Lecture 7: Statioary Perturbatio Theory I most practical applicatios the time idepedet Schrödiger equatio Hψ = Eψ (1) caot be solved exactly ad oe has to resort to some scheme of fidig approximate solutios, preferably by some method of iteratio that allows oe, at least i priciple, to fid the solutio with ay desired accuracy. Oe of the most widely used such methods is perturbatio theory. Perturbatio theory is applicable if there is a possibility of separatig the Hamiltoia H ito a sum of H 0, say, for which oe is able to fid the exact solutio of the correspodig Schrödiger equatio, ad a term H, the perturbatio. The latter term must be i some sese small for the method to work. The precise meaig of the smalless eeded to make the method useful has to be ivestigated i each case separately. However, i order to have a geeral picture of this meaig oe could thik of the eergy levels of a hydroge atom perturbed by some exteral force: if the shift of the eergy levels is small compared with their separatio i the uperturbed state, the perturbatio theory will be applicable. 1. The o-degeerate case Perturbatio theory proceeds somewhat differetly i the presece or i the absece of degeeracy. We shall cosider i this sectio oly the case of o-degeerate perturbatio theory. This will allow us to apply the method to most oe-dimesioal cases. Three-dimesioal systems will usually have some degree of degeeracy; this will be take ito accout i the ext sectio. Thus we assume that the Hamiltoia ca be writte i the form H = H 0 + λv (2) where I have writte λv for H. The parameter λ ca be thought of as some quatity which ca be varied at will ad ca be reduced to zero, i which case the perturbatio disappears ad all results must revert to the uperturbed case. If we cosider the case of a hydroge spectrum which we observe with the atom placed i a magetic field (Zeema effect) or i a electric field (Stark effect), the λ ca be thought of as characterisig the stregth of the field over which we have complete cotrol. The uperturbed Hamiltoia H 0 is supposed to have bee chose such that the solutio of the correspodig Schrödiger equatio is kow: 1 H 0 φ = E (0) φ (3) ad kow eigefuctios φ which form a complete set of or- with kow eigevalues E (0) thoormal fuctios: (φ m, φ ) = δ m. (4) 1 The otatio used here suggests the presece of discrete levels oly. The modificatios to iclude cotiuous states are straightforward (see, e.g. L.I. Schiff, Quatum Mechaics). 1
Before actually applyig perturbatio theory we shall rewrite the Schrödiger equatio i the form of a matrix equatio. This ca be doe by usig the completeess of the set {φ } to express the th eigefuctio of H i the form ψ = k c k φ k. (5) Here the expasio coefficiets are fuctios of λ: we must have c k = δ k for λ = 0 (6) to esure that ψ = φ for λ = 0. If we ow substitute (5) ito (1) ad take the scalar product with φ m, the we get (E E (0) m )c m = λ k c k V mk (7) where V mk = (φ m, V φ k ). Equatio (7) is the Schrödiger equatio i matrix form ad is equivalet to Eq. (1). Mathematically it is a liear simultaeous equatio with the expasio coefficiets c m ad the eergy eigevalue E as the ukow quatities. If the system uder cosideratio had oly a small umber of eergy levels oe could solve this equatio without ay approximatios by stadard methods of liear algebra. Exercise: Assumig that the system uder cosideratio has oly two eergy levels, show that the eigevalues are where H m = (φ m, Hφ ). E 1,2 = 1 2 (H 11 + H 22 ) ± 1 4 (H 11 + H 22 ) 2 + H 12 2 I most cases of practical iterest the umber of eigestates is large or eve