Second Order Exchange Energy of a d-Dimensional Electron Fluid

  • M. L. Glasser


A method is presented for reducing a 3d-fold integral occurring in higher order many-body integrals for a d-dimensional electron gas to a double integral. The result is applied to the second order exchange energy for a d-dimensional uniform electron fluid. The cases \(d=2,3\) are examined in detail.

24.1 Introduction

In their classic work on the ground-state energy of an interacting electron gas [1] Gell-Mann and Brueckner encountered the second order exchange term
$$\begin{aligned} E_{2b}=\frac{3p_F^3 e^4}{16\pi ^5}\int d\mathbf {p} \, d\mathbf {k} \, d\mathbf {q}\, \frac{f_{\mathbf {p}}f_{\mathbf {k}}(1-f_{\mathbf {p}+\mathbf {q}})(1-f_{\mathbf {k}+\mathbf {q}})}{q^2(\mathbf {p}+\mathbf {k}+\mathbf {q})^2\mathbf {q}\cdot (\mathbf {p}+\mathbf {k}+\mathbf {q})}, \end{aligned}$$
where \(f_p\) is the Fermi distribution function and \(p_F\) denotes the Fermi momentum. Gell-Mann’s assistant, H. Kahn, estimated by Monte-Carlo integration the value as \(-0.044\) and in a 1965 lecture in Istanbul [2] L. Onsager claimed that the exact value is \((\ln 2)/3-3\zeta (3)/2\pi ^2\), which remained unproven till eight years later when Onsager, Mittag and Stephen published a lengthy derivation [2]. In 1980, Ishihara and Ioriatti [3] evaluated the two-dimensional analogue of Eq. (24.1) and in 1984 the author published a note [4] indicating how such integrals might be handled in d-dimensions. But, due to a number of misprints [4] is difficult to follow and it seems appropriate to present a simplified and corrected version, particularly since the method has been found useful in other contexts [5] and, due to an oversight, it erroneously stated that the value given in [3] was confirmed. The dimension d will be treated as continuous by means of the expedient integration rule for an azimuthally symmetric integrand
$$\begin{aligned} \int d^d \mathbf {k}=\frac{2\pi ^{(d-1)/2}}{\varGamma \left[ \frac{1}{2}(d-1)\right] }\int _0^{\infty }dk \, k^{d-1}\int _0^{\pi }d\theta \sin ^{d-2}\theta . \end{aligned}$$
The following section covers the reduction of a basic 9d-dimensional integral to more manageable \(3d+2\)-dimensional form which, in Sect. 24.3, is applied to the second order exchange energy. The last section gives the results for \(d=2\) and \(d=3\).

24.2 Basic Integral Identity

The units \(\hbar =2m=1\), will be used along with the notation
$$\begin{aligned} f_p= & {} [1+\exp [\beta (p^2-p_F^2)]^{-1},\end{aligned}$$
$$\begin{aligned} Q(p)= & {} f_p(1-f_{\mathbf {p}+\mathbf {q}}),\end{aligned}$$
$$\begin{aligned} Q^\prime (p)= & {} f_{\mathbf {p}+\mathbf {q}}(1-f_p),\end{aligned}$$
$$\begin{aligned} \varDelta (p)= & {} f_{\mathbf {p}+\mathbf {q}}-f_{\mathbf {p}},\end{aligned}$$
$$\begin{aligned} \delta (p)= & {} (\mathbf {p}+\mathbf {q})^2-p^2. \end{aligned}$$
All vectors are d-dimensional, and vector integrals are over all space.

Lemma 24.1

In the zero temperature limit
$$\begin{aligned} \frac{Q(p)Q(k)-Q^\prime (p)Q^\prime (k)}{\mathbf {q}\cdot (\mathbf {p}+\mathbf {k}+\mathbf {q})}= -\frac{1}{\pi }\int _{-\infty }^{\infty }dz\, \frac{\varDelta (p)}{z-i\delta (p)}\frac{\varDelta (k)}{z+i\delta (k)}. \end{aligned}$$


The proof follows closely the derivation of a similar result in Appendix A of [3].

