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Chapter 3 Approximation Methods in QM Contents 3.1 Time independent PT (nondegenerate) . . . . . . . . . . . . . . . 52 3.2 Degenerate perturbation theory (PT) . . . . . . . . . . . . . . . . . 59 3.3 Time dependent PT and Fermi’s golden rule . . . . . . . . . . . 62 3.4 Variational method and helium atom . . . . . . . . . . . . . . . . . 67 Basic Questions A. What causes fine strucutre in hydrogen spectra? B. What is Fermi’s golden rule? C. What is the variational principle? 51 52 3.1 3.1.1 CHAPTER 3. APPROXIMATION METHODS IN QM Time-independent PT (nondegenerate) Introduction Two quantum problems can be solved exactly, they are harmonic oscillator and hydrogen (or hydrogen-like) atom. Most other quantum problems can not be solved exactly. One has to develop approximate method to solve such problem. Consider a Hamiltonian Ĥ, its Schrödinger eq. ĤΨp = Ep Ψp , p = quantum no. is difficult to solve. If Ĥ can be written as Ĥ = Ĥ0 + V̂ where V̂ is time-independent and ”small”, and the Schrödinger eq. of Ĥ0 has been solved Ĥ0 Φk = ek Φk → Ĥ0 |Φk i = ek |Φk i where k is the quantum number labelling the nondegenerate states. We wish to use the information of Ĥ0 to find the eigenstates and eigenvalues of Ĥ. Ĥ0 is usually referred to as unperturbed Hamiltonian, V̂ as perturbation potential (operator). In principle (the completeness principle), any state Ψp can be written as a linear combination of Φk as X |Ψp i = Cpk |Φk i k where {Cpk } are constants to be determined by eq. of Ĥ, together with eigenvalue Ep . Since the dimensionality (total number of the possible k) is in general infinite, the exact solution is not easy. However, since the perturbation V̂ is assumed small, it is not difficult to get approximate solution for |Ψp i and Ep . 3.1.2 Perturbation expansion Obviously, if V̂ is zero, p = k, |Ψk i = |Φk i, Ek = ek . Hence, for non-zero V̂ , this solution {p = k, |Ψk i = |Φk i , Ek = ek } is referred to as zero-order perturbation results. We need to go beyond this, and find nontrivial correction to this zero-order results. In order to do it in a systematic way, we introduce a parameter λ in the Hamiltonian Ĥ (λ) = Ĥ0 + λV̂ , 0 ≤ λ ≤ 1 so that as λ → 0, solutions for Ĥ (λ) is reduced to that of Ĥ0 and what we want is the results in the limit λ = 1. In perturbation theory, we assume p=k 3.1. TIME-INDEPENDENT PT (NONDEGENERATE) 53 i.e., the quantum number for the states of Ĥ is same as that of Ĥ0 . Therefore, we can expand each of the eigenvalues and eigenstates of Ĥ (λ) in terms of powers of λ as (0) (1) (2) (3) (0) (1) (2) (3) Ek = Ek + λEk + λ2 Ek + λ3 Ek + · · · Ψk = Ψk + λΨk + λ2 Ψk + λ3 Ψk + · · · (0) where Ek = ek is the zero-order approximation to the k-th eigenvalue of Ĥ, and (1) (2) (3) (0) Ek , Ek , Ek , · · · are the successive correction to this. Similarly, Ψk = Φk is the (1) (2) (3) zero-order appro. to k-th eigenstate of Ĥ and Ψk , Ψk , Ψk , · · · are the successive correction to this. (1) (2) (3) (1) (2) (3) In order to find Ek , Ek , Ek , · · · and Ψk , Ψk , Ψk , · · ·, we substitute the expansions into the original Schrödinger eq. ĤΨk = Ek Ψk or Ĥ0 + λV̂ = (0) (0) (1) (2) Ψk + λΨk + λ2 Ψk + · · · (1) (2) Ek + λEk + λ2 Ek + · · · (0) (1) (2) Ψk + λΨk + λ2 Ψk + · · · or, each side is arranged according to the powers of λ (0) (0) (1) Ĥ0 Ψk + λ V̂ Ψk + Ĥ0 Ψk (0) (0) (1) (0) (1) (2) + λ2 V̂ Ψk + Ĥ0 Ψk (0) (1) = E k Ψk + λ Ek Ψk + E k Ψk (2) (0) +··· (1) (1) (0) (2) + λ 2 Ek Ψk + E k Ψk + E k Ψk +··· Equating the coefficients of λ on both sides of the eq. (consider λ as an arbitrary number, this must be true), we have (0) (0) (0) Ĥ0 Ψk = Ek Ψk → (0) (0) Ψk = Φk , Ek = e k as zero-order result, and (0) (1) (1) (0) (0) (1) V̂ Ψk + Ĥ0 Ψk = Ek Ψk + Ek Ψk (1) (1) as first-order eq. for Ek and Ψk and (1) (2) (2) (0) (1) (1) (0) (2) V̂ Ψk + Ĥ0 Ψk = Ek Ψk + Ek Ψk + Ek Ψk (2) (2) as second-order eq. for Ek and Ψk etc. 54 3.1.3 CHAPTER 3. APPROXIMATION METHODS IN QM Solution for perturbation eq.: 1st-order Now consider the solution to the above sequence of the perturbation eqs. The 1storder eq. is (0) (1) (1) (0) (0) (1) V̂ Ψk + Ĥ0 Ψk = Ek Ψk + Ek Ψk or, using the zero-order results (1) V̂ − Ek (1) Φk = ek − Ĥ0 Ψk (1) By the completeness principle, we can always express Ψk in terms of a linear combination of Φk as X (1) (1) Ckk0 Φk0 Ψk = k0 (1) where Ckk0 are coefficients to be determined and the equation becomes, using Ĥ0 Φk0 = ek 0 Φk 0 (1) V̂ − Ek V̂ − (1) Ek Φk = X k0 |Φk i = X k0 (1) (ek − ek0 ) Ckk0 Φk0 (1) (ek − ek0 ) Ckk0 |Φk0 i (1) this is the eq. for Ek and Ckk0 . The standard technique to solve this kind eq. is to take inner product with state Φk00 on both sides of the eq., i.e., multiplying Φ∗k00 and integrate Z Z X (1) (1) d3 rΦ∗k00 V̂ − Ek Φk = (ek − ek0 ) Ckk0 d3 rΦ∗k00 Φk0 k0 using the orthonormal relation Z d3 rΦ∗k Φk0 = δkk0 we have (1) Vk00 k − Ek δkk00 = Vk00 k ≡ X Zk 0 (1) d3 rΦ∗k00 V̂ Φk Choosing k 00 = k we have (1) (1) (ek − ek0 ) Ckk0 δk00 k0 = (ek − ek00 ) Ckk00 Ek = Vkk = Z d3 rΦ∗k V̂ Φk 3.1. TIME-INDEPENDENT PT (NONDEGENERATE) Choosing k 00 6= k we have 55 Vk00 k . ek − ek00 (1) Ckk00 = Note that Ckk is still undetermined and can be chosen as zero. The 1st-order results are, letting λ = 1, Ek ≈ e k + (1) Ek = Vkk = Z d3 rΦ∗k V̂ Φk for the energy and Ψk ≈ Φ k + Vk 0 k Φk 0 0 e − e 0 k k k 6=k X for the state. Note: if we choose Ckk 6= 0, the coefficient of Φk in the above formula will not be unity and Ψk is then not properly normalized in the limit Vk0 k → 0. (1) Note that the determine of the 1st-order energy Ek only depends on the zero(0) (1) order state function Ψk = Φk , and the 1st-order statefunction Ψk is not necessary. (2) However, in order to determine the 2nd -order energy correction Ek , the 1st-order (1) statefunction correction Ψk is required. This is generally true: nth-order energy correction requires the (n − 1)th-order statefunction correction. 3.1.4 Solution for perturbation eq.: 2st-order The 2nd-order equation is (1) (2) (2) (0) (1) (1) (0) (2) V̂ Ψk + Ĥ0 Ψk = Ek Ψk + Ek Ψk + Ek Ψk or (2) (0) (1) Ek Ψk = V̂ − Ek (0) (1) (0) (1) (1) (0) Ψk + Ĥ0 − Ek (2) Ψk (2) (2) where Ek , Ek , Ψk and Ψk are known. We only need to determine Ek and Ψk . (2) Again, Ψk can be written as a linear combination of Φk (2) Ψk = X (2) Ckk0 Φk0 k0 substituting this into the eq., using the zero- and 1st-order results, we have (2) Ek Φk = X k0 = X k0 (1) (1) Ckk0 V̂ − Ek (1) (1) Ckk0 V̂ − Ek Φk 0 + X k0 Φk 0 + X k0 (2) Ckk0 Ĥ0 − ek Φk0 (2) Ckk0 (ek0 − ek ) Φk0 56 CHAPTER 3. APPROXIMATION METHODS IN QM as before, taking inner product with state Φk00 on both sides of the eq. and using the orthonormality of Φk we have (2) Ek δk00 k = X k0 (1) (1) (2) (1) Ckk0 V̂k00 k0 − Ek δk00 k0 + Ckk00 (ek00 − ek ) . (2) Choosing k 00 = k, we obtain Ek (2) Ek = X k0 = X (1) Ckk0 Vkk0 − Ek δkk0 (1) Ckk0 Vkk0 , ← k0 = (1) Ckk = 0 X |Vkk0 |2 Vkk0 Vk0 k = 0 0 k 0 6=k ek − ek k 0 6=k ek − ek X (2) where we have assume V̂ is a Hermitian operator. The coefficient Ckk00 can be deter(2) mined by choosing k 00 6= k, as before. Again, to determine Ek we only need to know (1) (2) Ckk0 and Ckk00 is not necessary. The 2nd-order result for the energy is then given by (0) (1) (0) (1) (2) Ek ≈ Ek + Ek + Ek = ek + Vkk + Ψk ≈ Ψ k + Ψ k = Φ k + with 0 Vk0 k ≡ hk | V̂ |ki = 3.1.5 |Vkk0 |2 0 k 0 6=k ek − ek X Vk 0 k Φk 0 0 k 0 6=k ek − ek X Z d3 r Φ∗k0 V̂ Φk . Example Consider an anharmonic oscillator with Hamiltonian Ĥ = Ĥ0 + V̂ 1 h̄2 ∂ 2 + mωx2 , Ĥ0 = − 2 2m ∂x 2 where α is a small number (α h̄ω/x30 = h̄ω been solved, 1 Ĥ0 |ni = h̄ω n + |ni , 2 1 Ĥ0 = h̄ω ↠â + 2 mω 2h̄ 3/2 V̂ = αx3 ). The eigenequation of Ĥ0 has n = 0, 1, 2, · · ·, ∞ 3.1. TIME-INDEPENDENT PT (NONDEGENERATE) 57 and we know the effect of ↠and â on the normalized state |ni √ √ ↠|ni = n + 1 |n + 1i , â |ni = n |n − 1i for n > 0 . we hence need to express V̂ in terms of â and ↠. Using the definition â ≡ r we have x̂ = hence V̂ ip̂ mω x̂ + √ , 2h̄ 2mh̄ω â ≡ s 1 p̂ = i h̄ â + ↠, 2mω h̄ = α 2mω !3/2 h̄ = α 2mω !3/2 h̄ = α 2mω !3/2 r † 3 â + ↠â + ↠â + ↠ip̂ mω x̂ − √ 2h̄ 2mh̄ω s mωh̄ â − ↠2 â2 + â↠+ ↠â + â†2 â2 + 2↠â + 1 + â†2 . Up to 2nd-order approximation, we have En ≈ en + Vnn + where Vnn0 Vn0 n e − e n0 n0 6=n n X 1 en = h̄ω n + 2 and Vnn = hn| V̂ |ni h̄ = α 2mω !3/2 hn| â + ↠h̄ = α 2mω = 0 !3/2 hn| â + ↠i.e., the first-order energy correction is zero. â2 + 2↠â + 1 + â†2 |ni â2 + 2n + 1 + â†2 |ni 58 CHAPTER 3. APPROXIMATION METHODS IN QM To determine the 2nd-order energy correction, we need to evaluate Vn0 n . Consider â + ↠= = â + ↠â + â † â2 + 2↠â + 1 + â†2 |ni â2 + 2n + 1 + â†2 |ni q n (n − 1) |n − 2i + (2n + 1) |ni + q (n + 1) (n + 2) |n + 2i q √ n (n − 1) (n − 2) |n − 3i + (2n + 1) n |n − 1i + (n + 1) (n + 2)2 |n + 1i q q √ + n (n − 1)2 |n − 1i + (2n + 1) n + 1 |n + 1i + (n + 1) (n + 2) (n + 3) |n + 3i q q √ = n (n − 1) (n − 2) |n − 3i + (2n + 1) n + n (n − 1)2 |n − 1i + q q √ + (n + 1) (n + 2)2 + (2n + 1) n + 1 |n + 1i + (n + 1) (n + 2) (n + 3) |n + 3i = q = q n (n − 1) (n − 2) |n − 3i + 3n3/2 |n − 1i + 3 (n + 1)3/2 |n + 1i q + (n + 1) (n + 2) (n + 3) |n + 3i therefore, there are only four nonzero contribution to Vn0 n !3/2 q Vn−3,n h̄ = hn − 3| V̂ |ni = α 2mω Vn−1,n h̄ = hn − 1| V̂ |ni = α 2mω !3/2 3n3/2 Vn+1,n h̄ = hn + 1| V̂ |ni = α 2mω !3/2 3 (n + 1)3/2 Vn+3,n h̄ = hn + 3| V̂ |ni = α 2mω !3/2 q n (n − 1) (n − 2) (n + 1) (n + 2) (n + 3) therefore 2 2 2 2 Vn−1,n Vn+1,n Vn+3,n Vn−3,n Vnn0 Vn0 n + + + = e − e n0 en − en−3 en − en−1 en − en+1 en − en+3 n0 6=n n X h̄ 1 2 = α 3h̄ω 2mω !3 n (n − 1) (n − 2) h̄ 1 + α2 h̄ω 2mω !3 9n3 h̄ 1 − α2 h̄ω 2mω !3 9 (n + 1)3 3.2. DEGENERATE PT 59 h̄ 1 2 α − 3h̄ω 2mω 1 2 h̄ = α 3h̄ω 2mω !3 (n + 1) (n + 2) (n + 3) !3 h i × n (n − 1) (n − 2) + 27n3 − 27 (n + 1)3 − (n + 1) (n + 2) (n + 3) h̄ 1 = − α2 h̄ω 2mω !3 30n2 + 30n + 11 Hence, up to 2nd-order, the energy value is 1 1 2 h̄ En = h̄ω n + − α 2 h̄ω 2mω !3 30n2 + 30n + 11 and the energy interval between adjacent energy levels ∆ = En − En−1 60 2 h̄ = h̄ω − α h̄ω 2mω !3 n See figure 3.2 Degenerate PT In the above perturbation theory, the unperturbed states Φk are nondegenerate, ek0 6= ek for all k 0 6= k. Now we extend the theory to the degenerate case. Our Hamiltonian is still given by Ĥ = Ĥ0 + V̂ where the