Why is the magnetic Schrödinger operator positive? In the book _Schrödinger Operators_ by Cycon et al. they prove that the magnetic Schrödinger operator (as well as the Pauli operator) have essential spectrum $\sigma_{ess} = [0,\infty)$ if $B$ has decay at infinity and the potential $V$ is $-\Delta$-compact. Remember that the magnetic Schrödinger operator is given by $$H_{ms} := (-i\nabla -A)^2 +V $$ where $A$ is the vector potential. In their proof they claim that since $H(A,0)= (-i\nabla - A)^2$ is **positive** , the spectrum must be contained in $[0,\infty)$. A self-adjoint operator $T$ is positive means that $\langle Tx,x\rangle \geq 0$ for all $x\in X$ where $X$ is Banach. Now why is $H(A,0)$ positive? Is it because it is the square of an operator, specifically $-i\nabla -A$? If so is it trivial to see this?

If $X$ is a vector with Hermitian components, $$\langle\psi |X\cdot X|\psi\rangle=\sum_i\langle\psi |X_i^2| \psi\rangle=\sum_i\langle\psi |X_i^TX_i| \psi\rangle=\sum_i\Vert X_i|\psi\rangle\Vert^2\ge 0.$$