### Exercise 3.3.1 (p.61) Som diskutert på QIT-møtet [[12024-10-02]]. > Show that $T$ is unitary iff it is norm-preserving. > **Solution** > Unitarity $\Rightarrow$ norm-preserving is trivial: $||T \ket{\psi}|| = \sqrt{\bra{\psi} T^\dagger T \ket{\psi}} = \sqrt{\braket{\psi}{\psi}} = || \ket{\psi} ||$. > > For the other direction, first off, the spectral identity (provided as a hint in the exercise) yields $\braket{\psi}{\phi} = \braket{\phi}{\psi} = (\braket{\psi}{\phi})^*$, so we know that the inner product is a real number. In particular, this implies that $(\bra{\psi} T^\dagger T \ket{\phi})^* = \bra{\phi} T^\dagger T \ket{\psi} = \bra{\psi} T^\dagger T \ket{\phi}$ for all states $\ket{\psi}$, $\ket{\phi}$. > > Second, let $\ket{i}$ be the $i$th computational basis vector. Then the requirement $\forall i\ \bra{i} T^\dagger T \ket{i} = 1$ implies that the diagonals of the matrix $T^\dagger T$ must all be $1$. > > Finally, let $\ket{\psi} = \frac{1}{\sqrt{2}}(\ket{i} + \ket{j})$, $i \neq j$. Then, > > $\bra{\psi} T^\dagger T \ket{\psi} = \frac{1}{2}(\bra{i} + \bra{j})T^\dagger T(\ket{i} + \ket{j})$ > $= \frac{1}{2}(\bra{i}T^\dagger T\ket{i} + \bra{i}T^\dagger T \ket{j} + \bra{j} T^\dagger T \ket{i} + \bra{j} T^\dagger T \ket{j})$ > $= \frac{1}{2}(2 + \bra{i} T^\dagger T \ket{j} + \bra{j} T^\dagger T \ket{i}) = 1$, > $\Rightarrow \bra{i} T^\dagger T \ket{j} = -\bra{j} T^\dagger T \ket{i}$. > > But we already had that $\bra{i} T^\dagger T \ket{j} = \bra{j} T^\dagger T \ket{i}$, implying $\bra{i} T^\dagger T \ket{j} = - \bra{i} T^\dagger T \ket{j}$. The result is that $\bra{i} T^\dagger T \ket{j} = \bra{j} T^\dagger T \ket{i} = 0$ for all $i \neq j$, and so that all off-diagonal entries of the matrix $T^\dagger T$ must be $0$. > > Thus, $T^\dagger T = I$, i.e. $T$ is unitary.