Particle in a 1D Box

Before we can understand the complex, 3D structure of an atom that gives us the periodic table, we need to understand why quantum mechanics produces discrete, "chunked" energy levels in the first place.

To see this mathematically, we start by solving the simplest possible system: a single particle trapped on a one-dimensional line inside an inescapable box.

2.1 The Setup: A Trapped Particle

Imagine a particle of mass $m$ trapped on an $x$-axis between two impenetrable walls located at $x=0$ and $x=L$.

• Inside the box ($0<x<L$): The particle feels no forces, so its potential energy is zero ($V=0$).
• Outside the box: The walls are infinitely hard, meaning it requires infinite potential energy to exist outside ($V=\mathrm{\infty }$).

Because the particle cannot have infinite energy, the probability of finding it outside the box is exactly zero. Therefore, its wave function $\psi (x)$ must be exactly zero everywhere outside.

Why must it be zero at the walls ($x=0,L$)? 🤔

The Schrödinger equation involves second-order derivatives, which can only be calculated if the wave function is continuous. If $\psi (x)$ is zero outside the box, it cannot suddenly "jump" to a non-zero value the moment it enters the box.

Proof by Contradiction: Imagine if $\psi (x)$ were discontinuous—meaning it had a sharp, vertical jump at the wall.

1. The first derivative ($d\psi /dx$) at that jump would be an "infinitely sharp" spike (mathematically, a Dirac delta function).
2. The second derivative ($d^2\psi /d{x}^2$) would be the derivative of that spike, which is even more extreme.
3. If we look back at the Schrödinger equation, the kinetic energy is directly proportional to this second derivative. An infinite spike in the derivative would mean the particle has infinite energy.

Since no real physical system can have infinite energy, we must conclude that a jump is impossible. The wave function must be continuous, meeting the zero-value of the outside world smoothly right at the boundary:

$$\psi (0)=0\text{and}\psi (L)=0$$

These are our Boundary Conditions. In quantum mechanics, boundary conditions are the magic mathematical ingredient that creates quantization.

2.2 Solving the Schrödinger Equation

We start with the Time-Independent Schrödinger Equation (TISE) derived in Chapter 1, tailored for one dimension:

$$-\frac{{\hslash }^2}{2m}\frac{d^2\psi (x)}{d{x}^2}+V(x)\psi (x)=E\psi (x)$$

Since we are only looking inside the box, we set $V(x)=0$. The equation simplifies beautifully to:

$$-\frac{{\hslash }^2}{2m}\frac{d^2\psi }{d{x}^2}=E\psi$$

Rearranging this to isolate the derivative:

$$\frac{d^2\psi }{d{x}^2}=-\frac{2mE}{{\hslash }^2}\psi$$

To make the math cleaner, physicists group the constants on the right side into a single squared term, $k^2$ (where $k=\sqrt{2mE}/\hslash$).

$$\frac{d^2\psi }{d{x}^2}=-k^2\psi$$

This is a classic differential equation. It asks: What function, when differentiated twice, returns the negative of itself (multiplied by a constant)? The mathematical answer is a combination of trigonometric sine and cosine waves. The general solution is:

$$\psi (x)=A\sin (kx)+B\cos (kx)$$

(where $A$ and $B$ are constants we need to figure out).

2.3 Applying the Boundary Conditions

Now we apply our physical reality (the boundary conditions) to this mathematical abstraction.

1. The Left Wall ($x=0$): We know $\psi (0)=0$. Plugging $x=0$ into our general equation:

$$\psi (0)=A\sin (0)+B\cos (0)$$$$0=0+B(1)⟹B=0$$

Since $B=0$, the cosine term completely vanishes! Our wave function is now just:

$$\psi (x)=A\sin (kx)$$

2. The Right Wall ($x=L$): We know $\psi (L)=0$. Plugging $x=L$ into our updated equation:

$$\psi (L)=A\sin (kL)=0$$

For this equation to equal zero, either $A=0$ or $\sin (kL)=0$. If $A=0$, the entire wave function is zero everywhere—meaning the particle doesn't exist. Since the particle does exist, it must be true that:

$$\sin (kL)=0$$

A sine wave is mathematically zero only at integer multiples of $\pi$ (i.e., $0,\pi ,2\pi ,3\pi \dots$). Therefore:

$$kL=n\pi$$$$k=\frac{n\pi }{L}$$

(where $n=1,2,3\dots$)

What about $n=0$ (or $k=0$)? 🤔

If we set $n=0$, then $k$ becomes 0. If $k=0$, the wave function becomes $\psi (x)=A\sin (0)=0$ for all values of $x$. A wave function that is zero everywhere means the particle has a $0\mathrm{\%}$ chance of being anywhere—it essentially doesn't exist. Since our premise is that a particle is in the box, we must discard the $n=0$ case as a "trivial" (physically impossible) solution.

2.4 The Birth of the Quantum Number

Look closely at what just happened. The value $n$ emerged purely from the geometric requirement that the wave function must smoothly drop to zero at the walls. This integer $n$ is our very first Quantum Number.

Because $n$ can only be a whole integer, $k$ can only take on specific, discrete values. We have naturally "quantized" the system just by trapping it!

2.5 The Energy Levels

Earlier, we defined $k^2=2mE/{\hslash }^2$. Now that we know $k=n\pi /L$, we can substitute it back to find the allowed energy ($E$) of the particle:

$${\left(\frac{n\pi }{L}\right)}^2=\frac{2mE}{{\hslash }^2}$$$$\frac{n^2{\pi }^2}{L^2}=\frac{2mE}{{\hslash }^2}$$

Solving for $E$, we get the final energy equation for a particle in a 1D box:

$$E_n=\frac{n^2{\pi }^2{\hslash }^2}{2m{L}^2}$$

The Takeaway

This equation is profound for three reasons:

1. Energy is chunked: The allowed energy $E_n$ is directly proportional to $n^2$. The particle cannot have an energy of 1.5 units; it must jump strictly from $1$ to $4$ to $9$, etc.
2. Zero-Point Energy: The lowest possible energy is when $n=1$ (since $n=0$ means no wave, and thus no particle). This means a quantum particle can never have zero energy; it can never be completely still.
3. The Blueprint for Atoms: We just proved that restricting a wave in space forces its energy to become quantized. In the coming chapters, we will wrap these "walls" into a 3D sphere to form an atom, and this exact same mathematical phenomenon will create the electron shells of the periodic table.