Theorem 24.1

For real \(\mathbf {r}\) and \(t\ge 0\)
$$\begin{aligned} \int d\mathbf {p}\, e^{i[\mathbf {r}\cdot \mathbf {p}+\delta (p)t]}\varDelta (p)= -2i\left( \frac{2\pi p_F}{\xi }\right) ^{d/2}e^{-\frac{1}{2}i\mathbf {r}\cdot \mathbf {q}} \sin \left( \frac{1}{2}\mathbf {q}\cdot \mathbf {\xi }\right) J_{d/2}\left( p_F\xi \right) , \end{aligned}$$
where \(\mathbf {\xi }=\mathbf {r}+2t\mathbf {q}\).


First of all note that \(\varDelta (p)\) is simply a rectangular pulse with height 1 and width q, so has the inverse Laplace transform representation
$$\begin{aligned} \varDelta (p)=\int _{c-i\infty }^{c+i\infty }\frac{ds}{2\pi i s}e^{sp_F^2}[e^{-s(\mathbf {p}+\mathbf {q})^2}-e^{-s p^2}], \quad c>0. \end{aligned}$$
By substituting Eq. (24.6) into (24.5) one obtains the difference of two integrals. In the first make the change of variable \(\mathbf {p}\rightarrow -\mathbf {p}-\mathbf {q}\). This gives
$$\begin{aligned} \int _{c-i\infty }^{c+i\infty }\frac{ds}{2\pi is}e^{sp_F^2}\{e^{-i\mathbf {r}\cdot \mathbf {q}}e^{-itq^2} C(-\mathbf {r}-2t\mathbf {q})-e^{itq^2}C(\mathbf {r}+2t\mathbf {q})\} \end{aligned}$$
$$\begin{aligned} C(\mathbf {\xi })=\int d\mathbf {p}\, e^{i\mathbf {p}\cdot \mathbf {\xi }}e^{-sp^2}=\left( \frac{\pi }{s}\right) ^{d/2}e^{-\xi ^2/{4s}}. \end{aligned}$$
Next, one has
$$\begin{aligned} \int _{c-i\infty }^{c+i\infty }\frac{ds}{2\pi i}e^{p_F^2s}s^{-1-d/2}e^{-\xi ^2/{4s}}=\left( \frac{2p_F}{\xi }\right) ^{d/2}J_{d/2}(p_F\xi ), \end{aligned}$$
which gives for Eq. (24.7)
$$\begin{aligned} \left( \frac{2 p_F\pi }{\xi }\right) J_{d/2}(p_F\xi )e^{itq^2}[e^{-i\mathbf {q}\cdot \mathbf {\xi }}-1]= -2i\left( \frac{2p_F\pi }{\xi }\right) J_{d/2}\left( p_F\xi \right) e^{-\frac{1}{2}i\mathbf {r}\cdot \mathbf {q}} \sin \left( \frac{1}{2}\mathbf {q}\cdot \mathbf {\xi }\right) . \end{aligned}$$
Now, we choose, from among other possibilities,
$$\begin{aligned} \alpha (\mathbf {q})=\int \frac{e^{i\mathbf {r}\cdot \mathbf {q}}}{r}\, d\mathbf {r} \end{aligned}$$
and define
$$\begin{aligned} A(\mathbf {q})=\int d\mathbf {p} \, d\mathbf {k}\, \alpha (\mathbf {p}+\mathbf {k}+\mathbf {q})\, \frac{Q(p)Q(k)}{\mathbf {q}\cdot (\mathbf {p}+\mathbf {k}+\mathbf {q})}. \end{aligned}$$
By making the substitution \(\mathbf {p} \rightarrow -\mathbf {p}-\mathbf {q}\), \(\mathbf {k}\rightarrow -\mathbf {k}-\mathbf {q}\), and adding the result back to Eq. (24.12), we find, using the identity in Lemma 24.1,
$$\begin{aligned} A(q)= & {} -\frac{1}{2\pi }\int _{-\infty }^{\infty }dz\int d\mathbf {p} \, d\mathbf {k}\, \alpha (\mathbf {p}+\mathbf {k}+\mathbf {q})\frac{\varDelta (p)\varDelta (k)}{(z+i\delta (p))(z-i\delta (k))} \nonumber \\&=-\frac{1}{2\pi }\int \frac{d\mathbf {r}}{r} \int _{-\infty }^{\infty }dz \, B(\mathbf {r},z)B(-\mathbf {r},z) \end{aligned}$$
$$\begin{aligned} B(\mathbf {r},z)=\int d\mathbf {p}\, e^{i\mathbf {r}\cdot \mathbf {p}}\frac{\varDelta (p)}{z+i\delta (p)}=\int _0^{\infty }dt\, e^{-zt}\int d\mathbf {p}\, e^{i[\mathbf {r}\cdot \mathbf {p}+t\delta (p)]}\varDelta (p). \end{aligned}$$
By applying Theorem 24.1 and performing the elementary z integration, we have, after scaling q out of \(t_j\),

