8 Foldy-Wouthuysen TransformationWe now have the Dirac equation with interactions. For a given problem we cansolve for the spectrum and wavefunctions (ignoring the negative energy solutionsfor a moment), for instance, the hydrogen atom, We can compare the solutionsto those of the schr¨odinger equation and find out the relativistic corrections tothe spectrum and the wavefunctions. In fact, the problem of hydrogen atom canbe solved exactly. However, the exact solutions are problem-specific and involveunfamiliar special functions, hence they not very illuminating. You can find theexact solutions in many textbooks and also in Shulten’s notes. Instead, in thissection we will develop a systematic approximation method to solve a system inthe non-relativistic regime (E−m  m). It corresponds to take the approximationwe discussed in the previous section to higher orders in a systematic way. Thisallows a physical interpretation for each term in the approximation and tells us therelative importance of various effects. Such a method has more general applicationsfor different problems.In Foldy-Wouthuysen transformation, we look for a unitary transformation UFremoves operators which couple the large to the small components.Odd operators (off-diagonal in Pauli-Dirac basis): αi, γi, γ5, · · ·Even operators (diagonal in Pauli-Dirac basis): 1, β, Σ, · · ·ψ0= UFψ = eiSψ, S = hermitian (171)First consider the case of a free particle, H = α · p + βm not time-dependent.i∂ψ0∂t= eiSHψ = eiSHe−iSψ0= H0ψ0(172)We want to find S such that H0contains no odd operators. We can tryeiS= eβα·ˆpθ= cos θ + βα ·ˆp sin θ, whereˆp = p/|p|. (173)H0= (cos θ + βα ·ˆp sin θ) (α · p + βm) (cos θ − βα ·ˆp sin θ)= (α · p + βm) (cos θ − βα ·ˆp sin θ)2= (α · p + βm) exp (−2βα ·ˆpθ)= (α · p)cos 2θ −m|p|sin 2θ+ β (m cos 2θ + |p| sin 2θ) . (174)To eliminate (α · p) term we choose tan 2θ = |p|/m, thenH0= βpm2+ |p|2. (175)This is the same as the first hamilton we tried except for the β factor which alsogives rise to negative energy solutions. In practice, we need to expand the hamiltonfor |p|  m.26General case:H = α · (p − eA) + βm + eΦ= βm + O + E, (176)O = α · (p − eA), E = eΦ, βO = −Oβ, βE = Eβ (177)H time-dependent ⇒ S time-dependentWe can only construct S with a non-relativistic expansion of the transformedhamilton H0in a power series in 1/m.We’ll expand top4m3andp×(E, B)m2.Hψ = i∂∂te−iSψ0= e−iSi∂ψ0∂t+i∂∂te−iSψ0⇒ i∂ψ0∂t=eiSH − i∂∂te−iSψ0= H0ψ0(178)S is expanded in powers of 1/m and is “small” in the non-relativistic limit.eiSHe−iS= H + i[S, H] +i22![S, [S, H]] + · · · +inn![S, [S, · · · [S, H]]]. (179)S = O(1m) to the desired order of accuracyH0= H + i[S, H] −12[S, [S, H]] −i6[S, [S, [S, H]]] +124[S, [S, [S, [S, βm]]]]−˙S −i2[S,˙S] +16[S, [S,˙S]] (180)We will eliminate the odd operators order by order in 1/m and repeat until thedesired order is reached.First order [O(1)]:H0= βm + E + O + i[S, β]m. (181)To cancel O, we choose S = −iβO2m,i[S, H] = −O +β2m[O, E] =1mβO2(182)i22[S, [S, H]] = −βO22m−18m2[O, [O, E]] −12m2O3(183)i33![S, [S, [S, H]]] =O36m2−16m3βO4(184)i44![S, [S, [S, [S, H]]]] =βO424m3(185)−˙S =iβ˙O2m(186)−i2[S,˙S] = −i8m2[O,˙O] (187)27Collecting everything,H0= βm +O22m−O48m3+ E −18m2[O, [O, E]] −i8m2[O,˙O] (188)+β2m[O, E] −O33m2+iβ˙O2m= βm + E0+ O0(189)Now O0is O(1m), we can transform H0by S0to cancel O0,S0=−iβ2mO0=−iβ2m β2m[O, E] −O33m2+iβ˙O2m!