Unitary Matrices

Definition

A square matrix $U$ with complex entries is unitary if its conjugate transpose equals its inverse:

U^\dagger U = U U^\dagger = I.

Equivalently, the columns of $U$ form an orthonormal set in $\mathbb{C}^n$ : each column has unit norm and every pair of distinct columns is orthogonal. The same is true of the rows.

Two ways to verify unitarity

Method 1 — column norms and inner products. Extract the columns $\mathbf{c}_1, \mathbf{c}_2, \ldots$ and check:

$\|\mathbf{c}_k\| = 1$ for every $k$ (unit norm).
$\langle \mathbf{c}_j, \mathbf{c}_k \rangle = 0$ for every $j \neq k$ (mutual orthogonality).

Method 2 — direct product. Compute $U^\dagger U$ explicitly and check that it equals the identity matrix $I$ .

Both methods are equivalent; which is faster depends on the matrix.

Why unitarity is the right condition for quantum gates

In quantum mechanics a state $|\psi\rangle$ must remain normalised: $\langle \psi | \psi \rangle = 1$ . If evolution is described by a matrix $U$ , the evolved state is $|\psi'\rangle = U|\psi\rangle$ . Normalisation after evolution requires

\langle \psi' | \psi' \rangle = \langle \psi | U^\dagger U | \psi \rangle = 1

for every choice of $|\psi\rangle$ . The only matrices that satisfy this for all unit vectors $|\psi\rangle$ are the unitary matrices $U^\dagger U = I$ . Unitarity is therefore the algebraic expression of conservation of total probability.

Key properties of unitary matrices

From the definition several useful facts follow immediately.

Determinant has unit modulus. If $U^\dagger U = I$ then $\det(U^\dagger)\det(U) = 1$ , so $|\det(U)|^2 = 1$ and therefore $|\det(U)| = 1$ . The determinant lies on the unit circle in the complex plane but is not necessarily $1$ .

Eigenvalues have unit modulus. If $U\,|\psi\rangle = \lambda\,|\psi\rangle$ then $\|\lambda\,|\psi\rangle\| = \|U\,|\psi\rangle\| = \||\psi\rangle\|$ , which forces $|\lambda| = 1$ .

Unitary matrices form a group. The product of two unitary matrices is unitary: $(UV)^\dagger(UV) = V^\dagger U^\dagger UV = V^\dagger V = I$ . The inverse of a unitary matrix is its conjugate transpose, also unitary. This group structure matches the physical requirement that composing two valid quantum operations gives another valid quantum operation.

A worked example — the Hadamard matrix

The Hadamard matrix is

H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}.

Its entries are real, so $H^\dagger = H^T$ . Let us compute $H^\dagger H$ directly.

Reading off the columns, the first column is $\mathbf{c}_1 = \tfrac{1}{\sqrt{2}}\binom{1}{1}$ and the second is $\mathbf{c}_2 = \tfrac{1}{\sqrt{2}}\binom{1}{-1}$ .

Column norms:

\|\mathbf{c}_1\|^2 = \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} + \frac{1}{2} = 1, \qquad \|\mathbf{c}_2\|^2 = \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{-1}{\sqrt{2}}\right)^2 = \frac{1}{2} + \frac{1}{2} = 1.

Cross inner product:

\langle \mathbf{c}_1, \mathbf{c}_2 \rangle = \frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\cdot\frac{-1}{\sqrt{2}} = \frac{1}{2} - \frac{1}{2} = 0.

Both columns are unit vectors and they are orthogonal, so $H$ is unitary: $H^\dagger H = I$ . Equivalently, since $H$ is also symmetric ( $H^T = H$ ) and $H^\dagger H = I$ , we have $H = H^{-1}$ , confirming that $H^2 = I$ — applying the Hadamard gate twice returns any state to its original form.

Complex unitary matrices

The condition $U^\dagger U = I$ extends beyond real orthogonal matrices. Consider the phase matrix

P(\theta) = \begin{pmatrix} e^{i\theta} & 0 \\ 0 & e^{-i\theta} \end{pmatrix}.

Here $P(\theta)^\dagger = \begin{pmatrix} e^{-i\theta} & 0 \\ 0 & e^{i\theta} \end{pmatrix}$ , so

P(\theta)^\dagger P(\theta) = \begin{pmatrix} e^{-i\theta}e^{i\theta} & 0 \\ 0 & e^{i\theta}e^{-i\theta} \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I.

This matrix is unitary for every real $\theta$ even though its entries are in general not real. The standard quantum gates $S = \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix}$ and $T = \begin{pmatrix} 1 & 0 \\ 0 & e^{i\pi/4} \end{pmatrix}$ are diagonal unitaries of exactly this kind — each column is a unit-modulus phase times a standard basis vector — and they appear throughout quantum computing. They coincide with members of the family $P(\theta)$ only up to an overall global phase (for example $P(\pi/4) = e^{i\pi/4}\,\mathrm{diag}(1, e^{-i\pi/2})$ ), and such a global phase has no physical effect.

Try it

This is a numerical exercise — your code should return a number, not a circuit. The Hadamard matrix $H = \tfrac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$ is real, so $H^\dagger = H^T$ . Compute the $(0,0)$ entry of $H^\dagger H$ . A unitary matrix satisfies $H^\dagger H = I$ , so the result should be exactly $1$ .

Run your code to see the quantum state.