perturbed term V̂ is assumed small and the unperturbed Hamiltonian Ĥ0 has been solved Ĥ0 Φnd = en Φnd with quantum number d = 1, 2, · · ·, M , labelling the degeneracy. In the case of P P Hydrogen, we know for each n, we have degenerate qn d = (l, m), d = l,m and M 2 . For convenience we only need one letter for the notation. We expect that the 60 CHAPTER 3. APPROXIMATION METHODS IN QM introduction of perturbed term will lift the degeneracy in l, so the eigen equation for Ĥ is ĤΨnd = End Ψnd . But if we still carry out the similar analysis as before, we expect to encounter a factor such as Ĥ → Ĥ (λ) = Ĥ0 + λV̂ , and the expansion series (0) (1) (0) (1) End = End + λEnd + · · · Ψnd = Ψnd + λΨnd + · · · substituting these into the original Schrödinger eq. and equating the coefficients of λ on both sides of the eq. (consider λ as an arbitrary number, this must be true), we have (0) (0) (0) (0) (0) Ĥ0 Ψnd = End Ψn,d → Ψnd = Φnd , End = en as zero-order result, and (1) (0) (1) (0) (0) (1) V̂ Ψnd + Ĥ0 Ψnd = End Ψnd + End Ψnd as first-order eq. etc. All those formulae remain the same, only replacing single qn by double ones k → (n, d). However, there are divergence problem when we go beyond zeroth-order. Consider the 1st-order correction to the wavefunction (1) Ψnd = X (1) cnd,n0 d0 Φn0 d0 n 0 d0 = X d0 6=d (1) cnd,nd0 Φnd0 + other contributions from states with n0 6= n with the diverging coefficient (1) cnd,nd0 = Vnd,nd0 (0) (0) End − End0 = Vnd,nd0 . en − e n Solution: Recall the basic assumption in PT quantum number after perturbation = quantum number before perturbation. We need to modify this by introducing a new quantum number k as before perturbation : (n, d) after perturbation : (n, k) . This is equivalent to introduce a new basic set {Ωnk , k = 1, 2, · · ·, M }as Ωnk = M X d=1 bkd Φnd , k = 1, 2, · · ·, M 3.2. DEGENERATE PT 61 with coefficients bkd to be determined by the condition to avoid the divergence mentioned above. Now our new PT expansion series are (0) (1) (0) (1) Enk = Enk + λEnk + · · · Ψnk = Ψnk + λΨnk + · · · and the zeroth-order equation does not change much (0) Ĥ0 Ψnk = (0) (0) Enk Ψnk → (0) Enk (0) Ψnk = en , = Ωnk = M X bkd Φnd d=1 and the 1st-order eq. (1) (0) (0) (1) V̂ − Enk Ψnk = Enk − Ĥ0 Ψnk (1) where we expand Ψnk in terms of Ωnk (1) Ψnk = X (1) ckk0 Ωnk0 , k0 k = 1, 2, · · ·, M . As before, we substitute this into the 1st-order eq. and take inner product with state Ωnk00 on both sides, we have Z (1) Enk Vnk00 ,nk − (1) Enk δk00 k d 3 r Ω∗nk00 or V̂ − Take k 00 = k, we have Ωnk = = Z Z d3 r Ω∗nk00 d3 r Ω∗nk00 X k0 X k0 (1) (0) ckk0 Enk − Ĥ0 Ωnk0 (1) ckk0 (en − en ) Ωnk0 = 0 . (1) Enk = Vnk,nk and take k 00 6= k, we have Vnk00 ,nk = Z d3 r Ω∗nk00 V̂ Ωnk = 0, for all k 00 6= k . This last equation means operator V̂ is diagonal in the new basis. Therefore, the P coefficients {bkd } in Ωnk = M d=1 bkd Φnd is chosen so that V̂ is diagonal. Example: Spin-orbit coupling (or LS-coupling). An electron moving in a central potential V̂ = V (r) has Hamiltonian as Ĥ0 = − h̄2 2 ∇ + V (r) . 