Factorisation Method for the Sech-Squared Potential

Source: Cambridge Quantum Mechanics, Example Sheet 1, Question 8; accompanies David Tong’s notes.

Sketch the potential \[U(x) = -\frac{\hbar^2}{m}\operatorname{sech}^2 x\,. \tag{1.1}\]

Show that the time-independent Schrödinger equation for a particle in this potential can be written \[\hat{A}^\dagger \hat{A}\,\chi = (\mathcal{E}+1)\,\chi\,,\tag{1.2}\] where $\mathcal{E} = 2mE/\hbar^2$ and \[\hat{A} = \frac{d}{d x} + \tanh x\,, \qquad \hat{A}^\dagger = -\frac{d}{d x} + \tanh x\,.\tag{1.3}\]

Show, by integrating by parts, that for any normalised wavefunction $\chi$, \[\int_{-\infty}^{\infty} \chi^*\hat{A}^\dagger\hat{A}\,\chi\,dx \;=\; \int_{-\infty}^{\infty} (\hat{A}\chi)^*(\hat{A}\chi)\,dx\,,\tag{1.4}\] and deduce that the eigenvalues of $\hat{A}^\dagger\hat{A}$ are non-negative. Hence show that the ground state (with lowest energy) has $\mathcal{E}\geq -1$. Show that a wavefunction $\chi_0(x)$ is an energy eigenstate with $\mathcal{E}=-1$ iff \[\frac{d\chi_0}{d x} + \tanh x\,\chi_0 = 0\,.\tag{1.5}\]

Find and sketch $\chi_0(x)$.

Figure 1. The sech-squared potential \eqref{eq:sech-pot

The Hamiltonian can be written as \[\begin{aligned}H &= -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} -\frac{\hbar^2}{m}\operatorname{sech}^2 x\\ &= \frac{\hbar^2}{2m}\left(-\frac{d^2}{dx^2}-2+\tanh^2x\right)\\ &= \frac{\hbar^2}{2m}\left(\frac{d}{dx}+\tanh x\right)\left(-\frac{d}{dx}+\tanh x\right)-\frac{\hbar^2}{m}.\end{aligned}\tag{1.6}\] Thus, we have \[\hat{A}^\dagger \hat{A}\chi = (\mathcal{E}+1)\chi.\tag{1.7}\] Rather trivially following from integrating by part, we find \[\int dx \chi^*A^\dagger A\chi = \int dx (A\chi)^*(A\chi),\tag{1.8}\] justifying that $A^\dagger$ is indeed the Hermitian conjugate to $A$. Given an eigenstate of $A^\dagger A$, $|a\rangle$, with eigenvalue $a$. We have \[\langle a|A^\dagger A|a\rangle = a\langle a |a\rangle = ||A|a\rangle||^2 \geq 0 \implies a\geq0,\tag{1.9}\] which implies that $\mathcal{E}+1\geq0$. For the ground state with energy $\mathcal{E}=-1$, we have \[\hat{A}^\dagger \hat{A}\chi_0 = 0,\tag{1.10}\] Taking the inner product of the above with $\chi_0$, we find \[\langle \chi_0|A^\dagger A|\chi_0\rangle = ||A|\chi_0\rangle||^2 = 0 \implies A|\chi_0\rangle=0.\tag{1.11}\] Solving the above, we find \[\chi_0(x) = \frac{N}{\cosh x},\tag{1.12}\] where $N$ is a normalisation constant.

Figure 2. Ground state wavefunction $\chi_0(x) = N\operatorname{sech}x$, peaked at the origin and decaying exponentially as $x\to\pm\infty$.