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49. The Foldy-Wouthuysen Transformation

Introduction

In The Nonrelativistic Limit of the Dirac Equation we carried out a nonrelativistic approximation to the Dirac equation through order (v/c)2(v/c)^2, relative to the rest-mass energy mc2mc^2, obtaining the Pauli equation including the μ⋅B\muvec\cdot\Bvec term with g=2g=2. In these notes we will carry out the expansion through order (v/c)4(v/c)^4. To do this we will use a systematic procedure known as the Foldy-Wouthuysen transformation. This transformation block-diagonalizes the matrices that appear in the Dirac equation, thereby decoupling the upper and lower 2-component spinors of the 4-component Dirac spinor and producing two decoupled 2-component spinor equations. Our presentation of this subject is based on that of Bjorken and Drell, Relativistic Quantum Mechanics.

Units and Ordering

In these notes we use natural units, in which ℏ=c=1\hbar=c=1. In these units the fine structure constant becomes

α=e2ℏc=e2, \alpha = {e^2\over\hbar c} = e^2,

so that e\approx 1/\sqrt{137} (recall that in atomic units, e=1e=1).

We take the basic parameter of smallness to be v/c=vv/c=v, but the order of other quantities (for example, the magnetic field) in terms of this small quantity depends on the physical sitation. That is, we must choose a reference system to establish the order magnitude of of various physical quantities in terms of v/cv/c. For our reference system we will take the ground state of hydrogen. Thus our results will be valid for any other system whose basic orders of magnitude are comparable to those of hydrogen, such as other atoms of low ZZ near the ground state as well as many problems in molecular or condensed matter physics. For other problems such as those with strong external fields it would be advisable to rethink the ordering scheme.

Although we are mainly interested in the electron, the Dirac equation describes any particle of spin 12\frac{1}{2} and g≈2g\approx 2, such as the positron or various muons. Therefore we will leave the mass mm and charge qq of the particle as free parameters. This will not affect the ordering of terms in powers of v/cv/c. Other particles with anomalous gg-factors require a modified treatment. See Prob. Paragraph.

In the ground state of hydrogen v/cv/c is of the order of the fine structure constant α\alpha, so the expansion in powers of v/cv/c can be thought of as an expansion in powers of α\alpha. We shall, however, use a formal expansion parameter η\eta instead of α\alpha, partly to avoid confusion between the fine struction constant and the Dirac matrices α\alphavec. For example, we shall say that the velocity vv is of order η\eta. The only purpose of η\eta is to keep track of the order of various terms.

Table 1:Ordering of various quantities in powers of η∼v/c\eta \sim v/c. It is assumed that electric and magnetic fields are internally generated.

QuantitySymbolOrder
Electron mass\text{Electron mass}mmη0=1\eta^0 = 1
Velocity\text{Velocity}vv or uuη1=1\eta^1 = 1
Kinetic energy\text{Kinetic energy}(1/2)mv2(1/2)mv^2η2\eta^2
Potential energy\text{Potential energy}qΦq\Phiη2\eta^2
Kinetic momentum\text{Kinetic momentum}π=mv\pivec = m \vvecη\eta
Fine structure constant\text{Fine structure constant}α\alphaη\eta
Compton λ\text{Compton } \lambdaλC=1/m\lambda_C = 1/mη0=1\eta^0 = 1
Bohr radius\text{Bohr radius}a0=λC/α=1/ma2a_0 = \lambda_C / \alpha = 1/ma^2η−1\eta^{-1}
Orbital period\text{Orbital period}T0∼a0/v0∼1/mα2T_0 \sim a_0/v_0 \sim 1/m\alpha^2η−2\eta^{-2}
∇\nabla∼1/a0\sim 1/a_0η\eta
∂/∂t\partial/\partial t∼1/T0\sim 1/T_0η2\eta^2
qE∥q\Evec_\parallel−q∇Φ-q\nabla\Phiη3\eta^3
qAq\Avec∼v(qΦ)\sim v (q\Phi)η3\eta^3
qBq\Bvecq∇×A q\nabla \times \Avecη4\eta^4
qE⊥q\Evec_\perp−q(∂A/∂t)-q(\partial \Avec/\partial t)η5\eta^5