ifiite. The the exact solutio of Eq. (7) becomes either impractical or impossible. I these cases it may be possible to fid the solutio by iteratio usig perturbatio theory. The essece of perturbatio theory is to expad the eergy eigevalue ad the eigefuctio ito a power series i λ 2 E = E (0) + λe (1) + λ 2 E (2) +... (8a) Note that Eq. (8b) is equivalet with c k = δ k + λc (1) k + λ2 c (2) k +... (8b) ψ = φ + λψ (1) + λ2 ψ (2) +... (8c) 2 May authors, e.g. P.A.M. Dirac, The Priciples of Quatum Mechaics, do ot use such a expasio parameter but assume each successive term to be of smaller order tha the preceedig term. L.I. Schiff, Quatum Mechaics, uses a expasio parameter λ but does ot associate with it ay physical sigificace: it appears as a auxiliary variable which is set equal to oe at the ed of the calculatio. It is strogly recommeded to study the approaches take by differet authors to see how may differet ways lead to the same result. 2
where ψ (i) = k c (i) k φ k, i = 1, 2,... Substitutio of Eqs. (8) ito (7) yields E(0) m + λe(1) + λ2 E (2) +...) (δ m + λc (1) m + λ2 c (2) m +...) = λ k (δ k + λc (1) k + λ2 c (2) k +...)V mk ad comparig powers of λ we get the followig set of equatios: E(0) m )δ m = 0 (9a) E (0) m )c (1) m + E (1) δ m = V m (9b) E(0) m )c(2) m + E(1) c(1) m + E(2) δ m = k c (1) k V mk (9c) E(0) m )c(3) m + E(1) c(2) m + E(2) c(1) m + E(3) δ m = k c (2) k V mk (9d) etc. Equatio (9a) cotais o ew iformatio: it is the statemet that E m (0) From Eq. (9b) we get E (1) by puttig m = : ad we get c (1) m by puttig m : E(0) for m. E (1) = V (10) c (1) m = Vm E (0) E m (0) (m ) (11) The coefficiet c (1) remais udefied by Eqs. (9) ad will be chose by the requiremet that the fuctio ψ = φ + λψ (1) be ormalised to the 1st order i λ. Thus, usig Eqs. (8) ad (4), we get (ψ, ψ ) = (φ, φ ) + λ[(ψ (1), φ ) + (φ, ψ (1) )] + O(λ2 ) = 1 + λ(c + c ) + O(λ2 ) = 1 ad it is evidetly sufficiet to put c (1) = 0 (11 ) to esure the ormalisatio of ψ to 1st order i λ. This completes the solutio of Eq. (1) to first order of perturbatio theory. To fid the 2d order correctios we use Eq. (9c). Settig m = ad usig Eqs. (11) we get E (2) = k V kv k = E (0) E (0) k k V k 2 E (0) E (0) k where i the last step we have used the hermiticity of V. The 2d order correctio to the wave fuctio ca be foud by puttig m i (9c), ad havig foud the 2d order correctio oe ca proceed to fid the 3rd order correctio usig (9d), etc. So, at least i priciple the solutio ca be foud to ay desired accuracy by successively icreasig the umber of terms. A importat result of our derivatio is that the 1st order correctio E (1) perturbatio, the 2d order correctio E (2) (12) is liear i the is quadratic i the perturbatio ad so o. This 3
implies at least the possibility that for sufficietly small perturbatios the series will coverge. However, it happes that eve i cases where the perturbatio series is kow to diverge oe gets sometimes good agreemet with experimetal results if oe takes oly the lowest order term (cf. P.A.M. Dirac, Priciples of Quatum Mechaics, Sectio 42). We summarise our results by writig dow the eergy eigevalue E of the perturbed system up to the 2d order of perturbatio theory: E = E (0) + H + k H k 2 E (0) E (0) k (13) where H = λv is the perturbatio. Exercise: 1. A oe dimesioal harmoic oscillator with Hamiltoia Ĥ0 = ˆp2 2m + 1 2 m ω2 x 2 is perturbed by a additioal potetial eergy λx. Fid the eergy eigevalues to secod order i the perturbatio, give the eigevalues of the eergy of the uperturbed harmoic oscillator E = hω( + 1 ), ad 2 h ( (u, xu k ) = + 1 δk,+1 + ) δ k, 1 2mω where u (x) are the eigefuctios of the uperturbed harmoic oscillator. 