Positive operator

In mathematics (specifically linear algebra, operator theory, and functional analysis) as well as physics, a linear operator $A$ acting on an inner product space is called positive-semidefinite (or non-negative) if, for every $x\in \mathop {\text{Dom}} (A)$ , $\langle Ax,x\rangle \in \mathbb {R}$ and $\langle Ax,x\rangle \geq 0$ , where $\mathop {\text{Dom}} (A)$ is the domain of $A$ . Positive-semidefinite operators are denoted as $A\geq 0$ . The operator is said to be positive-definite, and written $A>0$ , if $\langle Ax,x\rangle >0,$ for all $x\in \mathop {\mathrm {Dom} } (A)\setminus \{0\}$ .

In physics (specifically quantum mechanics), such operators represent quantum states, via the density matrix formalism.

Cauchy–Schwarz inequality

If $A\geq 0,$ then

\left|\langle Ax,y\rangle \right|^{2}\leq \langle Ax,x\rangle \langle Ay,y\rangle .

Indeed, let $\varepsilon >0.$ Applying Cauchy–Schwarz inequality to the inner product

(x,y)_{\varepsilon }{\stackrel {\text{def}}{=}}\ \langle (A+\varepsilon \cdot \mathbf {1} )x,y\rangle

as $\varepsilon \downarrow 0$ proves the claim.

It follows that $\mathop {\text{Im}} A\perp \mathop {\text{Ker}} A.$ If $A$ is defined everywhere, and $\langle Ax,x\rangle =0,$ then $Ax=0.$

On H_C, if A ≥ 0 then A is symmetric

Without loss of generality, let the inner product $\langle \cdot ,\cdot \rangle$ be anti-linear on the first argument and linear on the second. (If the reverse is true, then we work with $\langle x,y\rangle _{\text{op}}{\stackrel {\text{def}}{=}}\ \langle y,x\rangle$ instead). For $x,y\in \mathop {\text{Dom}} A,$ the polarization identity

{\begin{aligned}\langle Ax,y\rangle ={}&\langle A(x+y),x+y\rangle -\langle A(x-y),x-y\rangle \\[1mm]&{}-i\langle A(x+iy),x+iy\rangle +i\langle A(x-iy),x-iy\rangle \end{aligned}}

and the fact that $\langle Ax,x\rangle =\langle x,Ax\rangle ,$ for positive operators, show that $\langle Ax,y\rangle =\langle x,Ay\rangle ,$ so $A$ is symmetric.

In contrast with the complex case, a positive-semidefinite operator on a real Hilbert space $H_{\mathbb {R} }$ may not be symmetric. As a counterexample, define $A:\mathbb {R} ^{2}\to \mathbb {R} ^{2}$ to be an operator of rotation by an acute angle $\varphi \in (-\pi /2,\pi /2).$ Then $\langle Ax,x\rangle =\|Ax\|\|x\|\cos \varphi >0,$ but $A^{*}=A^{-1}\neq A,$ so $A$ is not symmetric.

If A ≥ 0 and Dom A = H_C, then A is self-adjoint and bounded

The symmetry of $A$ implies that $\mathop {\text{Dom}} A\subseteq \mathop {\text{Dom}} A^{*}$ and $A=A^{*}|_{\mathop {\text{Dom}} (A)}.$ For $A$ to be self-adjoint, it is necessary that $\mathop {\text{Dom}} A=\mathop {\text{Dom}} A^{*}.$ In our case, the equality of domains holds because $H_{\mathbb {C} }=\mathop {\text{Dom}} A\subseteq \mathop {\text{Dom}} A^{*},$ so $A$ is indeed self-adjoint. The fact that $A$ is bounded now follows from the Hellinger–Toeplitz theorem.

This property does not hold on $H_{\mathbb {R} }.$

Application to physics: quantum states

The definition of a quantum system includes a complex separable Hilbert space $H_{\mathbb {C} }$ and a set ${\cal {S}}$ of positive trace-class operators $\rho$ on $H_{\mathbb {C} }$ for which $\mathop {\text{Trace}} \rho =1.$ The set ${\cal {S}}$ is the set of states. Every $\rho \in {\cal {S}}$ is called a state or a density operator. For $\psi \in H_{\mathbb {C} },$ where $\|\psi \|=1,$ the operator $P_{\psi }$ of projection onto the span of $\psi$ is called a pure state. (Since each pure state is identifiable with a unit vector $\psi \in H_{\mathbb {C} },$ some sources define pure states to be unit elements from $H_{\mathbb {C} }).$ States that are not pure are called mixed.