Positive operator (Hilbert space)

In mathematics (specifically linear algebra, operator theory, and functional analysis) as well as physics, a linear operator ${\displaystyle A}$ acting on an inner product space is called positive-semidefinite (or non-negative) if, for every ${\displaystyle x\in \mathop {\text{Dom}} (A)}$, ${\displaystyle \langle Ax,x\rangle \in \mathbb {R} }$ and ${\displaystyle \langle Ax,x\rangle \geq 0}$, where ${\displaystyle \mathop {\text{Dom}} (A)}$ is the domain of ${\displaystyle A}$. Positive-semidefinite operators are denoted as ${\displaystyle A\geq 0}$. The operator is said to be positive-definite, and written ${\displaystyle A>0}$, if ${\displaystyle \langle Ax,x\rangle >0,}$ for all ${\displaystyle x\in \mathop {\mathrm {Dom} } (A)\setminus \{0\}}$.^[1]

Many authors define a positive operator ${\displaystyle A}$ to be a self-adjoint (or at least symmetric) non-negative operator. We show below that for a complex Hilbert space the self adjointness follows automatically from non-negativity. For a real Hilbert space non-negativity does not imply self adjointness.

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

Cauchy–Schwarz inequality

Take the inner product ${\displaystyle \langle \cdot ,\cdot \rangle }$ to be anti-linear on the first argument and linear on the second and suppose that ${\displaystyle A}$ is positive and symmetric, the latter meaning that ${\displaystyle \langle Ax,y\rangle =\langle x,Ay\rangle }$. Then the non negativity of

${\displaystyle {\begin{aligned}\langle A(\lambda x+\mu y),\lambda x+\mu y\rangle =|\lambda |^{2}\langle Ax,x\rangle +\lambda ^{*}\mu \langle Ax,y\rangle +\lambda \mu ^{*}\langle Ay,x\rangle +|\mu |^{2}\langle Ay,y\rangle \\[1mm]=|\lambda |^{2}\langle Ax,x\rangle +\lambda ^{*}\mu \langle Ax,y\rangle +\lambda \mu ^{*}(\langle Ax,y\rangle )^{*}+|\mu |^{2}\langle Ay,y\rangle \end{aligned}}}$

for all complex ${\displaystyle \lambda }$ and ${\displaystyle \mu }$ shows that

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

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

On a complex Hilbert space, if an operator is non-negative then it is symmetric

For ${\displaystyle x,y\in \mathop {\text{Dom}} A,}$ the polarization identity

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

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

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

If an operator is non-negative and defined on the whole Hilbert space, then it is self-adjoint and bounded

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

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

Partial order of self-adjoint operators

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

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

It can be seen that a similar result as the Monotone convergence theorem holds for monotone increasing, bounded, self-adjoint operators on Hilbert spaces.^[2]

Application to physics: quantum states

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

References

• Conway, John B. (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

This page was last edited on 3 January 2024, at 23:28