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 $x\in \mathop {\text{Dom}} (A)$ , $\langle Ax,x\rangle \geq 0$ , 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 $x\in \mathrm {Dom} (A)$ .

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 $Ax=0.$

A≥0 ⇔ A^*≥0

Indeed, if $A\geq 0,$ then

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

The inverse is proved likewise.

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

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, and hence $\mathop {\text{Dom}} A\subseteq \mathop {\text{Dom}} A^{*}$ 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 follows either from the Hellinger–Toeplitz theorem or the following argument. The graph of a linear operator $T$ is defined as $G(T)\ {\stackrel {\text{def}}{=}}\ \{(x,Tx)\mid x\in \mathop {\text{Dom}} T\}.$ In our case, the reader can verify that the set $G(A)$ is topologically closed in $H_{\mathbb {C} }\times H_{\mathbb {C} }.$ The Closed graph theorem then guarantees that the operator $A$ is continuous and therefore bounded.

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.