Gaussian wave packet with a step potential

Question

In principle of quantum mechanics by Shankaar on page 170, while doing transmission and reflection index for a step potential for a Gaussian wave packet moving to the right. We come to this integration:

$$\langle \psi_k \vert \psi_I \rangle=\frac{1}{(2\pi)^{1/2}}\left \{ \int_{-\infty}^{\infty} \left[e^{-ik_1x/\hbar} +\left(\frac{B}{A}\right)^\ast e^{ik_1x/\hbar}\right] \theta(-x)\psi_I(x)dx +\int_{-\infty}^{\infty} \left(\frac{C}{A}\right)^\ast e^{-ik_2x/\hbar}\theta (x)\psi_I(x)dx] \right \}$$

The right integral is zero because $\theta (x)$ is non zero for $x>0$ and $\psi(x)$ is non zero for $x<0$, so the integral can't be anything but zero.

For the right part of the left integral I can't understand the argument presented. He says the function is peaked at $k=+k_0$ and is orthogonal to left moving momentum states. Why does that mean the integral is zero?

I have made screenshot of the relevant pages from the book.

Edit: https://i.sstatic.net/55WJZ.jpg , Thomas only embedded one image.

Answer 1

He says the function is peaked at $k=+k_0$ and is orthogonal to left moving momentum states. Why does that mean the integral is zero?

Well, two functions being orthogonal means that the dot product between them is zero. And if you split the integral of sum into a sum of integrals, then you'll get exactly that from the term in question — dot product of the left-going momentum eigenstate with the right-going wave packet.

Now, of course, they aren't exactly orthogonal. If you do actual integration, you'll find that the integral is proportional to $\exp(-(k+k_0)^2\eta)$ for some constant $\eta>0$, and this would be the smaller the larger the sum $k+k_0$. So for high enough momenta of the wave packet the orthogonality would indeed be a good approximation.