Approximation Methods in QM

Transcription

1 Chapter 3 Approximation Methods in QM Contents 3.1 Time independent PT (nondegenerate) Degenerate perturbation theory (PT) Time dependent PT and Fermi s golden rule Variational method and helium atom Basic Questions A. What causes fine strucutre in hydrogen spectra? B. What is Fermi s golden rule? C. What is the variational principle? 51

2 5 CHAPTER 3. APPROXIMATION METHODS IN QM 3.1 Time-independent PT (nondegenerate) Introduction Two quantum problems can be solved exactly, they are harmonic oscillator and hydrogen (or hydrogen-lie) 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. ĤΨ p E p Ψ p, is difficult to solve. If Ĥ can be written as Ĥ Ĥ0 + ˆV p quantum no. where ˆV is time-independent and small, and the Schrödinger eq. of Ĥ 0 has been solved Ĥ 0 Φ e Φ Ĥ0 Φ e Φ where is the quantum number labelling the nondegenerate states. We wish to use the information of Ĥ 0 to find the eigenstates and eigenvalues of Ĥ. Ĥ 0 is usually referred to as unperturbed Hamiltonian, ˆV as perturbation potential (operator). In principle (the completeness principle), any state Ψ p can be written as a linear combination of Φ as Ψ p C p Φ where {C p } are constants to be determined by eq. of Ĥ, together with eigenvalue E p. Since the dimensionality (total number of the possible ) is in general infinite, the exact solution is not easy. However, since the perturbation ˆV is assumed small, it is not difficult to get approximate solution for Ψ p and E p Perturbation expansion Obviously, if ˆV is zero, p, Ψ Φ, E e. Hence, for non-zero ˆV, this solution {p, Ψ Φ, E e } 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 + λ ˆV, 0 λ 1 so that as λ 0, solutions for Ĥ (λ) is reduced to that of Ĥ 0 and what we want is the results in the limit λ 1. In perturbation theory, we assume p

3 3.1. TIME-INDEPENDENT PT (NONDEGENERATE) 53 i.e., the quantum number for the states of Ĥ is same as that of Ĥ 0. Therefore, we can expand each of the eigenvalues and eigenstates of Ĥ (λ) in terms of powers of λ as E E (0) + λe (1) + λ E () + λ 3 E (3) + Ψ Ψ (0) + λψ (1) + λ Ψ () + λ 3 Ψ (3) + where E (0) e is the zero-order approximation to the -th eigenvalue of Ĥ, and E (1), E(), E(3), are the successive correction to this. Similarly, Ψ(0) Φ is the zero-order appro. to -th eigenstate of Ĥ and Ψ (1), Ψ(), Ψ(3), are the successive correction to this. In order to find E (1), E(), E(3), and Ψ(1), Ψ(), Ψ(3),, we substitute the expansions into the original Schrödinger eq. ĤΨ E Ψ or (Ĥ0 + λ ˆV ) ( Ψ (0) + λψ (1) + λ Ψ () + ) ( E (0) + λe (1) + λ E () + ) ( Ψ (0) + λψ (1) + λ Ψ () + ) or, each side is arranged according to the powers of λ Ĥ 0 Ψ (0) + λ ( ˆV Ψ (0) + E (0) Ψ(0) + λ ( E (1) Ψ(0) ) (1) Ĥ0Ψ ( + λ (1) ˆV Ψ + E (0) Ψ(1) + ) ( + λ E () Ψ(0) ) () Ĥ0Ψ + + E (1) Ψ(1) + E (0) ) Ψ() + Equating the coefficients of λ on both sides of the eq. (consider λ as an arbitrary number, this must be true), we have as zero-order result, and Ĥ 0 Ψ (0) E (0) Ψ(0) Ψ (0) Φ, E (0) e ˆV Ψ (0) (1) + Ĥ0Ψ E (1) as first-order eq. for E (1) and Ψ (1) and ˆV Ψ (1) () + Ĥ0Ψ E () as second-order eq. for E () and Ψ () etc. Ψ(0) Ψ(0) + E (1) + E (0) Ψ(1) Ψ(1) + E (0) Ψ()

