Finite perturbation theory for the relativistic Coulomb problem

We are motivated by the success of nonrelativistic quantum mechanics, in which the total Hamiltonian is written as the sum of a free part and a part responsible for interaction (a potential $\hat{V}$):

$$\hat{H}_{i}=\hat{H}_{f}+\hat{V}.$$

(1)

Here $f$ indicates free, $i$ indicates interacting, hats indicate operators.

Thus we propose for relativistic quantum mechanics the four equations ($\mu=0,1,2,3$)

$$\hat{P}_{i}^{\mu}=\hat{P}_{f}^{\mu}+\hat{V}^{\mu}.$$

(2)

Here the potential of nonrelativistic quantum mechanics becomes the zero component of what must transform as a four-vector, as do the components of $P_{f}^{\mu}$ and $P_{i}^{\mu},$ to guarantee relativistic covariance. This is

$$U^{\dagger}(L)\,V^{\mu}\,U(L)=L_{\phantom{\mu}\nu}^{\mu}\,V^{\nu},$$

(3)

where $U(L)$ is the unitary representation of a general Lorentz transformation, $L,$ and $L_{\phantom{\mu}\nu}^{\mu}$ is the corresponding coordinate transformation matrix. For processes involving photons, we assert that $V^{\mu}$ must be gauge invariant to arrive at physical results independent of the gauge.

For the scattering of two spinless “electrons”, we choose as free basis vectors $|\,P,l,m;f\,\rangle.$ In the CM frame, these are defined in terms of the individual momentum eigenvectors as

$$|\,(2\omega^{\mathrm{\scriptscriptstyle CM}},\boldsymbol{0}),l,m;f\,\rangle=\int d^{2}\hat{p}^{\mathrm{\scriptscriptstyle CM}}\,|\,+\boldsymbol{p}^{\mathrm{\scriptscriptstyle CM}},-\boldsymbol{p}^{\mathrm{\scriptscriptstyle CM}};f\,\rangle\,p^{\mathrm{\scriptscriptstyle CM}}\,Y_{lm}(\hat{\boldsymbol{p}}^{\mathrm{\scriptscriptstyle CM}}).$$

(4)

Then a boost takes them to general four-momentum

$$|\,P,l,m;f\,\rangle=U(\Lambda(\frac{\boldsymbol{P}}{E}))\,|\,(2\omega^{\mathrm{\scriptscriptstyle CM}},0),l,m;f\,\rangle.$$

(5)

We can ignore the total linear momentum, $\boldsymbol{P},$ by using the normalization

$$\langle\,+\boldsymbol{p}_{1}^{\mathrm{\scriptscriptstyle CM}},-\boldsymbol{p}_{1}^{\mathrm{\scriptscriptstyle CM}};f\,|\,+\boldsymbol{p}_{2}^{\mathrm{\scriptscriptstyle CM}},-\boldsymbol{p}_{2}^{\mathrm{\scriptscriptstyle CM}};f\,\rangle=\delta^{3}(\boldsymbol{p}_{1}^{\mathrm{\scriptscriptstyle CM}}-\boldsymbol{p}_{2}^{\mathrm{\scriptscriptstyle CM}}),$$

(6)

which has no consequences.

These are eigenvectors of total four-momentum $P^{\mu}=(E,\boldsymbol{P})^{\mu}.$ They are also eigenvectors of $\boldsymbol{L}^{2}$ and $L_{z},$ where $\boldsymbol{L}$ is the total angular momentum (which in this case is entirely orbital angular momentum) in the CM frame (where $\boldsymbol{P}=0$). In the more general case, where the two particles can be any combination of electrons, positrons and photons, the basis vectors take the form $|P,J,M,\lambda_{1},\lambda_{2};f\,\rangle,$ where $J$ and $M$ are the quantum numbers associated with the total CM-frame angular momentum, $\boldsymbol{J},$ and $\lambda_{1}$ and $\lambda_{2}$ are the CM-frame helicities of the two particles.

