Quantum Particle in a Box
The Particle in a Box model is one of the most important applications of quantum mechanics. It explains why the energy of microscopic particles becomes quantized.
1. Physical Setup
Consider a particle of mass \(m\) confined inside a one-dimensional box of length \(L\).
The particle can move only between:
\[ x=0 \quad \text{and} \quad x=L \]
The walls are infinitely rigid, so the particle cannot escape.
2. Potential Energy Function
The potential energy is defined as:
$$ x \le 0 \; \text{or} \; x \ge L $$
This means:
- No external force acts on the particle.
- The particle moves freely inside the box.
- The total energy is purely kinetic energy.
2. Outside the Box
For:
\[ x\le0 \quad \text{or} \quad x\ge L \]
the potential energy becomes:
\[ V(x)=\infty \]
Infinite potential means:
- The particle cannot penetrate the walls.
- The probability of finding the particle outside the box is zero.
- The wave function must vanish at the walls.
3. Schrödinger Equation
The time-independent Schrödinger equation is:
\[ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi=E\psi \]
Inside the box, \(V(x)=0\), therefore:
\[ -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi \]
Rearranging:
\[ \frac{d^2\psi}{dx^2}+\frac{2mE}{\hbar^2}\psi=0 \]
Let:
\[ k^2=\frac{2mE}{\hbar^2} \]
Then the equation becomes:
\[ \frac{d^2\psi}{dx^2}+k^2\psi=0 \]
4. General Solution
The general solution is:
\[ \psi(x)=A\sin(kx)+B\cos(kx) \]
where:
- \(A\) and \(B\) are constants
- \(k\) is the wave number
5. Boundary Conditions
Since the walls are impenetrable:
\[ \psi(0)=0 \]
\[ \psi(L)=0 \]
Condition at \(x=0\)
\[ \psi(0)=A\sin(0)+B\cos(0) \]
\[ \psi(0)=B \]
Since \(\psi(0)=0\):
\[ B=0 \]
Therefore:
\[ \psi(x)=A\sin(kx) \]
Condition at \(x=L\)
\[ \psi(L)=A\sin(kL)=0 \]
For non-zero \(A\):
\[ \sin(kL)=0 \]
Hence:
\[ kL=n\pi \]
where:
\[ n=1,2,3,\dots \]
Thus:
\[ k=\frac{n\pi}{L} \]
6. Quantized Energy
We know:
\[ k^2=\frac{2mE}{\hbar^2} \]
Substituting \(k=\frac{n\pi}{L}\):
\[ \left(\frac{n\pi}{L}\right)^2=\frac{2mE}{\hbar^2} \]
Solving for energy:
\[ \boxed{ E_n=\frac{n^2\pi^2\hbar^2}{2mL^2} } \]
This shows that energy is quantized.
7. Ground State Energy
For the lowest state:
\[ n=1 \]
Therefore:
\[ E_1=\frac{\pi^2\hbar^2}{2mL^2} \]
The particle can never have zero energy. This minimum energy is called the zero-point energy.
8. Normalization of Wave Function
The normalization condition is:
\[ \int_0^L |\psi(x)|^2 dx =1 \]
Substituting:
\[ \psi(x)=A\sin\left(\frac{n\pi x}{L}\right) \]
We obtain:
\[ A=\sqrt{\frac{2}{L}} \]
Hence the normalized wave function is:
\[ \boxed{ \psi_n(x)= \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right) } \]
9. Probability Density
The probability density is:
\[ |\psi_n(x)|^2 \]
It represents the probability of finding the particle at position \(x\).
10. Nodes
Nodes are points where:
\[ \psi(x)=0 \]
Number of nodes:
\[ n-1 \]
- \(n=1\) → 0 nodes
- \(n=2\) → 1 node
- \(n=3\) → 2 nodes
11. Standing Wave Nature
Allowed wavelengths satisfy:
\[ L=\frac{n\lambda}{2} \]
Therefore:
\[ \boxed{ \lambda_n=\frac{2L}{n} } \]
Only waves fitting perfectly inside the box are allowed.
12. Momentum of the Particle
Using de Broglie relation:
\[ p=\hbar k \]
Therefore:
\[ \boxed{ p_n=\frac{n\pi\hbar}{L} } \]
13. Energy Level Spacing
Difference between adjacent levels:
\[ \Delta E=E_{n+1}-E_n \]
\[ \boxed{ \Delta E= \frac{(2n+1)\pi^2\hbar^2}{2mL^2} } \]
14. Important Conclusions
- Energy is quantized
- The particle cannot have zero energy
- Wave functions form standing waves
- Only specific wavelengths are allowed
- Quantum confinement causes discrete energy levels
15. Final Important Equations
Energy Eigenvalues
\[ \boxed{ E_n=\frac{n^2\pi^2\hbar^2}{2mL^2} } \]
Wave Functions
\[ \boxed{ \psi_n(x)= \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right) } \]
Allowed Wavelengths
\[ \boxed{ \lambda_n=\frac{2L}{n} } \]
Momentum
\[ \boxed{ p_n=\frac{n\pi\hbar}{L} } \]