# Gaussian wave packet with a step potential

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://imgur.com/a/TSpjpZN , Thomas only embedded one image.

• Oy,yeah let me fix that. – gyzgyz123 Jul 25 at 20:28

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?
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.
• @gyzgyz123 Dot product of functions $f$ and $g$ is: $\langle f|g\rangle=\int_{\mathbb R} f(x)^*g(x)\,dx$. Note the complex conjugation. This changes $\exp(-ikx)$ into $\exp(ikx)$ in the integrand, which is what's likely confusing you. – Ruslan Jul 25 at 21:16