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Problems 1055–1072: Delta Functions, WKB & Miscellaneous 1D

Selected problems from Problems & Solutions in Quantum Mechanics (Lim, ed.)—covering the rest of Chapter 1: sech-squared potentials, periodic potentials, delta-function lattices, energy bands, operator algebra, and the Heisenberg picture.

← Problems 1032–1054
Problem 1055Buffalo

Sech-Squared Potential: S-Matrix and Bound State

Problem Statement

A Schrödinger equation in one dimension reads $$\left(-\frac{\partial^2}{\partial x^2} - 2\,\text{sech}^2 x\right)\psi = \varepsilon\,\psi$$ where we have set $\hbar = 1$ and $m = 1/2$.

(a) Show that $\exp(ikx)(\tanh x + \text{const})$ is a solution for a particular value of the constant. Calculate the S-matrix (transmission and reflection coefficients) for this problem.

(b) The wave function $\text{sech}\,x$ happens to satisfy the Schrödinger equation. Calculate the energy of the corresponding bound state and give a simple argument that it must be the ground state.

(c) Outline how you might proceed to estimate the ground-state energy if you did not know the wave function.

Step-by-Step Solution

Part (a): Scattering Solution and S-Matrix

Let the constant in the given solution be $K$, so we try $\psi = e^{ikx}(\tanh x + K)$. Substituting into the Schrödinger equation, we need to compute the second derivative. Since $\frac{d}{dx}\tanh x = \text{sech}^2 x$ and $\frac{d}{dx}\text{sech}^2 x = -2\,\text{sech}^2 x\,\tanh x$:

$$\psi = e^{ikx}(\tanh x + K)$$ $$\psi' = e^{ikx}\bigl[ik(\tanh x + K) + \text{sech}^2 x\bigr]$$ $$\psi'' = e^{ikx}\bigl[-k^2(\tanh x + K) + 2ik\,\text{sech}^2 x - 2\,\text{sech}^2 x\,\tanh x\bigr]$$

Substituting into $-\psi'' - 2\,\text{sech}^2 x\,\psi = \varepsilon\,\psi$:

$$k^2(\tanh x + K) - 2ik\,\text{sech}^2 x + 2\,\text{sech}^2 x\,\tanh x - 2\,\text{sech}^2 x(\tanh x + K) = \varepsilon(\tanh x + K)$$

The $\text{sech}^2 x\,\tanh x$ terms cancel. Collecting the $\text{sech}^2 x$ terms:

$$-2ik\,\text{sech}^2 x - 2K\,\text{sech}^2 x = 0 \implies K = -ik$$

The remaining terms give $\varepsilon = k^2$. Hence:

$$\psi = e^{ikx}(\tanh x - ik)$$

Now examine the asymptotic behavior. As $x \to +\infty$, $\tanh x \to 1$, so $\psi \to (1 - ik)e^{ikx}$. As $x \to -\infty$, $\tanh x \to -1$, so $\psi \to -(1 + ik)e^{ikx}$.

Since $V(x) \le 0$ and $\varepsilon > 0$, the transmission coefficient $T = 1$ and the reflection coefficient $R = 0$. The particle travels through the $\text{sech}^2$ potential without reflection. The S-matrix is:

$$S = \begin{pmatrix} 1 & -(1-ik)/(1+ik) \\ -(1-ik)/(1+ik) & 0 \end{pmatrix}$$

The reflectionless nature of the $\text{sech}^2$ potential is a well-known result in quantum mechanics.

Part (b): Bound State Energy

Letting $\psi = \text{sech}\,x$ and substituting into the Schrödinger equation, we use $\frac{d^2}{dx^2}\text{sech}\,x = \text{sech}\,x - 2\,\text{sech}^3 x = \text{sech}\,x(1 - 2\,\text{sech}^2 x)$:

$$-\text{sech}\,x(1 - 2\,\text{sech}^2 x) - 2\,\text{sech}^2 x \cdot \text{sech}\,x = \varepsilon\,\text{sech}\,x$$ $$-\text{sech}\,x + 2\,\text{sech}^3 x - 2\,\text{sech}^3 x = \varepsilon\,\text{sech}\,x$$ $$-\text{sech}\,x = \varepsilon\,\text{sech}\,x$$

Hence $\varepsilon = -1$. The bound state energy is:

$$\boxed{\varepsilon = -1}$$

Because $\text{sech}\,x > 0$ for all $x$ (it is a nodeless function in the whole coordinate space), it must be the ground state. A node theorem guarantees that the ground state wave function of a one-dimensional potential has no nodes.

Part (c): Variational Estimate

We might proceed by assuming a nodeless, even trial wave function with a variational parameter, such as $\psi = A\,e^{-\alpha x^2}$ or $\psi = A/(1 + \beta x^2)$, and minimize $\langle H \rangle$ with respect to the parameter to obtain an approximate value of the ground-state energy by the variational method.

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