7_Perturbation_Theory

7_Perturbation_Theory - VII. Time Independent Perturbation...

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Unformatted text preview: VII. Time Independent Perturbation Theory Griffiths Chapter 6 1 A. Nondegenerate Perturbation Theory 1. Formulation Suppose we have solved the time independent Schrödinger equation for some potential. 0 H 0 |n0 ￿ = En |n0 ￿ Therefore we know the eigenvalues En0 and the orthonormal eigenfunctions 〈n0|m0〉 = δnm. Now we perturb the potential slightly. What are the new eigenvalues and eigenfunctions? 2 A. Nondegenerate Perturbation Theory In general, we won’t be able to solve the new Schrödinger equation, but we can use the old eigenvalues and eigenfunctions to get an approximate solution. H |n￿ = En |n￿ new Hamiltonian Let’s write the new Hamiltonian, corresponding to the perturbed potential, in terms of the old Hamiltonian. H = H0 + H1 Next, we assume the new eigenvalues and eigenfunctions can be expanded about the old ones. The s uperscripts g ives the power of m atrix element of H1 t hat we expect t he term to come out proportional to. 0 1 2 E n = En + En + En + · · · |n￿ = |n0 ￿ + |n1 ￿ + |n2 ￿ + · · · 3 A. Nondegenerate Perturbation Theory 0 1 2 E n = En + En + En + · · · Substitute everything in... H = H0 + H1 H |n￿ = En |n￿ |n￿ = |n0 ￿ + |n1 ￿ + |n2 ￿ + · · · 2. Zeroth Order Theory 0 H 0 |n0 ￿ = En |n0 ￿ We recover the u nperturbed system 3. First Order Theory 1st order in the m atrix element of H1, as expected. See board VII.A.3 1 En = ￿n0 |H 1 |n0 ￿ The first order correction to the e nergy is the expectation value o f t he perturbation in the unperturbed s tate. ￿ ￿m0 |H 1 |n0 ￿ |n1 ￿ = |m0 ￿ 0 0 (En − Em ) m￿=n 4 A. Nondegenerate Perturbation Theory 4. An Example Griffiths Problem 6.1: Suppose we put a delta function bump in the centre of an infinite square well: H 1 = αδ (x − a/2) where α is a constant. (a) Find the first order correction to the allowed energies. Explain why the energies are not perturbed for even n. (b) Find the first three non-zero terms in the expansion of the correction to the ground state wave function. Solution: See board VII.A.4 5 A. Nondegenerate Perturbation Theory 5. Second Order Energies The second order energies are found, using the same method, to be ￿ |￿m0 |H 1 |n0 ￿|2 2 En = 0 0 En − Em See board VII.A.5 m￿=n Most books stop here (Shankar, Griffiths, for example) ignoring the 2nd order wave functions. We’ll stop here too. For those interested, Liboff gives the 2nd order wave functions. 6 A. Nondegenerate Perturbation Theory 6. Another Example Griffiths Problem 6.4: Suppose we put a delta function bump in the centre of an infinite square well: 1 H = αδ (x − a/2) where α is a constant. Find the second order correction to the energies. Solution: See board VII.A.6 7 B. Degenerate Perturbation Theory Following Griffiths, we’ll focus on energy corrections for this case. 1. The Problem We found the first order perturbation correction for the wave function was ￿ ￿m0 |H 1 |n0 ￿ 1 |n ￿ = |m0 ￿ 0 0 (En − Em ) m￿=n If two different states, |n0〉 and |m0〉 have the same energy, En0 = Em0, then this expression blows up. If the first order wave function expression is not reliable in this case, there’s no reason to trust the first order energy either. 8 B. Degenerate Perturbation Theory 2. The Solution Instead of using the basis set {|n0〉}, let’s transform to a different basis set where the energy denominator doesn’t blow up. In other words, we’ll find a basis set that is nondegenerate. Then, we can use non-degenerate perturbation theory without a problem. This is possible because perturbations lift the degeneracy of unperturbed states. 9 B. Degenerate Perturbation Theory This is possible because perturbations lift the degeneracy of unperturbed states. 10 B. Degenerate Perturbation Theory Operationa y: 1. Find the matrix elements of H1 with respect to the degenerate elements of {|n0〉}. • H1nm = 〈n0|H1|m0〉. 2. Diagonalize the matrix H1nm. • That is, find the eigenvalues and eigenvectors. 3. The eigenvalues are the first order energy corrections, En1. The eigenvectors are the “good” zeroth order wave functions. Justification: See board VII.B.2 11 B. Degenerate Perturbation Theory Mathematically, this corresponds to choosing a new basis set, built from the unperturbed set {|n0〉}, that diagonalizes the matrix H1mn. |n￿ = ¯ ￿ i ain |i0 ￿ 1 ￿n|H 1 |m￿ = Hnm δnm ¯ ¯ ¯¯ ¯¯ ￿m|n￿ = δmn ¯¯ ¯¯ As shown on board VII.B.2 This gives for the first order energy correction: 1 En = ￿n|H 1 |n￿ ¯ ¯ This is the same form of energy correction we had for nondegenerate perturbation theory. If we used the unperturbed basis set {|n0〉} however instead of {|n〉}, we would get the wrong answer. 