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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.

image

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

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

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  • $\begingroup$ If it's orthogonal to left moving states wouldn't that leave the right part of the sum and cancel the left part of the sum instead of the opposite which seems to happen? I'm also not sure where you got a dot product from. If we split the summation in to two integrals, from where does the dot product appear? $\endgroup$ – gyzgyz123 Jul 25 at 21:08
  • $\begingroup$ @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. $\endgroup$ – Ruslan Jul 25 at 21:16
  • $\begingroup$ Oops, yeah it seems obvious now.... Thanks for your help. $\endgroup$ – gyzgyz123 Jul 25 at 21:34

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