It is advantageous to work in momentum space, where the square roots in the single-particle energies, $\omega(\boldsymbol{p})=\sqrt{\boldsymbol{p}^{2}+m_{e}^{2}},$ pose no problem. At this point there is no need to introduce spacetime coordinates for the two particles. Note that to see a potential with spacetime dependence requires taking matrix elements between two state vectors, at least one of which has spacetime dependence.

Corresponding to each free basis vector is an interacting basis vector that solves the Schrödinger equations (eq. (2)). We will explore this correspondence shortly.

We take matrix elements

$$\langle\,P_{f},l_{f},m_{f};f\,|\,\hat{P}_{i}^{\mu}\,|\,P_{i},l_{i},m_{i};i\,\rangle=\langle\,P_{f},l_{f},m_{f};f\,|\,\hat{P}_{f}^{\mu}+\hat{V}^{\mu}\,|\,P_{i},l_{i},m_{i};i\,\rangle,$$

(7)

which reduce to

	$\displaystyle(P_{f}^{\mu}-P_{i}^{\mu})\,\langle\,P_{f},l_{f},m_{f};f\,|\,P_{i},l_{i},m_{i};i\,\rangle$	$\displaystyle=-\langle\,P_{f},l_{f},m_{f};f\,|\,\hat{V}^{\mu}\,|\,P_{i},l_{i},m_{i};i\,\rangle$
		$\displaystyle=-\int d^{4}P_{f}^{\prime}\sum_{l_{f}^{\prime}=0}^{\infty}\sum_{m_{f}^{\prime}=-l_{f}^{\prime}}^{l_{f}^{\prime}}\langle\,P_{f},l_{f},m_{f};f\,|\,\hat{V}^{\mu}\,|\,P_{f}^{\prime},l_{f}^{\prime},m_{f}^{\prime};f\,\rangle\langle\,P_{f}^{\prime},l_{f}^{\prime},m_{f}^{\prime};f\,|\,P_{i},l_{i},m_{i};i\,\rangle,$		(8)

a set of integral equations. We ensure that the free basis vectors are orthonormalized and complete so that we can insert the completeness relation as we have just done.

A simplification is possible. We will find below that it is possible to define a unitary transformation that relates free and interacting basis vectors with the same quantum numbers:

$$|\,P,l,m;i\,\rangle=U_{if}\,|\,P,l,m;f\,\rangle.$$

(9)

Note that in the quantum electrodynamics formalism (Weinberg [5]), the electron is given a free mass different from its mass after including electromagnetic interactions. But the difference of these masses is calculated to diverge and only the interacting mass is observable. As we will see, the correspondence proposed here produces no divergences.

The transformation of operators that gives the same expectation values is

$$\langle\,P,l,m;i\,|\,P_{i}^{\mu}\,|\,P,l,m;i\,\rangle=\langle\,P,l,m;f\,|\,P_{f}^{\mu}\,|\,P,l,m;f\,\rangle\Rightarrow P_{i}^{\mu}=U_{if}\,P_{f}^{\mu}\,U_{if}^{\dagger}.$$

(10)

Note that we will have similar relations for the transformation of rotation generators ($\boldsymbol{J}$) and boost generators ($\boldsymbol{K}$):

$$J_{i}^{k}=U_{if}\,J_{f}^{k}\,U_{if}^{\dagger}\quad\mathrm{and}\quad K_{i}^{k}=U_{if}\,K_{f}^{k}\,U_{if}^{\dagger}\quad\mathrm{for}\ k=1,2,3.$$

(11)

Consequently, if

$$\hat{\boldsymbol{P}}_{f}\,|\,P,l,m;f\,\rangle=\boldsymbol{0},$$

(12)

then

$$\hat{\boldsymbol{P}}_{i}\,|\,P,l,m;i\,\rangle=U_{if}\,\hat{\boldsymbol{P}}_{f}\,U_{if}^{\dagger}\,U_{if}\,|\,P,l,m;f\,\rangle=U_{if}\,\hat{\boldsymbol{P}}_{f}\,|\,P,l,m;f\,\rangle=\boldsymbol{0}.$$

