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

Main article: 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 a complex Hilbert space, 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

${\begin{aligned}\langle Ax,y\rangle ={\frac {1}{4}}({}&\langle A(x+y),x+y\rangle -\langle A(x-y),x-y\rangle \\[1mm]&{}-i\langle A(x+iy),x+iy\rangle +i\langle A(x-iy),x-iy\rangle )\end{aligned}}$

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 = $H_{\mathbb {C} }$, 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 $H_{\mathbb {R} }.$

Order in self-adjoint operators on $H_{\mathbb {C} }$

A natural ordering of self-adjoint operators arises from the definition of positive operators. Define $B\geq A$ if the following hold:

1. $A$ and $B$ are self-adjoint
2. $B-A\geq 0$

It can be seen that a similar result as the Monotone convergence theorem holds for monotone increasing, bounded, self-adjoint operators on Hilbert spaces.

Application to physics: quantum states

Main articles: Quantum state and Density operator

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.

References

• Conway, John (1990), Functional Analysis: An Introduction, Springer Verlag, ISBN 0-387-97245-5

• Roman, Stephen (2008), Advanced Linear Algebra, Graduate Texts in Mathematics (Third ed.), Springer, ISBN 978-0-387-72828-5