Let's solve the given problem step by step. --- ## **Problem Statement** Show that the first two wavefunctions of the (unshifted) particle in a box are orthogonal; that is, \[ 0 = \int_0^L B \sin\left(\frac{n_1 \pi x}{L}\right) \cdot B \sin\left(\frac{n_2 \pi x}{L}\right) dx \quad \text{where} \quad n_1 = 1, \; n_2 = 2 \] --- ## **Step 1: Write the Integral** \[ I = \int_0^L B \sin\left(\frac{n_1 \pi x}{L}\right) \cdot B \sin\left(\frac{n_2 \pi x}{L}\right) dx \] Plug in \( n_1 = 1 \), \( n_2 = 2 \): \[ I = \int_0^L B^2 \sin\left(\frac{\pi x}{L}\right) \sin\left(\frac{2\pi x}{L}\right) dx \] --- ## **Step 2: Use the Product-to-Sum Formula** Recall: \[ \sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)] \] So, \[ \sin\left(\frac{\pi x}{L}\right) \sin\left(\frac{2\pi x}{L}\right) = \frac{1}{2} \left[ \cos\left(\frac{\pi x}{L} - \frac{2\pi x}{L}\right) - \cos\left(\frac{\pi x}{L} + \frac{2\pi x}{L}\right) \right] \] \[ = \frac{1}{2} \left[ \cos\left(-\frac{\pi x}{L}\right) - \cos\left(\frac{3\pi x}{L}\right) \right] \] But \(\cos(-\theta) = \cos \theta\): \[ = \frac{1}{2} \left[ \cos\left(\frac{\pi x}{L}\right) - \cos\left(\frac{3\pi x}{L}\right) \right] \] --- ## **Step 3: Substitute Back into the Integral** \[ I = B^2 \int_0^L \frac{1}{2} \left[ \cos\left(\frac{\pi x}{L}\right) - \cos\left(\frac{3\pi x}{L}\right) \right] dx \] \[ = \frac{B^2}{2} \left( \int_0^L \cos\left(\frac{\pi x}{L}\right) dx - \int_0^L \cos\left(\frac{3\pi x}{L}\right) dx \right) \] --- ## **Step 4: Integrate Each Term** Recall: \[ \int \cos(ax) dx = \frac{1}{a} \sin(ax) + C \] - For \(\cos\left(\frac{\pi x}{L}\right)\): \[ \int_0^L \cos\left(\frac{\pi x}{L}\right) dx = \left[ \frac{L}{\pi} \sin\left(\frac{\pi x}{L}\right) \right]_0^L \] Plug in limits: - At \(x = L\): \(\sin\left(\frac{\pi L}{L}\right) = \sin(\pi) = 0\) - At \(x = 0\): \(\sin(0) = 0\) So, the integral is \(0 - 0 = 0\) --- - For \(\cos\left(\frac{3\pi x}{L}\right)\): \[ \int_0^L \cos\left(\frac{3\pi x}{L}\right) dx = \left[ \frac{L}{3\pi} \sin\left(\frac{3\pi x}{L}\right) \right]_0^L \] Plug in limits: - At \(x = L\): \(\sin\left(\frac{3\pi L}{L}\right) = \sin(3\pi) = 0\) - At \(x = 0\): \(\sin(0) = 0\) So, the integral is \(0 - 0 = 0\) --- ## **Step 5: Combine Results** \[ I = \frac{B^2}{2} (0 - 0) = 0 \] --- ## **Final Answer** \[ \boxed{0 = \int_0^L B \sin\left(\frac{\pi x}{L}\right) \cdot B \sin\left(\frac{2\pi x}{L}\right) dx} \] **This shows that the first two wavefunctions are orthogonal.**