Unitary Operators

A unitary operator is a surjective linear map that preserves inner products. Equivalently, its adjoint is its inverse. Unitary maps preserve the complete geometry of a Hilbert space while allowing vectors, bases, and relative phases to change.

Why Quantum Mechanics Needs Unitarity

Unitary operators appear in several logically distinct roles:

• reversible evolution of a closed system;
• active transformations of states and observables;
• passive changes between orthonormal coordinate bases;
• ideal quantum gates;
• representations of continuous and discrete symmetries.

The same matrix can serve more than one role, but the interpretation and transformation formulas must be declared. Unitarity is the shared inner-product-preserving structure.

Definition

Let $U:\mathcal H\to\mathcal H$ be linear on a complex Hilbert space. It is unitary when

$$U^\dagger U=UU^\dagger=I.$$

Therefore

$$U^{-1}=U^\dagger.$$

For all $\phi,\psi\in\mathcal H$,

$$\langle U\phi\vert U\psi\rangle = \langle\phi\vert U^\dagger U \vert\psi\rangle = \langle\phi\vert\psi\rangle.$$

Conversely, if a linear map is surjective and preserves every inner product, then $U^\dagger U=I$ and surjectivity gives $UU^\dagger=I$.

In a finite-dimensional space, the following conditions are equivalent:

• $U^\dagger U=UU^\dagger=I$;
• $U^{-1}=U^\dagger$;
• $U$ preserves all inner products;
• $U$ is linear and preserves all norms;
• $U$ maps one orthonormal basis to another;
• the columns of its matrix form an orthonormal basis.

The finite-dimensional implication $U^\dagger U=I\Rightarrow UU^\dagger=I$ uses equal finite dimensions. The infinite-dimensional caveat appears below.

Columns, Rows, and Matrix Tests

In an orthonormal basis, write the matrix entries as $U_{kj}$. The matrix product $U^\dagger U=I$ says that the columns are orthonormal:

$$\sum_k U_{ki}^*U_{kj} = \delta_{ij}.$$

The product $UU^\dagger=I$ says that the rows are orthonormal:

$$\sum_k U_{ik}U_{jk}^* = \delta_{ij}.$$

For a square finite matrix, either condition implies the other. A rectangular matrix can have orthonormal columns without being a unitary operator from a space onto itself.

All singular values of a unitary matrix equal one. Consequently,

$$\lVert U\rVert_{\mathrm{op}}=1$$

on a nonzero Hilbert space, and

$$\lvert\det U\rvert=1$$

in finite dimension. The determinant condition alone is not sufficient for unitarity.

What Unitary Maps Preserve

Because all inner products are preserved, so are:

• norms and normalization;
• orthogonality and angles;
• distances between vectors;
• linear independence;
• dimensions of subspaces;
• transition-amplitude magnitudes;
• spectra under unitary conjugation.

For example,

$$\lVert U\psi-U\phi\rVert = \lVert\psi-\phi\rVert.$$

If the same unitary acts on both vectors, their overlap is unchanged:

$$\langle U\phi\vert U\psi\rangle = \langle\phi\vert\psi\rangle.$$

Unitarity does not mean that every coefficient, phase, or measurement probability in a fixed basis stays unchanged. A unitary can rotate a state relative to the fixed measurement basis while preserving Hilbert-space geometry.

Group Structure

Unitary operators form a group:

• the identity is unitary;
• the product of unitaries is unitary;
• the inverse of a unitary is unitary;
• the adjoint of a unitary is unitary.

If $U$ and $V$ are unitary,

$$\begin{aligned} (UV)^\dagger(UV) &= V^\dagger U^\dagger UV \\ &= V^\dagger V \\ &=I. \end{aligned}$$

The finite-dimensional unitary group is denoted $U(d)$. Its subgroup of determinant-one matrices is $SU(d)$. Group actions by unitary operators are developed in Unitary Representations.

Eigenvalues and Spectral Form

Suppose

$$Uv=\lambda v, \qquad v\ne0.$$

Norm preservation gives

$$\lVert v\rVert = \lVert Uv\rVert = \lvert\lambda\rvert\lVert v\rVert,$$

so

$$\lvert\lambda\rvert=1.$$

Every eigenvalue can therefore be written as a phase $e^{i\theta}$. A unitary operator is normal, so in finite dimension it has an orthonormal eigenbasis and spectral form

$$U = \sum_\theta e^{i\theta}P_\theta.$$

Unitary does not imply Hermitian. A matrix that is both unitary and Hermitian has eigenvalues restricted to $\pm1$ and satisfies $U^2=I$.

Active Transformations

An active transformation changes the vector while the coordinate basis is held fixed:

$$\lvert\psi\rangle \longmapsto U\lvert\psi\rangle.$$

In fixed coordinates,

$$c\longmapsto Uc.$$

If a symmetry transformation carries both states and observables into symmetry-related objects, then

$$\lvert\psi\rangle \longmapsto U\lvert\psi\rangle, \qquad A \longmapsto UAU^\dagger.$$

The expectation value is unchanged under this paired transformation:

$$\langle U\psi\vert UAU^\dagger \vert U\psi\rangle = \langle\psi\vert A\vert\psi\rangle.$$

If the state is transformed but the physical measurement $A$ is held fixed, the expectation value can change. That change is often the point of applying an active gate or control pulse.