4 54 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 ˆV Ψ (0) (1) + Ĥ0Ψ E (1) Ψ(0) + E (0) Ψ(1) or, using the zero-order results ( ) (1) ˆV E Φ ( ) e Ĥ0 (1) Ψ in terms of a linear com- By the completeness principle, we can always express Ψ (1) bination of Φ as Ψ (1) C (1) Φ where C (1) are coefficients to be determined and the equation becomes, using Ĥ0Φ e Φ ( ) (1) ˆV E Φ ( ˆV E (1) ) Φ (e e ) C (1) Φ (e e ) C (1) Φ this is the eq. for E (1) and C. The standard technique to solve this ind eq. is to tae inner product with state Φ on both sides of the eq., i.e., multiplying Φ and integrate using the orthonormal relation we have ( ) d 3 rφ (1) ˆV E Φ (e e ) C (1) d 3 rφ Φ δ d 3 rφ Φ V E (1) δ (e e ) C (1) δ (e e ) C (1) V d 3 rφ ˆV Φ Choosing we have E (1) V d 3 rφ ˆV Φ

5 3.1. TIME-INDEPENDENT PT (NONDEGENERATE) 55 Choosing we have C (1) V e e Note that C is still undetermined and can be chosen as zero. The 1st-order results are, letting λ 1, E e + E (1) V. d 3 rφ ˆV Φ for the energy and V Ψ Φ + e e Φ for the state. Note: if we choose C 0, the coefficient of Φ in the above formula will not be unity and Ψ is then not properly normalized in the limit V 0. Note that the determine of the 1st-order energy E (1) only depends on the zeroorder state function Ψ (0) Φ, and the 1st-order statefunction Ψ (1) is not necessary. However, in order to determine the nd -order energy correction E (), the 1st-order statefunction correction Ψ (1) is required. This is generally true: nth-order energy correction requires the (n 1)th-order statefunction correction Solution for perturbation eq.: st-order The nd-order equation is ˆV Ψ (1) () + Ĥ0Ψ E () Ψ(0) + E (1) Ψ(1) + E (0) Ψ() or E () Ψ(0) ( ) (1) (1) ˆV E Ψ + ( Ĥ 0 E (0) ) () Ψ where E (0) Again, Ψ (), E(1), Ψ(0) and Ψ (1) are nown. We only need to determine E () and Ψ (). can be written as a linear combination of Φ Ψ () C () Φ substituting this into the eq., using the zero- and 1st-order results, we have E () Φ C (1) ( ) (1) ˆV E Φ + C () ) (Ĥ0 e Φ C (1) ( (1) ˆV E ) Φ + C () (e e ) Φ

6 56 CHAPTER 3. APPROXIMATION METHODS IN QM as before, taing inner product with state Φ on both sides of the eq. and using the orthonormality of Φ we have E () δ C (1) ( ˆV E(1) δ ) + C () (e e ). Choosing, we obtain E () E () C (1) ( V E (1) δ ) C (1) V, C(1) 0 V V e e V e e where we have assume ˆV is a Hermitian operator. The coefficient C () mined by choosing, as before. Again, to determine E () C (1) and C() is not necessary. The nd-order result for the energy is then given by E E (0) + E (1) + E () e + V + Ψ Ψ (0) + Ψ (1) Φ + V e e Φ can be deter- we only need to now V e e with V ˆV d 3 r Φ ˆV Φ Example Consider an anharmonic oscillator with Hamiltonian Ĥ Ĥ0 + ˆV Ĥ 0 h m x + 1 mωx, ˆV αx 3 where α is a small number (α hω/x 3 0 hω ( ) mω 3/). h The eigenequation of Ĥ 0 has been solved, ( Ĥ 0 n hω n + 1 ) n, n 0, 1,,, ( Ĥ 0 hω â â + 1 )

7 3.1. TIME-INDEPENDENT PT (NONDEGENERATE) 57 and we now the effect of â and â on the normalized state n â n n + 1 n + 1, â n n n 1 for n > 0. we hence need to express ˆV in terms of â and â. Using the definition â mω h ˆx + iˆp, m hω â mω h ˆx iˆp m hω we have hence ˆx h (â + â ), ˆp 1 i mω h (â â ) ˆV ( ) 3/ h (â α + â 3 ( ) 3/ h (â α + â â + ââ + â â + â ) ( ) 3/ h (â α + â â + â â â ). Up to nd-order approximation, we have E n e n + V nn + n n V nn V n n e n e n where ( e n hω n + 1 ) and V nn n ˆV n ( ) 3/ h α n ( â + â ) ( â + â â â ) n ( ) 3/ h α n ( â + â ) ( â + n â ) n 0 i.e., the first-order energy correction is zero.