(13)

So, from eq. (), we see the vanishing of the matrix elements of the three-vector part of the potential, $\hat{\boldsymbol{V}},$ in the one CM frame that is the same in the free and interacting theories:

$$\langle\,(E_{f},\boldsymbol{0}),l_{f},m_{f};f\,|\,\hat{\boldsymbol{V}}\,|\,(E_{i},\boldsymbol{0}),l_{i},m_{i};i\,\rangle=\boldsymbol{0}.$$

(14)

In what follows, we will perform calculations only in the CM frame, knowing that we could transform differential cross sections to other frames, if necessary.

We define $\hat{V}=\hat{V}^{0}.$ We know

$$\langle\,P_{f},l_{f},m_{f};f\,|\,P_{i},l_{i},m_{i};i\,\rangle\propto\delta^{3}(\boldsymbol{P}_{f}-\boldsymbol{P}_{i})$$

(15)

by translational invariance, so we can remove all reference to the three-momenta and arrive at the simplified equation in the CM frame (understanding $\boldsymbol{P}_{f}=\boldsymbol{P}_{i}=0$ throughout)

$$(E_{f}-E_{i})\,\langle\,E_{f},l_{f},m_{f};f\,|\,E_{i},l_{i},m_{i};i\,\rangle=-\int_{2m_{e}}^{\infty}dE_{f}^{\prime}\sum_{l_{f}^{\prime}=0}^{\infty}\sum_{m_{f}^{\prime}=-l_{f}^{\prime}}^{l_{f}^{\prime}}\langle\,E_{f},l_{f},m_{f};f\,|\,\hat{V}\,|\,E_{f}^{\prime},l_{f}^{\prime},m_{f}^{\prime};f\,\rangle\langle\,E_{f}^{\prime},l_{f}^{\prime},m_{f}^{\prime};f\,|\,E_{i},l_{i},m_{i};i\,\rangle.$$

(16)

We are considering the case of a potential that is rotationally invariant ($[\boldsymbol{J}_{i},\hat{V}]=0$). Thus, by the Wigner-Eckhart theorem (Messiah [6]), the matrix elements take the form

$$\langle\,E_{f},l_{f},m_{f};f\,|\,\hat{V}\,|\,E_{f}^{\prime},l_{f}^{\prime},m_{f}^{\prime};f\,\rangle\propto\delta_{l_{f}l_{f}^{\prime}}\delta_{m_{f}m_{f}^{\prime}}.$$

(17)

Furthermore, if the potential is rotationally invariant, we expect that the free to interacting transformation will also be rotationally invariant, preserving the quantum numbers $l$ and $m$. A further reduction, ignoring $m,$ gives

$$(E_{f}-E_{i})\,\langle\,E_{f},l;f\,|\,E_{i},l;i\,\rangle=-\int_{2m_{e}}^{\infty}dE_{f}^{\prime}\langle\,E_{f},l;f\,|\,\hat{V}\,|\,E_{f}^{\prime},l;f\,\rangle\langle\,E_{f}^{\prime},l;f\,|\,E_{i},l;i\,\rangle.$$

(18)

The phase shifts that we want to calculate come from the asymptotic behaviour of interacting position wavefunctions as the separation, $r,$ of the two particles increases without bound. Thus we need to construct eigenvectors of the operator associated with this separation. Note that we cannot here use the techniques of nonrelativistic quantum mechanics, where the kinetic energy depends quadratically on the momentum, to form the total linear momentum and the relative momentum.