(190)After transformation with S0,H00= eiS0H0− i∂∂te−iS0= βm + E0+β2m[O0, E0] +iβ˙O02m(191)= βm + E0+ O00, (192)where O00is O(1m2), which can be cancelled by a third transformation, S00=−iβO002mH000= eiS00H00− i∂∂te−iS00= βm + E0(193)= βm +O22m−O48m3+ E −18m2[O, [O, E]] −i8m2[O,˙O] (194)Evaluating the operator products to the desired order of accuracy,O22m=(α · (p − eA))22m=(p − eA)22m−e2mΣ · B (195)18m2[O, E] + i˙O=e8m2(−iα · ∇Φ − iα ·˙A) =ie8m2α · E (196)O,ie8m2α · E=ie8m2[α · p, α · E]=ie8m2Xi,jαiαj−i∂Ej∂xi+e4m2Σ · E × p (197)=e8m2(∇ · E) +ie8m2Σ · (∇ × E) +e4m2Σ · E × pSo, the effective hamiltonial to the desired order isH000= βm +(p − eA)22m−p48m3+ eΦ −e2mβΣ · B−ie8m2Σ · (∇ × E) −e4m2Σ · E × p −e8m2(∇ · E) (198)28The individual terms have a direct physical interpretation.The first term in the parentheses is the expansion ofp(p − eA)2+ m2(199)and −p4/(8m3) is the leading relativistic corrections to the kinetic energy.The two terms−ie8m2Σ · (∇ × E) −e4m2Σ · E × p (200)together are the spin-orbit energy. In a spherically symmetric static potential, theytake a very familar form. In this case ∇ × E = 0,Σ · E × p = −1r∂Φ∂rΣ · r × p = −1r∂Φ∂rΣ · L, (201)and this term reduces toHspin−orbit=e4m21r∂Φ∂rΣ · L. (202)The last term is known as the Darwin term. In a coulomb potential of a nucleuswith charge Z|e|, it takes the form−e8m2(∇ · E) = −e8m2Z|e|δ3(r) =Ze28m2δ3(r) =Zαπ2m2δ3(r), (203)so it can only affect the S (l = 0) states whose wavefunctions are nonzero at theorigin.For a Hydrogen-like (single electron) atom,eΦ = −Ze24πr, A = 0. (204)The shifts in energies of various states due to these correction terms can be com-puted by taking the expectation values of these terms with the correspondingwavefunctions.Darwin term (only for S (l = 0) states):ψnsZαπ2m2δ3(r)ψns=Zαπ2m2|ψns(0)|2=Z4α4m2n3. (205)Spin-orbit term (nonzero only for l 6= 0):Zα4m21r3σ · r × p=Z4α4m4n3[j(j + 1) − l(l + 1) − s(s + 1)]l(l + 1)(l +12). (206)29Relativistic corrections:−p48m3=Z4α4m2n434−nl +12. (207)We find∆E(l = 0) =Z4α4m2n434− n(208)= ∆E(l = 1, j =12), (209)so 2S1/2and 2P1/2remain degenerate at this level. They are split by Lamb shift(2S1/2> 2P1/2) which can be calculated after you learn radiative corrections inQED. The 2P1/2and 2P3/2are split by the spin-orbit interaction (fine structure)which you should have seen before.∆E(l = 1, j =32) − ∆E(l = 1, j =12) =Z4α4m4n3(210)9 Klein Paradox and the Hole TheorySo far we have ignored the negative solutions. However, the negative energy solu-tions are required together with the