Positive operator

In mathematics (specifically in linear algebra and functional analysis) as well as physics, a linear operator $A$ acting on an inner product space is called positive (also non-negative, positive-semidefinite; written $A\geq 0$ ) if, for every $x\in \mathop {\text{Dom}} A,$ $\langle Ax,x\rangle \geq 0.$ In physics, such operators represent quantum states.

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 $Ax=0.$

If A is positive, so is A^*

Indeed,

\langle A^{*}x,x\rangle ={\overline {\langle x,A^{*}x\rangle }}={\overline {\langle Ax,x\rangle }}=\langle Ax,x\rangle \geq 0.

Positive everywhere-defined operator on H_C is self-adjoint

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). 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. For $A$ to be self-adjoint, it is necessary that $\mathop {\text{Dom}} A=\mathop {\text{Dom}} A^{*}.$ Recall that, for an arbitrary operator $T,$ $\mathop {\text{Dom}} T\subseteq \mathop {\text{Dom}} T^{*}.$ In our case, the equality of domains holds because $\mathop {\text{Dom}} A=H_{\mathbb {C} }:$ $H_{\mathbb {C} }=\mathop {\text{Dom}} A\subseteq \mathop {\text{Dom}} A^{*},$ so $A$ is indeed self-adjoint.

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.