We take the reference energy to be the electron rest mass-energy mc2=mmc^2=m, which we regard as of order η0=1\eta^0=1, as indicated in Table 1. The velocity vv is of order η\eta; in the hydrogen ground state we have denoted the velocity by v0v_0 (see, for example, Table 1), which in natural units is α\alpha (the fine structure constant). Thus the kinetic energy of the electron mv2/2mv^2/2 is of order η2\eta^2, as shown in the table. Assuming that kinetic and potential energy trade off between themselves as is typical in a bound state problem, the potential energy qΦq\Phi must also be of order η2\eta^2. Also, the kinetic momentum,

π=p−qA=mv\pivec=\pvec-q\Avec = m\vvec

is of order η\eta, as shown in the table.

The unit of length in natural units is the Compton wavelength, λC=ℏ/mc=1/m\lambda_C =\hbar/mc = 1/m, which is regarded as of order η0=1\eta^0=1. The Bohr radius a0a_0 is a factor of 1/α1/\alpha times larger than this, a0=λC/α=1/mαa_0=\lambda_C/\alpha = 1/m\alpha, which is therefore of order η−1\eta^{-1}. The scale of the atom is large in the units we are using. Likewise, the orbital period T0T_0 in the atom is of order a0/v0=1/mα2a_0/v_0=1/m\alpha^2, which is of order η−2\eta^{-2}. This is also large in the units we are using. (The natural unit of time is ℏ/mc2=1/m\hbar/mc^2 = 1/m; the orbital period in hydrogen is a factor of 1/α21/\alpha^2 times larger than this.)

We will order all electric and magnetic fields as if they were internally generated. This does not mean that they have to be internally generated, it just means that they will be ordered as if they were. (For example, in 26. The Zeeman Effect in Hydrogen and Alkali Atoms{} it was noted that strong laboratory magnetic fields are roughly of the same order of magnitude as those generated by electron orbital and spin currents in an atom.) For example, we will treat the electrostatic potential as of order Φ∼e/a0\Phi\sim e/a_0, so that qΦ∼e2/a0=α/a0=α2/mq\Phi \sim e^2/a_0 = \alpha/a_0 = \alpha^2/m. This is another way of seeing that qΦq\Phi is of order η2\eta^2.

We assume we are working in Coulomb gauge, so the longitudinal electric field is given by E∥=−∇Φ\Evec_\parallel=-\del\Phi. But since the atomic scale length a0a_0 is of order η−1\eta^{-1}, the operator ∇\del must be treated as of order 1/a0∼η1/a_0\sim\eta. Thus we find that qE∥q\Evec_\parallel is of order η3\eta^3, as indicatated in the table.

The scalar and vector potentials Φ\Phi and A\Avec produced by a point charge qq with position r\rvec and velocity v\vvec are obtained from (5) and (9), with ρ(x)=q δ(x−r)\rho(\xvec)=q\,\delta(\xvec-\rvec) and J(x)=qv δ(x−r)\Jvec(\xvec)=q\vvec\,\delta(\xvec-\rvec). These show that A∼(v/c)Φ\Avec\sim (v/c)\Phi, so that qAq\Avec is of order η3\eta^3. This means that qB=q∇×Aq\Bvec=q\del\cross\Avec is of order η4\eta^4. Also, the time derivative ∂/∂t\partial/\partial t must be of order 1/T01/T_0, that is, order η2\eta^2. Therefore the quantity qE⊥=−q(∂A/∂t)q\Evec_\perp=-q(\partial\Avec/\partial t), where E⊥\Evec_\perp is the transverse electric field, is of order η5\eta^5. These orderings are summarized in Table 1.

Transforming the Hamiltonian

The Hamiltonian for a Dirac particle interacting with an electromagnetic field is

η0η1η2H=mβ+α⋅π⃗+qΦ,\begin{array}{ccccccc} && \eta^0 && \eta^1 && \eta^2 \\ H &=& m\boldsymbol{\beta} &+& \boldsymbol{\alpha}\cdot\vec{\pi} &+& q\Phi, \end{array}

where the terms have been arranged in increasing order in η\eta, as shown (that is, in decreasing order of magnitude). We work with the Dirac-Pauli representation of the Dirac matrices.