2m 62 CHAPTER 3. APPROXIMATION METHODS IN QM In general the eigenvalues Enl of Ĥ0 are 2 (2l + 1)-fold degenerate (for hydrogen, Enl = En , and degeneracy is 2n2 -fold). One of relativistic effects (Dirac) is the addition of a potential, the so-called spin-orbit coupling (or LS-coupling) V̂LS = A (r) L̂ · Ŝ . We can treat this V̂LS as perturbation potential and write 1 2 V̂LS = A (r) Ĵ2 − L̂2 −Ŝ , 2 Ĵ ≡ L̂ + Ŝ As we know, the eigenstates of Ĥ0 can be represented by the product of orbital wavefunctions and spin wavefunctions, or |nlmi |sms i = |nlml sms i. Within the first-order degenerate PT, we need to use these |nlml sms i to diagonalize the perturbation operator V̂LS and to obtain the zero-order wavefunction and first-order energy correction. We have already done so in the previous chapter. The zeroth-order wavefunction are angular momentum states |nlsjmi which are eigenstates of Ĵ2 and Jˆz with angular momentum quantum number, using s = 1/2 1 j = l ± , m = −j, −j + 1, · · ·, j − 1, j . 2 Hence, the energy level shift by LS-coupling is (1) ELS 3.3 3.3.1 3 h̄2 j (j + 1) − l (l + 1) − hA (r)inl . = hnlsjm| V̂LS |nlsjmi = 2 4 Time-dependent perturbation theory Newton’s equation As we have seen before, the time-dependent Schrödinger eq. is ih̄ ∂ Ψ (r, t) = ĤΨ (r, t) , ∂t −ih̄ ∂ ∗ Ψ (r, t) = ĤΨ∗ (r, t) . ∂t The expectation value of an arbitrary operator  =  (t), assumed Ψ (r, t) is normalized, Z D E  = hΨ|  |Ψi = d3 r Ψ∗ (r, t)  (t) Ψ (r, t) has the following evolution equation Z i d D E ∂ h ∗  = d3 r Ψ (r, t)  (t) Ψ (r, t) dt ∂t Z Z 1 ∂ Â Ψ (r, t) + = d3 r Ψ∗ (r, t) d3 r Ψ∗ (r, t) ÂĤ − Ĥ Â Ψ (r, t) ∂t ih̄ 3.3. TIME-DEPENDENT PERTURBATION THEORY or d D E  = dt * ∂  ∂t + + 63 iE 1 Dh Â, Ĥ . ih̄ Using this formula, Newton’s equation becomes, using Ĥ = p̂2 /2m + V̂ (r) D E d hp̂i = − ∇V̂ . dt 3.3.2 Time-dependent expansion Consider now the following Hamiltonian Ĥ = Ĥ0 + V̂ (t) where Ĥ0 is time-independent with eigen eqs. Ĥ0 Φn (r) = En Φn (r) , n = 1, 2, · · · and the time dependent part of the Hamiltonian is V̂ (t). The general time-dependent wavefunction Ψ (r, t) can be expanded in the following general form as Ψ (r, t) = X dn (t) Φn (r) = n X cn (t) Φn (r) e−iEn t/h̄ n and after substituting into the Schrödinger eq., we have X n ! X dcn Φn (r) e−iEn t/h̄ = En + V̂ cn Φn (r) e−iEn t/h̄ En cn + ih̄ dt n or ih̄ X n0 X dcn0 Φn0 (r) e−iEn0 t/h̄ = cn0 V̂ Φn0 (r) e−iEn0 t/h̄ . dt n0 Taking inner product with Φn (r) on both sides of eq., we have the time-evolution equation for cn (t) ih̄ dcn X = Vnn0 cn0 eiωnn0 t , dt 0 n with Vnn0 ≡ hΦn | V̂ |Φn0 i = Z d3 r Φ∗n V̂ Φn0 , n = 1, 2, · · · h̄ωnn0 = En − En0 . these are a set of first-order, coupled equations for {cn , n = 1, 2, · · ·}. Normally difficult to solve. For some special case, exact solution can be found. See Mandl, Quantum Mechanics, Section 9.2, P 198. 64 CHAPTER 3. APPROXIMATION METHODS IN QM Consider a typical initial condition: at t = 0, the system is in one of the eigenstate, say, |1i, namely cn (0) = δ1n . For V̂ (t) = 0, the amplitudes cn (t) retain this initial value at all times and Ψ (r, t) = X cn (t) Φn (r) e−iEn t/h̄ = Φ1 (r) e−iE1 t/h̄ n the time-dependent part is trivial. 3.3.3 Perturbation Theory Now we treat V̂ as perturbation potential and write the Hamiltonian as before Ĥ = Ĥ0 + λV̂ (t) and correspondingly (1) 2 (2) cn = c(0) n + λcn + λ cn + · · · the evolution equation for cn becomes ih̄ X (0) d (0) (1) (2) 2 (2) cn + λc(1) cn0 + λcn0 + λ2 cn0 + · · · λVnn0 eiωnn0 t . n + λ cn + · · · = dt n0 Setting the coefficients of each power of λ equal on both sides we have a series successive eqs., d (0) c = 0, c(0) n (t) = const. dt n X (0) d ih̄ c(1) = cn0 Vnn0 eiωnn0 t n dt n0 X d (1) = cn0 Vnn0 eiωnn0 t ih̄ c(2) n dt n0 ··· . ih̄ Consider the typical initial condition: applying the perturbation V̂ (t) after t > 0, namely ( = 0, t≤0 V̂ (t) 6= 0, t>0 and assume for t ≤ 0, the system is in one of the eigenstates of Ĥ0 , say |1i, cn (0) = c(0) n (t) = δ1n 3.3. TIME-DEPENDENT PERTURBATION THEORY 65 Hence the first-order equation becomes ih̄ d (1) X (0) cn = cn0 Vnn0 eiωnn0 t = Vn1 eiωn1 t dt n0 Integrating these equations from t = 0 to t > 0, subject to the condition that cn (0) = δ1n , we have (1) c1 (t) = 1 + c(1) n (t) = (1) 1 ih̄ Z t 0 V̂11 (t0 ) dt0 1 Zt 0 V̂n1 (t0 ) eiωn1 t dt0 , ih̄ 0 2 ωn1 = En − E 1 , h̄ n 6= 1 where c1 (t) is the probability (in 1st-order approximation) that the system at 2 t > 0 stays in the original state |1i, and c(1) n (t) is the probability (in 1st-order approximation) that the system at t > 0 is in another state |ni (n 6= 1). Example. A hydrogen atom is placed in a spatially homogeneous time-dependent electric field in the z-axis ( 0, t≤0 ε= −t/τ ε0 e , t>0 with τ > 0. At time t = 0, the atom is in the (1s) ground state with wavefunction and energy 1 −r/a0 4π0h̄2 e2 φ1s = q , a0 = . , E1s = − e 8π0 a0 me2 πa30 Find the probability, in the 1st order perturbation theory, that after a sufficiently long time the atom is in the (2p) excited state with wavefunction and energy φ2p = q 1 32πa50 r cos θ e−r/2a0 , E2p = E1s . 4 Solution: According to the 1st-order perturbation theory, the probability amplitude for the transition (1s) → (2p) is 1 c (t) = ih̄ with ω2p,1s = Z t 0 0 V̂2p,1s (t0 ) eiω2p,1s t dt0 E2p − E1s 3E1s =− h̄ 4h̄ and 0 0 V̂ (t0 ) = eε0 e−t /τ z = eε0 e−t /τ r cos θ . 66 CHAPTER 3. APPROXIMATION METHODS IN QM The matrix element V̂2p,1s (t0 ) is 0 V̂2p,1s (t0 ) = h2p| V̂ (t0 ) |1si = eε0 e−t /τ h2p| r cos θ |1si = eε0 e Z −t0 /τ 1 q r cos θ e−r/2a0 · q 32πa50 Z ∞ 1 πa30 Z π e−r/a0 r 2 sin θdrdθdφ 2π 1 2 4 −3r/2a0 √ dφ cos θ sin θdθ r e dr = eε0 e 0 0 4 2πa40 0 1 4!25 a50 2 215/2 −t0 /τ −t/τ √ = eε0 e · · · 2π = Ce , C ≡ 5 eε0 a0 35 3 3 4 2πa40 −t0 /τ Z and the amplitude C c (t) = ih̄ Z t 0 0 0 eiω2p,1s t −t /τ dt0 = C eiω2p,1s t−t/τ − 1 ih̄ iω2p,1s − 1/τ and the transition probability is |c(t)|2 = C 2 1 + e−2t/τ − 2e−t/τ cos(ω2p,1s t) , 2 ω2p,1s + 1/τ 2 h̄2 and after a sufficiently long time, t = ∞, the probability is |c (∞)|2 = 3.3.4 (eε0 a0 )2 215 . 310 (3E1s /4)2 + (h̄/τ )2 Fermi golden rule Now we consider the following simpler case, a step function perturbation potential V̂ (t) = ( 0, V̂ , t≤0 t>0 where V̂ is independent of time. What is the transition probability per unit time for the system originally in state |1i at t ≤ 0 to the state |2i at t > 0? Solution: From the above 1sr-order perturbation theory, the time-integration in the transition probability amplitude can be carried out, (1) c2←1 (t) = 1 ih̄ Z t 0 V2,1 1 − eiω2,1 t h̄ω2,1 E2 − E 1 = h̄ 0 V2,1 (t0 ) eiω2,1 t dt0 = V2,1 = h2| V̂ |1i , ω2,1 3.4. VARIATIONAL METHOD 67 and the probability is, using sin y = (eiy − e−iy )/2, 4 |V2,1 |2 sin2 (ω2,1 t/2) 2πt = 2 |V2,1 |2 D (ω2,1 , t) 2 2 ω2,1 h̄ h̄ 2 sin2 (xt/2) D (x, t) = , x = ω2,1 πt x2 2 (1) P2←1 (t) = c2←1 (t) = where function D (x, t) has a simple integral Z 2 D (x, t) dx = πt −∞ ∞ Z ∞ −∞ sin2 (xt/2) dx = 1, x2 using Z ∞ −∞ sin2 (ax) dx = πa, x2 similar to the continuous delta function δ (x). In the limit t → ∞, the two become identical. So we have the transition probability given by, for large t P2←1 (t) = 2πt 2πt 2 |V2,1 |2 δ (E2 − E1 ) 2 |V2,1 | δ (ω2,1 ) = h̄ h̄ and the