We consider two different individual particle translations applied to a free individual momentum eigenvector:

$$U(T_{1}(+\boldsymbol{r}/2))\,U(T_{2}(-\boldsymbol{r}/2))\,|\,\frac{1}{2}\boldsymbol{P}+\boldsymbol{p},\frac{1}{2}\boldsymbol{P}-\boldsymbol{p};f\,\rangle=|\,\frac{1}{2}\boldsymbol{P}+\boldsymbol{p},\frac{1}{2}\boldsymbol{P}-\boldsymbol{p}\,\rangle\,e^{-i\boldsymbol{p}\cdot\boldsymbol{r}}.$$

(19)

This transformation changes the separation of the two particles by $\boldsymbol{r}$ but leaves the average position unchanged. So we define simultaneous eigenvectors of separation and total three-momentum as

$$|\,\boldsymbol{r},\boldsymbol{P};f\,\rangle=\int\frac{d^{3}p}{(2\pi)^{\frac{3}{2}}}\,|\,\frac{1}{2}\boldsymbol{P}+\boldsymbol{p},\frac{1}{2}\boldsymbol{P}-\boldsymbol{p};f\,\rangle\,e^{-i\boldsymbol{p}\cdot\boldsymbol{r}},$$

(20)

noting that

$$\hat{\boldsymbol{r}}=i\frac{\partial}{\partial\boldsymbol{p}}=i\frac{\partial}{\partial\boldsymbol{p}_{1}}-i\frac{\partial}{\partial\boldsymbol{p}_{2}}$$

(21)

and $\hat{\boldsymbol{P}}$ commute.

Then we form eigenvectors of the magnitude of the separation and of orbital angular momentum,

$$|\,r,l,m,\boldsymbol{P};f\,\rangle=\int d^{2}\hat{r}\,|\,\boldsymbol{r},\boldsymbol{P};f\,\rangle\,r\,Y_{lm}(\hat{\boldsymbol{r}}),$$

(22)

where the argument of the spherical harmonic is here a unit vector, not an operator. We note that $\hat{\boldsymbol{L}}=\hat{\boldsymbol{r}}\times\hat{\boldsymbol{p}}$ defined in the CM frame also commutes with $\boldsymbol{P}.$

Note that we will only be requiring these basis vectors for the special case $\boldsymbol{P}=\boldsymbol{0},$ where $r$ is the separation and $l,m$ label the orbital angular momentum, both in the CM frame. Note also that these basis vectors will have calculable but very complicated Lorentz transformation properties, as with any position eigenvectors.

If we choose the orthonormalization

$$\langle\,\frac{1}{2}\boldsymbol{P}_{a}+\boldsymbol{p}_{a},\frac{1}{2}\boldsymbol{P}_{a}-\boldsymbol{p}_{a};f\,|\,\frac{1}{2}\boldsymbol{P}_{b}+\boldsymbol{p}_{b},\frac{1}{2}\boldsymbol{P}_{b}-\boldsymbol{p}_{b};f\,\rangle=\delta^{3}(\boldsymbol{P}_{a}-\boldsymbol{P}_{b})\,\delta^{3}(\boldsymbol{p}_{a}-\boldsymbol{p}_{b}),$$

(23)

then we find

$$\langle\,r_{a},l_{a},m_{a},\boldsymbol{P}_{a};f\,|\,r_{b},l_{b},m_{b},\boldsymbol{P}_{b};f\,\rangle=\delta^{3}(\boldsymbol{P}_{a}-\boldsymbol{P}_{b})\,\delta(r_{a}-r_{b})\,\delta_{l_{a}l_{b}}\delta_{m_{a}m_{b}}.$$

(24)

We find the relation between eigenvectors of momentum magnitude and eigenvectors of separation magnitude:

$$\langle\,r,l_{a},m_{a},\boldsymbol{P}_{a};f\,|\,p,l_{b},m_{b},\boldsymbol{P}_{b};f\,\rangle=\delta^{3}(\boldsymbol{P}_{a}-\boldsymbol{P}_{b})\,\delta_{l_{a}l_{b}}\delta_{m_{a}m_{b}}\,\sqrt{\frac{2}{\pi}}\,pr\,j_{l_{a}}(pr)=\delta^{3}(\boldsymbol{P}_{a}-\boldsymbol{P}_{b})\,\delta_{l_{a}l_{b}}\delta_{m_{a}m_{b}}\,y_{l_{a}}^{(f)}(r,p),$$