Passive Changes of Orthonormal Basis

A passive basis change leaves the abstract vector and operator fixed and changes only their coordinates. Let

$$\lvert f_j\rangle = U\lvert e_j\rangle,$$

where the columns of $U$ are the new basis vectors expressed in the old orthonormal basis. Then

$$c_{\mathcal F} = U^\dagger c_{\mathcal E}$$

and

$$A_{\mathcal F} = U^\dagger A_{\mathcal E}U.$$

These are passive coordinate formulas. They do not describe a state being physically driven. A different convention may name $U^\dagger$ as the basis-change matrix, moving daggers to the opposite side. The canonical coordinate convention is Change of Basis, and the physics-facing bookkeeping is Change of Basis.

Worked Example: Hadamard Matrix

The Hadamard matrix

$$H = \frac{1}{\sqrt2} \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix}$$

is real and symmetric, so $H^\dagger=H$. Direct multiplication gives

$$H^\dagger H = H^2 = I.$$

Actively,

$$H\lvert0\rangle=\lvert+\rangle, \qquad H\lvert1\rangle=\lvert-\rangle.$$

Passively, the same array can convert coordinates between the computational basis and the $\lvert\pm\rangle$ basis, but the direction of the coordinate map must follow the declared convention.

The Hadamard matrix is both unitary and Hermitian. Its eigenvalues are $\pm1$. This is a special property, not the definition of a unitary.

Worked Example: A Relative Phase

The phase gate

$$P(\varphi) = \begin{pmatrix} 1&0\\ 0&e^{i\varphi} \end{pmatrix}$$

is unitary because each diagonal entry has unit modulus. Acting on

$$\lvert+\rangle = \frac{\lvert0\rangle+\lvert1\rangle}{\sqrt2}$$

gives

$$P(\varphi)\lvert+\rangle = \frac{ \lvert0\rangle +e^{i\varphi}\lvert1\rangle }{\sqrt2}.$$

The norm is unchanged, but the relative phase changes interference in later bases. Unitarity preserves overlaps between two vectors transformed by the same operator; it does not freeze the relationship between one transformed state and a fixed reference basis.

Multiplying the entire operator by a phase,

$$U\longmapsto e^{i\chi}U,$$

does not change its action on isolated pure-state rays. That phase can become relative and observable when the operation occurs conditionally in one branch of a larger controlled process. See Rays and Global Phase.

Hermitian Generators

If $H=H^\dagger$, then

$$U(t) = \exp\left( -\frac{iHt}{\hbar} \right)$$

is unitary. Conversely, let $U(t)$ be a differentiable one-parameter family with

$$U(0)=I, \qquad U(t)^\dagger U(t)=I.$$

Differentiating at $t=0$ gives

$$\dot U(0)^\dagger+\dot U(0)=0.$$

Thus $\dot U(0)$ is skew-Hermitian, and

$$H = i\hbar\dot U(0)$$

is Hermitian. Finite-dimensional continuous unitary motion therefore has a Hermitian infinitesimal generator.

The exponential construction belongs to Matrix Functions and Exponentials. The physical closed-system statement is Unitary Time Evolution.

Unitary, Hermitian, Projective, and Antiunitary

Unitary operations are stable under the standard constructions used for composite and block-structured systems:

$$U_A\otimes U_B \quad\text{and}\quad U_1\oplus U_2$$

are unitary whenever their factors are. Conjugation preserves products, commutators, spectra, trace, and positivity:

$$(UAU^\dagger)(UBU^\dagger)=UABU^\dagger,$$ $$[UAU^\dagger,UBU^\dagger]=U[A,B]U^\dagger.$$

For a density operator,

$$\rho\longmapsto U\rho U^\dagger$$

preserves its eigenvalues, purity, and von Neumann entropy. This statement applies to closed evolution on the whole modeled system; reduced subsystem dynamics need not be unitary.

A frequently used two-level exponential is

$$e^{-i\theta\mathbf n\cdot\boldsymbol\sigma/2} =\cos\frac\theta2\,I -i\sin\frac\theta2\,\mathbf n\cdot\boldsymbol\sigma, \qquad \lVert\mathbf n\rVert=1,$$

which follows from $(\mathbf n\cdot\boldsymbol\sigma)^2=I$.

These operator classes answer different questions:

Class	Defining relation	Typical role
Unitary	$U^\dagger U=UU^\dagger=I$	reversible geometry-preserving map
Hermitian	$A^\dagger=A$	real spectral quantity or generator
Orthogonal projector	$P^\dagger=P=P^2$	selection of a subspace
Antiunitary	conjugate-linear and norm preserving	time-reversal-type symmetry

A nontrivial projector is not unitary because it discards components orthogonal to its range. An antiunitary map preserves transition-probability magnitudes but is conjugate-linear rather than linear; see Antiunitary Symmetries, First Look.

