# Inequality for expectation-value of commutator

While working on the derivation of the Heisenberg Uncertainty Principle, I'm getting stuck on showing that the following inequality holds true for the Hermitian operators $$A$$ and $$B$$ and the arbitrary quantum state $$| \psi \rangle$$: $$|\langle \psi |[A,B]| \psi \rangle|^2 \leq 4\langle \psi |A^2| \psi \rangle\langle \psi |B^2| \psi \rangle$$ Following (in)equalities are already given/proven by this point: $$|\langle \psi |[A,B]| \psi \rangle|^2 + |\langle \psi |\{A,B\}| \psi \rangle|^2 = 4|\langle \psi |AB| \psi \rangle|^2$$ $$\text{Cauchy-Schwarz:}\quad|\langle \psi |AB| \psi \rangle|^2 \leq \langle \psi |A^2| \psi \rangle\langle \psi |B^2| \psi \rangle$$ What is it that I'm missing? I've made it to $$0 \leq \langle \psi |A^2| \psi \rangle\langle \psi |B^2| \psi \rangle + Re(\langle \psi |AB| \psi \rangle^2)$$ which to me seems like a dead end. I've been looking into this way to long, so it might be very obvious and I just made a silly mistake... Thanks in advance.

From

$$|\langle \psi |[A,B]| \psi \rangle|^2 + |\langle \psi |\{A,B\}| \psi \rangle|^2 = 4|\langle \psi |AB| \psi \rangle|^2$$

You have

$$|\langle \psi |[A,B]| \psi \rangle|^2 =- |\langle \psi |\{A,B\}| \psi \rangle|^2 + 4|\langle \psi |AB| \psi \rangle|^2$$

Which means that

$$|\langle \psi |[A,B]| \psi \rangle|^2 \leq 4|\langle \psi |AB| \psi \rangle|^2$$

Now, Using Cauchy-Schwarz the identity you wanted to prove is proven:

$$|\langle \psi |[A,B]| \psi \rangle|^2 \leq 4|\langle \psi |AB| \psi \rangle|^2 \leq 4\langle \psi |A^2| \psi \rangle\langle \psi |B^2| \psi \rangle$$

• Wow, that's so obvious... After having looked at it so meticulously I didn't see the big picture anymore... Well, thanks :D Jun 10, 2020 at 22:06
• It often happens that we don't see something like that, especially when the proof is easy. Jun 11, 2020 at 6:33