Theorem 24.2

$$\begin{aligned} A(q)= \frac{2}{\pi q} \left( 2\pi p_F\right) ^d \int \frac{d\mathbf {r}}{r}\; \int _0^{\infty }\int _0^{\infty }\frac{dt_1 dt_2}{t_1+t_2}\frac{\sin \left( \frac{1}{2}q\xi _1\right) \sin \left( \frac{1}{2}q\xi _2\right) }{(\xi _1\xi _2)^{d/2}}J_{d/2}(p_F\xi _1)J_{d/2}(p_F\xi _2), \end{aligned}$$
where \(\mathbf { \xi _1}=\mathbf {r}+2t_1\hat{q}\), \(\mathbf {\xi _2}=\mathbf {r}-2t_2\hat{q}\).

24.3 Application to Second Order Exchange

For our choice of Coulomb interaction
$$\begin{aligned} \alpha (q)=e^2\frac{(4\pi )^{(d-1)/2}}{q^{d-1}}\varGamma \left( \frac{d-1}{2}\right) \end{aligned}$$
which requires \(d>1\), the second order exchange contribution to the ground-state energy per unit volume of a d-dimensional electron fluid is
$$\begin{aligned} E_{2x}= \frac{1}{(2\pi )^{2d}}\int \alpha (q)A(q)d\mathbf {q}. \end{aligned}$$
For \(d>2\) we take the polar axis as the \(\hat{q}\)-direction and apply Theorem 24.2. The q integration is elementary and we have
$$\begin{aligned} E_{2x}=K_d\int d\varOmega _q\int \frac{d\mathbf {r}}{r}\int _0^{\infty }\int _0^{\infty }\frac{dt_1 dt_2}{t_1+t_2}\ln \left| \frac{\xi _1+\xi _2}{\xi 1-\xi _2}\right| \frac{J_{d/2}(p_F\xi _1)J_{d/2}(p_F\xi _2)}{(\xi _1\xi _2)^{d/2}}, \end{aligned}$$
where \(K_d\) collects all the numerical prefactors and powers of \(p_F\) (for \(d=2\) \(\int d\varOmega _q=2\pi \)) and will be made explicit in the final result. Now set \(t_2=ut_1\) and \(\mathbf {r}\rightarrow t_1\mathbf {r}\), so
$$\begin{aligned} E_{2x}=K_d\int d\varOmega _q \int _0^{\infty }\frac{du}{u+1}\int \frac{d\mathbf {r}}{r(\eta _1\eta _2)^{d/2}}\ln \left| \frac{\eta _1+\eta _2}{\eta _1-\eta _2}\right| \int _0^{\infty }\frac{dt}{t}J_{d/2}(p_Ft\eta _1)J_{d/2}(p_Ft\eta _2), \end{aligned}$$
where \(\eta _1=|\mathbf {r}+2\hat{q}|\), \(\eta _2=|\mathbf {r}-2u\hat{q}|\). The \(\theta ,t-\) integrals can be done next, yielding
$$\begin{aligned} E_{2x}=K_d\int _0^{\infty }\frac{du}{u+1}\int \frac{d\mathbf {r}}{r \eta _>^d}\ln \left| \frac{\eta _1+\eta _2}{\eta _1-\eta _2}\right| ,\quad (d>2). \end{aligned}$$
For \(d>2\) we can switch to \(d-\)dimensional cylindrical coordinates with axis along \(\hat{q}\). Since the integrand is independent of the azimuthal angle
$$\begin{aligned} \int d\mathbf {r}=\frac{2\pi ^{(d-1)/2}}{\varGamma [\frac{1}{2}(d+1)]}\int _{-\infty }^{\infty }dz\int _0^{\infty }\rho ^{d-2}d\rho . \end{aligned}$$
Next, after the successive transformations \(t=(z-1)/(z+1)\) and \(\rho =2s/(1-t)\) we have
$$\begin{aligned} E_{2x}= & {} K_d\int _{-1}^1\frac{dt}{1-t}\ln \left( \frac{1+t}{1-t}\right) F(t), \nonumber \\ F(t)= & {} \int _0^{\infty }\frac{s^{d-2}ds}{(s^2+1)^{d/2}}\int _t^1\frac{dy}{\sqrt{y^2+s^2}}. \end{aligned}$$
Carrying out the s integration, we come to
$$\begin{aligned} F(t)=\frac{1}{d-1}\int _t^1\frac{dy}{|y|}\;_2F_1\left( \frac{1}{2}, \frac{1}{2}(d-1);\frac{1}{2}(d+1);1-y^{-2}\right) . \end{aligned}$$
By integrating by parts and noting that
$$\;_2F_1\left( \frac{1}{2},a;a+1;1-y^{-2}\right) =\frac{2|y|}{1+|y|}\; \;_2F_1\left( 1,1-a;a+1;\frac{1-|y|}{1+|y|}\right) ,$$
we arrive at the principal result

