Let's break down the solution into **two steps** as requested, with explanations at each step: --- ### **Step 1: Prove the Inequality Using Cauchy-Schwarz** **Statement:** For any two vectors \( |\mu\rangle \) and \( |\nu\rangle \), and a Hermitian operator \( H \) with positive eigenvalues, \[ |\langle \mu | H | \nu \rangle|^2 \leq \langle \mu | H | \mu \rangle \langle \nu | H | \nu \rangle \] **Proof & Explanation:** - Define a new inner product: \( (\phi, \psi)_H = \langle \phi | H | \psi \rangle \). - Since \( H \) is Hermitian and has positive eigenvalues, this is a valid inner product. - By the **Cauchy-Schwarz inequality** for inner products: \[ |(\mu, \nu)_H|^2 \leq (\mu, \mu)_H (\nu, \nu)_H \] which gives: \[ |\langle \mu | H | \nu \rangle|^2 \leq \langle \mu | H | \mu \rangle \langle \nu | H | \nu \rangle \] **Explanation:** This follows directly from the Cauchy-Schwarz inequality, as the positivity of \( H \) ensures the inner product is well-defined. --- ### **Step 2: Prove \(\mathrm{Tr}(H) > 0\)** **Proof & Explanation:** - The trace of an operator in an orthonormal eigenbasis is the sum of its eigenvalues. - Let the eigenvalues of \( H \) be \( \lambda_i > 0 \). - Therefore, \[ \mathrm{Tr}(H) = \sum_i \lambda_i \] - Since all \( \lambda_i \) are positive, the sum is **strictly greater than zero**. **Explanation:** The trace is positive because it is a sum of positive numbers (the positive eigenvalues). --- ### **Final Answers:** 1. **Inequality:** \[ |\langle \mu | H | \nu \rangle|^2 \leq \langle \mu | H | \mu \rangle \langle \nu | H | \nu \rangle \] (Follows from Cauchy-Schwarz for the positive-definite inner product defined by \( H \).) 2. **Trace:** \[ \mathrm{Tr}(H) > 0 \] (Because all eigenvalues of \( H \) are positive.)