Notes on Perturbation Theory, Lecture notes of Physics

Download Notes on Perturbation Theory and more Lecture notes Physics in PDF only on Docsity! Lecture 17 Perturbation Theory 147 148 LECTURE 17. PERTURBATION THEORY 17.1 Introduction So far we have concentrated on systems for which we could find exactly the eigenvalues and eigenfunctions of the Hamiltonian, like e.g. the harmonic oscillator, the quantum rotator, or the hydrogen atom. However the vast majority of systems in Nature cannot be solved exactly, and we need to develop appropriate tools to deal with them. Perturbation theory is extremely successful in dealing with those cases that can be mod- elled as a “small deformation” of a system that we can solve exactly. Let us translate the above statement into a precise mathematical framework. We are going to consider systems that have an Hamiltonian: Ĥ = Ĥ0 + ￿V̂ , (17.1) where Ĥ0 is the Hamiltonian of the unperturbed system, ￿ is a small parameter, and V̂ is the potential describing the perturbation. We shall assume that the perturbation V is independent of time. Let us also assume that we can solve the time-independent Schrödinger equation for Ĥ0, i.e. that we know its eigenvalues and eigenfunctions: Ĥ0ψ (n)(x) = E(n)ψ(n)(x) . (17.2) For simplicity we start by considering the case where all the unperturbed levels E(n) are not degenerate. 17.2 Perturbative solution Let us discuss the solution of the time-independent Schrödinger equation for the full Hamil- tonian H. The eigenvalue equation reads: Ĥψ(x) = Eψ(x) . (17.3) Since ￿ is a small parameter, we shall expand the solution of Eq. (17.3) as a Taylor series in ￿: ψ(x) = ψ0(x) + ￿ψ1(x) + ￿ 2ψ2(x) + . . . , (17.4) E = E0 + ￿E1 + ￿ 2E2 + . . . . (17.5) Plugging Eqs. (17.4) and (17.5) into Eq. (17.3), we obtain: ￿ Ĥ0 + ￿V̂ ￿￿ ψ0(x)+￿ψ1(x) + ￿ 2ψ2(x) + . . . ￿ = ￿ E0 + ￿E1 + ￿ 2E2 + . . . ￿￿ ψ0(x) + ￿ψ1(x) + ￿ 2ψ2(x) + . . . ￿ . (17.6) We can now solve Eq. (17.6) order by order in ￿. 17.3. DEGENERATE LEVELS 151 which yields the correction of order ￿L to the unperturbed energy level. Following the computation above we also obtain: ￿ψ(m)|ψL￿ = ￿ψ(m)|V |ψL−1￿ E(n) − E(m) − 1 E(n) − E(m) L−1￿ K=1 EK￿ψ (m) |ψL−K￿ . (17.21) Using Eq. (17.21) for L = 2 we find the second-order correction to the n-th energy level: E2 = ￿ψ0|V̂ |ψ1￿ = ￿ m ￿=n ￿ψ(n)|V̂ |ψ(m)￿￿ψ(m)|V̂ |ψ(n)￿ E(n) − E(m) . (17.22) 17.3 Degenerate levels Equation (17.15) shows that the correction to the energy eigenfunctions at first order in perturbation theory is small only if ￿ψ(m)|V̂ |ψ(n)￿ E(n) − E(m) ￿ 1 . (17.23) If the energy splitting between the unperturbed levels is small compared to the matrix element in the numerator, then the perturbation becomes large, and the approximation breaks down. In particular, if there are degenerate levels, the denominator is singular, and the solution is not applicable. Let us see how we can deal with a g0-fold degenerate level of the unperturbed Hamiltonian. We shall denote P the projector onto such level, and Q the projector orthogonal to this level. The first-order equation: ￿ Ĥ0 − E0 ￿ ψ1 + ￿ V̂ − E1 ￿ ψ0 = 0 (17.24) can be projected using P onto the space spun by the degenerate states: P ￿ V̂ − E1 ￿ ψ0 = 0 . (17.25) Choosing a basis for the space of degenerate levels, we can write ψ0 as: ψ0 = g0￿ i=1 ciφi , (17.26) and then rewrite Eq. (17.25): ￿φi|V̂ |φj￿cj = E1ci , (17.27) i.e. E1 is an eigenvalue of the matrix Vij = ￿φi|V̂ |φj￿. This equation has g0 roots (not necessarily distinct), and generalizes Eq. (17.10) to the case of degenerate levels. If the eigenvalues are indeed all distinct, then the degeneracy is completed lifted. If some of the eigenvalues are equal, the degeneracy is only partially lifted. 