(25)

in terms of the free spherical waves, $y_{l_{a}}^{(f)}(r,p),$ familiar from partial wave analysis (Messiah [6]). This motivates our definition for the interacting case,

$$\langle\,r,l_{a},m_{a},P_{a};f\,|\,p,l_{b},m_{b},P_{b};i\,\rangle\equiv\delta^{3}(\boldsymbol{P}_{a}-\boldsymbol{P}_{b})\,\delta_{l_{a}l_{b}}\delta_{m_{a}m_{b}}\,y_{l_{a}}^{(i)}(r,p),$$

(26)

in terms of the interacting wavefunctions, $y_{l_{a}}^{(i)}(r,p).$

Note that the position eigenvectors are common to the free and interacting theories. We do not define interacting position eigenvectors.

A unitary transformation connecting the free and interacting theories, in the nonrelativistic case, was introduced in Hoffmann [3], and a perturbation series was developed to second order in the coupling strength. Only minor changes are needed to treat the relativistic case.

In the CM frame, the total energies are

$$E(p)=2\sqrt{p^{2}+m_{e}^{2}},$$

(27)

where $p$ is the magnitude of both single-particle momenta. Then a change of quantum numbers to eq. (18) gives

$$(E(k)-E(p))\,\langle\,k,l;f\,|\,p,l;i\,\rangle=-\int_{0}^{\infty}dk^{\prime}\,\langle\,k,l;f\,|\,\hat{V}\,|\,k^{\prime},l;f\,\rangle\langle\,k^{\prime},l;f\,|\,p,l;i\,\rangle.$$

(28)

The normalizations are

$$\langle\,E_{1},l;f/i\,|\,E_{2},l;f/i\,\rangle=\delta(E_{1}-E_{2})\Leftrightarrow\langle\,p_{1},l;f/i\,|\,p_{2},l;f/i\,\rangle=\delta(p_{1}-p_{2}).$$

(29)

We expand the unitary transformation to second order in $\alpha$:

$$U_{if}=e^{-i\Theta}\cong 1-i\alpha\,\Theta^{(1)}-\frac{i}{2}\alpha^{2}\Theta^{(2)}-\frac{1}{2}\alpha^{2}\Theta^{(1)2},$$

(30)

where

$$\Theta\cong\alpha\,\Theta^{(1)}+\frac{1}{2}\alpha^{2}\,\Theta^{(2)}.$$

(31)

Then we solve the equation

$$H_{i}+V=U_{if}\,H_{f}\,U_{if}^{\dagger}$$

(32)

order by order, with $V=\mathcal{O}(\alpha)$. The results are

$$i[H_{f},\alpha\,\Theta^{(1)}]=V\quad\mathrm{and}\quad[H_{f},\alpha^{2}\,\Theta^{(2)}]=[\alpha\,\Theta^{(1)},V].$$

(33)

The solutions for the matrix elements of $\Theta^{(1)}$ and $\Theta^{(2)}$ are

$$\langle\,k_{1},l;f\,|\,\alpha\,\Theta^{(1)}\,|\,k_{2},l;f\,\rangle=-iP\frac{\mathcal{V}_{l}(k_{1},k_{2})}{E(k_{1})-E(k_{2})}$$

(34)

and

$$\langle\,k_{1},l;f\,|\,\alpha^{2}\,\Theta^{(2)}\,|\,k_{2},l;f\,\rangle=-iP\,\frac{1}{E(k_{1})-E(k_{2})}\int_{0}^{\infty}dk^{\prime}\,\left\{\frac{\mathcal{V}_{l}(k_{1},k^{\prime})\mathcal{V}_{l}(k^{\prime},k_{2})}{E(k_{1})-E(k^{\prime})}-\frac{\mathcal{V}_{l}(k_{1},k^{\prime})\mathcal{V}_{l}(k^{\prime},k_{2})}{E(k^{\prime})-E(k_{2})}\right\},$$

