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- 1. WELCOME TO YOU ALL PHYSICAL CHEMISTRY PRESENTATION M.SC.-II – (SEM. III ) 2013– 2014 PERTURBATION Presented by :– Dharmendra R. Prajapati RAMNIRANJAN JHUNJHUNWALA COLLEGE
- 2. Perturbation: Background Algebraic Differential Equations
- 3. Perturbation Original Equation X 2 − 25 = 0 Y = X 2 + ε X − 25 Perturbed equation X 2 + ε X − 25 = 0 Y vs X 15 10 0 ≤ ε <1 5 0 -8 -6 -4 -2 0 2 4 6 8 -5 Y Epsilon=0.8 -10 Epsilon=0.5 Epsilon=0.0 -15 -20 -25 -30 X
- 4. Perturbation Perturbed equation X 2 + ε X − 25 = 0 Change in result (absolute values) vs Change in equation Perturbation in the result Perturbation in the result (root) Root -1 -2 -0.15 -0.15 Simple (Regular) Perturbation 0.06 ε=0.1 ε=0.1 0.05 ε=-0.1 ε=-0.1 0.04 0.03 Answer can be in the form 0.02 -0.1 -0.1 0.01 0.01 ε=-0.01 ε=-0.01 0 0 -0.05 0 -0.05 0 ε=0.01 ε=0.01 0.05 0.05 Epsilon (perturbation) Epsilon (perturbation) 0.1 0.1 0.15 0.15 X = X 0 + ε φ ( X 0 ) + ε 2 ...
- 5. Y = εX + X − 5 2 Y vs X Perturbation Original Equation X −5 = 0 Perturbed equation ε X 2 + X −5 = 0 40 35 30 25 Epsilon=0 Two roots instead 15 Y 20 Epsilon=0.8 one 10 Epsilon=1 5 0 -6 -4 -2 -5 0 -10 X 2 4 6 Roots are not close to the original root of
- 6. Perturbation Y = εX + X − 5 Change in result (absolute values) vs Change in equation 2 Other root varies from the original root dramatically, as epsilon approaches zero! Root-1 Root-2 3.5 1200 Perturbation Perturbation in Resultin Result 3 2.5 1000 2 800 1.5 600 1 0.5 400 0 200 0 0.2 0.4 0.6 0.8 1 1.2 Epsilon 0 0 0.2 0.4 0.6 Epsilon 0.8 1 1.2 Singular perturbation Answer may NOT be in the form X = X 0 + ε φ ( X 0 ) + ε 2 ...
- 7. dy =a dx Solution Differential Equations a <1 y =ax Perturbation-1 Solution y = 0, at x = 0 dy = a +ε y dx ( y = 0, at x = 0 ) a εx a ε 2 x2 y = e −1 = 1 + ε x + + ... − 1 ε ε 2 a ε 2 x2 = ε x + + ... ε 2 ε → 0, y → ax εx = ax1 + + ... 2 Regular Perturbation
- 8. § Time-Independent Perturbation Theory FIRST ORDER PERTURBATION:• We are often interested in systems for which we could solve the Schreodinger equation if the potential energy were slightly different. • Consider a one-dimensional example for which we can write the actual potential energy as Vactual(x) = V(x) + υ(x) (12.1) • where υ(x) is a small perturbation added to the unperturbed potential V(x).
- 9. • The Hamiltonian operator of the “unperturbed” (i.e. exactly solvable) system is • and the Schreodinger equation of that system therefore is H0ψl = Elψl (12.3) with a known set of eigenvalues El and eigenfunctions ψl. • The Schreodinger equation of the “perturbed” system contains the additional term υ(x) in the Hamiltonian:
- 10. [H0 + υ(x)]ψn’ = En’ψn’ (12.4) • • which may make it impossible to solve directly for the eigenfunctions ψn’ and eigenvalues En’. • However, Postulate 3 tells us that any acceptable wave function may be expanded in a series of eigenfunctions of the unperturbed Hamiltonian. Therefore we may write each function ψn’ as a • series of the functions ψl with constant coefficients: • (The letter l is simply an index, having nothing to do with the angular momentum quantum number; this
- 11. • Substitution into Eq.(12.4) gives • The technique used in finding coefficients in a Fourier series may be employed here. We multiply each side of Eq.(12.6) by ψm*—the • complex conjugate of a particular unperturbed eigenfunction ψm. We then integrate both sides over all x, to obtain ∞ ∞ ∞ ∞ −∞ l =1 −∞ l =1 * ' * ψ m [ H 0 + v( x)] ∑ anlψ l dx = ∫ Enψ m ∑ anlψ l dx ∫
- 12. • Integrating each side of Eq.(12.7) term by term and removing the space-independent factors anl and En’ from the integrals. we obtain • Because the functions ψ are normalized and orthogonal, the right-hand side of Eq.(12.8) reduces to the single term anmEn’ for which l = m, and we have • We can rewrite the left-hand side of Eq.(12.8) by using the fact that H0ψl = Elψl [Eq.(12.3)]; this
- 13. • • ∞ ∞ l =1 −∞ * ' anl ∫ψ m [ El + v( x)]ψ l dx = anm En ∑ or ∞ ∞ ∞ ∞ * * ' anl El ∫ψ mψ l dx + ∑ anl ∫ψ m [ v( x)]ψ l dx = anm En ∑ • • (12.9) Again, because the functions ψ are normalized and orthogonal, the first term on the left reduces to anmEm. • The second term can be written in abbreviated form as ∑∞l=1 anlυml, where υml is an abbreviation for the integral ∫∞-∞ψ*mυ(x)ψl dx, called the matrix element of the perturbing potential υ(x) between the states m and l. l =1 −∞ l =1 −∞
- 14. • (This term can also be written in Dirac notation as <m|υ(x)|l>.) • Substitution into Eq.(12.9) now yields • and after rearranging terms, • Equation (12.10) is exact. No approximations have been used in deriving it. However, it contains too many unknown quantities to permit an exact solution in most cases.
- 15. • The most important of the unknown quantities is the perturbed energy of the nth level, En’,or more precisely,Δ En = En’ - En . So we let m=n in Eq.(12.10) to obtain ∞ ' anl vnl = ann ( En − En ) • ∑ (12.10a) l =0 • To find a first approximation to this energy difference, we make the arbitrary assumption that anl = 1 if l = n and anl = 0 if l ≠ n. In that case, the lefthand side of Eq.(12.10a) collapses to a single term: vnn. • Thus Eq.(12.10) is reduced to the approximation
- 16. • This is a first-order approximation to the perturbed energy En’. You can recognize this integral from the formula for the expectation value of an operator. • This formula is not exact because the integral contains the wave function of the unperturbed system rather than the actual wave function.
- 17. Applications of perturbation theory Perturbation theory is an important tool for describing real quantum systems, as it turns out to be very difficult to find exact solutions to the Schrödinger equation for Hamiltonians of even moderate complexity. The Hamiltonians to which we know exact solutions, such as the hydrogen atom, the quantum harmonic oscillator and the particle in a box, are too idealized to adequately describe most systems. Using perturbation theory, we can use the known solutions of these simple Hamiltonians to generate solutions for a range of more complicated systems.
- 18. RefeRenceS Reference books: phySical chemiStRy - Skoog , holleR phySical chemiStRy (3 Rd edition) =Silbey & albeRty phySical chemiStRy(6 th edition) =atkinS p.w.
- 19. • • The End Thank You for Your Attention!