Particle in a One‑Dimensional Box

1. Introduction

Consider a particle of mass \(m\) trapped in a one‑dimensional box of length \(a\). The particle can move only along the \(x\)-axis. The potential inside the box is zero, and outside it is infinite:

We will solve the time‑independent Schrödinger equation using the postulates of quantum mechanics. The Hamiltonian operator for one dimension is:

2. Schrödinger Wave Equation

The time‑independent Schrödinger equation \(\hat{H}\psi = E\psi\) gives:

Multiplying by \(-\frac{2m}{\hbar^{2}}\) and rearranging:

Using \(\hbar = h/(2\pi)\), \(\hbar^{2} = h^{2}/(4\pi^{2})\). Then \(\frac{2m}{\hbar^{2}} = \frac{8\pi^{2}m}{h^{2}}\). Hence (4) becomes:

3. Solution Outside the Box (\(V=\infty\))

Outside the box, \(V=\infty\). Substituting into (5):

Since \(E\) is finite, \(E-\infty \to -\infty\). The term \(\frac{8\pi^{2}m}{h^{2}}(E-\infty)\psi\) dominates, forcing:

Thus the particle cannot exist outside the box; the wavefunction is zero there.

4. Solution Inside the Box (\(V=0\))

Inside the box (\(0 < x < a\)), \(V=0\). Equation (5) reduces to:

Define the constant:

Then equation (8) becomes a simple harmonic oscillator equation:

The general solution is:

where \(A\) and \(B\) are constants to be determined by boundary conditions and normalization.

5. Boundary Conditions

The wavefunction must be continuous, so it must vanish at the walls where the potential is infinite:

• At \(x = 0\): \(\psi(0)=0\)
• At \(x = a\): \(\psi(a)=0\)

First boundary condition (\(x=0\))

Thus \(B=0\) and the wavefunction reduces to:

Second boundary condition (\(x=a\))

Since \(A \neq 0\) (otherwise \(\psi\) is trivial), we require \(\sin(ka)=0\). Therefore:

Note: \(n=0\) would give \(\psi(x)=0\) everywhere (trivial, not a valid state). So \(n\) is a positive integer.

From (15):

Thus the wavefunction becomes:

6. Normalization

The wavefunction must satisfy the normalization condition:

Substituting (17):

Using the identity \(\sin^{2}\theta = \frac{1}{2}(1-\cos2\theta)\), we get:

The sine term vanishes at both limits, leaving:

Choosing the positive root (phase convention), we obtain the normalized wavefunction:

7. Energy Eigenvalues

From equations (9) and (16):

Solving for \(E\):

Thus the energy is quantized. The ground state (lowest energy) corresponds to \(n=1\):

Higher states have energies proportional to \(n^{2}\). The integer \(n\) is called the quantum number.

8. Explicit Wavefunctions for Low States

Using equation (23):

Higher states are obtained similarly. Note that \(n=0\) would give \(\psi=0\) everywhere and is not a valid physical state.

First three wavefunctions and their probability densities (n=1,2,3).

9. Probability Density

The probability density is given by \(|\psi_{n}(x)|^{2}\):

It is positive and normalized to 1 over the box. The probability of finding the particle in an interval \([x, x+dx]\) is \(|\psi|^{2}dx\).

10. Video Lecture (Urdu/Hindi)

11. Summary of Key Results

• Normalized wavefunction: \(\displaystyle \psi_{n}(x) = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)\), \(n=1,2,3,\dots\)
• Energy eigenvalues: \(\displaystyle E_{n} = \frac{n^{2}h^{2}}{8ma^{2}}\)
• Boundary conditions force quantization of both wavefunction and energy.
• The ground state has non‑zero energy (zero‑point energy).
• Probability density vanishes at the walls and has \(n-1\) nodes inside.
• The wavefunctions form a complete orthonormal set.

These results are the foundation for understanding quantum wells, molecular orbitals, and many other confined quantum systems.