2. Show that the same result ca be obtaied without the use of perturbatio theory by solvig the eigevalue problem of the perturbed harmoic oscillator exactly. 3. (i) If the harmoic oscillator is perturbed by a additioal potetial eergy V 1 = g x 3, discuss the physical justificatio of usig statioary perturbatio theory to fid the perturbed eergy levels. What is the lowest value of the total eergy at which perturbatio theory is certaily goig to fail? (ii) Assumig that the use of perturbatio theory is justified, explai the procedure of calculatig the perturbative correctio, to which the x 3 term gives rise, i the lowest o-vaishig order. 4. Explai why the use of statioary perturbatio theory is justified i the case of a perturbatio of the harmoic oscillator by V (x) = ax 3 + bx 4 ad calculate the correctio to the eergy i the lowest o-vaishig order. Aswer. The eergy shift i the lowest o-vaishig order is 3b 4 ( ) 2 h (2 2 + 2 + 1) 15 h 2 ( mω 4 a2 2 + 11 ) m 3 ω 4 30 Note that this result implies that the correctio due to the x 3 term is of 2d order whereas that due to the x 4 term is of 1st order. 4
5. Calculate the eergy shift of the groud state eergy of a hydroge-like atom (or io) caused by the fiite size of the ucleus, assumig that the charge of the ucleus is uiformly distributed over its volume. Aswer. The P.E. of the electro i the field of a ucleus of charge Ze ad radius R, assumig uiform charge distributio, is give by V (r) = Ze2 2R Ze2 r ( 3 r 2 R 2 ) r R r R ad if we choose the uperturbed Hamiltoia to be H 0 = ˆT + V c (r), with V c (r) = Ze2, r the the groud state eigefuctio of H 0 is the well kow hydroge wave fuctio ψ 0 = 1 πa 3 e r/a, where a = a Bohr /Z ad a Bohr = 5.3 10 11 m is the Bohr radius. The correspodig groud state eergy is E (0) 1 = 1 2 (Zα)2 m e c 2. The perturbatio the becomes { V (r) H Vc (r) for r R = 0 for r R The 1st order correctio is give by E (1) 1 = (ψ 0, H ψ 0 ) ad we ote that the itegratio over the polar agles gives just a factor of 4π leavig us with the itegral over r. This itegral ca be simplified if oe realises that the upper limit of itegratio is the uclear radius R which eve for the heaviest uclei does ot exceed 10 14 m. This implies that over the etire rage of r the expoetial fuctio is to a good approximatio equal to oe. The remaiig itegratio is straightforward ad yields ( ) R 2 E (1) 1 = 0.8 E (0) 1 a Thus for hydroge oe fids E (1) 1 / E (0) 1 10 9, for oxyge (Z = 8) oe gets 4 10 7, ad for uraium (Z = 92) 3 10 4, where for the order-of-magitude estimates we have used a well kow result from uclear physics, R = R 0 A 1/3, R 0 = 1.3 10 15 m, together with the approximate relatio A 2Z betwee mass umber A ad atomic umber Z. 6. Calculate the shift of the groud state eergy of the hydroge atom i a uiform electric field (Stark effect). Aswer. The perturbatio to the Hamiltoia of the hydroge atom is i this case H = eez if E is the electric field stregth ad we assume that the field is i the z directio. The 1st order correctio vaishes by symmetry. The 2d order correctio is give by E (2) 1 = >1,lm (φ lm, H φ 100 ) 2 E (0) 1 E (0) Recall that the sum has to be exteded over the complete set of eigefuctios. I the preset case the complete set of eigefuctios icludes the eigefuctios of the cotiuum states for which the summatio over must be replaced by a itegratio 5