152 LECTURE 17. PERTURBATION THEORY Example A well-known example of degenerate perturbation theory is the Stark effect, i.e. the separation of levels in the H atom due to the presence of an electric field. Let us consider the n = 2 level, which has a 4-fold degeneracy: |2s￿, |2p, 0￿, |2p,+1￿, |2p,−1￿ . (17.28) The electric field is chosen in the z-direction, hence the perturbation can be written as: V = −ezE , (17.29) where E is the magnitude of the electric field. We need to compute the matrix Vij in the subspace of the unperturbed states of the H atom with n = 2. This is a 4× 4 Hermitean matrix. Note that the perturbation V is odd under parity, and therefore it has non-vanishing matrix elements only between states of opposite parity. Since the eigenstates of the H atom are eigenstates of L2 and Lz, we find that only the matrix elements between s and p states can be different from zero. Moreover, V commutes with Lz and therefore only matrix elements between states with the same value of Lz are different from zero. So we have proved that the only non-vanishing matrix elements are ￿2s|V̂ |2p, 0￿ and its Hermitean conjugate. Hence the matrix V is given by:   0 3eEa0 0 0 3eEa0 0 0 0 0 0 0 0 0 0 0 0   , (17.30) where a0 is the Bohr radius. We see that the external field only removes the degeneracy between the |2s￿, and the |2p, 0￿ states; the states |2p,±1￿ are left unchanged. The two other levels are split: E = E2 ± 3ea0E . (17.31) 17.4 Applications There are numerous applications of perturbation theory, which has proven to be a very effective tool to gain quantitative information on the dynamics of a system whenever a small expansion parameter can be identified. Here we discuss briefly two examples. 17.4. APPLICATIONS 153 17.4.1 Ground state of Helium We can now attempt to incorporate the effect of the inter-electron Coulomb repulsion by treating it as a perturbation. We write the Hamiltonian as Ĥ = Ĥ0 + Ĥ ￿ where Ĥ0 = Ĥ1 + Ĥ2 and Ĥ ￿ = e2 4π￿0|r1 − r2| The ground state wavefunction that we wrote down earlier is an eigenfunction of the unperturbed Hamiltonian, Ĥ0; Ψ(ground state) = u100(r1)u100(r2)χ0,0. To compute the first order correction to the ground state energy, we have to evaluate the expectation value of the perturbation, Ĥ ￿, with respect to this wavefunction; ∆E1 = e2 4π￿0 ￿ u∗100(r1)u ∗ 100(r2)χ ∗ 0,0 1 r12 u100(r1)u100(r2)χ0,0 dτ1dτ2 The scalar product of χ0,0 with its conjugate = 1, since it is normalised. Putting in the explicit form of the hydrogenic wavefunction from Lecture 10 u100(r) = 1 √ π (Z/a0) 3/2 exp(−Zr/a0) thus yields the expression ∆E1 = e2 4π￿0 ￿ Z3 πa30 ￿2 ￿ 1 r12 exp{−2Z(r1 + r2)/a0} dτ1dτ2 Amazingly, this integral can be evaluated analytically. See, for example, Bransden and Joachain, Introduction to Quantum Mechanics, pp 465-466. The result is ∆E1 = 5 4 Z Ry = 5 2 Ry = 34 eV giving for the first-order estimate of the ground state energy E1 = −108.8 + 34 eV = −74.8 eV = −5.5 Ry to be compared with the experimentally-measured value of −78.957 eV .