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I'm having a problem in section 5.7 of Susskind's "Quantum Mechanics, The Theoretical Minimum". Specifically, I'm trying to derive eq. 5.11, $$ 2\sqrt{ \langle \mathbf{A}^2 \rangle \langle \mathbf{B}^2 \rangle} = |\langle \psi | \mathbf{AB} | \psi \rangle - \langle \psi | \mathbf{BA} | \psi \rangle | \tag{5.11} $$ where $\mathbf{A}$ and $\mathbf{B}$ and observables (and, hence, Hermitian), by following the text and substituting eqs. 5.10, $$ | X \rangle = \mathbf{A} | \psi \rangle \\ | Y \rangle = i\mathbf{B} | \psi \rangle \tag{5.10} $$ into the Cauchy-Schwarz inequality, eq. 5.9, $$ 2|X||Y| \ge | \langle X | Y \rangle + \langle Y | X \rangle | \tag{5.9} $$ But I'm having a problem.

I can derive the left side of eq. 5.9 using, $$ |X| = \sqrt{ \langle X | X \rangle }\\ = \sqrt{\langle \psi | \mathbf{A}^\dagger \mathbf{A} | \psi \rangle }\\ = \sqrt{\langle \psi | \mathbf{A} \mathbf{A} | \psi \rangle} \\ = \sqrt{\langle \psi | \mathbf{A}^2 | \psi \rangle} \\ = \sqrt{\langle \mathbf{A}^2 \rangle} $$ And, $$|Y| = \sqrt{\langle Y | Y \rangle} \\ = \sqrt{\langle \psi | -i \mathbf{B}^\dagger i \mathbf{B} | \psi \rangle} \\ = \sqrt{\langle \psi | -i^2 \mathbf{B} \mathbf{B} | \psi \rangle} \\ = \sqrt{\langle \psi | \mathbf{B}^2 | \psi \rangle} \\ = \sqrt{\langle \mathbf{B}^2 \rangle} $$ So, $$2|X||Y| = 2\sqrt{\langle \mathbf{A}^2 \rangle} \sqrt{\langle \mathbf{B}^2 \rangle} = 2\sqrt{\langle \mathbf{A}^2 \rangle \langle \mathbf{B}^2 \rangle} $$ So far so good.

But when I try to derive the right side of eq. 5.9 I end out with an extra $i$. Using,

$$ \langle X | Y \rangle = \langle \psi | \mathbf{A}^\dagger i \mathbf{B} | \psi \rangle \\ = \langle \psi | i \mathbf{AB} | \psi \rangle \\ = i \langle \psi | \mathbf{AB} | \psi \rangle $$ And, $$\langle Y | X \rangle = \langle \psi | -i \mathbf{B}^\dagger \mathbf{A} | \psi \rangle \\ = \langle \psi | -i \mathbf{BA} | \psi \rangle \\ = -i \langle \psi | \mathbf{BA} | \psi \rangle $$ Which means, $$ | \langle X | Y \rangle + \langle Y | X \rangle | = | i \langle \psi | \mathbf{AB} | \psi \rangle -i \langle \psi | \mathbf{BA} | \psi \rangle | $$ Which is almost the same as eq. 5.11, but I've got extra $i$'s on the right hand side. What am I doing wrong?

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You are correct, but there's one last simplifying step. The norm of the product is the product of the norms. For two complex numbers, $z_1$ and $z_2$:

\begin{equation} |z_1\cdot z_2| = |z_1|\cdot|z_2| \end{equation}

So the right hand side of your last equation is

\begin{align} |i\langle\psi|\mathbf{AB}|\psi\rangle - i\langle\psi|\bf{BA}|\psi\rangle| &= |i(\langle\psi|\mathbf{AB}|\psi\rangle - \langle\psi|\mathbf{BA}|\psi\rangle)| \\ &=|i|\cdot|\langle\psi|\bf{AB}|\psi\rangle - \langle\psi|\bf{BA}|\psi\rangle| \\ &= |\langle\psi|\bf{AB}|\psi\rangle - \langle\psi|\bf{BA}|\psi\rangle| \end{align}

which is the desired equation.

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