12 B. Degenerate Perturbation Theory 3. An Example Griffiths Example 6.2: Consider the three dimensional infinite cubic well. Let’s introduce the perturbation H= 1 ￿ V0 0 , , if 0 < x < a/2 and 0 < y < a/2; otherwise. If we calculate the first excited state energy corrections using non-degenerate perturbation theory, we get V0/4 for all three states. Using degenerate perturbation theory, we find, correctly, that the three states all have different energies. Solution: See board VII.B.3 13 B. Degenerate Perturbation Theory 4. An Easier Way Choose a Hermitian operator A that commutes with both H0 and H1 (if you can find one). Use simultaneous eigenfunctions of A and H0 as the basis set |n〉. This set will automatically diagonalize H1. The energies are then easily found by using 1 En = ￿n|H 1 |n￿ ¯ ¯ Proof: See board VII.B.4 Example: The Fine Structure of Hydrogen. 14 C. The Fine Structure of Hydrogen The Bohr Formula gave us the spectrum of hydrogen. ￿ ￿ 2 ￿2 ￿ 2 E1 mc e 1 En = 2 = − , n = 1, 2, 3, · · · 2c2 4π￿ 2 n 2￿ n 0 on the order of α2mc2 e2 1 α≡ ≈ 4π￿0 ￿c 137 The Bohr energy is on the order of α2mc2. 15 C. The Fine Structure of Hydrogen The Bohr energy is on the order of α2mc2. Relativistic corrections to the electron in the hydrogen atom are on the order of α4mc2. Spin-orbit coupling corrections are also on the order of α4mc2. Taken together, these two corrections give the fine structure of the hydrogen atom. Higher order corrections, that we will not study, include the Lamb shift on the order of α5mc2 and the hyperfine structure, on the order of (m/mp)α4mc2. 16 C. The Fine Structure of Hydrogen 1. The Relativistic Correction The Hamiltonian for the electron in the hydrogen atom is written, as always, as the kinetic energy plus the potential energy: H = T+V. P2 non-relativistic kinetic energy T= 2m Correcting this kinetic energy for the electron using special relativity gives, to lowest order, P2 P4 T= − + ··· 3 c2 2m 8m with special relativity Therefore our first order perturbation can be taken as P4 1 denotes correction due to relativity Hr = − 3 2 8m c See board VII.C.1.a 17 C. The Fine Structure of Hydrogen We can put this perturbation into our perturbation theory formula: 1 En = ￿n|H 1 |n￿ ¯ ¯ 1 Hr P4 =− 3 2 8m c What basis set should we use? Instead of diagonalizing the matrix, we can note that Lz and L2 are both operators that commute with Hr1 and with the unperturbed hydrogen Hamiltonian H0 that we previously derived. Therefore the eigenfunctions of L2 and Lz will automatically diagonalize Hr1 and these form our “good” basis set. We found these eigenfunctions when we solved the unperturbed hydrogen atom. They are just |nlm〉 or ψnlm(r). Thus the unperturbed wave functions of the hydrogen atom form a “good” basis set, which is to say, n, l, m are good quantum numbers for this problem. See board VII.C.1.b C. The Fine Structure of Hydrogen The first order, relativistic, correction to the Bohr energies is 1 Er = − (En ) 2mc2 2 ￿ 4n l + 1 /2 ￿ −3 See board VII.C.1.c 19 C. The Fine Structure of Hydrogen 2. Spin-Orbit Coupling Imagine an electron “orbiting” a proton. From the electron frame, the proton is orbiting the electron. The moving positive charge sets up a magnetic field that the “spinning” electron feels as a torque. This tends to align the magnetic moment μ of the electron in the direction of the magnetic field B. The Hamiltonian of the electron due to this effect will be H = −µ · B 20 C. The Fine Structure of Hydrogen H = −µ · B The magnetic field created by the proton is found using the Biot-Savard law. It is: 1 e L B= See board VII.C.2.a 2 r3 4π￿0 mc orbital angular momentum of the electron The magnetic dipole moment of the electron is e µ=− S m See board VII.C.2.b spin angular momentum of the electron Combining these gives us the first order perturbation correction for the hydrogen atom electron due to spin-orbit coupling. ￿ 2￿ denotes correction due e 1 1 Hso = S · L t o spin-orbit coupling 2 c2 r 3 8π￿0 m 21 C. The Fine Structure of Hydrogen We can put this perturbation into our perturbation theory formula: 1 Hso = ￿ 2 e 8π￿0 ￿ 1 S·L 2 c2 r 3 m 1 En = ￿n|H 1 |n￿ ¯ ¯ What basis set should we use? L2 also commutes with this spin-orbit H1so, (so l is a good quantum number), but Lz does not commute, so m (or ml) is not a good quantum number. S2 commutes, as does J2 