8 58 CHAPTER 3. APPROXIMATION METHODS IN QM To determine the nd-order energy correction, we need to evaluate V n n. Consider (â + â ) ( â + â â â ) n ( â + â ) ( â + n â ) n ( â + â ) ( ) n (n 1) n + (n + 1) n + (n + 1) (n + ) n + n (n 1) (n ) n 3 + (n + 1) n n 1 + (n + 1) (n + ) n n (n 1) n 1 + (n + 1) n + 1 n (n + 1) (n + ) (n + 3) n + 3 [ n (n 1) (n ) n 3 + (n + 1) ] n + n (n 1) n 1 + [ + (n + 1) (n + ) + (n + 1) ] n + 1 n (n + 1) (n + ) (n + 3) n + 3 n (n 1) (n ) n 3 + 3n 3/ n (n + 1) 3/ n (n + 1) (n + ) (n + 3) n + 3 therefore, there are only four nonzero contribution to V n n therefore n n V nn V n n e n e n V n 3,n ( ) 3/ n 3 ˆV h n α n (n 1) (n ) V n 1,n ( ) 3/ n 1 ˆV h n α 3n 3/ V n+1,n ( ) 3/ n + 1 ˆV h n α 3 (n + 1) 3/ V n+3,n ( ) 3/ n + 3 ˆV h n α (n + 1) (n + ) (n + 3) V n 3,n e n e n hω α + 1 hω α 1 hω α V n 1,n + V n+1,n + V n+3,n e n e n 1 e n e n+1 e n e n+3 ( ) 3 h n (n 1) (n ) ( ) 3 h 9n 3 ( ) 3 h 9 (n + 1) 3

9 3.. DEGENERATE PT 59 1 ( h 3 hω α 1 3 hω α ( h ) 3 ) 3 (n + 1) (n + ) (n + 3) [ n (n 1) (n ) + 7n 3 7 (n + 1) 3 (n + 1) (n + ) (n + 3) ] 1 ( ) 3 h ( hω α 30n + 30n + 11 ) Hence, up to nd-order, the energy value is ( E n hω n + 1 ) 1 ( ) 3 h ( hω α 30n + 30n + 11 ) and the energy interval between adjacent energy levels See figure E n E n 1 hω 60 ( ) 3 h hω α n 3. Degenerate PT In the above perturbation theory, the unperturbed states Φ are nondegenerate, e e for all. Now we extend the theory to the degenerate case. Our Hamiltonian is still given by Ĥ Ĥ0 + ˆV where the perturbed term ˆV is assumed small and the unperturbed Hamiltonian Ĥ0 has been solved Ĥ 0 Φ nd e n Φ nd with quantum number d 1,,, M, labelling the degeneracy. In the case of Hydrogen, we now for each n, we have degenerate qn d (l, m), d l,m and M. For convenience we only need one letter for the notation. We expect that the

10 60 CHAPTER 3. APPROXIMATION METHODS IN QM introduction of perturbed term will lift the degeneracy in l, so the eigen equation for Ĥ is ĤΨ nd E nd Ψ nd. But if we still carry out the similar analysis as before, we expect to encounter a factor such as Ĥ Ĥ (λ) Ĥ0 + λ ˆV, and the expansion series E nd E (0) nd + λe(1) nd + Ψ nd Ψ (0) nd + λψ(1) nd + 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 Ĥ 0 Ψ (0) nd E(0) nd Ψ(0) n,d Ψ (0) nd Φ nd, E (0) nd e n as zero-order result, and ˆV Ψ (0) (1) nd + Ĥ0Ψ nd E(1) nd Ψ(0) nd + E(0) nd Ψ(1) nd as first-order eq. etc. All those formulae remain the same, only replacing single qn by double ones (n, d). However, there are divergence problem when we go beyond zeroth-order. Consider the 1st-order correction to the wavefunction Ψ (1) nd n d c (1) nd,n d Φ n d c (1) nd,nd Φ nd + other contributions from states with n n d d with the diverging coefficient c (1) nd,nd V nd,nd E (0) nd E(0) nd Solution: Recall the basic assumption in PT V nd,nd e n e n. quantum number after perturbation quantum number before perturbation. We need to modify this by introducing a new quantum number as before perturbation : (n, d) after perturbation : (n, ). This is equivalent to introduce a new basic set {Ω n, 1,,, M}as M Ω n b d Φ nd, d1 1,,, M

