Perturbation Theory According to Rayleigh Schrodinger - Lecture Notes | PH 451, Study notes of Quantum Mechanics

Download Perturbation Theory According to Rayleigh Schrodinger - Lecture Notes | PH 451 and more Study notes Quantum Mechanics in PDF only on Docsity! 478 XV. PERTURBATION THEORY ACCORDING TO RAYLEIGH SCHRÖDINGER a) without degeneracy Only few problems in QM are exactly solvable. Therefore, approximation methods are extremely important. The procedure presented here was already used by Rayleigh for acoustical problems end of the 19th century., and was introduced by Schroedinger into QM. One assumes that the problem is exactly solvable, if one neglects certain terms, which are relatively small. These small terms are then treated as perturbation, making use of the expansion in a complete system of the zeroth order solution. (15.1) $ $ $ ...... ...... ( ) (1) ( ) ( ) (1) ( ) H H H E E E E = + = + + + = + + + 0 1 0 2 0 2 ψ ψ ψ ψ The upper index indicates here, that the quantity is small in 1st, 2nd,........nth order . The exact Schroedinger equation is: $H Eψ ψ= For the unperturbed Hamilton operator we have $ ( ) ( ) ( ) ( )H E0 0 0 0ψ ψ= We substitute the expansion of E and R in the exact Schroedinger equation and obtain 479 (15.2) E E E E H H H H E H E H E H E H E H E H n n n ( ) (1) ( ) ( ) ( ) (1) ( ) ( ) ( ) (1) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (1) (1) (1) ( ) ( ) ( ) ( ) (1) (1) (1) ( ) ( ) ( ....... $ $ $ ..... $ ........ $ $ $ .... $ $ $ 0 2 0 2 0 2 0 0 0 0 0 0 0 0 2 2 2 0 + + + + − − − − ⋅ + + + = − + − + − + + − + − + − ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ d i d i d i d i d i d i ) ...+ = 0 Here we have compiled terms of the same order. We start from the non-degenerated energy term and obtain in zeroth approximationEn ( )0 (15.3)E H or E n H nn n n ( ) ( ) ( ) ( ) ( ) ( ) ( )$0 0 0 0 0 0 00− = =c hψ First approximation: (15.4)E H E Hn n n n (1) (1) ( ) ( ) ( ) (1)$ $− + − =d i d iψ ψ0 0 0 0 We expand and substitute it into (15.4) :ψ ψn nm m m c(1) ( )= ∑ 0 (15.5)E H E H cn n n nm m m (1) (1) ( ) ( ) ( ) ( )$ $− + − =∑d i d iψ ψ0 0 0 0 0 We multiply from the left with and integrateψ n ( )0 ∗ E n H n c E En nm n m mn This term is always zero (1) ( ) (1) ( ) ( ) ( )$ ( )− + − =0 0 0 0 0δ1 2444 3444 482 Then we can calculate E(2) : (15.15)ψ ψ ψ ψn n n n n n nE H E H E H ( ) ( ) ( ) ( ) (1) (1) (1) ( ) ( ) ( )$ $ $0 0 0 2 2 2 0 0⋅ − + − + − = −∞ ∞z d i d i d i ψ ψn nm m mc ( ) ( ) ( )2 2 0= ∑ c E E E n H m c Enm m n m nm n nm nm n nn ( ) ( ) ( ) (1) ( ) (1) ( ) (1) ( )$2 0 0 0 0 2 0∑ − + − + =c h e jδ δ δ (15.16)⇒ = −≠ ∑E n H m m H n E En n mm n ( ) ( ) (1) ( ) ( ) (1) ( ) ( ) ( ) $ $ 2 0 0 0 0 0 0 Examples for perturbation theory to first order: He-atom : Z =2 (15.17)$H Ze r Ze r e r r = − ∇ − ∇ − − + − h h r r2 1 2 2 2 2 2 1 2 2 2 1 22 2µ µ 1 4 1 0πε = The perturbation operator is 483 $ (1)H e r r = − 2 1 2 r r We want to calculate the energy of the ground state $H Eψ ψ0 0 0= In the following I set h, ,µ and e = 1 (15.19)ψ φ φ φ π0 1 1 1 3 2 1 2 = ⋅ = F HG I KJ − s s s rwith Z e The factor arises from the fact that I do the integration over the whole volume (dJ) 1 1 2 π F HG I KJ and not only over dr for the normalization. (or in the other notation ) consists of 2 hydrogen- like Hamilton operators$H0 $ ( )H 0 − ∇ − F HG I KJ = 1 2 1 1 0 12 2 Z r E s Total energy sφ φ } 484 (Virial Theorem) E Z0 2 2 2 = − (Here I have used, that the potential energy of the 1s- state is -Z/a0 . Then I have divided by 2 and multiplied by 2 and set a0 =1, because I have set and .h, ,µ and e = 1 a e0 2 2= h / µ Since Z = 2 , it leads to the above result ) E0 = - Z 2 (15.20) E r r d d0 0 2 1 2 1 2 1(1) = −zz ψ τ τr r (15.21)E Z e r r d d Z Z r r 0 6 2 2 1 2 1 2 1 2 1 2 5 8 (1) ( ) = − = = ⋅ − −zzπ τ τ ψ φ φ 1 r r (15.22)E Z Z0 2 1 5 8 = − −FHG I KJ 487 The selection rule is a consequence of the fact that the perturbing interaction commutes with .