(35)

where

$$\mathcal{V}_{l}(k_{1},k_{2})=\langle\,k_{1},l;f\,|\,V\,|\,k_{2},l;f\,\rangle.$$

(36)

Also

$$\langle\,k_{1},l;f\,|\,\alpha^{2}\,\Theta^{(1)2}\,|\,k_{2},l;f\,\rangle=-P\int_{0}^{\infty}dk^{\prime}\,\frac{\mathcal{V}_{l}(k_{1},k^{\prime})}{E(k_{1})-E(k^{\prime})}\frac{\mathcal{V}_{l}(k^{\prime},k_{2})}{E(k^{\prime})-E(k_{2})}.$$

(37)

The symbol $P$ indicates that when these expressions are integrated over momentum, the principal part prescription (also known as finding the Cauchy principal value) must be used. We have seen in (Hoffmann [3]) that use of this prescription eliminates divergences and give results in agreement with an exactly solvable model.

Then the interacting position separation wavefunction to second order is

	$\displaystyle y_{l}^{(2)}(r,p)$	$\displaystyle=\langle\,r,l;f\,|\,U_{if}^{(2)}\,|\,p,l;f\,\rangle$
		$\displaystyle=y_{l}^{(f)}(r,p)-P\int_{0}^{\infty}dk\,y_{l}^{(f)}(r,k)\,\frac{\mathcal{V}_{l}(k,p)}{E(k)-E(p)}+\frac{1}{2}P\int_{0}^{\infty}dk\,y_{l}^{(f)}(r,k)\int_{0}^{\infty}dk^{\prime}\,\frac{\mathcal{V}_{l}(k,k^{\prime})\mathcal{V}_{l}(k^{\prime},p)}{(E(k)-E(k^{\prime}))(E(k^{\prime})-E(p))}$
		$\displaystyle\quad-\frac{1}{2}P\int_{0}^{\infty}dk\,y_{l}^{(f)}(r,k)\int_{0}^{\infty}dk^{\prime}\,\left\{\frac{\mathcal{V}_{l}(k,k^{\prime})\mathcal{V}_{l}(k^{\prime},p)}{(E(k)-E(p))(E(k)-E(k^{\prime}))}-\frac{\mathcal{V}_{l}(k,k^{\prime})\mathcal{V}_{l}(k^{\prime},p)}{(E(k)-E(p))(E(k^{\prime})-E(p))}\right\}.$		(38)

For the Coulomb potential in the CM frame, the matrix elements are

	$\displaystyle\mathcal{V}_{l}(k_{1},k_{2})$	$\displaystyle=\int_{0}^{\infty}dr\,y_{l}^{(f)}(r,k_{1})\,\frac{\alpha}{r}\,y_{l}^{(f)}(r,k_{2})$
		$\displaystyle=\frac{\alpha}{\sqrt{\pi}}\,\frac{l!}{\Gamma(l+\frac{3}{2})}\,\begin{cases}\left(\rho\right)^{l+1}\,_{2}F_{1}(\frac{1}{2},l+1;l+\frac{3}{2};\rho^{2})&\rho<1,\\ \left(\frac{1}{\rho}\right)^{l+1}\,_{2}F_{1}(\frac{1}{2},l+1;l+\frac{3}{2};\frac{1}{\rho^{2}})&\rho>1,\end{cases}$		(39)

with $\rho=k_{1}/k_{2}.$ Both sections of this function are singular at $k_{1}=k_{2}.$ Close to the singularity, with $k_{1}=k_{2}(1+x)$ for small $x,$ we find the approximation (Olver et al. [7], their eq. (15.8.10))

$$\mathcal{V}_{l}(k_{2}(1+x),k_{2})\rightarrow\frac{\alpha}{\pi}(C_{l}-\ln|x|)+\mathcal{O}(x),$$

(40)

with

$$C_{l}=\ln 2-\boldsymbol{C}-\psi(l+1)$$

(41)

and

$$\boldsymbol{C}=0.577215\dots$$

(42)

known as Euler’s constant. Here $\psi(z)$ is the logarithmic derivative of the gamma function (Gradsteyn and Ryzhik [8], their section 8.36).