11 3.. DEGENERATE PT 61 with coefficients b d to be determined by the condition to avoid the divergence mentioned above. Now our new PT expansion series are E n E (0) n + λe(1) n + Ψ n Ψ (0) n + λψ(1) n + and the zeroth-order equation does not change much and the 1st-order eq. Ĥ 0 Ψ (0) n E(0) n Ψ(0) n E(0) n e n, Ψ (0) n Ω n where we expand Ψ (1) n in terms of Ω n ( ) (1) (0) ˆV E n Ψ n ( E (0) n ) Ĥ0 (1) Ψ n Ψ (1) n c (1) Ω n, 1,,, M. M b d Φ nd d1 As before, we substitute this into the 1st-order eq. and tae inner product with state Ω n on both sides, we have ( ) d 3 r Ω (1) n ˆV E n Ωn d 3 r Ω ( (0) n E n Ĥ0) Ωn or Tae, we have V n,n E (1) n δ d 3 r Ω n c (1) E (1) n V n,n c (1) (e n e n ) Ω n 0. and tae, we have V n,n d 3 r Ω ˆV n Ω n 0, for all. This last equation means operator ˆV is diagonal in the new basis. Therefore, the coefficients {b d } in Ω n M d1 b d Φ nd is chosen so that ˆV is diagonal. Example: Spin-orbit coupling (or LS-coupling). An electron moving in a central potential ˆV V (r) has Hamiltonian as Ĥ 0 h m + V (r).

12 6 CHAPTER 3. APPROXIMATION METHODS IN QM In general the eigenvalues E nl of Ĥ 0 are (l + 1)-fold degenerate (for hydrogen, E nl E n, and degeneracy is n -fold). One of relativistic effects (Dirac) is the addition of a potential, the so-called spin-orbit coupling (or LS-coupling) ˆV LS A (r) ˆL Ŝ. We can treat this ˆV LS as perturbation potential and write ˆV LS 1 A (r) (Ĵ ˆL Ŝ ), Ĵ ˆL + Ŝ As we now, the eigenstates of Ĥ0 can be represented by the product of orbital wavefunctions and spin wavefunctions, or nlm sm s nlm l sm s. Within the first-order degenerate PT, we need to use these nlm l sm s to diagonalize the perturbation operator ˆV LS 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 nlsjm which are eigenstates of Ĵ and Ĵz with angular momentum quantum number, using s 1/ j l ± 1, m j, j + 1,, j 1, j. Hence, the energy level shift by LS-coupling is E (1) LS nlsjm ˆV [ LS nlsjm h j (j + 1) l (l + 1) 3 ] A (r) 4 nl. 3.3 Time-dependent perturbation theory Newton s equation As we have seen before, the time-dependent Schrödinger eq. is i h t Ψ (r, t) ĤΨ (r, t), i h t Ψ (r, t) ĤΨ (r, t). The expectation value of an arbitrary operator   (t), assumed Ψ (r, t) is normalized, Â Ψ Â Ψ d 3 r Ψ (r, t)  (t) Ψ (r, t) has the following evolution equation d  d 3 r dt t [ Ψ (r, t)  (t) Ψ (r, t)] d 3 r Ψ (r, t)  1 Ψ (r, t) + t i h d 3 r Ψ (r, t) ( ÂĤ ĤÂ) Ψ (r, t)

13 3.3. TIME-DEPENDENT PERTURBATION THEORY 63 or d dt   + 1 [Â, ] Ĥ. t i h Using this formula, Newton s equation becomes, using Ĥ ˆp /m + ˆV (r) d dt ˆp ˆV Time-dependent expansion Consider now the following Hamiltonian Ĥ Ĥ0 + ˆV (t) where Ĥ0 is time-independent with eigen eqs. Ĥ 0 Φ n (r) E n Φ n (r), n 1,, and the time dependent part of the Hamiltonian is ˆV (t). The general time-dependent wavefunction Ψ (r, t) can be expanded in the following general form as Ψ (r, t) n d n (t) Φ n (r) n c n (t) Φ n (r) e ient/ h and after substituting into the Schrödinger eq., we have ( E n c n + i h dc ) n Φ n (r) e ient/ h ( En + dt ˆV ) c n Φ n (r) e ient/ h n or n i h n dc n dt Φ n (r) e ie n t/ h n c n ˆV Φn (r) e ie n t/ h. Taing inner product with Φ n (r) on both sides of eq., we have the time-evolution equation for c n (t) i h dc n dt n V nn c n e iω nn t, n 1,, with V nn Φ n ˆV Φ n d 3 r Φ n ˆV Φ n, hω nn E n E n. these are a set of first-order, coupled equations for {c n, n 1,, }. Normally difficult to solve. For some special case, exact solution can be found. See Mandl, Quantum Mechanics, Section 9., P 198.

