Let’s solve **step by step:** --- ### 1. **Setting up the Problem** - **Potential:** - \( V(x) = 0 \) for \( 0 < x < L \) - \( V(x) = \infty \) elsewhere - **Wave function:** \( \Psi(x) = 0 \) for \( x \leq 0 \) and \( x \geq L \) --- ### 2. **Time-Independent Schrödinger Equation (TISE)** For \( 0 < x < L \), the TISE is: \[ -\frac{\hbar^2}{2m} \frac{d^2 \Psi(x)}{dx^2} = E \Psi(x) \] or \[ \frac{d^2 \Psi(x)}{dx^2} + k^2 \Psi(x) = 0 \] where \( k^2 = \frac{2mE}{\hbar^2} \). --- ### 3. **General Solution** The general solution of the above equation is: \[ \Psi(x) = A \sin(kx) + B \cos(kx) \] --- ### 4. **Applying Boundary Conditions** #### **Boundary 1: \( x = 0 \), \( \Psi(0) = 0 \)** \[ \Psi(0) = A \sin(0) + B \cos(0) = B = 0 \] So \( B = 0 \). #### **Boundary 2: \( x = L \), \( \Psi(L) = 0 \)** \[ \Psi(L) = A \sin(kL) = 0 \] For non-trivial solution (\(A \neq 0\)), \(\sin(kL) = 0\) ⇒ \(kL = n\pi\), \(n = 1,2,3,\dots\). So, \[ k = \frac{n\pi}{L} \] --- ### 5. **Normalized Wave Functions** So, \[ \Psi_n(x) = A \sin\left(\frac{n\pi x}{L}\right) \] for \( n = 1, 2, 3, \ldots \) **Normalization:** Require \[ \int_0^L |\Psi_n(x)|^2 dx = 1 \] \[ A^2 \int_0^L \sin^2\left(\frac{n\pi x}{L}\right) dx = 1 \] But \[ \int_0^L \sin^2\left(\frac{n\pi x}{L}\right) dx = \frac{L}{2} \] So \[ A^2 \cdot \frac{L}{2} = 1 \implies A = \sqrt{\frac{2}{L}} \] **Final normalized wave function:** \[ \boxed{ \Psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right), \quad n=1,2,3,\ldots,\quad 0 < x < L } \] --- ### 6. **Probability Density** \[ |\Psi_n(x)|^2 = \left(\sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)\right)^2 = \frac{2}{L} \sin^2\left(\frac{n\pi x}{L}\right) \] --- ### 7. **Summary & Boundary Conditions** - **Boundary conditions:** \( \Psi(0) = 0, \Psi(L) = 0 \) - **Normalized wave functions:** \[ \boxed{ \Psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right) } \] - **Probability density:** \[ \boxed{ |\Psi_n(x)|^2 = \frac{2}{L} \sin^2\left(\frac{n\pi x}{L}\right) } \] - Valid for \( 0 < x < L \), and zero elsewhere. --- Let me know if you want the energy levels as well!