These functions $\mathcal{V}_{l}(k_{1},k_{2})$ have the property that they have a singular point but are integrable across the singularity. This will be sufficient for the principal part integrals to converge to finite values.

To find the asymptotic form of $y_{l}^{(2)}(r,p)$ in eq. (38), we first use the asymptotic forms of the free spherical waves (Messiah [6])

$$y_{l}^{(f)}(r,k)\rightarrow\sqrt{\frac{2}{\pi}}\,\sin(kr-l\frac{\pi}{2}).$$

(43)

For the first-order term (with $k=p(1+x)$)

$$I^{(1)}\rightarrow-P\int_{-1}^{\infty}dx\,\sqrt{\frac{2}{\pi}}\,\{\sin(pr-l\frac{\pi}{2})\cos(prx)+\cos(pr-l\frac{\pi}{2})\sin(prx)\}\,\frac{\mathcal{V}_{l}(p(1+x),p)}{x}\left\{\frac{E(p(1+x))+E(p)}{4p(2+x)}\right\}.$$

(44)

We review the rules we learned from the similar case in Hoffmann [3]:

• •

  For the $\cos(prx)$ and $\sin(prx)$ integrals, the contributions from $x>1$ will vanish asymptotically like $1/pr,$ since the integrand has no singularities on that region and the sinusoids oscillate rapidly.

• •

  The factor $\sin(prx)/x\rightarrow\pi\,\delta(x)$ acts as a delta function. When combined with the singularity of $\mathcal{V}_{l}(p(1+x),p),$ we have (with $z=prx$)

  $$-\frac{\alpha}{\pi}\int_{-\infty}^{\infty}dz\,\mathrm{sinc}(z)(C_{l}-\ln|z/pr|)=\alpha(-C_{l}-\ln(pr)-\boldsymbol{C})=\alpha(-\ln(2pr)+\psi(l+1)).$$

  (45)

• •

  There is no delta function in the integral containing $\cos(prx).$ On $x\in[-1,1],$ only integrands even in $x$ will survive the principal part integration. In this integrand, there will be factors of the form $[A]_{-}[B]_{+}/x+[A]_{+}[B]_{-}/x,$ with $A=\mathcal{V}_{l}(p(1+x),p)$ and $B=\{E(p(1+x))+E(p)\}/p(2+x)$ and $[f(x)]_{\pm}=\frac{1}{2}\{f(x)\pm f(-x)\}.$ We note that the first of these terms and $[B]_{-}/x$ are finite at the origin. So we consider integrals of the form

  $$\int dx\,\ln x\,\cos(prx)=x\,\ln x\,\mathrm{sinc}(prx)-\frac{\mathrm{Si}(prx)}{pr},$$

  (46)

  where $\mathrm{Si}(x)=\int_{0}^{x}dt\,\mathrm{sinc}\,t.$ This integral vanishes at $x=0$ and decreases like $1/pr$ for $x\neq 0.$

Thus we find (with $B(0)=E(p)/2p=1/2\beta,$ where $\beta=p/\sqrt{p^{2}+m_{e}^{2}}$ is the relativistic velocity magnitude for either of the particles)

$$I^{(1)}=\sqrt{\frac{2}{\pi}}\,\cos(pr-l\frac{\pi}{2})\,\eta\,\{-\ln(2pr)+\psi(l+1)\}\equiv\sqrt{\frac{2}{\pi}}\,\cos(pr-l\frac{\pi}{2})\,\{\delta_{l}^{(1)}(p)\},$$

