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? 
 A: Let $\mathcal{H}$ denote the Hilbert space of wave functions $\psi:\mathbb{R}^3\to\mathbb{C}$.  Recall that
$$\langle u|v\rangle =\iiint_{\mathbb{R}^3}\bar{u}(x,y,z)\ v(x,y,z)\ dx\ dy\ dz$$
for all $u,v\in \mathcal{H}$.  Write $h$ for the operator $(-i\nabla -A)^2$.  Observe that
\begin{align}\langle hu|v\rangle &=\iiint_{\mathbb{R}^3}\overline{hu}(x,y,z)\ v(x,y,z)\ dx\ dy\ dz
\\&=\iiint_{\mathbb{R}^3}(i\nabla -A)^2\bar{u}(x,y,z)\ v(x,y,z)\ dx\ dy\ dz
\\&=\iiint_{\mathbb{R}^3}(i\nabla -A)\cdot \overline{\Phi u}(x,y,z)\ v(x,y,z)\ dx\ dy\ dz,
\end{align}
where $\Phi=-i\nabla-A$.  That is,
\begin{align}\langle hu|v\rangle &=i\iiint_{\mathbb{R}^3}\nabla\cdot \overline{\Phi u}(x,y,z)\ v(x,y,z)\ dx\ dy\ dz-\iiint_{\mathbb{R}^3} \overline{\Phi u}(x,y,z)\cdot Av(x,y,z)\ dx\ dy\ dz
\\&=-i\iiint_{\mathbb{R}^3}\overline{\Phi u}(x,y,z)\cdot \nabla v(x,y,z)\ dx\ dy\ dz
-\iiint_{\mathbb{R}^3}\overline{\Phi u}(x,y,z)\cdot Av(x,y,z)\ dx\ dy\ dz,\end{align}
where we apply integration by parts in higher dimension, assuming that $u(x,y,z)$ and $v(x,y,z)$ vanish quickly when $(x,y,z)$ is large.  That is,
\begin{align}\langle hu|v\rangle &=\iiint_{\mathbb{R}^3}\overline{\Phi u}(x,y,z)\cdot(-i\nabla-A)v(x,y,z)\ dx\ dy\ dz
\\&=\iiint_{\mathbb{R}^3}\overline{\Phi u}(x,y,z)\cdot\Phi v(x,y,z)\ dx\ dy\ dz.
\end{align}
In particular,
\begin{align}\langle hu|u\rangle &=\int_{\mathbb{R}^3}\overline{\Phi u}(x,y,z)\cdot\Phi u(x,y,z)\ dx\ dy\ dz\\&=\int_{\mathbb{R}^3}\big\Vert\Phi u(x,y,z)\big\Vert^2\ dx\ dy\ dz\geq 0\end{align}
A: If $X$ is a vector with Hermitian components, $$\langle\psi |X\cdot X|\psi\rangle=\sum_i\langle\psi |X_i^2| \psi\rangle=\sum_i\langle\psi |X_i^TX_i| \psi\rangle=\sum_i\Vert X_i|\psi\rangle\Vert^2\ge 0.$$
