Ground State and Zero-Point Energy

Energy levels of the infinite square well

For a particle of mass $m$ confined between impenetrable walls at $x = 0$ and $x = L$ , the Schrödinger equation inside the box admits only standing-wave solutions that vanish at both walls. This boundary condition forces the allowed wavelengths to satisfy $\lambda_n = 2L/n$ for positive integers $n$ , giving momenta $p_n = h/\lambda_n = n\pi\hbar/L$ . The kinetic energy of state $n$ is then

E_n = \frac{p_n^2}{2m} = \frac{n^2 \pi^2 \hbar^2}{2mL^2}, \qquad n = 1, 2, 3, \ldots

Energy is discretized — the allowed values are the integer-square multiples $E_n = n^2 E_1$ of the base unit $E_1$ .

The ground state

The ground state is the state of lowest possible energy, reached when $n = 1$ :

E_1 = \frac{\pi^2 \hbar^2}{2mL^2}.

For an electron ( $m_e = 9.109 \times 10^{-31}\,\text{kg}$ ) in a box of width $L = 1\,\text{nm}$ , inserting numbers gives

E_1 = \frac{(3.1416)^2 \times (1.055 \times 10^{-34})^2}{2 \times (9.109 \times 10^{-31}) \times (10^{-9})^2} = 6.030 \times 10^{-20}\,\text{J} \approx 0.376\,\text{eV}.

Halving the box width to $L = 0.5\,\text{nm}$ multiplies $E_1$ by four (it scales as $1/L^2$ ), giving roughly $1.50\,\text{eV}$ — the value you will compute in the exercise.

Why zero-point energy cannot be zero

A classical particle at rest has zero kinetic energy. Quantum mechanics forbids this inside a box. The Heisenberg uncertainty principle requires $\Delta x \, \Delta p \geq \hbar/2$ . A particle in the box has position uncertainty at most $\Delta x \approx L$ . Consequently its momentum uncertainty satisfies $\Delta p \gtrsim \hbar/(2L)$ , and the corresponding kinetic energy is at minimum on the order of $\hbar^2/(8mL^2)$ — non-zero, and of the same form as $E_1$ .

This irreducible energy floor is called the zero-point energy. It has real physical consequences: liquid helium remains a fluid at atmospheric pressure all the way to absolute zero because the zero-point kinetic energy of the light helium atoms is large enough to prevent solidification.

Summary

| Quantity | Expression | |---|---| | Allowed energies | $E_n = n^2 \pi^2 \hbar^2 / (2mL^2)$ | | Ground state ( $n = 1$ ) | $E_1 = \pi^2 \hbar^2 / (2mL^2)$ | | Zero-point energy | $E_1 \neq 0$ — forbidden by the uncertainty principle | | Box-width scaling | $E_1 \propto 1/L^2$ |

Try it

This is a numerical exercise — your code should return a number. Compute the ground-state energy in eV for an electron confined in a box of width $L = 0.5\,\text{nm}$ .

Run your code to see the quantum state.