 5/4/04::9:56:8 6 D n, = n H n = S c nk n H k n k = S ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ nh k kh n ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ n k E n -E k = S ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ SnH kw ÅÅÅÅÅÅ n k E n -E k Again, this is consistent with the two state system, where the second order shift for the non-degenerate case was given by ÅÅÅÅÅÅÅ d D. ü Higher order terms The second order expansion coefficients can be determined by operating on the l -equation with kw. The procedure can be iterated, but the algebra gets increasingly cumbersome. It becomes useful to develop a more formal approach to the perturbation series. This is done in Sakurai 5., or Merzbacher 8.-. If time allows I'll add some notes at the end to discuss this development. The development is valueable as it has features similar to the development of scatterng theory. ü Normalization As noted earlier, the state Sn' = Sn 0 + l Sn... = Sn + Sdn is not normalized to unity. Accordingly, one "renormalizes" by defining Sn è = Z n I Sn' so that n è n è = Z n n' n' = or Z n - = n' n' = n n + Ren dn + dn dn = + l n n +... or, taking the inverse, Z n = - l n n +... - l S c * nk c nl k l kl = - l S Sc nk W k = - l kh S n ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ k E n -E k Note: this behavior is typical. The loss of overlap is second order in the perturbation. ü Comment: sudden vs adiabatic ü Degenerate perturbation theory For the non-degenerate case, the first three results are

apr_0-may_5.nb: 5/4/04::9:56:8 7 D = n H n Sn = k n S Sk c nk = k n S Sk ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ kh n kh D = S n ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ k E n -E k E n -E k The first order energy shift is always valid, but if the energy eigenvalues are nearly degenerate E n - E k / k H n then the perturbative series will not converge, or at least not quickly, i.e. c nk is of order unity, and D is of the same order as D. In this case, one must take a different approach to dealing with the degenerate states. Hopefully the number of degenerate states is small, in which case it may be practical to solve the degenerate subspace exactly. Let the set of degenerate states be identified as Sd, and define the projection operators P d = S d Sd dw, which projects onto the degenerate subspace, and P êê d = - P d, which projects out of the degenerate subspace. Then the simplest perturbative procedure is to diagonalize H within the degenerate subspace, and then use the new eigenstates Sd ', along with the original non-degenerate eigenstates as the basis for constructing a perturbative series to determine Sdn and D. ü Degenerate perturbation theory: Algebra Consider H = H 0 + H where H 0 has a degenerate subspace Sd. Then P d Sd = Sd, P d Sd êê = 0, P êê d Sd = 0, and êê Pd Sd êê = Sd êê. Further, the unperturbed H 0 has H 0 Sd = E d Sd H 0 Sk = E k Sk where E d is the degenerate energy, and where Sk is a non-degenerate state. Then, a partial solution to the perturbation problem is to solve within the P d subspace first, H Sd ' = E d ' Sd ', where Sd ' = c d d' Sd is found be diagonalizing P d H P d. Of course this will mix up the matrix elements k H d. Specifically, there will be a unitary matrix U such that Sd ' = U Sd, and Sk ' = U Sk = Sk for states not in the subspace, i.e. P d Sk = 0. Once U has been determined, the problem can be rephrased using the basis Sd ' and the operator H ' = U H U = UH 0 + H U = H 0 + U H U and solved using non-degenerate techniques. This can be depicted pictorially in matrix form. For convenience suppose the states are ordered so that the Sdcome first. In the original basis Ed + H, d d H, k d H, d k E k + H, k k After transforming E d' H, k d' H, d' k E k + H, k k Where E d' = E d + e d' is now diagonal, and the non-degenerate technique can be applied. Note that the non-degenerate part H, k k is not modified, but that the mixed terms are. One might worry that the expansion is not well defined since kh n ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ has a small denominator for n, k both within the degenerate subspace; however, in the diagonalized basis Sd ' E n -E k the matrix element in the numerator vanishes and so everything is, in fact, well behaved.