We organize the Dirac matrices into those that are “even”, that is, block-diagonal,

1=(1001),β=(100−1),Σ=(σ00σ),\begin{aligned} 1=\begin{pmatrix}1 & 0 \\ 0 & 1 \\\end{pmatrix}, \qquad \beta=\begin{pmatrix}1 & 0 \\ 0 & -1 \\\end{pmatrix}, \qquad \Sigmavec=\begin{pmatrix}\sigmavec & 0 \\ 0 & \sigmavec\\\end{pmatrix}, \end{aligned}

and those that are “odd”, that is, off-diagonal,

α=(0σσ0),\begin{aligned} \alphavec=\begin{pmatrix}0 & \sigmavec \\ \sigmavec & 0\\\end{pmatrix}, \end{aligned}

where all 4×44\times4 Dirac matrices are partitioned into four 2×22\times2 matrices. The only odd matrices we will need are the three components of α\alphavec, but γ5\gamma_5 is also odd and would be needed in other problems. We see that the terms of the Hamiltonian (3) that are even (odd) as Dirac matrices are also of even (odd) order in η\eta. We will also write the Hamiltonian (3) as

η0η1η2H=mβ ⁣+ ⁣O ⁣+ ⁣E,\begin{array}{ccccccc} && \eta^0 && \eta^1 && \eta^2 \\[2pt] H &=& m\boldsymbol{\beta} &\!+\!& \odd &\!+\!& \even, \end{array}

where O\odd and E\even are the odd and even terms at order η\eta and η2\eta^2, respectively,

O=π⋅α,E=qΦ.\odd=\pivec\cdot\alphavec, \qquad \even=q\Phi.

We shall subject the Hamiltonian (3) to a unitary transformation that eliminates the odd Dirac matrices, thereby transforming it into a new Hamiltonian which is block-diagonal. The new Dirac equation will consists of two 2-component equations that are decoupled from one another, one of which will turn out to be the Pauli equation with relativistic corrections. The transformation will be carried out as a power series expansion in powers of η\eta, that is, v/cv/c.

We write the original Dirac equation as

i∂ψ∂t=Hψ,i\frac{\partial \psi}{\partial t} = H\psi,

and the new one as

i∂ψ′∂t=H′ψ′,i\frac{\partial \psi'}{\partial t} =H'\psi',

that is, with primes, where the old and new wave functions are related by

ψ′=eSψ.\psi' = e^S \psi.

We refer to SS as the “generator” of the transformation; we choose it to be anti-Hermitian, so that eSe^S will be unitary. From (10) we obtain

i∂ψ′∂t=i∂eS∂t ψ+eSi∂ψ∂t=(i∂eS∂t e−S+eSHe−S)ψ′,i\frac{\partial \psi'}{\partial t} = i\frac{\partial e^S}{\partial t}\,\psi + e^S i\frac{\partial \psi}{\partial t} =\Bigl(i\frac{\partial e^S}{\partial t}\,e^{-S} + e^S H e^{-S}\Bigr) \psi',

so that

H′=eSHe−S+i∂eS∂t e−S.H'=e^S H e^{-S} + i\frac{\partial e^S}{\partial t}\,e^{-S}.

As we shall see, SS is small, that is, of positive order in η\eta, so that we can expand the exponentials and carry out the transformation in terms of power series.

