[NOTES/QCQI-01004] Positive Operators

$\newcommand{\average}[2]{\langle#1|#2|#1\rangle}$
$\newcommand{\innerproduct}[2]{\langle#1|#2\rsngle}$

1. Positive Operators

Definition : An operator $P$, in a complex inner product space, is called positive operator if
\begin{equation}\label{eq01}
\average{f}{A}\ge0\,.
\end{equation}
holds for all vectors $\ket{f}$ in the vector space.


2. Properties of positive operators

• The eigenvalues of a positive operator are positive. Proof: If $P$ is a positive operator and $\alpha$ is an eigenvalue with eigenvector $\ket{\alpha}$ \begin{equation}\label{eq02} P\ket{\alpha} = \alpha\ket{\alpha} \Rightarrow {\bra{\alpha}P\ket{\alpha}\over \langle\alpha\ket{\alpha}}=\alpha \end{equation} Therefore $\alpha\ge0$. Thus the eigenvalues of a positive operator are non negative.
• A positive operator in an inner product space is hermitian.Proof: We shall use the fact that $\average{f}{X}=0$ for all vectors implies that the operator $X=0$ \begin{equation}\label{eq03} \average{f}{X} = 0\Rightarrow X =0\,. \end{equation}For a positive operator $\average{f}{P^\dagger} = \average{f}{P}^*=\average{f}{P}$. The last step follows from positivity property of operator $P$. Therefore \begin{equation}\label{eq04} \average{f}{(P-P^\dagger)} = \average{f}{P^\dagger}-\average{f}{P}=0 \end{equation} holds for every vector in $\ket{f}$ in the vector space. This implies that $P-P^\dagger=0$ or $P$ is a hermitian operator.

3.Proof of (3):

For completeness we  write out proof of (3). This result is a consequence of polarization
like identity
\begin{eqnarray}\label{eq05}
4i\matrixelement{f}{X}{g} &=& \average{(f+g)}{X} -
\average{(f-g)}{X}\nonumber\\
&&+i\average{(f+ig)}{X}-i\average{(f-ig)}{X}
\end{eqnarray}
If $\average{f}{X}$ vanishes for all $f$, the r.h.s. of \eqref{eq05}
is zero giving $\matrixelement{f}{X}{g}=0$ for all vectors $\ket{f}$
and $\ket{g}$. This in turn implies that the operator $X$ itself is
zero.

4. Example:

An operator $A$ having
positive eigenvalues need not be a positive operator. To see this, consider
$A^\prime=SAS^{-1}$ has positive eigenvalues, for every invertible
operator $S$. If $A$ is hermitian, $A^\prime$ need not be hermitian.
Since positive operators are necessarily hermitian, we conclude that
$A$ need not be positive operator.

Question for you:

In a finite dimensional complex vector space, give an example of an operator $X$and vector $\ket{f}$
such that the eigenvalues of $X$ are positive but $\average{f}{X}$ is negative.