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- Jan 26, 2012

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*Quantum Computation and Quantum Information*asks the reader to show that a positive operator is necessarily Hermitian. There is a hint given; namely, that you first show an arbitrary operator can be written $A=B+iC$, where $B$ and $C$ are Hermitian. N.B., Nielsen and Chuang are pretty much always working in finite-dimensional Hilbert spaces. Recall that a positive operator $A$ is one such that $\langle x|A|x\rangle\ge 0$ for all vectors $|x\rangle$. An Hermitian operator $A$ is one such that $A=A^{\dagger}$. Here is my solution to the problem:

I claim that $A=B+iC$, where

\begin{align*}

B&=\frac{A+A^{\dagger}}{2} \\

C&=\frac{A-A^{\dagger}}{2i}.

\end{align*}

We can easily verify that $A=B+iC$. Note that

\begin{align*}

B^{\dagger}&=\frac{A^{\dagger}+A}{2}=B \\

C^{\dagger}&=-\frac{A^{\dagger}}{2i}+\frac{A}{2i}=C.

\end{align*}

Hence, $B$ and $C$ are both Hermitian. Now we assume that $A$ is positive, and that $B$ and $C$ are defined as above. For a positive operator, we must have $\langle x|A|x\rangle\ge 0$ for all vectors $|x\rangle$. Since $C$ is Hermitian, it is normal, and hence is diagonalizable. That is, it has a representation

$$C=\sum_i \lambda_i|i\rangle\langle i|,$$

where the $\{|i\rangle\}$ is an orthonormal basis of the space $V$. If it is an orthonormal basis, then we can write

$$|x\rangle=\sum_jx_j|j\rangle.$$

Hence,

\begin{align*}

C|x\rangle&=\sum_i \lambda_i|i\rangle\langle i|\sum_jx_j|j\rangle \\

&=\sum_{i,j}\lambda_ix_j|i\rangle\langle i|j\rangle \\

&=\sum_i\lambda_i x_i|i\rangle.

\end{align*}

Since

$$\langle x|=\sum_j x_j^*\langle j|,$$

we have that

\begin{align*}

\langle x|C|x\rangle&=\sum_j x_j^*\langle j|\sum_i\lambda_i x_i|i\rangle \\

&=\sum_{j,i}\lambda_ix_j^*x_i\langle j|i\rangle \\

&=\sum_i\lambda_i|x_i|^2.

\end{align*}

Since $C$ is Hermitian, its eigenvalues are real. Hence, $\langle x|C|x\rangle$ is real. By the same token, $\langle x|B|x\rangle$ is real. In order to be able even to write

$$\langle x|A|x\rangle=\langle x|(B+iC)|x\rangle=\langle x|B|x\rangle+i\langle x|C|x\rangle\ge 0,$$

the portion $\langle x|C|x\rangle$ must be either pure imaginary or zero. It is not pure imaginary. Hence, it must be zero. Therefore, $A$ is Hermitian.

Here is the commentary thread.