Expanding the New Hamiltonian

The expansion of the two terms in (12) in power series was explored in Prob. Paragraph. The solution involves iterated commutators of SS with other operators, which are conveniently expressed by the notation,

LSX=[S,X],LS2X=[S,[S,X]],LS3X=[S,[S,[S,X]]],L_SX = [S,X], \quad L_S^2X=[S,[S,X]], \quad L_S^3X=[S,[S,[S,X]]],

etc., where XX is any operator and where LS0X=XL_S^0X=X. In this notation the two terms of (12) can be written

eSHe−S=eLSH=∑n=0∞1n!LSnH=H+LSH+12!LS2H+…,e^S H e^{-S} = e^{L_S} H = \sum_{n=0}^\infty {1\over n!} L_S^n H = H + L_S H + {1\over 2!}L_S^2 H + \ldots,

and

i∂eS∂te−S=i∑n=0∞1(n+1)!LSnS˙=i(S˙+12!LSS˙+13!LS2S˙+…),i\frac{\partial e^S}{\partial t}e^{-S} = i\sum_{n=0}^\infty {1\over (n+1)!} L_S^n {\dot S}=i\Bigl( {\dot S} + {1\over 2!}L_S{\dot S} +{1\over 3!}L_S^2{\dot S} +\ldots\Bigr),

where

S˙=∂S∂t.{\dot S}=\frac{\partial S}{\partial t}.

We now apply the series (14) to the three terms (6), arranging the terms so that (12) can be written

η0η1η2η3η4H′=mβ ⁣+ ⁣LS(mβ) ⁣+ ⁣12LS2(mβ) ⁣+ ⁣16LS3(mβ) ⁣+ ⁣124LS4(mβ) ⁣+ ⁣… ⁣+ ⁣O ⁣+ ⁣LSO ⁣+ ⁣12LS2O ⁣+ ⁣16LS3O ⁣+ ⁣… ⁣+ ⁣E ⁣+ ⁣LSE ⁣+ ⁣12LS2E ⁣+ ⁣… ⁣+ ⁣terms in S˙,\begin{array}{ccccccccccccc} && \eta^0 && \eta^1 && \eta^2 && \eta^3 && \eta^4 && \\[4pt] H' &=& m\boldsymbol{\beta} &\!+\!& L_S(m\boldsymbol{\beta}) &\!+\!& \frac{1}{2}L_S^2(m\boldsymbol{\beta}) &\!+\!& \frac{1}{6}L_S^3(m\boldsymbol{\beta}) &\!+\!& \frac{1}{24}L_S^4(m\boldsymbol{\beta}) &\!+\!& \ldots \\[4pt] &&&\!+\!& \odd &\!+\!& L_S\odd &\!+\!& \frac{1}{2}L_S^2\odd &\!+\!& \frac{1}{6}L_S^3\odd &\!+\!& \ldots \\[4pt] &&&&&\!+\!& \even &\!+\!& L_S\even &\!+\!& \frac{1}{2}L_S^2\even &\!+\!& \ldots \\[4pt] &&&&&&&\!+\!& \mathrlap{\text{terms in $\dot{S}$}}, &&&& \end{array}

The terms in S˙{\dot S} that are referred to come from the second term of (12), expanded according to (15).

The Generator SS

Our strategy will be to choose the generator SS so that

LS(mβ)=m[S,β]=−O,L_S(m\beta)=m[S,\beta]=-\odd,

thereby killing the two terms in the η1\eta^1 column of (17). Notice that this makes SS of order η\eta, confirming that powers of LSL_S correspond to increasing powers of η\eta, as indicated by the columns of terms in (17). By writing out SS as a 4×44\times4 matrix with four unknown 2×22\times2 submatrices and computing LS(mβ)L_S(m\beta) it is easy to solve for SS. We find

S=12mβO,S={1\over 2m}\beta\odd,

which is anti-Hermitian, as required. In manipulating these expressions it is worth noting that π\pivec is purely a spatial operator that commutes with the Dirac matrices α\alphavec, β\beta and Σ\Sigmavec, which are purely spin operators. The anticommutator

{β,O}=0\{\beta,\odd\}=0

is also useful.

Next, from

S=12mβ(α⋅π)=12mβα⋅(p−qA)S={1\over2m}\beta(\alphavec\cdot\pivec)= {1\over 2m}\beta\alphavec\cdot(\pvec-q\Avec)

we obtain

S˙=−q2mβα⋅∂A∂t,{\dot S} = -{q\over 2m}\beta\alphavec\cdot\frac{\partial \Avec}{\partial t},

which, by Table 1, is of order η5\eta^5 and thus off the charts in (17). We see that through fourth order, we can ignore all terms in S˙{\dot S}.