apr_0-may_5.nb: 5/4/04::9:56:8 8 ü Comment: no level crossing - states repel each other and exchange identities. Return for a moment to the two state system, H = H 0 + H, with H 0 = E 0 0 and H 0 E = +D 0 d d -D and consider the behavior of the system as D changes. Specifically, consider the trajectory of the energy eigenvalues as a function of D for the case that D begins with a value less than 0, passes through zero, and becomes positive. 4 Eigenvalue trajectories for /0., E 0.5.5 E.5 0.5 - -0.5 0 0.5 The energy eigenvalues during this process are E = E 0 H D + d As long as d is non-zero the energy eigenvalues do not cross. The states are sometimes said to "repel" each other. Also note that for D 5 d, one can use perturbation theory, but the meaning of the unperturbed state is inverted from that for D + -d. One may the states are exchanged. This result is fairly general. If as a result of a change in strength of a term of the Hamiltonian it appears that the trajectories for two energy eigenvalues are going to cross, then in fact they repel each other and exchange identities. Examples ü Stark effect The Stark effect concerns the shift in energy levels of an atom under the influence of an electric field. Taking the field to be in the z` direction, the potential energy for an electron is V = -e z. The problem is then to find the atomic states for H = H 0 - e z

apr_0-may_5.nb: 5/4/04::9:56:8 9 with H 0 being the unperturbed atomic hamiltonian. The simplest case is the ideal hydrogen atom. The unperturbed states are characterized by Sn l m. The principle quantum number must obey n > l, the bound state energies are given by E n = - a m ÅÅÅÅÅÅÅÅÅÅ n = E 0 ÅÅÅÅÅÅ n, where E 0 =.6 ev is the ground state energy (one Rydberg), a = ÅÅÅÅÅÅ e c ÅÅÅÅÅÅÅ 7 is the "fine structure constant" and m = m e m ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ p is the reduced mass of the electron in the hydrogen atom. m e +m p r q f n l m = R n l r Y l m q, f. There is considerable degeneracy, since the energy only depends on n, and l < n, the number of states N n with energy E n is N n = n- l=0 l + = n. This counting neglects spin. In the presence of a perturbation, the new states will be linear combinations of the hydrogenic states. The main effect of perturbations will be to mix up states within the degenerate subspaces. The only non-degenerate state is the ground state, which is interesting in its own right. ü D and parity As a starting point, note that in the unperturbed basis, the diagonal matrix elements vanish, i.e. D = -e n l m z n l m = 0. This can be shown by direct integration for small values of l and m. More generally, this is a consequence of parity. Parity, often denoted by P, is the discrete operation of reflection. Acting on a position-ket P Sx = S-x Amongst its other properties, P =, reflection twice is the identity operation. As a consequence the allowed eigenvalues for eigenstates of parity are p =. Note that the only position-ket which can be an eigenstate of parity is S0. Inspection of the spherical harmonics shows that Y lm -ǹ = - l Y lm ǹ At the same time Y lm -ǹ = -ǹ lm = -ǹ P - P lm = ǹ P lm = - l ǹ lm. It follows that the spherical harmonics are eigenstates of parity with P l m = - l l m( As with other operations, one can examine the behavior of operators under parity, O Ø O' = POP -. If O = O' then O is considered to be an even parity operator. Similarly, if O = -O' then O is odd. If neither is true then one says that O does not have definite parity. Now consider the case of a matrix element of some operator O with parity p O between two states Sn and Sm with parity p n and p m. Then under parity m O n Ø m' O' n' = p m p o p n m O n At the same time m O n = V y m * x Ox y n x Ø V y m * -x O-x y n -x = -V y m * x Ox y n x = V y m * x Ox y n x = m O n

