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Problems 1011–1031: Basic Principles & 1D Motions

From Problems & Solutions in Quantum Mechanics (Lim, ed.)—covering uncertainty principle, infinite/finite square wells, delta potentials, step potentials, harmonic oscillator, and fundamental theorems.

Problem 1011Columbia

Uncertainty Principle: Position and Momentum Estimates

Problem Statement

Use the uncertainty principle to estimate the ground-state energy of a particle of mass $m$ in the following potentials:

(a) $V(x) = \frac{1}{2}m\omega^2 x^2$ (harmonic oscillator).

(b) $V(x) = \lambda |x|$ (linear potential, $\lambda > 0$).

(c) $V(x) = -V_0\,\delta(x)$ (attractive delta function potential, $V_0 > 0$).

Step-by-Step Solution

Part (a): Harmonic Oscillator

By the uncertainty principle, $\Delta x \,\Delta p \geq \hbar/2$. Taking $\Delta x \sim a$ and $\Delta p \sim \hbar/(2a)$ as the characteristic position and momentum spreads, the total energy is approximately:

$$E \sim \frac{(\Delta p)^2}{2m} + \frac{1}{2}m\omega^2(\Delta x)^2 = \frac{\hbar^2}{8ma^2} + \frac{1}{2}m\omega^2 a^2$$

Minimizing with respect to $a$: $\frac{dE}{da} = -\frac{\hbar^2}{4ma^3} + m\omega^2 a = 0$, giving:

$$a^2 = \frac{\hbar}{2m\omega} \quad\Longrightarrow\quad E_{\min} = \frac{1}{2}\hbar\omega$$

This matches the exact ground-state energy. The harmonic oscillator is the unique case where the uncertainty-principle estimate gives the exact result.

Part (b): Linear Potential

With $V = \lambda|x|$, the energy estimate is:

$$E \sim \frac{\hbar^2}{8ma^2} + \lambda a$$

Minimizing: $-\frac{\hbar^2}{4ma^3} + \lambda = 0$, so $a^3 = \frac{\hbar^2}{4m\lambda}$, giving $a = \left(\frac{\hbar^2}{4m\lambda}\right)^{1/3}$. Substituting:

$$E_{\min} \sim \frac{3}{2}\left(\frac{\hbar^2\lambda^2}{4m}\right)^{1/3} \sim \left(\frac{\hbar^2\lambda^2}{m}\right)^{1/3}$$

The exact Airy-function solution gives $E_0 = 2.338\left(\frac{\hbar^2\lambda^2}{2m}\right)^{1/3}$, so the estimate captures the correct scaling.

Part (c): Delta Function Potential

For the delta potential, the particle is localized within a region of size $a$, so $\langle V \rangle \sim -V_0/a$ (since $\delta(x)$ samples the wave function at $x=0$ and $|\psi(0)|^2 \sim 1/a$). The energy is:

$$E \sim \frac{\hbar^2}{2ma^2} - \frac{V_0}{a}$$

Minimizing: $-\frac{\hbar^2}{ma^3} + \frac{V_0}{a^2} = 0$, so $a = \frac{\hbar^2}{mV_0}$. Substituting:

$$E_{\min} = -\frac{mV_0^2}{2\hbar^2}$$

This matches the exact bound-state energy of the attractive delta function potential.

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