over the eergy E. But such a direct approach leads to a itractable mathematical problem. A method that leads to the exact result is described i L.I. Schiff, Quatum Mechaics. Evaluatio of the above sum gives the result E (2) 1 = 2a 3 Bohr E 2 2 2 1 f() where f() = 28 7 ( 1) 2 5. Covergece of the above series is poor ad oe has to 3 ( + 1) 2+5 take a large umber of terms for a reasoable approximatio. Thus oe fids =2 E (2) 1 = 1.832a 3 BohrE 2 which is about 20% too small o accout of the eglected cotiuum states. I spite of the umerical deficiecy of our result it is ot without iterest physically: it tells us that the hydroge atom i a electric field acquires a iduced dipole momet d that is proportioal to the exteral electric field: d = E(2) 1 E 4a3 Bohr E 2. Perturbatio theory i the presece of Degeeracy A problem additioal to the oe solved i the previous sectio arises if the eergy level, for which we wat to calculate the effect of the perturbatio, is degeerate. Let us express the degeeracy of the th level of the uperturbed system by writig the wave fuctios which correspod to the eergy eigevalue E (0) i the form φ α α = 1,..., f (14) The label α has to be uderstood i a geeric sese. A well kow example of degeeracy is foud i the statioary states of the hydroge atom which are characterised by three quatum umbers: the pricipal quatum umber ad the agular mometum quatum umbers l ad m. I this case α represets both l ad m. With the above otatio the Schrödiger equatio of the uperturbed system is H 0 φ α = E (0) φ α (15) ad the eigefuctios form a complete set of orthoormal fuctios: (φ mα, φ β ) = δ m δ αβ. (16) The characteristic ovel feature that arises i the case of perturbatio of a degeerate level is that the perturbatio must be expected to have a differet effect o each of the f differet states φ α. Therefore the perturbed eergies must also be distiguished by the label α. We 6
take this ito accout by writig the perturbatio expasios correspodig to Eq. (8) i the form E α = E (0) 1α + λ 2 a 2α +... ψ α = ψ α (0) α + λ2 ψ α (2) +... (17) where for simplicity we have suppressed the label o the correctios a. Now, if the perturbatio is switched off (λ 0) the by virtue of Eq. (17) the eergy correctly takes o the correspodig value of the uperturbed system, but the wave fuctio becomes ψ α, (0) ad ab iitio we do ot kow whether this will be ay oe of the uperturbed states φ α or possibly some liear combiatio of these states. We must therefore cosider the most geeral possibility, amely that ψ (0) α = β c αβ φ β. (18) Thus, i the case of degeeracy we have the additioal problem to fid the superpositio coefficiets c αβ. We shall ow show that this problem is solved automatically together with the problem of fidig the 1st order correctios a 1α. Let us write dow the eigevalue problem for the Hamiltoia H = H 0 + λv i the form ad substitute the expasios (17). The we get (E α H 0 )ψ α = λv ψ α (19) (E (0) H 0 + λa 1α +...)(ψ α (0) + λψ(1) α +...) = λv (ψ(0) α + λψ(1) α +...) (20) ad comparig powers of λ we get a set of equatios similar to Eqs. (9). The first oe of these is agai the trivial statemet ad eed ot be metioed ay more. Of particular iterest i the preset case is oly the equatio that arises from the terms liear i λ: H 0)ψ (1) α + a 1αψ (0) α = V ψ(0) α. (21) Takig the scalar product with φ γ ad usig Eq. (18) yields a 1α c αγ = f β=1 c αβv () γβ (22) where V () γβ = (φ γ, V φ β ), ad sice α is fixed ad γ ca take o f differet values Eq. (22) represets a system of f simultaeous liear homogeeous equatios for the f ukow coefficiets c αβ. The 1st order eergy shift a 1α here plays the role of a eigevalue. The characteristic equatio of (22) is a polyomial of order f i a 1α so that we ca expect up to f differet roots, each root correspodig to oe value of α. If all roots are differet oe says that the degeeracy is completely removed i 1st order. Ofte the degeeracy is oly partly removed i 1st order, i.e. ot all roots of the characteristic equatio are differet. Let us cosider i detail the particular case of two-fold degeeracy, f = 2. The Eq. (22) takes o the form a 1α c α1 = V 11 c α1 + V 12 c α2 a 1α c α2 = V 21 c α1 + V 22 c α2 (23) 7