Theorem 24.3

The second order exchange contribution to the ground-state energy of a \(d>2-\)dimensional electron fluid is
$$\begin{aligned} E_{2x}= & {} K_dG(d) , \end{aligned}$$
$$\begin{aligned} G(d)= & {} \int _0^1\frac{dy}{y+1}\left[ \frac{\pi ^2}{3}-\ln ^2y\right] \;_2F_1\left[ 1,\frac{1}{2}(3-d);\frac{1}{2}(1 +d);y\right] .\qquad \qquad \qquad \end{aligned}$$

24.4 Discussion

Equation (24.24a) is as far as one can proceed without specifying the dimensionality. For \(d=3\), we find, since the hypergeometric function reduces to unity,
$$\begin{aligned} E_{2x}=K_3G(3)=\frac{e^4p_F^3}{4\pi ^2} \int _0^1\frac{dy}{1+y}\left( \frac{\pi ^2}{3}-\ln ^2 y\right) =\frac{1}{6}[\pi ^2\ln 4-9\zeta (3)]. \end{aligned}$$
which is exactly the Onsager–Stephen–Mittag value, since they have \(e^2=2\) and \(p_F=1\).
For the case \(d=2\) we take the limit of Eq. (24.24a) which gives
$$\begin{aligned} G(2)=2\int _0^1\left( \frac{\pi ^2}{3}-4\ln ^2y\right) \frac{\tan ^{-1}y}{y^2+1}dy , \end{aligned}$$
which, unlike the corresponding integral in [3] does not seem to be analytically evaluable. This gives
$$\begin{aligned} E_{2x}=K_2 G(2)=18.0586 \cdot \frac{p_F^2 e^4}{32\pi ^4} , \end{aligned}$$
about 30% less than the value \(28.3664 \cdot (p_F^2 e^4/32\pi ^4)\) in [3]. A possible reason is that in Eq. (14) of Ref. [3] the argument of the second Bessel function is \(|\mathbf {r}-2\hat{u}t|\) and after making the substitution \(\mathbf {r}\rightarrow (t+x)\mathbf {r}+\hat{u}(x-t)\), in Eq. (16) of Ref. [3] the authors present it as \((x+t)|\mathbf {r}-\hat{u}|\), which is incorrect. It is this error which renders the remainder of the evaluation analytically tractable. An attempt to continue the calculation after correcting this was stymied by a further difficulty in Eq. (14) of Ref. [3]; the factor of 2 in the numerator of the argument of the logarithm means that, as \(x\rightarrow \infty \) this argument tends to 2, rather than unity as required for convergence at the upper limit of the x integration.



Financial support of MINECO (Project MTM2014-57129-C2-1-P) and Junta de Castilla y León (VA057U16) is acknowledged.


  1. 1.
    M. Gell-Mann, K.A. Brueckner, Phys. Rev. 106, 364 (1957).
  2. 2.
    L. Onsager, L. Mittag, M.J. Stephen, Ann. Phys. 473(1–2), 71 (1966).
  3. 3.
    A. Isihara, L. Ioriatti, Phys. Rev. B 22, 214 (1980).
  4. 4.
    M.L. Glasser, J. Comp. Appl. Math. 10(3), 293 (1984).
  5. 5.
    M.L. Glasser, G. Lamb, J. Phys. A: Math. Theor. 40(6), 1215 (2007).

Copyright information

© Springer International Publishing AG, part of Springer Nature 2018

Authors and Affiliations

  1. 1.Department of PhysicsClarkson UniversityPotsdamUSA
  2. 2.Department of Theoretical PhysicsUniversity of ValladolidValladolidSpain
  3. 3.Donostia International Physics CenterSan SebastiánSpain

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