# Why first-order Born Approximation doesn't satisfy optical theorem?

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.

Therefore:

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

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?

• Where is $\theta$ in $f(\theta)$? Commented Dec 25, 2018 at 11:58
• $\theta$ 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.
– tak
Commented Dec 25, 2018 at 12:32

$$\let\th=\theta \let\dag=\dagger \def\bk{\mathbf k} \def\br{\mathbf r} \def\st{\sigma_{\mathrm{tot}}} \def\Im{\mathrm{Im}}$$ 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.