which has the characteristic equatio a 2 1α T a 1α + = 0 where T ad are the trace ad the determiat of the matrix V αβ, respectively: T = (V 11 + V 22 ), = V 11 V 22 V 12 V 21 The two roots of the characteristic equatio are a 1α = 1 2 T ± (1 2 T ) 2. Thus the degeeracy is removed i 1st order uless T 2 4 = 0. The separatio of the split eergy levels is give by a 1 a 1+ a 1 = (V 11 V 22 ) 2 + 4 V 12 2. The split eergy levels correspod to the zeroth-order wave fuctios ψ ± = c 1± φ 1 + c 2± φ 2 where the coefficiets must be foud from the simultaeous equatios together with the ormalisatio coditio (V 11 a 1 )c 1 + V 12 c 2 = 0 V 12 c 1 + (V 22 a 1 )c 2 = 0 c 1 2 + c 2 2 = 1. After straightforward if somewhat tedious algebra oe ca get the coefficiets i the followig form: c 1± = 1/2 V 12 V 11 V 22 1 ± 2 V 12 (V 11 V 22 ) 2 + 4 V 12 2 c 2± = 1/2 V 21 V 11 V 22 ± 1 2 V 12 (V 11 V 22 ) 2 + 4 V 12 2 Note that oly the relative phase betwee c 1± ad c 2± is fixed by the simultaeous equatios; the overall phase is arbitrary ad was chose such as to give the resultig formulæ a particularly symmetric form. Exercise: Calculate the splittig of the 1st excited state of the hydroge atom i a uiform electric field i 1st order of degeerate perturbatio theory (Stark effect). Aswer. The 1st excited state correspods to the pricipal quatum umber = 2; there are four degeerate states with agular mometum quatum umbers l = 0, 1, m = l, l+1,..., l. 8
The characteristic equatio is a quartic equatio i the 1st order eergy correctio which must be computed from a 4 4 determiat. This seems a rather tedious task. However, with a few very geeral cosideratios oe ca see that there is oly oe o-zero matrix elemet: V 01 = 3eEa Bohr, ad that the characteristic equatio reduces to the simple form a 2 1(a 2 1 V 01 2 ) = 0 whose four roots are 0, 0, ad ± V 01. Thus there is oe two-fold degeerate ushifted level ad two o-degeerate levels shifted by equal amouts up ad dow. The wave fuctio correspodig to the upper level is (φ 200 φ 210 )/ 2, ad the wave fuctio of the lower level is (φ 200 + φ 210 )/ 2; the ushifted state remais some liear superpositio of the states φ 211 ad φ 21 1. To derive these results cosider the geeral matrix elemet (φ 2lm, V φ 2l m ) where the perturbatio Hamiltoia is V = eez. This Hamiltoia has cylidrical symmetry about the z axis ad therefore commutes with the agular mometum operator L z. 3 Thus i.e. Next we ote that (φ 2lm, [L z, V ]φ 2l m ) = h(m m )(φ 2lm, V φ 2l m ) = 0, (φ 2lm, V φ 2l m ) = 0 uless m m = 0. (φ 2lm, V φ 2lm ) = 0 by parity, which leaves oly the matrix elemets (φ 2lm, V φ 2l m) with l l to calculate. But there are oly two matrix elemets: V 01 ad V 10, ad they are related by Hermiticity of the Hamiltoia: V 01 = V10, i.e. effectively we eed to evaluate oly the matrix elemet V 01. To do this we must substitute the hydroge wave fuctios ψ 200 (r, θ, φ) = R 20 (r)y 00 (θ, φ) ad ψ 210 (r, θ, φ) = R 21 (r)y 10 (θ, φ) ad carry out the itegratios, hece V 01 = ee d 3 rr 20 (r)y 00 (θ, φ)zr 21 (r)y 10 (θ, φ) = 3eEa Bohr. 3 Formally oe ca see this by otig that z = r cos θ ad L z = i h / φ. 9