$ $L and Jz z If we calculate the matrix elements with we find that they are except for an n J M J= , , constant the same as the matrix elements of .$J z n L M z n L M n L z n L M n L M z n L M n L z n L L M n L M z n L M n L z n L L M , , , , , , , , , , , , , , , , , , ′ = ′ ⋅ ′ − = ′ − ⋅ − − ′ = − ′ ⋅ − 1 1 1 1 2 2 2 2 Landau Lifschitz: Quantum Mechanics If one sets the sum m m E E const n mm n n( ) ( ) ( )0 0− = ≠ ∑ one obtains (15.25)E a b M Fn n n J ( ) ( )2 2 2= + 488 Fig. 15. 1 Stark-effect for the hydrogen atom. The dashed lines represent the linear Stark effect, the full lines the quadratic corrections. These corrections are magnified by a factor 10 compared with the linear splitting., i.e. in reality much smaller. The ground state and l = 1, m = ± 1 show only a quadratic Stark-effect . 489 b) Perturbation theory with degeneracy The procedures presented before are based on non degenerated energy levels. This restriction can be lifted, if one uses for the wavefunction of a degenerated state an appropriate linear combination. How does one find it, is explained in the following example: (15.25)E n H nn ( ) ( ) (1) ( )$0 0 0= is f-fold degenerated .n( )0 (15.26)n n n n f1 0 2 0 3 0 0( ) ( ) ( ) ( ), , ,......., We assume that we would have found the right linear combination. The perturbation will$ (1)H lift the degeneracy partly (15.27)E n H nn k kkk (1) ( ) (1) ( )$= 0 0 is a linear combination of all f vectors of the subspace n(0) .nk ( )0 (15.28)n n n nk i i i k ( ) ( ) ( ) ( )0 0 0 0= ∑ We obtain then (15.29)E n n n H n n nn k i i j i j j kkk (1) ( ) ( ) ( ) (1) ( ) , ( ) ( )$= ∑ 0 0 0 0 0 0 492 H H A H d dn n m m12 21 1 21 2 1 2 (1) (1) (1)( ) ( ) $ ( ) ( )= = = ∗ ∗zzφ φ φ φ τ τ K E A A K E − − = (1) (1) 0 K E A K E A E K AI II − − = − = ± = (1) (1) , (1) c h2 2 0 ∓ The equations are (see ** on page 490) E K K A c AcI I II (1): ( )− + + = 0 c c I II = 1 Normalization: c c c c symmetric eigenfunctionI II I II 2 2 1 1 2 + = = = 493 E K K A c AcII I II (1): ( )− − + = 0 c c I II = −1 Normalization: c c c c symmetric eigenfunction I II I II 2 2 1 1 2 + = = − = Thus we have found the new eigenfunctions, which diagonalize .$ (1)H (15.32) ψ ψ φ φ φ φ ψ ψ φ φ φ φ s I n m n m n m a II n m n m n m r r E E E K A r r E E E K A = = + = + + + = = − = + + − ( , ) ( ) ( ) ( ) ( ) ( , ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) r r r r 1 2 0 0 1 2 0 0 1 2 1 2 2 1 1 2 1 2 2 1 l q l q 494 Fig. 15. 2 Diagram of the exchange splitting of the Helium levels is the symmetric function (position space) ; since the total wavefunction is antisymmetricψ I the spin function has to be antisymmetric; therefore belongs to the singulet state (Antiparallelψ I spins : S = 0 ; Para-Helium). The ground state in helium has a symmetric position function and this is reflected in our result. The lowest level, E1, belongs to one wavefunction (not degenerated) . In order to form the lowest level of the Helium atom, we have onlyφ100 1( ) r r one possibility, namely to bring the second electron into the same state ( Pauli principle demands then antiparallel spins) In the lowest state N1 = N2 and Ra = 0 . Therefore we have for the ground state as single solution ψ φ φs r r E E K A = = + + 100 1 100 2 12 ( ) ( ) r r According to (15.32) the energy difference between the para- and ortho state is 2A. The level diagram decomposes into two energetically different systems of para- and ortho Helium. Each level En + Em of the Helium atom , which was obtained without considering the electron- electron interaction, corresponds to two levels , if it is taken into account, the level of the para- Helium, En + Em + K + A , and the level of the ortho-Helium, En + Em + K - A . If one elctron is in the sate wit E1, the second one in the state with E2 (energy E1 + E2) , then we obtain two levels again, taking into account the electron-electron interaction .