KAHIBARO
Table of Contents
A First Quantum Model
The particle in a box is one of the simplest and most important models in quantum mechanics. It describes a particle that is free to move inside a limited region of space, but cannot exist outside it. Even though it is an idealized model, it teaches several central quantum ideas very clearly.
In classical physics, a particle trapped between two walls can have any position inside the box and can move with any speed. In quantum physics, the situation is very different. The allowed states are restricted, the energy cannot take arbitrary values, and the particle is described by a wave function rather than a definite path.
This model is sometimes called the infinite square well. The word infinite means that the walls are assumed to be impossibly high, so the particle can never escape. The word well refers to the potential energy shape.
The Physical Setup
Consider a one dimensional box extending from $x = 0$ to $x = L$. Inside the box, the potential energy is zero. Outside the box, the potential energy is infinite. This is written as
$$
V(x) =
\begin{cases}
0, & 0 < x < L \\
\infty, & x \le 0 \text{ or } x \ge L
\end{cases}
$$
Because the potential is infinite outside the box, the particle cannot be found there. Therefore the wave function must vanish outside, and also at the walls.
For a particle in an infinite box,
$$
\psi(0) = 0, \qquad \psi(L) = 0
$$
These are the boundary conditions.
The Potential Well Picture
A simple sketch helps show the idea.
Inside the interval $0 < x < L$, the particle experiences no force because the potential is constant there. But the walls force the wave function to fit exactly into the box.
Solving the Schrödinger Equation
Inside the box, where $V(x)=0$, the time independent Schrödinger equation becomes
$$
-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} = E\psi
$$
or equivalently,
$$
\frac{d^2\psi}{dx^2} + k^2 \psi = 0
$$
where
$$
k^2 = \frac{2mE}{\hbar^2}
$$
The general solution of this differential equation is
$$
\psi(x) = A\sin(kx) + B\cos(kx)
$$
Now apply the boundary conditions.
From $\psi(0)=0$, we get
$$
\psi(0) = A\sin(0) + B\cos(0) = B = 0
$$
So the wave function becomes
$$
\psi(x) = A\sin(kx)
$$
From $\psi(L)=0$, we get
$$
A\sin(kL)=0
$$
For a nonzero wave function, this requires
$$
\sin(kL)=0
$$
which happens when
$$
kL = n\pi
$$
with
$$
n = 1,2,3,\dots
$$
Thus the allowed values of $k$ are
$$
k_n = \frac{n\pi}{L}
$$
Quantized Energy Levels
Using $k^2 = 2mE/\hbar^2$, the allowed energies are
$$
E_n = \frac{\hbar^2 k_n^2}{2m}
= \frac{\hbar^2}{2m}\left(\frac{n\pi}{L}\right)^2
$$
So
$$
E_n = \frac{n^2\pi^2\hbar^2}{2mL^2}
$$
This means the energy is quantized. Only specific discrete values are allowed.
For a particle in a one dimensional infinite box of length $L$,
$$
E_n = \frac{n^2\pi^2\hbar^2}{2mL^2}, \qquad n=1,2,3,\dots
$$
The energy cannot take arbitrary values.
A very important result is that the lowest energy is not zero. For $n=1$,
$$
E_1 = \frac{\pi^2\hbar^2}{2mL^2}
$$
This is called the ground state energy. The particle can never be completely at rest in the box.
Allowed Wave Functions
The normalized wave functions are
$$
\psi_n(x) =
\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right),
\qquad 0 < x < L
$$
and
$$
\psi_n(x)=0
\qquad \text{outside the box}
$$
The factor $\sqrt{2/L}$ is chosen so that the total probability of finding the particle somewhere in the box is 1.
The normalized stationary states for the infinite well are
$$
\psi_n(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)
$$
for $n=1,2,3,\dots$
Why Normalization Matters
The wave function itself is not directly observable. What matters physically is $|\psi(x)|^2$, which gives the probability density.
For normalization,
$$
\int_0^L |\psi_n(x)|^2 \, dx = 1
$$
This means the particle must be somewhere inside the box.
Probability Density
For the state $n$, the probability density is
$$
|\psi_n(x)|^2 = \frac{2}{L}\sin^2\left(\frac{n\pi x}{L}\right)
$$
This tells us where the particle is most likely to be found in a measurement.
For the ground state, the probability density is largest in the middle of the box and zero at the walls. For higher states, the probability density has more peaks and more zeros inside the box.
Nodes and State Number
A node is a point where the wave function is zero. In the infinite box, the walls are always nodes. Higher energy states also have internal nodes.
For the state with quantum number $n$, the number of internal nodes is
$$
n-1
$$
So the first few states look like this:
| Quantum number $n$ | Energy | Internal nodes |
|---|---|---|
| 1 | $E_1$ | 0 |
| 2 | $4E_1$ | 1 |
| 3 | $9E_1$ | 2 |
| 4 | $16E_1$ | 3 |
The energy increases like $n^2$, not linearly.
Ground State and Excited States
The state $n=1$ is the ground state, the lowest possible energy state. States with $n=2,3,4,\dots$ are excited states.
A key quantum feature is that there is no $n=0$ solution. If we tried $n=0$, the wave function would be zero everywhere, and that does not represent a physical state.
There is no allowed state with $n=0$ in the infinite box.
Therefore the minimum energy is not zero.
Dependence on Mass and Box Size
The energy formula shows how the allowed energies depend on the particle mass $m$ and box length $L$:
$$
E_n \propto \frac{1}{mL^2}
$$
This means that a lighter particle has larger energy spacing, and a smaller box also has larger energy spacing.
So confinement matters. If a particle is squeezed into a smaller region, the quantum energies spread farther apart.
Momentum in the Box
The wave functions are standing waves. They are not simple single direction traveling waves. Because of this, a state in the box is not a state of definite momentum in the simplest sense.
However, the allowed wave numbers are fixed by
$$
k_n = \frac{n\pi}{L}
$$
and the corresponding momentum scale is
$$
p_n = \hbar k_n = \frac{n\pi\hbar}{L}
$$
This helps connect the box solution to the idea of matter waves.
A Standing Wave Interpretation
The allowed wave functions fit into the box like standing waves on a string fixed at both ends. Only certain wavelengths are possible:
$$
L = \frac{n\lambda_n}{2}
$$
so
$$
\lambda_n = \frac{2L}{n}
$$
This immediately gives
$$
k_n = \frac{2\pi}{\lambda_n} = \frac{n\pi}{L}
$$
This standing wave picture is a simple way to understand why the energies are quantized.
Summary of the Model
The particle in a box is simple, but it reveals several foundational ideas of quantum mechanics very clearly.
| Feature | Result |
|---|---|
| Region of motion | $0 < x < L$ |
| Potential inside | $V=0$ |
| Potential outside | $V=\infty$ |
| Boundary conditions | $\psi(0)=\psi(L)=0$ |
| Allowed wave functions | $\psi_n(x)=\sqrt{2/L}\sin(n\pi x/L)$ |
| Allowed energies | $E_n=\dfrac{n^2\pi^2\hbar^2}{2mL^2}$ |
| Lowest state | $n=1$ |
| Ground state energy | Nonzero |
The particle in a box shows three major quantum ideas at once:
$$
\text{discrete states, discrete energies, and probability waves}
$$
These are among the most important differences between classical and quantum physics.