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

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).

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

• You are correct, I think Shankar is a bit cavalier. The "normalization" $1/\sqrt{2\pi}$ is the one needed for a single plane wave to have $\langle \Psi_{k'}| \Psi_{k} \rangle = \delta(k'-k)$ but does not work here. But the main point is that this function is not normalizable, these are scattering states.
– lcv
Jun 6, 2020 at 15:03
• @lcv Thanks for comment - by normalizable I meant $\langle \Psi_{k'} | \Psi_{k} \rangle = \delta(k-k')$ and this is crucial in Shankar as this allows to obtain coefficients of wavepacket in terms of initial conditions. Jun 6, 2020 at 16:02
• I think that information about $B(k)$ is missing Jun 6, 2020 at 22:07
• @Elborito that is a good point - I added expressions for $B(k)$ and $C(k)$, maybe that can help Jun 8, 2020 at 21:07

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

• hi! Thanks for your answer. Yes, this is exactly what I tried but I couldn't make it till the end. Because then what you have is terms that are proportional to $B(k)$, $B(k')$ or $B(k)B(k')$ but these terms are not equal to zero. On the other hand, normalization constant $A = 1/ \sqrt{2\pi}$ does not depend on $B(k)$ or $B(k')$. Maybe I could update my question Jun 6, 2020 at 2:03
• in that case, I suppose that is not normalizable. But I am not sure Jun 6, 2020 at 2:16

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.

• That is not a wave packet but a linear combination of two waves.
– lcv
Jun 6, 2020 at 8:04
• @lcv, you are correct that the questioner writes down a linear combination of two waves. What I meant in my answer is that Shankar uses a wave packet solution in the book, which is built from this plane wave solution that the questioner refers to. I agree with you that my original phrasing did not make this clear, so I made an edit. Jun 6, 2020 at 8:06
• @ProfM Thanks for your answer! By normalizable I meant $\langle \Psi_{k'} | \Psi_{k} \rangle = \delta(k-k')$ and this is crucial in Shankar as this allows to obtain coefficients of wavepacket that you mentioned in terms of initial conditions. Jun 6, 2020 at 16:02