We are asked to find the matrix elements of the operator (\hat{\Omega}) in the given two-dimensional orthonormal basis (|\phi_1\rangle, |\phi_2\rangle). Let’s solve it step by step numerically. The operator is: [ \hat{\Omega} = 3 |\phi_1\rangle \langle \phi_1| + \sqrt{2} |\phi_1\rangle \langle \phi_2| + \sqrt{2} |\phi_2\rangle \langle \phi_1| + 2 |\phi_2\rangle \langle \phi_2| ] The matrix elements of (\hat{\Omega}) in this basis are: [ \Omega_{ij} = \langle \phi_i | \hat{\Omega} | \phi_j \rangle \quad \text{for } i,j=1,2 ] --- ### Step 1: Compute (\Omega_{11} = \langle \phi_1 | \hat{\Omega} | \phi_1 \rangle) [ \begin{aligned} \Omega_{11} &= \langle \phi_1 | \big( 3 |\phi_1\rangle \langle \phi_1| + \sqrt{2} |\phi_1\rangle \langle \phi_2| + \sqrt{2} |\phi_2\rangle \langle \phi_1| + 2 |\phi_2\rangle \langle \phi_2| \big) | \phi_1 \rangle \ &= 3 \langle \phi_1 | \phi_1 \rangle \langle \phi_1 | \phi_1 \rangle + \sqrt{2} \langle \phi_1 | \phi_1 \rangle \langle \phi_2 | \phi_1 \rangle + \sqrt{2} \langle \phi_1 | \phi_2 \rangle \langle \phi_1 | \phi_1 \rangle + 2 \langle \phi_1 | \phi_2 \rangle \langle \phi_2 | \phi_1 \rangle \end{aligned} ] Since the basis is orthonormal: (\langle \phi_1 | \phi_1 \rangle = 1), (\langle \phi_1 | \phi_2 \rangle = 0) [ \Omega_{11} = 3 \cdot 1 + \sqrt{2} \cdot 1 \cdot 0 + \sqrt{2} \cdot 0