In fact, you can repeat the argument before with an arbitrary initial state $\vert \psi_0\rangle$. The conditional probabilities $p(\mathbf{y}\vert \psi_0) = \vert \langle\mathbf{y}\vert \psi_0\rangle\vert ^2$ must add up to $1$, and hence $$ \langle \psi_0\vert U^\dagger U\vert \psi_0\rangle = 1. $$ As you can check below, this implies that $$ \langle \mathbf{x}\vert U^\dagger U\vert \mathbf{x}'\rangle = \delta_{\mathbf{x}\mathbf{x}'} \quad \text{or equivalently} \quad U^\dagger U = I. $$ In other words, the adjoint of $U$ is its inverse, $U^\dagger = U^{-1}$.

This is the defining property of a **unitary** matrix. By thinking about probability and black boxes, we have rediscovered that the black boxes of quantum mechanics are unitary operators! --- ***Exercise H.2.2.*** For a system of $n$ qubits, how many real, independent parameters are required to specify the unitary matrix $U$? Compare to the number of parameters in a classical random system, $2^{2n} - 2^n$. More parameters means more types of behaviour are possible, and in theory, more powerful computers!

Solution.

For $n$ qubits, the state vector $\vert \psi\rangle$ has $2^n$ components. Thus, $U$ is a $2^n \times 2^n$ matrix of complex entries, each of which is specified by two real numbers. This gives a total of $2 \cdot 2^n \cdot 2^n = 2 \cdot 2^{2n}$ parameters. The equation $U^\dagger U = I$ gives us a whole $2^n \times 2^n$ grid of real conditions to satisfy, so the total number of independent real parameters is $$ 2 \cdot 2^{2n} - 2^{2n} = 2^{2n}. $$ This is a factor of $2^n$ more than the number of parameters in a classical random matrix. Much more is possible in the quantum realm! ▢

--- ***Exercise H.2.3.*** Let $\mathbf{x}\neq\mathbf{x}'$ be distinct computational basis states, and define $$ \vert \psi_0\rangle = \alpha\vert \mathbf{x}\rangle+\beta\vert \mathbf{x}'\rangle, \quad \vert \alpha\vert ^2 + \vert \beta\vert ^2 = 1. $$ Assuming the condition $\langle \psi_0\vert U^\dagger U\vert \psi_0\rangle = 1$, this sequence of exercises will guide you to the conclusion that $U^\dagger U = I$. (a) Show that $\langle \psi_0\vert U^\dagger U\vert \psi_0\rangle = 1$ implies $$ \Re[\alpha^*\beta \langle\mathbf{x}\vert U^\dagger U\vert \mathbf{x}'\rangle] = 0, $$ where $\Re$ denotes the real component of a complex number. (b) By choosing $\alpha$ and $\beta$ judiciously, argue that $\langle\mathbf{x}\vert U^\dagger U\vert \mathbf{x}'\rangle = 0$. (c) Conclude that $\langle\mathbf{x}\vert U^\dagger U\vert \mathbf{x}'\rangle = \delta_{\mathbf{x}\mathbf{x}'}$, and thus $U^\dagger U = I$. Here, $\delta_{\mathbf{x}\mathbf{x}'}$ is the *Kronecker delta*, defined by $$ \delta_{\mathbf{x}\mathbf{x}'} = \begin{cases} 1 & \mathbf{x} = \mathbf{x}' \\ 0 & \text{otherwise.} \end{cases} $$

Solution.

(a) A little algebra shows that $$ \langle \psi_0\vert U^\dagger U\vert \psi_0\rangle = \vert \alpha\vert ^2\langle \mathbf{x}\vert U^\dagger U\vert \mathbf{x}\rangle + \vert \beta\vert ^2\langle \mathbf{x}'\vert U^\dagger U\vert \mathbf{x}'\rangle + \alpha^*\beta\langle \mathbf{x}\vert U^\dagger U\vert \mathbf{x}'\rangle + (\alpha^*\beta\langle \mathbf{x}\vert U^\dagger U\vert \mathbf{x}'\rangle)^\dagger. $$ From $(1)$, the first term in brackets gives $\vert \alpha\vert ^2$ and the second $\vert \beta\vert ^2$, which sum to $1$ since $\vert \psi_0\rangle$ is normalized. Hence $$ \langle \psi_0\vert U^\dagger U\vert \psi_0\rangle = 1 + \alpha^*\beta\langle \mathbf{x}\vert U^\dagger U\vert \mathbf{x}'\rangle + (\alpha^*\beta\langle \mathbf{x}\vert U^\dagger U\vert \mathbf{x}'\rangle)^\dagger = 1 + 2\Re [\alpha^*\beta\langle \mathbf{x}\vert U^\dagger U\vert \mathbf{x}'\rangle], $$ since the sum of a complex number and its conjugate is twice the real part. Assuming $\langle \psi_0\vert U^\dagger U\vert \psi_0\rangle = 1$ then implies that the real part of $\alpha^*\beta \langle \mathbf{x}\vert U^\dagger U\vert \mathbf{x}'\rangle$ must vanish. (b) We can now simply choose $\alpha$ and $\beta$ to ensure that the expression is real! Take for instance $\vert \alpha\vert = \vert \beta\vert = 1/\sqrt{2}$, and choose the phases so that $\alpha^*\beta = e^{-i\theta}/2$ cancels the phase $e^{i\theta}$ of $\langle\mathbf{x}\vert U^\dagger U\vert \mathbf{x}'\rangle$. Then $$ \Re [\alpha^*\beta\langle\mathbf{x}\vert U^\dagger U\vert \mathbf{x}'\rangle] = \frac{1}{2}e^{-i\theta}\langle\mathbf{x}\vert U^\dagger U\vert \mathbf{x}'\rangle = 0 \quad \Longrightarrow \quad \langle\mathbf{x}\vert U^\dagger U\vert \mathbf{x}'\rangle = 0. $$ (c) We know from $(1)$ that $\langle\mathbf{x}\vert U^\dagger U\vert \mathbf{x}\rangle = 1$ for all $\mathbf{x}$, and hence $\langle\mathbf{x}\vert U^\dagger U\vert \mathbf{x}'\rangle = \delta_{\mathbf{x}\mathbf{x}'}$. This is just the component form of the statement $U^\dagger U = I$, so $U$ is unitary as claimed! ▢

--- The discussion above explain *why* quantum gates are unitaries. This is important because, in order to actually make a quantum computer, we need to harness the laws of nature in a way that implements exactly the unitary gate we want. There is no one single way to do this, it depends on the specific type of quantum computer we choose to build. Is everything we can do on quantum system modelled by unitary matrices? Mostly yes but there's one big exception... Projective measurements aren't unitary! This is one of the biggest and unsolved mysteries in physics, and it's known as the measurement problem.

Codercise H.2.1 - Unitary or not?

Codercise H.2.2 - Only good unitaries allowed

Conserving total probabilities

Why unitaries?