one-dimensional box
\begin{equation} V(x) = \begin{cases} 0 & \text{for } 0 < x < L \ \infty & \text{otherwise} \end{cases} \end{equation}
This means that the particle is free to move within the interval (0, L) but cannot exist outside this region. Schrödinger Equation To find the allowed energy levels of the particle, we solve the time-independent Schrödinger equation:\begin{equation} -\frac{\hbar^2}{2m} \frac{d^2\psi(x)}{dx^2} = E\psi(x) \end{equation}
where \hbar is the reduced Planck's constant, m is the mass of the particle, E is the energy, and \psi(x) is the wave function of the particle. Boundary Conditions The boundary conditions for the wave function are:\begin{equation} \psi(0) = 0 \quad \text{and} \quad \psi(L) = 0 \end{equation}
These conditions arise from the fact that the wave function must vanish at the walls of the box, where the potential is infinite. General Solution The general solution to the Schrödinger equation can be expressed as:\begin{equation} \psi(x) = A \sin\left(kx\right) + B \cos\left(kx\right) \end{equation}
where k = \frac{\sqrt{2mE}}{\hbar}, and A and B are constants determined by the boundary conditions. Applying Boundary Conditions Applying the boundary condition \psi(0) = 0 gives:\begin{equation} \psi(0) = B = 0 \end{equation}
Thus, the wave function simplifies to:\begin{equation} \psi(x) = A \sin\left(kx\right) \end{equation}
Next, applying the second boundary condition \psi(L) = 0 leads to:\begin{equation} \psi(L) = A \sin(kL) = 0 \end{equation}
This condition is satisfied if:\begin{equation} kL = n\pi \quad (n = 1, 2, 3, \ldots) \end{equation}
From this, we can express k as:\begin{equation} k = \frac{n\pi}{L} \end{equation}
Wave Function Substituting k back into the wave function gives:\begin{equation} \psi_n(x) = A \sin\left(\frac{n\pi x}{L}\right) \end{equation}
The normalization condition requires that:\begin{equation} \int_0^L |\psi_n(x)|^2 dx = 1 \end{equation}
Calculating this integral leads to:\begin{equation} A^2 \int_0^L \sin^2\left(\frac{n\pi x}{L}\right) dx = 1 \end{equation}
Using the integral result:\begin{equation} \int_0^L \sin^2\left(\frac{n\pi x}{L}\right) dx = \frac{L}{2} \end{equation}
we find:\begin{equation} A^2 \cdot \frac{L}{2} = 1 \quad \Rightarrow \quad A = \sqrt{\frac{2}{L}} \end{equation}
Thus, the normalized wave function is:\begin{equation} \psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right) \end{equation}
Energy Levels The energy levels corresponding to these wave functions can be found using the relation for k:\begin{equation} E_n = \frac{\hbar^2 k^2}{2m} = \frac{\hbar^2}{2m} \left(\frac{n\pi}{L}\right)^2 \end{equation}
This simplifies to:\begin{equation} E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2} \end{equation}
where n = 1, 2, 3, \ldots represents the quantum number. Conclusion The one-dimensional box model illustrates the quantization of energy levels in quantum mechanics. The allowed energy levels are discrete, and the wave functions are sinusoidal, reflecting the confinement of the particle within the box. This model serves as a foundational example in quantum mechanics, providing insights into more complex systems.Listen to this page as a podcast (about 1 hour, generated with AI).
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