14 64 CHAPTER 3. APPROXIMATION METHODS IN QM Consider a typical initial condition: at t 0, the system is in one of the eigenstate, say, 1, namely c n (0) δ 1n. For ˆV (t) 0, the amplitudes c n (t) retain this initial value at all times and Ψ (r, t) n c n (t) Φ n (r) e ient/ h Φ 1 (r) e ie 1t/ h the time-dependent part is trivial Perturbation Theory Now we treat ˆV as perturbation potential and write the Hamiltonian as before and correspondingly c n c (0) n the evolution equation for c n becomes i h d dt Ĥ Ĥ0 + λ ˆV (t) + λc(1) n + λ c () n + ( c (0) n + λc (1) n + λ c () n + ) ( (0) c n + λc(1) n + λ c () n + ) λv nn e iω nn t. n Setting the coefficients of each power of λ equal on both sides we have a series successive eqs., i h d dt c(0) n 0, c (0) n (t) const. i h d dt c(1) n n c (0) n V nn eiω nn t i h d dt c() n n c (1) n V nn eiω nn t. Consider the typical initial condition: applying the perturbation ˆV (t) after t > 0, namely ˆV (t) { 0, t 0 0, t > 0 and assume for t 0, the system is in one of the eigenstates of Ĥ0, say 1, c n (0) c (0) n (t) δ 1n

15 3.3. TIME-DEPENDENT PERTURBATION THEORY 65 Hence the first-order equation becomes i h d dt c(1) n n c (0) n V nn eiω nn t V n1 e iω n1t Integrating these equations from t 0 to t > 0, subject to the condition that c n (0) δ 1n, we have c (1) 1 (t) i h c (1) n (t) 1 i h t t 0 ˆV 11 (t ) dt ˆV n1 (t ) e iω n1t dt, ω n1 E n E 1, n 1 h 0 where c (1) 1 (t) is the probability (in 1st-order approximation) that the system at t > 0 stays in the original state 1, and c (1) n (t) is the probability (in 1st-order approximation) that the system at t > 0 is in another state n (n 1). Example. A hydrogen atom is placed in a spatially homogeneous time-dependent electric field in the z-axis ε { 0, t 0 ε 0 e t/τ, t > 0 with τ > 0. At time t 0, the atom is in the (1s) ground state with wavefunction and energy φ 1s 1 e r/a 0, E 1s e, a 0 4πɛ 0 h. πa 3 8πɛ 0 a 0 me 0 Find the probability, in the 1st order perturbation theory, that after a sufficiently long time the atom is in the (p) excited state with wavefunction and energy φ p 1 r cos θ e r/a 0, 3πa 5 0 E p E 1s 4. Solution: According to the 1st-order perturbation theory, the probability amplitude for the transition (1s) (p) is with and c (t) 1 t ˆV p,1s (t ) e iω p,1st dt i h 0 ω p,1s E p E 1s h 3E 1s 4 h ˆV (t ) eε 0 e t /τ z eε 0 e t /τ r cos θ.

16 66 CHAPTER 3. APPROXIMATION METHODS IN QM The matrix element ˆV p,1s (t ) is ˆV p,1s (t ) p ˆV (t ) 1s eε 0 e t /τ p r cos θ 1s and the amplitude eε 0 e t /τ 1 r cos θ e r/a0 3πa 5 0 ( eε 0 e t /τ 1 4 πa πa 3 0 e r/a 0 ) ( π r 4 e 3r/a 0 dr cos θ sin θdθ 0 r sin θdrdθdφ ) ( π eε 0 e t /τ 1 4 4!5 a 5 0 πa π Ce t/τ, C 15/ 3 eε 0a 5 0 c (t) C i h t and the transition probability is 0 e iω p,1st t /τ dt C e iωp,1st t/τ 1 i h iω p,1s 1/τ c(t) C h 1 + e t/τ e t/τ cos(ω p,1s t) ω p,1s + 1/τ, and after a sufficiently long time, t, the probability is Fermi golden rule c ( ) (eε 0 a 0 ) (3E 1s /4) + ( h/τ). 0 ) dφ Now we consider the following simpler case, a step function perturbation potential ˆV (t) { 0, t 0 ˆV, t > 0 where ˆV is independent of time. What is the transition probability per unit time for the system originally in state 1 at t 0 to the state at t > 0? Solution: From the above 1sr-order perturbation theory, the time-integration in the transition probability amplitude can be carried out, c (1) 1 (t) 1 t V,1 (t ) e iω,1t dt V (,1 ) 1 e iω,1t i h 0 hω,1 V,1 ˆV 1, ω,1 E E 1 h