apr_0-may_5.nb: 5/4/04::9:56:8 0 where in going from the first to the second line a coordinate change x Ø -x is performed. Since the integral is over all space, one can use -V = V. It follows that for operators and states of definite parity that m O n = p m p o p n m O n. Either p m p o p n = or m O n = 0. Returning to the case of D for the ground state Stark effect, the parity in this case is p D = - l p z - l = p z. Since z Ø -z under a reflection, p z = -. It follows that p D = -, and therefore D = 0. Note: It might appear that this argument is independent of the value of l, however, for the hydrogen atom there is the accidental degeneracy between different l-states with the same n. In this case one must use degenerate perturbation theory, which will replace the pure l-states with linear combinations that will not have definite parity, and the above argument can't be applied. ü D (n=, ground state) Since the first-order term vanishes, the leading energy shift is given by D = e Snlmz00W n S ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ E n -E 0 where E 0 =.6 ev is the ground state energy, and the sum is over all states not in the ground state. In general, sums like this may be difficult. In the preent case, progress can be made by approximating the denominator for each term in the sum by E - E 0 ÅÅÅÅ 4 - E 0 = - ÅÅÅÅ 4 E 0. A constant denominator can be taken out of the integral, D e ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ E -E 0 S n Snlm z 00W = 4 ÅÅÅÅ e ÅÅÅÅÅÅÅÅÅÅÅ Å E 0 S n Snlm z 00W Now, since the ground state itself has a vanishing matrix element it can be added to the sum without changing the result. This produces a complete sum over states, which can be evaluated D 4 ÅÅÅÅ e ÅÅÅÅÅÅÅÅÅÅÅ Å E 0 S nlm Snlm z 00W = ÅÅÅÅ 4 = ÅÅÅÅ 4 e ÅÅÅÅÅÅÅÅÅÅÅ Å E 0 S nlm 00 z nlm nlm z 00 e ÅÅÅÅÅÅÅÅÅÅÅ E 0 Å 00 z 00 Using 00 z 00 = a 0 (for example, Appendix A of Sakurai has wave functions) yields D 4 ÅÅÅÅ e ÅÅÅÅÅÅÅÅÅÅÅ E 0 Å a 0 ü n=, degenerate case The n = states include both an l = 0 singlet and an l = triplet. The unperturbed states are degenerate, so one must pursue degenerate perturbation theory, diagonalizing the perturbation within the n = subspace before considering other states. The z operator behaves as a T 0 tensor operator. As such it can't change the m-value of a state. The m = states are unaffected by the perturbation and remain degenerate. The m = 0 states, however, are degenerate and may mix. Similarly, the parity selection rule requires that the diagonal elements for the m = 0 states vanish. This leaves just the matrix element

apr_0-may_5.nb: 5/4/04::9:56:8 D = 0 z 00 = e a 0 The full mixing matrix is lm 00 0-00 0 D 0 0 0 D 0 0 0 0 0 0 0-0 0 0 0 After diagonalizing, the two eigenstates are S = ÅÅÅÅÅÅÅÅ S00 S0 H with energies E = E D From this point, nd order perturbation theory must be applied to the m = 0, states separately. Each m-value will mix with other states of different n, l but only with states of the same m, due to the q = 0 nature of z as a tensor operator. Thus nd order perturbation theory can be directly applied to the m = states. ü Spin-Orbit Coupling 5/4/04 - This is a brief overview of the issues in lieu of longer discussion. There is a coupling between the spinand orbital magnetic moments of an atom. This results in an interation hamiltonian of the form H I = A L ÿ S where A is an operator which acts on the radial part of the state. The simple hydrogen atom has states defined in an angular basis Slsm l m s. H I is not diagonal in this basis. Given the degeneracies of the H atom, a perturbative analysis would require determine all the matrix components of H I, and then diagonalizing. A simpler procedure is to recognize L ÿ S = ÅÅÅÅ J - L - S which is diagonal in the Sls jm basis, and that one can go back and forth between the two basis sets by using Clebsch Gordon coefficients. Including just the spin-orbit interaction leads to a small splitting of the degeneracy in different n-levels of the hydrogen atom. ü Zeeman effect An alternative problem is to consider a simple hydrogen atom in an external B-field. In this case the interaction hamiltonian is eb H B = - ÅÅÅÅÅÅÅÅÅÅÅÅÅ m e c L z + S z

apr_0-may_5.nb: 5/4/04::9:56:8 This H B is diagonal in the Slsm l m s basis, and D can be calculated directly. ü SO+Zeeman This is really a problem in perturbation theory. How one proceeds depends on the strength of the external B-field. For a strong field one should use the Slsm l m s basis for the unperturbed states, applying H = H 0 + H B as the unperturbed hamiltonian. The spin-orbit term is then treated as a perturbation. Where H B already splits the degeneracy, the spin-orbit term can be treated with non-degenerate methods. If H B leaves degeneracy then degenerate methods must be used. This can be done conveniently by appropriate use of L ÿ S = L z S z + ÅÅÅÅ L + S - + L - S + to connect different states. For a weak field. Begin in the Sls jmbasis and solve the spin-orbit problem first, treating H B as a weak perturbation. A more formal approach ü Use of projection operator ü Evaluate P - ü Stuff