probability per unit time Ṗ2←1 (t) = 2π |V2,1 |2 δ (E2 − E1 ) . h̄ This is the famous Fermi Golden rule. 3.4 3.4.1 Variational method Variational principle Variational principle (Rayleigh-Ritz): The expectation value of a given Hamiltonian Ĥ in any state Ψ is always greater or equal to the exact ground-state energy E 0 , hΨ| Ĥ |Ψi ≥ E0 hΨ |Ψi the equality holds if |Ψi is the exact ground state. Proof: Suppose Ĥ |Φn i = En |Φn i , n = 0, 1, 2, · · · E0 < En , for n > 0 68 CHAPTER 3. APPROXIMATION METHODS IN QM Write |Ψi as a linear combination of |Φn i ∞ X |Ψi = n=0 Cn |Φn i = C0 |Φ0 i + 2 hΨ |Ψi = |C0 | + ∞ X n=1 ∞ X n=1 Cn |Φn i |Cn |2 Therefore Ĥ |Ψi = Ĥ C0 |Φ0 i + = E0 C0 |Φ0 i + ∞ X n=1 ∞ X n=1 ! Cn |Φn i Cn En |Φn i and 2 hΨ| Ĥ |Ψi = E0 |C0 | + 2 ∞ X n=1 > E0 |C0 | + E0 so En |Cn |2 ∞ X n=1 2 2 |Cn | = E0 |C0 | + ∞ X n=1 2 |Cn | ! = E0 hΨ |Ψi hΨ| Ĥ |Ψi ≥ E0 . hΨ |Ψi In general, we do not know the exact ground state energy E0 . We wish to find an approximation for its value. By the variational principle, our approximation hĤi using any trial wavefunction |Ψi will always be greater than the exact value E0 . Hence, the lower the expectation value hĤi (i.e., the closer to the exact value), the better the trial wavefunction. A typical varitaional calculation is as follows: we first choose a trial wavefunction Ψ with a few variational parameters α1 , α2 , · · · Ψ (α1 , α2 , · · ·) and calculate the expectation value E (α1 , α2 , · · ·) = hΨ| Ĥ |Ψi . hΨ |Ψi Next, we minimize this E (α1 , α2 , · · ·) through variational equations ∂ E (α1 , α2 , · · ·) = 0 ∂α1 ∂ E (α1 , α2 , · · ·) = 0 ∂α2 ··· 3.4. VARIATIONAL METHOD 69 The solution for (α1 , α2 , · · ·) of these equations then put back to E (α1 , α2 , · · ·) and this minimum E represents the best estimate for the trial function of the form Ψ (α1 , α2 , · · ·). Clearly, the success of this nonperturbative method will depend crucially on the form of the trial wave functions we choose to start with. 3.4.2 Examples Hydrogen atom. The trial function for the ground state of Hydrogen is choose as Φα = Ce−αr , C= s α3 π the Hamiltonian is h̄2 2 e2 Ĥ = − ∇ − , 2m 4π0 r ∂ 1 ∂ r2 ∇ = 2 r ∂r ∂r 2 ! + 1  (θφ) r2 and the expectation value is Eα = hΦα | Ĥ |Φα i = e2 α h̄2 α2 − 2m 4π0 hence ∂Eα e2 m 1 =0 → α= 2 = ∂α a0 4π0h̄ where a0 is the Bohr radius. So the best estimate for the ground state energy is h̄2 1 1 e2 e2 α 1 Eα = = − − 2m a20 4π0 a0 2 4π0 a0 and for the ground state is Φ= q 1 a30 π e−r/a0 both are actually exact. This is not surprising because the trial function has the correct form. Helium atom. In this case we do not know the exact result and we are dealing with two identical particles. Consider in general Helium-like ions with the following Hamiltonian describing two electrons interacting each other and with a nucleus containing Z protons h̄2 2 h̄2 2 Ze2 Ĥ = − ∇1 − ∇2 − 2m 2m 4π0 1 1 e2 1 + + . r1 r2 4π0 r12 70 CHAPTER 3. APPROXIMATION METHODS IN QM We want to approximate the ground state of this system by a simple trial wavefunction (actually, PT), using the knowledge from hydrogen-like ions. Ze2 1 h̄2 ∇2 − 4π are We know the eigenstates φnlm (r) for a single electron Ĥ = − 2m 0 r s Z 3 −Zr/a0 e πa30 s Zr −Zr/a0 Z3 1 − φ2s (r) = R20 (r) Y00 (θ, φ) = e 8πa30 2a0 ... φ1s (r) = R10 (r) Y00 (θ, φ) = with φ1s (r) as the ground state, and eigenvalues Enl = En = − e2 Z 2 , 8π0 a0 n2 n = 1, 2, · · · . We also know electrons