Now (18) allows us to combine the first two rows of (17), for example,

12LS2(mβ)+LSO=12LSO.{1\over2}L_S^2(m\beta) + L_S\odd = {1\over2}L_S\odd.

Then (17) becomes

η0η2η3η4H′=mβ ⁣+ ⁣12LSO ⁣+ ⁣13LS2O ⁣+ ⁣18LS3O ⁣+ ⁣… ⁣+ ⁣E ⁣+ ⁣LSE ⁣+ ⁣12LS2E ⁣+ ⁣…,\begin{array}{ccccccccccc} && \eta^0 && \eta^2 && \eta^3 && \eta^4 && \\[2pt] H' &=& m\boldsymbol{\beta} &\!+\!& \frac{1}{2}L_S\odd &\!+\!& \frac{1}{3}L_S^2\odd &\!+\!& \frac{1}{8}L_S^3\odd &\!+\!& \ldots \\[4pt] &&&\!+\!& \even &\!+\!& L_S\even &\!+\!& \frac{1}{2}L_S^2\even &\!+\!& \ldots, \end{array}

in which the odd column at order η\eta has disappeared.

This shows that we will need powers of LSL_S applied to O\odd. The first of these is

LSO=12m[βO,O]=12m(βO2−OβO)=1mβO2,L_S\odd ={1\over2m}[\beta\odd,\odd]={1\over2m} (\beta\odd^2-\odd\beta\odd)={1\over m}\beta\odd^2,

where we have used the anticommutator (20). In a similar manner we find

LS2O=−1m2O3,LS3O=−1m3βO4.L_S^2\odd = -{1\over m^2}\odd^3, \qquad L_S^3\odd = -{1\over m^3}\beta\odd^4.

The New Hamiltonian Through Second Order

Equations~(24) and (25) show that the second order Hamiltonian involves the operator O2\odd^2, which we put into a useful form as follows. First, noting that all components of π\pivec commute with all components of α\alphavec, we have

O2=(α⋅π)2=αiπiαjπj=αiαjπiπj.\odd^2 = (\alphavec\cdot\pivec)^2 = \alpha_i\pi_i \alpha_j\pi_j = \alpha_i\alpha_j\pi_i\pi_j.

Next, by using the definitions (4) and (5) we verify the useful identity,

αiαj=δij+iϵijk Σk,\alpha_i\alpha_j = \delta_{ij} + i\epsilon_{ijk}\, \Sigma_k,

which shows that

O2=π2+iϵijk Σkπiπj.\odd^2 = \pivec^2 + i\epsilon_{ijk}\,\Sigma_k \pi_i\pi_j.

In the final term in this equation we note that ϵijk\epsilon_{ijk} is antisymmetric in (ij)(ij), so this term can be written

i2ϵijk Σk(πiπj−πjπi)=i2ϵijk Σk [πi,πj].{i\over2}\epsilon_{ijk}\,\Sigma_k(\pi_i\pi_j-\pi_j\pi_i) ={i\over2}\epsilon_{ijk}\,\Sigma_k\,[\pi_i,\pi_j].

But it follows from the definition (2) that

[πi,πj]=iq ϵijℓ Bℓ,[\pi_i,\pi_j]=iq\,\epsilon_{ij\ell}\, B_\ell,

which are essentially the velocity commutators previously derived as (5). These commutators also appeared in Prob. Paragraph. Thus the expression (30) can be written

−q2ϵijk Σk ϵijℓ Bℓ=−q δkℓΣkBℓ=−qΣ⋅B,-{q\over 2} \epsilon_{ijk}\, \Sigma_k \, \epsilon_{ij\ell}\, B_\ell = -q\,\delta_{k\ell} \Sigma_k B_\ell = -q\Sigmavec\cdot\Bvec,

where we use (54). Altogether, we obtain

O2=π2−qΣ⋅B.\odd^2 = \pivec^2 -q\Sigmavec\cdot\Bvec.