and Jz. Taken together, J2, Jz, L2, S2 form a complete set of commuting observables, that is, they share eigenfunctions. They also all commute with H0. Therefore the common eigenfunction of J2, Jz, L2, S2, |jmjls〉 are our basis set. Happily, we know these from solving the L2 eigenvalue equation. See board VII.C.2.c 22 C. The Fine Structure of Hydrogen The first order, spin-orbit, correction to the Bohr energies is 1 Eso = (En ) mc2 2 ￿ n[j (j + 1) − l(l + 1) − 3/4] l(l + 1/2)(l + 1) ￿ See board VII.C.2.d Combining this with the relativistic first order correction gives ￿ ￿ denotes total fine 2 (En ) 4n 1 s tructure correction Ef s = 3− 2mc2 j + 1/2 See Griffiths problem 6.17 Combining this with the Bohr formula gives the corrected energy levels for the hydrogen atom. ￿ ￿ ￿￿ 2 13.6eV α n 3 Enj = − 1+ 2 − See board 2 n n j + 1/2 4 23 VII.C.2.e C. The Fine Structure of Hydrogen Enj ￿ ￿ ￿￿ 2 13.6eV α n 3 =− 1+ 2 − 2 n n j + 1/2 4 The fine structure breaks the degeneracy in l so that for a given n, different l no longer all have the same energy. It preserves degeneracy in j. The good quantum numbers are n, l, s, j, and mj. 24 D. The Zeeman Effect The energy levels of an atom placed in an external magnetic field Bext will be shifted. This is called the Zeeman Eect. This is represented by the Hamiltonian HZ = −(µl + µs ) · Bext magnetic dipole e µl = − L moment due to the 2m e lectron orbital motion magnetic dipole moment due to the e lectron spin The Hamiltonian therefore becomes e HZ = (L + 2S) · Bext 2m 25 µs = − e S m D. The Zeeman Effect e HZ = (L + 2S) · Bext 2m The nature of the Zeeman effect for a hydrogen atom in a magnetic field depends on the relative strengths of the external field compared with the internal magnetic field. 1 e B= 4π￿0 mc2 r3 L For Bext << Bint, fine structure dominates and the Zeeman effect can be treated as a perturbation on the fine structure results. For Bext >> Bint , then the Zeeman effect dominates and fine structure should be treated as a perturbation on the Zeeman results. 26 D. The Zeeman Effect 1. Weak-Field Zeeman Eect In the weak field, the Zeeman Hamiltonian can be treated as a perturbation on the solved fine structure hydrogen atom. The set of operators, J2, Jz, L2, S2 from the the fine structure hydrogen atom commute with the Zeeman Hamiltonian, so the appropriate wave functions for our perturbation theory are |jmjls〉. See Griffiths problem 6.16 The formula for the first order corrections to the energy of the fine structure hydrogen atom due to a weak external magnetic field would be added n to match Griffiths 1 1 EZ = ￿nljmj |HZ |nljmj ￿ left off s as a spectator variable 27 1 HZ D. The Zeeman Effect e = (L + 2S) · Bext 2m 1 1 EZ = ￿nljmj |HZ |nljmj ￿ e = Bext · ￿nljmj |L + 2S|nljmj ￿ 2m Solving this, we find an energy correction of choosing z-axis along 1 EZ 1 EZ where and = µB gJ Bext mj t he direction of the m agnetic field e￿ µB = = 5.788 × 10−5 eV/T 2m j (j + 1) − l(l + 1) + 3/4 gJ = 1 + 2j (j + 1) See board VII.D.1.a 28 Bohr magneton Landé g-factor D. The Zeeman Effect Adding this energy correction to the total, fine structure corrected, Bohr energy formula, gives ￿ ￿ ￿￿ 2 −13.6eV α n 3 Enjmj = 1+ 2 − + µB gJ Bext mj 2 n n j + 1/2 4 The non-degenerate ground state of the hydrogen atom splits into two when placed in such a magnetic field. m = +1/2 j ￿ E1,1/2,±1/2 = −13.6eV 1 + 2 α 4 ￿ ± µB Bext mj = -1/2 See board VII.D.1.b 29 D. The Zeeman Effect 2. Strong-Field Zeeman Eect Let’s assume that Bext >> Bint. In fact, let’s begin by taking Bint = 0. The Zeeman Hamiltonian e HZ = (L + 2S) · Bext 2m becomes just eBext HZ = (Lz + 2Sz ) 2m if we let Bext lie along the z-axis. The energy levels of the hydrogen atom due to this PaschenBack Eect, ignoring fine structure, are Perturbation theory Enml ms 13.6eV =− + µB Bext (ml + 2ms ) 2 n 30 n ot used here! See board VII.D.2.a D. The Zeeman Effect Griffiths then adds fine structure as a perturbation. He doesn’t fully justify his use of basis set – he proceeds as if Lz and Sz commute with H1so. He does this in an average sense. He is using “almost good” quantum numbers. Note his typo however in equation 6.81. See board VII.D.2.b Overall, Griffiths gets 1 Efs = 13.6eV 2 α 3 n ￿ ￿ 3 l(l + 1) − ml ms − 4n l(l + 1/2)(l + 1) ￿￿ See Griffiths problem 6.22 31 ...
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This note was uploaded on 09/28/2011 for the course PHYS 334 taught by Professor Resch during the Spring '08 term at Waterloo.

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