have spin degree of freedom, indicated by two states |↑i or |↓i. We want to construct the simplest two-body wavefunction from these single-body state. There is a (anti)symmetry requirement we need to impose. The following is a trial wavefunction satisfying this requirement, Ψ0 (1, 2) = φ1s (r1 ) φ1s (r2 ) χ12 , 1 χ12 = √ (|↑i1 |↓i2 − |↓i1 |↑i2 ) 2 hence Ψ (1, 2) = −Ψ (2, 1). This is a spin-singlet state. Another possible trial state is the spin-triplet, 1 Ψ00 (1, 2) = √ [φ1s (r1 ) φ2s (r2 ) − φ2s (r1 ) φ1s (r2 )] τ12 , 2 1 τ12 = |↑i1 |↑i2 , √ (|↑i1 |↓i2 + |↓i1 |↑i2 ) , |↓i1 |↓i2 2 but this will give higher energy due to the present of the higher level state φ2s (r1 ). (It is quite easy to test experimentally if the ground state Helium is singlet or triplet.) Using the singlet state Ψt (1, 2), we evaluate the energy expectation value D Ĥ E = = = = e2 1 Ze2 1 h̄2 2 + ∇1 − 2· 2m 4π0 r1 4π0 r12 ! 2 2 2 Z e 5 Ze 2· − + 8π0 a0 4 8π0 a0 5 e2 −2Z 2 + Z Ry, Ry = 4 8π0 a0 −5.50 Ry, Z = 2 * + which is only 6% above the experimental result of −5.81 Ry. 3.4. VARIATIONAL METHOD Math Note: Evaluate 1 r12 D 1 r12 1 = 2 π E 71 : Z a0 6 Z d 3 r1 d 3 r1 1 e−2Z(r1 +r2 )/a0 . |r1 − r2 | First we use the expression (Fourier transformation) 1 = |r1 − r2 | hence 1 r12 1 = 2 π Z a0 6 Z d3 k ik·(r1 −r2 ) 4π e k2 (2π)3 Z d3 k 4π (2π)3 k 2 now Z and finally d3 r1 eik·r1 −2Zr1 /a0 = h 1 r12 4Z = πa0 Z ∞ 0 (x2 Z 3 d r1 e 2 ik·r1 −2Zr1 /a0 16πZ/a0 k 2 + (2Z/a0 )2 i2 dx 5Z . 4 = 8 a0 + 1) 72 CHAPTER 3. APPROXIMATION METHODS IN QM We can improve our wavefunction by introducing a variational parameter Z 0 s φ1s (r) = Z 03 −Z 0 r/a0 e , πa30 Ψ0 (1, 2) = φ1s (r1 ) φ1s (r2 ) χ12 and repeat our calculation as before, see example one, D Ĥ E Z0 Ze2 1 h̄2 2 ∇1 − = 2· 2m 4π0 r1 * 0 2 02 2 + !Z 0 e2 + 4π0 1 r12 Z0 0 2 ZZ e 5 Ze Z e − + 8π0 a0 4π0 a0 4 8π0 a0 5 e2 = 2 Z 02 − 2Z 0 Z + Z 0 Ry, Ry = 8 8π0 a0 = 2· and variational eq. ∂ D E 5 Ĥ 0 = 0 → Z 0 = Z − 0 Z ∂Z 16 and the estimated ground-state energy D Ĥ E Z 0 =Z−5/16 = −2 Z − = − 5 16 2 Ry, Z=2 2 · 272 Ry = −5.70 Ry 162 which is closer to the experimental result of −5.81 Ry. (higher no more than 2%). Physically, the variational parameter Z 0 represents the mutual screening by the two electrons. Now how does one go from here? Can we further improve our estimate and how? A more sophisticated variational wavefunction (Hylleraas) is Ψ (1, 2) = P (s, t, u) Ψ0 (1, 2) Z0 Z0 s ≡ (r1 + r2 ) , t ≡ (r1 − r2 ) , a0 a0 where P (s, t, u) is expanded in power series of s, t and u P (s, t, u) = X cn,2l,m sn t2l um l,n,m and variational eqs. are ∂ ∂cn,2l,m D E Ĥ = 0, ∂ D E Ĥ = 0 ∂Z 0 u≡ Z0 r12 a0 3.4. VARIATIONAL METHOD 73 and by including 10 cn,2l,m parameters, the accuracy can reach to 6-7 digits. See Intermediate QM, 1964, H.A. Bethe. Variational method and diagonalization. For a given basis set (truncated or untruncated), {|ni , n = 1, 2, · · ·, N }, the trial wavefunction can be written as |Ψi = N X n=1 cn |ni , hΨ| = N X n=1 hn| c∗n where {c∗n , cn ; n = 1, 2, · · ·, N } are variational parameters determined by δ D E δ D E Ĥ = Ĥ = 0, δc∗n δcn or hΨ |Ψi or PN n0 =1 N X n0 =1 D E Ĥ = Hnn0 cn0 − hΨ| Ĥ |Ψi cn = 0, hΨ |Ψi2 D E Hnn0 cn0 = Ĥ cn , hΨ| Ĥ |Ψi hΨ |Ψi Hnn0 ≡ hn| Ĥ |n0 i n = 1, 2, · · ·, N . This is precisely D E the eigenequation for the Hamiltonian matrix {Hnn0 } with the energy eigenvalue Ĥ . Therefore, the variational method and diagonalization is equivalent.