Unitary Matrices

mathReading-focused

10 min read

The adjoint, the unitary condition, and why every quantum gate (but not measurement) has to satisfy it.

Suggested before this lesson

Not every matrix is a valid quantum gate. This lesson covers the one condition that decides which matrices are: unitarity. It's the formal version of something the algorithms track will lean on directly, that a quantum gate has to be reversible.

The Adjoint

The adjoint (or conjugate transpose, or "dagger") of a matrix $A$ , written $A^\dagger$ , is formed by transposing $A$ and conjugating every entry:

$A = \begin{pmatrix}a & b \\ c & d\end{pmatrix} \quad\Longrightarrow\quad A^\dagger = \begin{pmatrix}a^* & c^* \\ b^* & d^*\end{pmatrix}$

For a matrix with only real entries, like $X$ , $Z$ , or $H$ , conjugating does nothing, so $A^\dagger$ is just the ordinary transpose.

The Unitary Condition

A matrix $U$ is unitary if $U^\dagger U = I$ (equivalently $UU^\dagger = I$ too, for square matrices), where $I$ is the identity. This single condition says two things at once:

$U^\dagger$ is $U$ 's inverse. Undoing $U$ is always possible, and it's always $U^\dagger$ , nothing fancier is needed.
$U$ preserves inner products: $\langle U\psi | U\phi\rangle = \langle\psi|\phi\rangle$ for any $\psi,\phi$ (this follows directly from $U^\dagger U = I$ ). In particular, taking $\psi=\phi$ , $U$ preserves norm.

That second point is exactly why quantum gates must be unitary: a state's norm being 1 is what makes its squared amplitudes valid probabilities that sum to 1. A gate that didn't preserve norm would turn a valid state into an invalid one, probabilities that don't add up to 100%. Unitary gates are precisely the linear maps that never let that happen.

Checking H

$H^\dagger H = \begin{pmatrix} \tfrac{1}{\sqrt2} & \tfrac{1}{\sqrt2} \\ \tfrac{1}{\sqrt2} & -\tfrac{1}{\sqrt2} \end{pmatrix}\begin{pmatrix} \tfrac{1}{\sqrt2} & \tfrac{1}{\sqrt2} \\ \tfrac{1}{\sqrt2} & -\tfrac{1}{\sqrt2} \end{pmatrix} = \begin{pmatrix} \tfrac12+\tfrac12 & \tfrac12-\tfrac12 \\ \tfrac12-\tfrac12 & \tfrac12+\tfrac12 \end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix} = I$

( $H$ is real and symmetric, so $H^\dagger = H$ here, meaning $H$ is its own inverse, which checks out: applying qc.h(0) twice in a row returns a qubit to where it started.)

Unitary Eigenvalues Always Have Modulus 1

Here's the fact Quantum Phase Estimation is eventually built on: every eigenvalue $\lambda$ of a unitary matrix satisfies $|\lambda| = 1$ , meaning it can always be written $\lambda = e^{i\theta}$ for some real $\theta$ . This follows straight from norm preservation: if $Uv = \lambda v$ for a unit vector $v$ , then $\|Uv\| = \|v\| = 1$ (unitarity), but also $\|Uv\| = \|\lambda v\| = |\lambda|\,\|v\| = |\lambda|$ , so $|\lambda|=1$ . A unitary gate can never scale an eigenvector up or down, only rotate its phase, which is exactly the $\pm1$ pattern you already saw for $Z$ and $X$ in the last lesson ( $e^{i\pi}=-1$ , $e^{i0}=1$ ), just the special real-valued case of a much more general fact.

What Isn't Unitary

Measurement. Collapsing a superposition down to a single definite outcome is fundamentally irreversible: it's a many-to-one process, many different pre-measurement states can collapse to the same post-measurement outcome, and a many-to-one map can never have an inverse. This is also exactly the reversibility argument the oracle lesson makes about classical functions: a generic $f$ isn't invertible either, which is why an oracle has to be built as an XOR-into-an-ancilla unitary instead of "compute $f(x)$ and overwrite the answer directly."

Watch norm preservation happen (or fail)

The violet loop traces $Mv$ for every direction $v$ . For $X$ or $H$ it stays a perfect circle: norm preserved, exactly the unitary condition. Switch to the non-unitary matrix from the quiz below and watch the loop distort.

Vector transform

$Angle of v$

20°

$v$

(0.94, 0.34)

$M v$

(0.34, 0.94)

$∥ M v ∥$

1.000

The violet loop traces every possible output $Mv$ as $v$ sweeps a full circle. A circle in, circle out means $M$ preserves length for every direction, exactly the unitary condition; a squashed or stretched loop means it doesn't.

Check your understanding

M = \begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}

unitary?

