Positive operator

Cauchy–Schwarz inequality

Main article: Cauchy–Schwarz inequality

Take the inner product \langle \cdot, \cdot \rangle to be anti-linear on the first argument and linear on the second and suppose that A is positive and symmetric, the latter meaning that \langle Ax,y \rangle= \langle x,Ay \rangle . Then the non negativity of : \begin{align} \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{align} for all complex \lambda and \mu shows that :\left|\langle Ax,y\rangle \right|^2 \leq \langle Ax,x\rangle \langle Ay,y\rangle. 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 an operator is non-negative then it is symmetric

For x,y \in \operatorname{Dom}A, the polarization identity

: \begin{align} \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{align} 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 an operator is non-negative and defined on the whole complex Hilbert space, then it is self-adjoint and [[bounded operator|bounded]]

The symmetry of A implies that \operatorname{Dom}A \subseteq \operatorname{Dom}A^* and A = A^|_{\operatorname{Dom}(A)}. For A to be self-adjoint, it is necessary that \operatorname{Dom}A = \operatorname{Dom}A^. In our case, the equality of domains holds because H_\mathbb{C} = \operatorname{Dom}A \subseteq \operatorname{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}.

Partial order of self-adjoint operators

A natural partial 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 article: Quantum state, 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

• {{citation | last=Roman | first=Stephen

References

1. {{harvnb. Roman. 2008
2. Eidelman, Yuli, Vitali D. Milman, and Antonis Tsolomitis. 2004. Functional analysis: an introduction. Providence (R.I.): American mathematical Society.