# Problem Deriving "The General Uncertainty Principle" in Section 5.7 of Susskind's "Quantum Mechanics"

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?

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 $$z_2$$:
$$$$|z_1\cdot z_2| = |z_1|\cdot|z_2|$$$$
\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}