(47)

with

$$\delta_{l}^{(1)}(p)=-\eta\ln(2pr)+\eta\,\psi(l+1).$$

(48)

Here $\eta=\alpha/2\beta$ corresponds to the parameter defined in the nonrelativistic case, $\eta_{\mathrm{NR}}=\alpha/(p/\mu),$ with $\mu=m_{e}/2$ the reduced mass in this case. We will find that our perturbation series is in powers of $\eta$ rather than of $\alpha.$

We use similar methods for the second-order terms, employing the four rules

	$\displaystyle\int_{0}^{\infty}dk\,\frac{\sin(kr-l\frac{\pi}{2})}{E(k)-E(p)}\,f(k)$	$\displaystyle\rightarrow\cos(pr-l\frac{\pi}{2})\,\frac{\pi}{2\beta}\,f(p),$
	$\displaystyle\int_{0}^{\infty}dx\,\frac{\cos(kr-l\frac{\pi}{2})}{E(k)-E(p)}\,f(k)$	$\displaystyle\rightarrow-\sin(pr-l\frac{\pi}{2})\,\frac{\pi}{2\beta}\,f(p),$
	$\displaystyle\int_{-1}^{\infty}dx\,\frac{\sin(kr-l\frac{\pi}{2})}{E(k)-E(p)}\,\mathcal{V}_{l}(p(1+x),p)\,f(x)$	$\displaystyle\rightarrow\cos(pr-l\frac{\pi}{2})\,\delta_{l}^{(1)}(p)\,f(0)$
	$\displaystyle\int_{-1}^{\infty}dx\,\frac{\cos(kr-l\frac{\pi}{2})}{E(k)-E(p)}\,\mathcal{V}_{l}(p(1+x),p)\,f(x)$	$\displaystyle\rightarrow-\sin(pr-l\frac{\pi}{2})\,\delta_{l}^{(1)}(p)\,f(0),$		(49)

all for $f(k)$ having no singularities. We evaluate the terms in eq. (38) as written, performing each $k^{\prime}$ integral before the $k$ integral. We find that changing the order of the double momentum integrals changes some of the terms, but the total is invariant. An ambiguity remains. Using partial fractions to cancel some terms gives a different final result. We suspect this is because doing this changes the set of poles in the integrands. This matter is under further investigation.

We find the result

$$y_{l}^{(2)}(r,p)\rightarrow\sqrt{\frac{2}{\pi}}\,\sin(pr-l\frac{\pi}{2})\,\left\{-\frac{1}{2}\delta_{l}^{(1)}(p)^{2}\right\}+\sqrt{\frac{2}{\pi}}\,\cos(pr-l\frac{\pi}{2})\,\left\{\delta_{l}^{(1)}(p)+\frac{1}{2}\,\frac{\pi}{2\beta}\,P\int_{0}^{\infty}dk^{\prime}\,\frac{\mathcal{V}_{l}(k^{\prime},p)^{2}}{E(k^{\prime})-E(p)}\right\}.$$

(50)

We note that the term proportional to $\sin(pr-l\frac{\pi}{2})$ has the correct form required by unitarity at this order. The last term is

$$\delta_{l}^{(2)}(p)=\frac{1}{2}\,\frac{\pi}{2\beta}\,P\int_{0}^{\infty}dk^{\prime}\,\frac{\mathcal{V}_{l}(k^{\prime},p)^{2}}{E(k^{\prime})-E(p)}=\frac{1}{2}\,\frac{\pi}{(2\beta)^{2}}\,P\int_{-1}^{\infty}dx\,\frac{\mathcal{V}_{l}(p(1+x),p)^{2}}{x}\,g(p,x),$$

(51)

where

$$g(p,x)=2\beta\,\frac{E(p(1+x))+E(p)}{4p(2+x)}\rightarrow 1$$

(52)

at $x=0.$