17 3.4. VARIATIONAL METHOD 67 and the probability is, using sin y (e iy e iy )/, P 1 (t) (1) c 1 (t) 4 V,1 sin (ω,1 t/) h πt ω,1 h V,1 D (ω,1, t) D (x, t) sin (xt/), x ω πt x,1 where function D (x, t) has a simple integral D (x, t) dx πt sin (xt/) x dx 1, using sin (ax) x dx πa, 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 P 1 (t) πt h and the probability per unit time This is the famous Fermi Golden rule. V,1 δ (ω,1 ) πt h V,1 δ (E E 1 ) P 1 (t) π h V,1 δ (E E 1 ). 3.4 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, Ψ Ĥ Ψ Ψ Ψ E 0 the equality holds if Ψ is the exact ground state. Proof: Suppose Ĥ Φ n E n Φ n, n 0, 1,, E 0 < E n, for n > 0

18 68 CHAPTER 3. APPROXIMATION METHODS IN QM Write Ψ as a linear combination of Φ n Therefore and so Ψ C n Φ n C 0 Φ 0 + C n Φ n n0 n1 Ψ Ψ C 0 + C n n1 ( ) Ĥ Ψ Ĥ C 0 Φ 0 + C n Φ n n1 E 0 C 0 Φ 0 + C n E n Φ n n1 Ψ Ĥ Ψ E 0 C 0 + E n C n n1 > E 0 C 0 ) + E 0 C n E 0 ( C 0 + C n E 0 Ψ Ψ n1 n1 Ψ Ĥ Ψ E 0. Ψ Ψ In general, we do not now the exact ground state energy E 0. We wish to find an approximation for its value. By the variational principle, our approximation Ĥ using any trial wavefunction Ψ will always be greater than the exact value E 0. Hence, the lower the expectation value Ĥ (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 α 1, α, and calculate the expectation value Ψ (α 1, α, ) E (α 1, α, ) Ψ Ĥ Ψ Ψ Ψ. Next, we minimize this E (α 1, α, ) through variational equations E (α 1, α, ) α 1 0 E (α 1, α, ) α 0

19 3.4. VARIATIONAL METHOD 69 The solution for (α 1, α, ) of these equations then put bac to E (α 1, α, ) and this minimum E represents the best estimate for the trial function of the form Ψ (α 1, α, ). Clearly, the success of this nonperturbative method will depend crucially on the form of the trial wave functions we choose to start with Examples Hydrogen atom. The trial function for the ground state of Hydrogen is choose as α Φ α Ce αr 3, C π the Hamiltonian is Ĥ h m e 4πɛ 0 r, and the expectation value is 1 ( r ) + 1 r r r r  (θφ) E α Φ α Ĥ Φ α h α m e α 4πɛ 0 hence E α α 0 α e m 4πɛ 0 h 1 a 0 where a 0 is the Bohr radius. So the best estimate for the ground state energy is and for the ground state is E α h m 1 a 0 e α 1 1 4πɛ 0 a 0 Φ 1 a 3 0π e r/a 0 e 4πɛ 0 a 0 both are actually exact. This is not surprising because the trial function has the correct form. Helium atom. In this case we do not now the exact result and we are dealing with two identical particles. Consider in general Helium-lie ions with the following Hamiltonian describing two electrons interacting each other and with a nucleus containing Z protons ( Ĥ h m 1 h m Ze ) + e 1. 4πɛ 0 r1 r 4πɛ 0 r 1