This work allows us to write the new Hamiltonian through second order as

H′=mβ+12mβ(π2−qΣ⋅B)+qΦ+O(η3).H' = m\beta + {1\over2m}\beta(\pivec^2-q\Sigmavec\cdot\Bvec) +q\Phi + O(\eta^3).

To second order in η\eta the Hamiltonian has been block-diagonalized. The upper 2-components of the transformed Dirac equation through second order have now become the usual Pauli equation (1), with g=2g=2 and the constant term mc2mc^2 added to the Hamiltonian. The calculation thus far is equivalent to what we did in an ad hoc manner in The Nonrelativistic Limit of the Dirac Equation, but we have all the higher order terms and can continue with the expansion.

A Second Foldy-Wouthuysen Transformation

The transformation so far has eliminated the odd contribution to the Hamiltonian at order η\eta, but the odd term at order η3\eta^3 remains. It is

H3′=13LS2O+LSE.H'_3 = {1\over3}L_S^2\odd + L_S\even.

To eliminate this term we must carry out a second Foldy-Wouthousen transformation, this time with a generator S′S' that takes H′H' into a new Hamiltonian H′′H''. That is, we write

H′′=eS′H′e−S′+i∂eS′∂te−S′,H''=e^{S'}H'e^{-S'} + i\pop{e^{S'}}{t}e^{-S'},

a version of (12), taken to the next step. Expanding the exponentials in power series as before, we find that to eliminate H3′H'_3 we must find S′S' such that

LS′(mβ)=m[S′,β]=−H3′.L_{S'}(m\beta)=m[S',\beta]=-H'_3.

But we do not need to solve this equation explicitly, we only need to note that a solution exists, and that it gives S′S' which is of order η3\eta^3.

The New Hamiltonian at Fourth Order

With this knowledge we can see that all other commutators and correction terms in (36) are of order η5\eta^5 and higher. Thus we find

H4′′=H4′=18LS3O+12LS2E=−18m3βO4+12LS2E,H''_4 = H'_4 = {1\over8}L_S^3\odd + {1\over2}L_S^2\even = -{1\over 8m^3}\beta\odd^4 + {1\over2}L_S^2\even,

where we use (26). As for O4\odd^4, if we only want this to fourth order in η\eta we can ignore the correction term in (33), which by Table 1 is of order η4\eta^4. For the same reason we can ignore the correction term in Eq.(2), which according to the table is of order η3\eta^3. Thus

O4=(π2−qΣ⋅B)2→π4=(p−qA)4→p4,\odd^4 = (\pivec^2-q\Sigmavec\cdot\Bvec)^2 \to \pivec^4 = (\pvec-q\Avec)^4 \to p^4,

where →\to means dropping terms of order η5\eta^5 or higher. Altogether, the first term in H4′′H''_4 in (38) becomes

18LS3O→−βp48m3.{1\over8}L_S^3\odd \to -\beta{p^4\over 8m^3}.

To compute the second term in (38) we start with

LSE=12m[βO,E]=q2m[βαiπi,Φ]=q2mβαi[πi,Φ],L_S\even = {1\over 2m}[\beta\odd,\even] ={q\over2m}[\beta\alpha_i\pi_i,\Phi] ={q\over2m}\beta\alpha_i[\pi_i,\Phi],

since β\beta and αi\alpha_i are spin operators and πi\pi_i and Φ\Phi are spatial. But

[πi,Φ]=[pi,Φ]=−i∂Φ∂xi→iEi,[\pi_i,\Phi] = [p_i,\Phi] = -i\frac{\partial \Phi}{\partial x_i} \to i E_i,

where in the last step we drop the transverse electric field which is of higher order, as noted in Table 1. Thus we find

LSE=iq2mβ(α⋅E),L_S\even = {iq\over 2m}\beta(\alphavec\cdot\Evec),

to the order required.

