First-order Born Approximation in Quantum Mechanics states that scattering amplitude is a Fourier transform of potential:

$$ f(\theta) = \int d^3 r^{\prime} e^{-i (\bf k - k_i)r^{\prime}} V(r^{\prime}) $$

where $\bf k $ is a wave vector of incident wave and $ \bf k_i$ is a wave vector of scattered wave. $\theta$ is defined as the angle between $\bf k$ and $\bf k_i$ (i.e. scattering angle). $|\bf k|$ is equal to $|\bf k_i|$ in elastic scattering, so $\bf k - k_i $$=0$ for $\theta = 0$ as far as we consider elastic scattering.


$$ f(0) = \int d^3 r^{\prime} V(r^{\prime}) $$

However, this fails to satisfy the optical theorem:

$$ \mbox{Im}f(0) = \frac{k}{4\pi} \sigma_{tot} $$

because $f(0)$ is obviously real. Why is it happened?

  • $\begingroup$ Where is $\theta$ in $f(\theta)$? $\endgroup$ Commented Dec 25, 2018 at 11:58
  • $\begingroup$ $\theta$ is defined as the angle between $\bf k$ and $\bf k_i$ (i.e. scattering angle). Then, $\bf k - k_i$ becomes zero when $\theta=0$ because $|\bf k|$ is equal to $|\bf k_i|$ in elastic scattering. $\endgroup$
    – tak
    Commented Dec 25, 2018 at 12:32

1 Answer 1


$\let\th=\theta \let\dag=\dagger \def\bk{\mathbf k} \def\br{\mathbf r} \def\st{\sigma_{\mathrm{tot}}} \def\Im{\mathrm{Im}}$ Shortly, because it's a first-order perturbative approximation. Let me recall how optical theorem is proven. There are two ways, equivalent at their root:

  • flux conservation
  • unitarity of $S$-matrix.

By flux conservation I mean the following. Start from $$u(\br) = e^{i\bk\cdot\br} + f(\th)\,{e^{ikr} \over r}$$ as the asymptotic form of solution for a scattering problem. We expect that asymptotically outgoing flux equates ingoing one, i.e. that total asymptotic flux vanishes. Let's write shortly $$u = u_1 + u_2.$$ Computing flux, which is quadratic in $u$, we shall find three terms: $$\Phi(u_1) + \Phi(u_2) + \hbox{interference term}.$$

$\Phi(u_1)$ is obviuosly zero. $\Phi(u_2)$ is proportional to $\st$, and the interference term gives $\Im f(0)$. Result is the optical theorem.

Now note that $f(\th)$ is of the first order in $V(r)$, whereas $\st$ is of second order. A first-order perturbative approach neglects second-order terms, thus giving $\Im f(0) = 0$.

Unitarity of $S$-matrix means $SS^\dag=I$. Define $S = I + T$. Then $$I = SS^\dag = I + T + T^\dag + TT^\dag.$$ Again $T$ expresses outgoing wave, i.e. scattering amplitude (with an $i$ multiplier, so that $T+T^\dag$ is $\Im f$). $TT^\dag$ gives $\st$, and the above argument may be repeated.


Your Answer

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