1D quantum scattering from step potential - how to prove orthonormality of eigenstates?

Question

In R. Shankar book "Principles of quantum mechanics" $2$nd edition on page $170$ he states that for a step potential the following is energy eigenfunction, where $k = \sqrt{2mE/\hbar^2}$.

$$ \Psi_k (x) = A \left( \left( e^{ikx} + B(k) e^{-ikx} \right)\theta(-x) + C(k) e^{i\sqrt{k^2 - q^2}x} \theta(x) \right) $$

Here, $B(k)$ and $C(k)$ are some given functions only of variable $k$, and $q^2$ is a fixed real number. $\theta(x)$ is Heaviside step function which is non zero for $x>0$ where it has value $1$.

Then, he says that for this function to be normalized we must choose $A = 1/\sqrt{2\pi}$.

Question: How do I verify that with this choice of $A$ this eigenfunction is normalized? I tried computing inner product where I separated parts $x<0$ and $x>0$, but then I am unable to simplify expressions to get something proportional to $\delta(k-k')$.

My attempt is the following.

$$ \langle \Psi_{k'}(x) | \Psi_{k} (x) \rangle = \int_{-\infty}^{0} \Psi_{k'}^*(x) \Psi_{k'}(x) \mathrm{d}x + \int_{0}^{\infty} \Psi_{k'}^*(x) \Psi_{k'}(x) \mathrm{d}x $$ $$ = |A|^2 \int_{-\infty}^{0} (e^{ik'x} + B(k')e^{-ik'x})^* (e^{ikx} + B(k) e^{-ikx}) \mathrm{d}x $$ $$ + |A|^2 \int_0^{\infty} \left(C(k')e^{i\sqrt{k'^2-q^2}x}\right)^* \left(C(k)e^{-i\sqrt{k^2-q^2}x} \right) \mathrm{d}x $$

For example, consider the first integral. I am not sure how to proceed further.

$$ |A|^2 \int_{-\infty}^{0} (e^{ik'x} + B(k')e^{-ik'x})^* (e^{ikx} + B(k) e^{-ikx}) \mathrm{d}x $$ $$ = |A|^2 \Bigg( \int_{-\infty}^{0} e^{i(k-k')x}\mathrm{d}x + \int_{-\infty}^{0} B(k) e^{i(-k-k')x} \mathrm{d}x $$ $$ + \int_{-\infty}^{0}B(k') e^{i(k'+ k)x} \mathrm{d}x + \int_{-\infty}^{0} B(k)B(k') e^{i(k'-k)x} \mathrm{d}x \Bigg) $$ $$ = |A|^2 \Bigg( \pi \delta(k-k') + \frac{i}{k'-k} + B(k) \left(\pi \delta(k+k') + \frac{i}{k+k'} \right) $$ $$ + B(k') \left(\pi \delta(k+k') - \frac{i}{k+k'} \right) + B(k)B(k') \left( \pi \delta(k-k') + \frac{i}{k-k'} \right)\Bigg)$$

Here, $B(k)$ and $C(k)$ are the following functions solved from the boundary conditions (taken from Shankar equations 5.4.9, 5.4.10).

$$ B(k) = \frac{k-\sqrt{k^2-q^2}}{k+\sqrt{k^2-q^2}} $$ $$ C(k) = \frac{2k}{k+\sqrt{k^2-q^2}} $$

Answer 1

Hint: You have to use the Fourier transform. You can use the definition:

$$ \mathcal{F}\{f(x),k\}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(x)e^{-ikx}\,dx $$

So, for example, the first integral is:

\begin{align} \int_{-\infty}^0 e^{i(k-k')x}\,dx &= \int_{-\infty}^\infty \theta(-x)e^{i(k-k')x}\,dx \\ &= \int_{-\infty}^\infty \theta(x)e^{i(k'-k)x}\,dx \\ &= \sqrt{2\pi}\,\mathcal{F}\{\theta(x),k-k'\}=\sqrt{2\pi}\sqrt{\frac{\pi}{2}}\left(\frac{1}{i\pi(k-k')}+\delta(k-k')\right) \\ &=\frac{1}{i(k-k')}+\pi\delta(k-k') \\ \end{align}

In fact, you can use the same trick for all integrals, because $B(k),B(k')$ are constants into the integral

Answer 2

A plane wave $Ae^{ikx}$ is not a normalizable wave function, as attempting to calculate the integral of its absolute value squared gives: $$ |A|^2\int_{-\infty}^{\infty}1dx, $$ and the integral is not finite. This in turn implies that a free particle of a definite momentum, which is represented by a plane wave, is not a valid solution to the system. What Shankar is doing in their book is to contruct a different, normalizable, solution: a wave packet. Shankar constructs the wave packet using the energy eigenstates that you write in your question as a superposition of two plane waves. This wave packet (chosen to be Gaussian for mathematical simplicity), peaks around some momentum $p_0=\hbar k_0$, and the width around this momentum is chosen to be small (but as we have just argued, it cannot be exactly zero). This means that the wave packet represents a particle of momentum $p\simeq p_0$, and this is the best you can do in terms of defining the momentum of a particle. What Shankar is then doing is normalizing the wave packet, and this is something that you can do in the usual way without any infinities.

A similar situation arises for "position eigenstates". In that case, the position wave function $\delta(x-x_0)$ of a particle at $x_0$ is also non-normalizable, so instead you must consider a normalizable wave function that peaks at $x_0$ but still has a finite width.