We must now apply LSL_S to this. We obtain

LS2E=iq4m2[βO,β(α⋅E)]=iq4m2[βOβ(α⋅E)−β(α⋅E)βO]=−iq4m2[O,α⋅E]→−iq4m2[α⋅p,α⋅E],\begin{aligned} L_S^2\even &= {iq\over 4m^2}[\beta\odd, \beta(\alphavec\cdot\Evec)] = {iq\over 4m^2} [\beta\odd\beta(\alpha\cdot\Evec)- \beta(\alphavec\cdot\Evec)\beta\odd] =-{iq\over 4m^2}[\odd,\alphavec\cdot\Evec] \\ &\to -{iq\over 4m^2}[\alphavec\cdot\pvec,\alphavec\cdot\Evec], \\ \end{aligned}

where we use the anticommutator {β,αi}=0\{\beta,\alpha_i\}=0 and drop the correction term in (2). The remaining commutator is

[α⋅p,α⋅E]=αipiαjEj−αiEiαjpj=αiαj(piEj−Eipj)=(δij+iϵijk Σk)(piEj−Eipj)=p⋅E−E⋅p+iΣ⋅(p×E−E×p),\begin{aligned} [\alphavec\cdot\pvec,\alphavec\cdot\Evec] &= \alpha_ip_i\alpha_jE_j-\alpha_iE_i\alpha_jp_j =\alpha_i\alpha_j(p_iE_j-E_ip_j) \\ &=(\delta_{ij}+i\epsilon_{ijk}\,\Sigma_k)(p_iE_j-E_ip_j) = \pvec\cdot\Evec-\Evec\cdot\pvec +i\Sigmavec\cdot(\pvec\cross\Evec-\Evec\cross\pvec), \\ \end{aligned}

where we have used the identity (28). This can be simplified with

p⋅E−E⋅p=[pi,Ei]=−i∇⋅E.\pvec\cdot\Evec-\Evec\cdot\pvec = [p_i,E_i] = -i\del\cdot\Evec.

Now putting the pieces together we obtain the total fourth order Hamiltonian as

H4′′=18LS3O+12LS2E=−βp48m3−q8m2∇⋅E+q8m2Σ⋅(p×E−E×p),H''_4 = {1\over8}L_S^3\odd+{1\over2}L_S^2\even = -\beta{p^4\over 8m^3} -{q\over8m^2}\del\cdot\Evec + {q\over8m^2} \Sigmavec\cdot(\pvec\cross\Evec-\Evec\cross\pvec),

to the order required. The Hamiltonian is now block-diagonalized through fourth order in v/cv/c. The total Hamiltonian through fourth order is obtained by adding this term to (34).

The Upper Two Components

To obtain the effective 2-component Hamiltonian for the upper two components of the transformed Dirac spinor ψ′′\psi'' we just replace the Dirac matrices β\beta and Σ\Sigmavec by their upper 2×22\times2 blocks, which are 1 and σ\sigmavec, respectively. Writing simply HH for this Hamiltonian, it is

H=m+π22m−q2mσ⋅B+qΦ−p48m3−q8m2∇⋅E+q8m2σ⋅(p×E−E×p).H=m + {\pivec^2\over 2m} -{q\over2m}\sigmavec\cdot\Bvec +q\Phi -{p^4\over 8m^3} -{q\over8m^2}\del\cdot\Evec +{q\over8m^2}\sigmavec\cdot(\pvec\cross\Evec-\Evec\cross\pvec).

We restore ordinary units to this, obtaining

H=mc2+12m(p−qcA)2−gqℏ2mcS⋅B+qΦ−p48m3c2−qℏ28m2c2∇⋅E−q4m2c2S⋅(p×E−E×p),\begin{aligned} H & = mc^2 + {1\over2m}\Bigl(\pvec-{q\over c}\Avec\Bigr)^2 -g{q\hbar\over2mc}\Svec\cdot\Bvec + q\Phi -{p^4\over 8m^3c^2} -{q\hbar^2\over8m^2c^2}\del\cdot\Evec \\ & \qquad-{q\over4m^2c^2}\Svec\cdot(\pvec \cross\Evec-\Evec \cross\pvec), \end{aligned}

where we write π=p−(q/c)A\pivec=\pvec-(q/c)\Avec, S=(ℏ/2)σ\Svec=(\hbar/2)\sigmavec and g=2g=2.