20 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 nowledge from hydrogen-lie ions. We now the eigenstates φ nlm (r) for a single electron Ĥ h m Ze 1 are 4πɛ 0 r Z 3 φ 1s (r) R 10 (r) Y 00 (θ, φ) e Zr/a 0 φ s (r) R 0 (r) Y 00 (θ, φ)... πa 3 0 Z 3 with φ 1s (r) as the ground state, and eigenvalues 8πa 3 0 ( 1 Zr ) e Zr/a 0 a 0 E nl E n e Z, n 1,,. 8πɛ 0 a 0 n We also now electrons have spin degree of freedom, indicated by two states or. 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, Ψ 0 (1, ) φ 1s (r 1 ) φ 1s (r ) χ 1, χ 1 1 ( 1 1 ) hence Ψ (1, ) Ψ (, 1). This is a spin-singlet state. Another possible trial state is the spin-triplet, Ψ 0 (1, ) 1 [φ 1s (r 1 ) φ s (r ) φ s (r 1 ) φ 1s (r )] τ 1, τ 1 1, 1 ( ), 1 but this will give higher energy due to the present of the higher level state φ s (r 1 ). (It is quite easy to test experimentally if the ground state Helium is singlet or triplet.) Using the singlet state Ψ t (1, ), we evaluate the energy expectation value Ĥ h m 1 Ze 1 4πɛ 0 r 1 ( Z e ) + 5 Ze 8πɛ 0 a 0 4 8πɛ 0 a 0 ( Z + 5 ) 4 Z 5.50 Ry, Z + e 1 4πɛ 0 r 1 Ry, Ry e 8πɛ 0 a 0 which is only 6% above the experimental result of 5.81 Ry.

21 3.4. VARIATIONAL METHOD 71 Math Note: Evaluate 1 r 1 : 1 r 1 1 π ( Z a0 ) 6 d 3 r 1 d 3 r 1 1 r 1 r e Z(r 1+r )/a 0. First we use the expression (Fourier transformation) 1 r 1 r d 3 ei (r 1 r ) 4π (π) 3 hence now and finally 1 r 1 1 π ( Z a0 ) 6 d 3 r 1 e i r 1 Zr 1 /a 0 1 r 1 4Z πa 0 0 d 3 4π (π) 3 d 3 r 1 e i r 1 Zr 1 /a 0 16πZ/a 0 [ + (Z/a 0 ) ] dx (x + 1) 4 5 Z. 8 a 0

22 7 CHAPTER 3. APPROXIMATION METHODS IN QM We can improve our wavefunction by introducing a variational parameter Z Z 3 φ 1s (r) e Z r/a 0, Ψ πa 3 0 (1, ) φ 1s (r 1 ) φ 1s (r ) χ 1 0 and repeat our calculation as before, see example one, Ĥ h Z m 1 Ze 1 + e 1 4πɛ 0 r 1 4πɛ Z 0 r 1 Z ( Z e ZZ e ) + 5 Z e 8πɛ 0 a 0 4πɛ 0 a 0 4 8πɛ 0 a 0 (Z Z Z + 5 ) 8 Z Ry, Ry e 8πɛ 0 a 0 and variational eq. Ĥ Z Z 0 Z Z 5 16 and the estimated ground-state energy Ĥ Z Z 5/16 ( Z 16) 5 Ry, Z 7 16 Ry 5.70 Ry which is closer to the experimental result of 5.81 Ry. (higher no more than %). Physically, the variational parameter Z 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, ) P (s, t, u) Ψ 0 (1, ) s Z a 0 (r 1 + r ), t Z a 0 (r 1 r ), u Z a 0 r 1 where P (s, t, u) is expanded in power series of s, t and u P (s, t, u) c n,l,m s n t l u m l,n,m and variational eqs. are c n,l,m Ĥ 0, Z Ĥ 0

23 3.4. VARIATIONAL METHOD 73 and by including 10 c n,l,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), { n, n 1,,, N}, the trial wavefunction can be written as N N Ψ c n n, Ψ n c n n1 n1 where {c n, c n; n 1,,, N} are variational parameters determined by or or δ δc n Ĥ δ δc n Ĥ 0, Ĥ Ψ Ĥ Ψ Ψ Ψ Ψ Ψ N n 1 H nn c n Ψ Ĥ Ψ c n Ψ Ψ 0, H nn n Ĥ n N H nn c n Ĥ c n, n 1,,, N. n 1 This is precisely the eigenequation for the Hamiltonian matrix {H nn } with the energy eigenvalue Ĥ. Therefore, the variational method and diagonalization is equivalent.