Yes, since it's a valid 2×2 matrix.

No:

M^\dagger M = \begin{pmatrix}1&0\\1&1\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix} = \begin{pmatrix}1&1\\1&2\end{pmatrix} \neq I

Only if it's also symmetric.

Yes, because its determinant is 1.

Check your understanding

Why can't measurement be represented as a unitary matrix?

Measurement matrices haven't been discovered yet.

Measurement is many-to-one (many different states can collapse to the same outcome), and a many-to-one map can't have an inverse, but every unitary matrix is invertible (its own adjoint).

Measurement only applies to single qubits, and unitary matrices need at least 2 qubits.

It can; Qiskit just chooses not to represent it that way.

The Adjoint

The adjoint (or conjugate transpose, or "dagger") of a matrix

A

, written

A^\dagger

, is formed by transposing

A

and conjugating every entry:

A = \begin{pmatrix}a & b \\ c & d\end{pmatrix} \quad\Longrightarrow\quad A^\dagger = \begin{pmatrix}a^* & c^* \\ b^* & d^*\end{pmatrix}

For a matrix with only real entries, like

X

Z

, or

H

, conjugating does nothing, so

A^\dagger

is just the ordinary transpose.

The Unitary Condition

A matrix

U

is unitary if

U^\dagger U = I

(equivalently

UU^\dagger = I

too, for square matrices), where

I

is the identity. This single condition says two things at once:

U^\dagger

U

's inverse. Undoing

U

is always possible, and it's always

U^\dagger

, nothing fancier is needed.

U

preserves inner products:

\langle U\psi | U\phi\rangle = \langle\psi|\phi\rangle

for any

\psi,\phi

(this follows directly from

U^\dagger U = I

). In particular, taking

\psi=\phi

U

preserves norm.

Checking H

H^\dagger H = \begin{pmatrix} \tfrac{1}{\sqrt2} & \tfrac{1}{\sqrt2} \\ \tfrac{1}{\sqrt2} & -\tfrac{1}{\sqrt2} \end{pmatrix}\begin{pmatrix} \tfrac{1}{\sqrt2} & \tfrac{1}{\sqrt2} \\ \tfrac{1}{\sqrt2} & -\tfrac{1}{\sqrt2} \end{pmatrix} = \begin{pmatrix} \tfrac12+\tfrac12 & \tfrac12-\tfrac12 \\ \tfrac12-\tfrac12 & \tfrac12+\tfrac12 \end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix} = I

(

H

is real and symmetric, so

H^\dagger = H

here, meaning

H

is its own inverse, which checks out: applying qc.h(0) twice in a row returns a qubit to where it started.)

Unitary Eigenvalues Always Have Modulus 1

Here's the fact Quantum Phase Estimation is eventually built on: every eigenvalue

\lambda

of a unitary matrix satisfies

|\lambda| = 1

, meaning it can always be written

\lambda = e^{i\theta}

for some real

\theta

. This follows straight from norm preservation: if

Uv = \lambda v

for a unit vector

v

, then

\|Uv\| = \|v\| = 1

(unitarity), but also

\|Uv\| = \|\lambda v\| = |\lambda|\,\|v\| = |\lambda|

, so

|\lambda|=1

. A unitary gate can never scale an eigenvector up or down, only rotate its phase, which is exactly the

\pm1

pattern you already saw for

Z

and

X

in the last lesson (

e^{i\pi}=-1

e^{i0}=1

), just the special real-valued case of a much more general fact.

What Isn't Unitary

f

isn't invertible either, which is why an oracle has to be built as an XOR-into-an-ancilla unitary instead of "compute

f(x)

and overwrite the answer directly."

Watch norm preservation happen (or fail)

Vector transform

$Angle of v$

20°

$v$

(0.94, 0.34)

$M v$

(0.34, 0.94)

$∥ M v ∥$

1.000

Check your understanding

M = \begin{pmatrix}1 & 1\\0 & 1\end{pmatrix}

unitary?

Yes, since it's a valid 2×2 matrix.

No:

M^\dagger M = \begin{pmatrix}1&0\\1&1\end{pmatrix}\begin{pmatrix}1&1\\0&1\end{pmatrix} = \begin{pmatrix}1&1\\1&2\end{pmatrix} \neq I

Only if it's also symmetric.

Yes, because its determinant is 1.

Check your understanding

Why can't measurement be represented as a unitary matrix?

Measurement matrices haven't been discovered yet.

Measurement is many-to-one (many different states can collapse to the same outcome), and a many-to-one map can't have an inverse, but every unitary matrix is invertible (its own adjoint).

Measurement only applies to single qubits, and unitary matrices need at least 2 qubits.

It can; Qiskit just chooses not to represent it that way.