2D Infinite Well Quantum Wave Function Plotter

Last Updated on September 16, 2024 by Max

Derivation of Wave Functions in a 2D Infinite Well

To derive the wave function for a particle in a 2D infinite potential well, we start by considering a quantum mechanical particle confined within a rectangular region of space with infinite potential walls.

This means that the particle is trapped in a box where it cannot escape, and its wave function must satisfy certain boundary conditions imposed by the walls of the well.

Dimensions of the Infinite Well

• The potential well has a length $L_x$ in the $x$-direction and a length $L_y$ in the $y$-direction.
• Inside the well (for $0 \leq x \leq L_x$ and $0 \leq y \leq L_y$), the potential $V(x, y) = 0$.
• Outside the well, the potential $V(x, y) = \infty$.

Schrödinger Equation:

The time-independent Schrödinger equation for a particle of mass $m$ confined in a 2D infinite potential well is given by:

$$-\frac{\hbar^2}{2m} \left( \frac{\partial^2 \psi(x, y)}{\partial x^2} + \frac{\partial^2 \psi(x, y)}{\partial y^2} \right) = E \psi(x, y),$$

where $\hbar$ is the reduced Planck’s constant, $\psi(x, y)$ is the wave function of the particle, and $E$ is the energy of the particle.

Separation of Variables

To solve the Schrödinger equation, we use the method of separation of variables. We assume the solution can be written as a product of two functions, one depending only on $x$ and the other only on $y$:

$$\psi(x, y) = X(x) Y(y).$$

Substituting this into the Schrödinger equation gives:

$$-\frac{\hbar^2}{2m} \left( \frac{d^2 X(x)}{dx^2} Y(y) + X(x) \frac{d^2 Y(y)}{dy^2} \right) = E X(x) Y(y).$$

Dividing both sides by $X(x) Y(y)$, we obtain:

$$-\frac{\hbar^2}{2m} \left( \frac{1}{X(x)} \frac{d^2 X(x)}{dx^2} + \frac{1}{Y(y)} \frac{d^2 Y(y)}{dy^2} \right) = E.$$

Since the left side is a sum of two terms, one depending only on $x$ and the other only on $y$, each term must be equal to a constant. Let’s denote these separation constants by $E_x$ and $E_y$ such that:

$$\frac{1}{X(x)} \frac{d^2 X(x)}{dx^2} = -\frac{2mE_x}{\hbar^2}, \quad \frac{1}{Y(y)} \frac{d^2 Y(y)}{dy^2} = -\frac{2mE_y}{\hbar^2}.$$

Thus, the total energy $E$ of the system is:

$$E = E_x + E_y.$$

Solution in the $x$-Direction

The differential equation for $X(x)$ is:

$$\frac{d^2 X(x)}{dx^2} + \frac{2mE_x}{\hbar^2} X(x) = 0.$$

This is a standard form of the second-order differential equation whose solutions are sinusoidal functions:

$$X(x) = A \sin\left(k_x x\right) + B \cos\left(k_x x\right),$$

where $k_x = \sqrt{\frac{2mE_x}{\hbar^2}}$.

To satisfy the boundary conditions:

$\psi(x, y) = 0$ at $x = 0$ and $x = L_x$,

We find:

1. At $x = 0$: $X(0) = A \sin(0) + B \cos(0) = B = 0 \Rightarrow B = 0$.
2. At $x = L_x$: $X(L_x) = A \sin(k_x L_x) = 0$.

For a non-trivial solution, $A \neq 0$, so:

$$k_x L_x = n_x \pi, \quad \text{where } n_x = 1, 2, 3, \ldots$$

Thus, $k_x = \frac{n_x \pi}{L_x}$, and the solution for $X(x)$ becomes:

$$X(x) = A \sin\left( \frac{n_x \pi x}{L_x} \right).$$

Solution in the $y$-Direction

Similarly, for the $y$-direction, the differential equation for $Y(y)$ is:

$$\frac{d^2 Y(y)}{dy^2} + \frac{2mE_y}{\hbar^2} Y(y) = 0,$$

with solutions:

$$Y(y) = C \sin\left(k_y y\right) + D \cos\left(k_y y\right),$$

where $k_y = \sqrt{\frac{2mE_y}{\hbar^2}}$.

Applying the boundary conditions:

$\psi(x, y) = 0$ at $y = 0$ and $y = L_y$,

We find:

1. At $y = 0$: $Y(0) = C \sin(0) + D \cos(0) = D = 0 \Rightarrow D = 0$.
2. At $y = L_y$: $Y(L_y) = C \sin(k_y L_y) = 0$.

For a non-trivial solution, $C \neq 0$, so:

$$k_y L_y = n_y \pi, \quad \text{where } n_y = 1, 2, 3, \ldots$$

Thus, $k_y = \frac{n_y \pi}{L_y}$, and the solution for $Y(y)$ becomes:

$$Y(y) = C \sin\left( \frac{n_y \pi y}{L_y} \right).$$

The General Solution for the Wave Function

The general solution for the wave function $\psi(x, y)$ is the product of the solutions in the $x$- and $y$-directions:

$$\psi(x, y) = X(x) Y(y) = A C \sin\left( \frac{n_x \pi x}{L_x} \right) \sin\left( \frac{n_y \pi y}{L_y} \right).$$

Normalization of the Wave Function

The normalization condition requires that the total probability of finding the particle in the well is 1:

$$\int_0^{L_x} \int_0^{L_y} |\psi(x, y)|^2 \, dx \, dy = 1.$$

Substitute the wave function:

$$\int_0^{L_x} \int_0^{L_y} A^2 C^2 \sin^2\left( \frac{n_x \pi x}{L_x} \right) \sin^2\left( \frac{n_y \pi y}{L_y} \right) \, dx \, dy = 1.$$

Separate the integrals:

$$A^2 C^2 \left( \int_0^{L_x} \sin^2\left( \frac{n_x \pi x}{L_x} \right) \, dx \right) \left( \int_0^{L_y} \sin^2\left( \frac{n_y \pi y}{L_y} \right) \, dy \right) = 1.$$

Each integral evaluates to $\frac{L_x}{2}$ and $\frac{L_y}{2}$ respectively, so:

$$A^2 C^2 \left( \frac{L_x}{2} \right) \left( \frac{L_y}{2} \right) = 1.$$

Solving for $A C$:

$$A C = \frac{2}{\sqrt{L_x L_y}}.$$

Thus, the normalized wave function is:

$$\psi(x, y) = \frac{2}{\sqrt{L_x L_y}} \sin\left( \frac{n_x \pi x}{L_x} \right) \sin\left( \frac{n_y \pi y}{L_y} \right).$$

Energy Levels

The energy levels of the particle in the 2D infinite well are given by:

$$E_{n_x, n_y} = \frac{\hbar^2 \pi^2}{2m} \left( \frac{n_x^2}{L_x^2} + \frac{n_y^2}{L_y^2} \right), \quad \text{where } n_x, n_y = 1, 2, 3, \ldots$$

Conclusion

The wave function $\psi(x, y)$ describes the quantum state of a particle confined in a 2D infinite potential well, and the energy levels are quantized based on the quantum numbers $n_x$ and $n_y$. The derived wave function shows the sinusoidal nature of the probability amplitude, with nodes determined by the quantum numbers, reflecting the particle’s confinement within the well.