This is a step from Appendix A in Szabo and Ostlund's Quantum Chemistry. In short, the question is how to get from \ref{A.20} to \ref{A.21}. We can factor out some constants $K$ in the expressions to get:

\begin{align} (A| -Z_C/r_{1C} |B) &= K \left(\frac{1}{2\pi^2}\right) \int \mathrm{d}\mathbf{k} \; e^{-k^2/4p} k^{-2} \exp\big(-i\,\mathbf{k}\cdot(\mathbf{R}_P - \mathbf{R}_C)\big) \tag{A.20}\label{A.20}\\ &= K \left(\frac{2}{\pi|\mathbf{R}_P-\mathbf{R}_C|}\right) \int\limits_0^\infty \mathrm{d}k \; e^{-k^2/4p} k^{-1} \sin(k|\mathbf{R}_P-\mathbf{R}_C|) \tag{A.21}\label{A.21} \end{align}

Part of the note between these steps is that (paraphrasing) we can let $\mathbf{R}_P-\mathbf{R}_C$ lie on the $z$-axis and use that $k\cdot (\mathbf{R}_P-\mathbf{R}_C)= k|\mathbf{R}_P-\mathbf{R}_C|\cos\theta$, "then we can easily perform the angular part of the integration over $\boldsymbol{k}$ to obtain [\ref{A.21}]."

What I have done: It seems that Euler's formula must be applied to the second exponent in \ref{A.20} but using the hint we'd have something like $$\cos(k|\mathbf{R}_P-\mathbf{R}_C|\cos\theta) + i\sin \dots \text{etc}. \tag{???}$$ I guess that the sine in \ref{A.21} follows integration of the (real) cosine in Euler's expansion. In line with this, \ref{A.21} is multiplied by a factor of $2\pi$.

I will keep at this but I think if it puzzles me, then others may have stumbled over it so I am posting.
Hopefully removal of the common factors ($K$) makes this easier to read but the problem is on pages 413-414 of Szabo and Ostlunds book.

  • 2
    $\begingroup$ Don't have my Szabo at hand (heresy, I know), will look into it today/tomorrow... $\endgroup$
    – user37142
    Aug 21, 2018 at 14:05
  • 2
    $\begingroup$ There is an integration 'missing' effectively $\int \exp(-a\cos(\theta) ) d\theta$. Substituting $z=\sin(\theta)$ and some fiddling about makes the integral easier. $\endgroup$
    – porphyrin
    Aug 21, 2018 at 19:12
  • $\begingroup$ Using porphyrin's substitution and Euler's relation I get an answer which is off by a factor of $2\pi ~k^2.$ If a spherical volume was intended by the original vector notation then this seems to be OK. Would still like an authoritative answer... $\endgroup$
    – daniel
    Aug 24, 2018 at 18:33

1 Answer 1


As commented, do note that the integral (A.20) is taken over all $\mathbf{k}$-space, whereas the integral (A.21) is one-dimensional. The explicit calculation of this integral follows as

$$\begin{align} \int\text{d}\mathbf{k}\,\Psi(k)e^{-i\mathbf{k\cdot r}} &=\int_0^\infty\text{d}k\,k^2\int_0^\pi\text{d}\theta\int_0^{2\pi}\text{d}\varphi\,\sin\theta\Psi(k)e^{-ikr\cos\theta}\\ &= 2\pi\int_0^\infty\text{d}k\,k^2\Psi(k)\int_0^\pi\text{d}\theta\,\sin\theta e^{-ikr\cos\theta}\\ &= 2\pi\int_0^\infty\text{d}k\,k^2\Psi(k)\left[\frac{e^{ikru}}{ikr}\right]_{u=-1}^1\\ &= \frac{4\pi}{r}\int_0^\infty\text{d}k\,k\Psi(k)\sin(kr).\end{align}$$

In the first equality we convert this integral from Cartesian to spherical coordinates, in the second equality we evaluate the integral over $\varphi$, in the third equality we substitute $u = -\cos\theta$, and in the fourth equality we make use of Euler's formula to get the exponential representation of $\sin(kr)$.

  • $\begingroup$ This looks right to me...$\Psi(k)$ has absorbed the $k^{-2}$ in A.20, right? $\endgroup$
    – daniel
    Aug 28, 2018 at 18:42
  • $\begingroup$ @daniel, yes.$\ $ $\endgroup$ Aug 28, 2018 at 18:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.