Welcome to our community

Be a part of something great, join today!

Quantum Computing Series: Positive Operators are Hermitian

  • Thread starter
  • Admin
  • #1

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,183
Exercise 2.24 on page 71 of Nielsen and Chuang's 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.