The last three terms of this Hamiltonian are the correction terms at order (v/c)4(v/c)^4. The first of these is the relativistic kinetic energy correction, denoted HRKEH_RKE in (5). In the second of these we write E=−∇Φ\Evec=-\del\Phi and V=qΦV=q\Phi, so that it becomes

ℏ28m2c2∇2V,{\hbar^2\over 8m^2c^2}\del^2V,

which is precisely the Darwin term given previously in (23). The final term in (49) is the spin-orbit term, which which we have previously considered only in the case of a central force potential. To specialize to that case, we assume Φ=Φ(r)\Phi=\Phi(r) so that

qE=−q∇Φ(r)=−qxrdΦdr=−xrdVdr.q\Evec=-q\del\Phi(r) = -q{\xvec\over r}\dod{\Phi}{r} = -{\xvec\over r} \dod{V}{r}.

Then

qp×E=−(p×x)1rdVdr=1rdVdrL=−qE×p,q\pvec\cross\Evec=-(\pvec\cross\xvec){1\over r}\dod{V}{r} = {1\over r}\dod{V}{r}\Lvec = -q\Evec\cross\pvec,

so that for central force potentials the final term of (49) can be written

12m2c21rdVdrL⋅S,{1\over 2m^2c^2}{1\over r}\dod{V}{r} \Lvec\cdot\Svec,

which is the spin-orbit term seen previously in (13).

Discussion

In the pseudo-derivation of the spin-orbit term given in 25. Fine Structure in Hydrogen and Alkali Atoms{} we glossed over the derivation of the extra factor of 1/21/2 due to Thomas precession, because it would have been a substantial excursion to go into it in detail. Nevertheless, it is interesting to see how rotations can be generated by the composition of boosts, and the physical consequences of these. The Dirac equation, however, implicitly knows about Thomas precession, and about how the extra rotation effectively cancels out one half of the precession of the spin in the electron’s rest frame. The Dirac equation also implicitly knows that the gg factor of the electron is 2 (very nearly).

In the case of hydrogen the Dirac equation gives results that are in agreement with spectroscopic data to high precision, the main discrepancies being the hyperfine structure and the Lamb shift, both of which involve extra physics not incorporated into the Dirac equation. The same can be said for the corrections to the gg-factor, which along with the Lamb shift are due to interactions with the electromagnetic field (they are so-called “radiative corrections”). Of course there are also corrections due to the finite mass of the nucleus, which are mostly accounted for by replacing the electron mass mm by the reduced mass μ\mu. To account for these in a manner that is fully satisfactory from a fundamental standpoint, however, is more challenging.

Perhaps the most important point of these results concerns our narrative on the development of our understanding of the Dirac equation. The ingredients in the Dirac equation are the simplest that can be conceived for a relativistic wave equation for a particle with spin. The implementation of the Dirac algebra is the simplest, and the coupling to the electromagnetic field is minimal. The fact that such a large amount of physically correct detail, in excellent agreement with experiment, emerges from these assumptions is impressive evidence that the Dirac equation is an important and largely correct step in the merging of quantum mechanics with special relativity. Nothing we have said so far, however, sheds any light on the meaning of the negative energy solutions of the Dirac equation, which are the main conceptual difficulty that must be solved before we have a fully satisfactory framework for understanding relativistic quantum mechanics. The next few sets of notes outline the resolution of this problem.

Problems

Problem 1.

Equation~(\eqr\cn.49) gives the effective 2-component Hamiltonian for the upper two components of the transformed Dirac wave function ψ′′\psi''. In a similar manner write out the effective 2-component Hamiltonian for the lower two components. Is it identical with or related to any realistic 2-component wave equation of the Pauli type?

Problem 2.

Specialize Eq.~(\eqr\covariance.116) to the case of the neutron, as an example of a problem with an anomalous magnetic moment. Remember that q=0q=0, which simplifies the calculation. Carry out a Foldy-Wouthuysen transformation to fourth order, as in these notes.