Quantum Country exercise 9

This is one in a series of posts working through the exercises in the Quantum Country online introduction to quantum computing and related topics. The exercises in the original document are not numbered; I have added my own numbers for convenience in referring to them.

Exercise 9. Show that for any matrix $M$ and vector $\vert \psi \rangle$ the following identity holds: $\Vert M \vert \psi \rangle \Vert^2 = \langle \psi \vert M^\dagger M \vert \psi \rangle$.

Answer: By definition we have $\Vert M \vert \psi \rangle \Vert^2 = \left( M \vert \psi \rangle \right)^\dagger M \vert \psi \rangle$. It was shown in the text that $\left( M \vert \psi \rangle \right)^\dagger = \vert \psi \rangle^\dagger M^\dagger$. By definition we also have $\langle \psi \vert = \vert \psi \rangle^\dagger$.

We then have

$\Vert M \vert \psi \rangle \Vert^2 = \left( M \vert \psi \rangle \right)^\dagger M \vert \psi \rangle = = \vert \psi \rangle^\dagger M^\dagger M \vert \psi \rangle = \langle \psi \vert M^\dagger M \vert \psi \rangle$

So we have shown that $\Vert M \vert \psi \rangle \Vert^2 = \langle \psi \vert M^\dagger M \vert \psi \rangle$ for any matrix $M$ and vector $\vert \psi \rangle$.

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