# Trouble with Hermitian operators?

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1. Nov 17, 2014

### 21joanna12

I am looking at the derivation of the Heisenberg Uncertainty Principle presented here: http://socrates.berkeley.edu/~jemoore/p137a/uncertaintynotes.pdf [Broken]

and am confused about line (21)...

I do not understand why $AB$ and $BA$ are complex conjugates of each other... (I'm still in high school so I don't really have much of a background in the algebra of operators).

If I assume what is written in line 21, I think I get the following lines- if $AB$ and $BA$ are complex conjugates, then $|\langle AB-BA \rangle |^2 = |2\langle AB\rangle |^2 = 4|\langle AB\rangle |^2$ I think... The expected value signs are what is confusing me most...

Thank you for any help!

Last edited by a moderator: May 7, 2017
2. Nov 17, 2014

### Staff: Mentor

A hermitian operator has the property $A^\dagger = A$, where the $\dagger$ symbol refers to hermitian conjugation, which is the extension to operators (or matrices) of complex conjugation. Also, when applying hermitian conjugation to a product of operators, you have to invert their order:
$$(A B)^\dagger = B^\dagger A^\dagger$$
So, if $A$ and $B$ are hermitian, you get
$$(A B)^\dagger = B^\dagger A^\dagger = BA$$
The complex conjugate of a bracket is given by
$$\left( \langle \psi | A | \phi \rangle \right)^* = \langle \phi | A^\dagger | \psi \rangle$$
When you combine these things together, you find
\begin{align*} \left( \langle AB \rangle \right)^* &= \left( \langle \psi | AB | \psi \rangle \right)^* \\ &= \langle \psi | (A B)^\dagger | \psi \rangle \\ &= \langle \psi | BA | \psi \rangle \\ &= \langle B A \rangle \end{align*}
So, as for all complex numbers $|z| = |z^*|$, you have that $|\langle AB \rangle| = |\langle BA \rangle|$.

You can only write an inequality here, not equalities.