MHB Quantum Computing: Positive Operators are Hermitian

Ackbach
Gold Member
MHB
Messages
4,148
Reaction score
93
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.
 
Last edited by a moderator:
Physics news on Phys.org
You don't need to introduce a basis. Since ##\hat{C}=\hat{C}^{\dagger}## you have
$$\langle x|\hat{C} x \rangle=\langle \hat{C}^{\dagger} x| x\rangle = \langle \hat{C} x|x \rangle=\langle x|\hat{C} x \rangle^* \; \Rightarrow \; \langle x|\hat{C} x \rangle \in \mathbb{R}.$$
Since by assumption ##\hat{A}## is positive semidefinite, from that you must have ##\langle x|\hat{C} x \rangle=0## for all ##|x \rangle \in \mathcal{H}##, and thus ##\hat{C}=0##.
 
  • Like
Likes PeroK and Ackbach
I read Hanbury Brown and Twiss's experiment is using one beam but split into two to test their correlation. It said the traditional correlation test were using two beams........ This confused me, sorry. All the correlation tests I learnt such as Stern-Gerlash are using one beam? (Sorry if I am wrong) I was also told traditional interferometers are concerning about amplitude but Hanbury Brown and Twiss were concerning about intensity? Isn't the square of amplitude is the intensity? Please...
I am not sure if this belongs in the biology section, but it appears more of a quantum physics question. Mike Wiest, Associate Professor of Neuroscience at Wellesley College in the US. In 2024 he published the results of an experiment on anaesthesia which purported to point to a role of quantum processes in consciousness; here is a popular exposition: https://neurosciencenews.com/quantum-process-consciousness-27624/ As my expertise in neuroscience doesn't reach up to an ant's ear...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Back
Top