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.