Processes That Are Not Unitary

Not every quantum process acting on a subsystem is a unitary map on that subsystem. Examples include:

• projective state update;
• postselection;
• decoherence and dissipation;
• discarding an environment;
• noisy quantum channels.

Such processes can preserve total probability without preserving all inner products or being reversible. Many arise from a unitary interaction on a larger system followed by ignoring or conditioning on part of that system. The canonical framework is Quantum Channels and Noise.

Reversible Computation applies unitary invertibility to logical maps, reversible embeddings, workspace, and uncomputation without treating measurement, reset, or discard as unitary operations.

Infinite-Dimensional Isometry Warning

In finite dimension,

$$U^\dagger U=I$$

for a square operator implies surjectivity and therefore $UU^\dagger=I$. This implication fails in infinite dimension.

On $\ell^2(\mathbb N)$, define the unilateral shift

$$S(c_1,c_2,c_3,\ldots) = (0,c_1,c_2,\ldots).$$

It preserves norms and satisfies

$$S^\dagger S=I.$$

But its range consists only of sequences whose first component is zero, so it is not surjective. In fact,

$$SS^\dagger = I-\lvert e_1\rangle\langle e_1\vert \ne I.$$

The shift is an isometry, not a unitary operator. On an infinite-dimensional Hilbert space, require both adjoint identities or explicitly require a surjective isometry.

Every unitary is a bounded operator of operator norm one.

Numerical Checks

For a computed matrix $\widetilde U$, inspect a unitarity defect such as

$$\epsilon_U = \lVert \widetilde U^\dagger\widetilde U-I \rVert.$$

Also check the opposite product for rectangular or infinite-dimensional approximations. A determinant of unit modulus is only a necessary condition.

If theory guarantees an exact unitary but roundoff produces a nearby nonsingular matrix $M$, the unitary polar factor is

$$U = M(M^\dagger M)^{-1/2}.$$

Applying this correction is appropriate only when projection back to the unitary group matches the intended model. It should not be used to erase genuine loss, gain, or nonunitary dynamics.

Common Mistakes

• Confusing unitary with Hermitian. Unitary means inverse equals adjoint; Hermitian means operator equals adjoint.
• Checking only the determinant. $\lvert\det U\rvert=1$ does not imply orthonormal columns.
• Assuming norm preservation without linearity implies unitarity. Unitary operators are linear.
• Switching between active and passive uses. Declare whether vectors move or coordinates change.
• Assuming all probabilities in a fixed basis remain unchanged. A unitary can rotate a state relative to that basis.
• Treating a global operator phase as always irrelevant. It can become a relative phase in a controlled or interferometric setting.
• Calling an infinite-dimensional isometry unitary without checking surjectivity.
• Modeling measurement update, noise, or subsystem dynamics as a unitary on the subsystem alone.

Exercises

1. Let

$$U = \frac{1}{\sqrt2} \begin{pmatrix} 1&i\\ i&1 \end{pmatrix}.$$

Verify unitarity and find its action on the computational basis vectors.

Solution

The adjoint is

$$U^\dagger = \frac{1}{\sqrt2} \begin{pmatrix} 1&-i\\ -i&1 \end{pmatrix}.$$

Multiplication gives

$$U^\dagger U = \frac12 \begin{pmatrix} 2&0\\ 0&2 \end{pmatrix} =I.$$

Its action is

$$\begin{aligned} U\lvert0\rangle &= \frac{ \lvert0\rangle+i\lvert1\rangle }{\sqrt2}, \\ U\lvert1\rangle &= \frac{ i\lvert0\rangle+\lvert1\rangle }{\sqrt2}. \end{aligned}$$

The images form an orthonormal basis.

2. Prove that every eigenvalue of a finite-dimensional unitary matrix has unit modulus. What does this imply for its determinant?

Solution

If $Uv=\lambda v$ with $v\ne0$, then

$$\lVert v\rVert = \lVert Uv\rVert = \lvert\lambda\rvert\lVert v\rVert.$$

Therefore $\lvert\lambda\rvert=1$. A unitary matrix is normal and has a full orthonormal eigenbasis, so its determinant is the product of its eigenvalues:

$$\lvert\det U\rvert = \prod_j\lvert\lambda_j\rvert =1.$$

3. Let the columns of a unitary matrix $U$ be a new orthonormal basis written in old-basis coordinates. Derive the passive coordinate rule for a vector and the matching rule for an operator.

Solution

The old and new coordinate columns obey

$$c_{\mathrm{old}} = U c_{\mathrm{new}},$$

because the columns of $U$ are the new basis vectors in old coordinates. Multiplying by $U^\dagger=U^{-1}$ gives

$$c_{\mathrm{new}} = U^\dagger c_{\mathrm{old}}.$$

For an operator,

$$\begin{aligned} A_{\mathrm{new}}c_{\mathrm{new}} &= U^\dagger A_{\mathrm{old}} c_{\mathrm{old}} \\ &= U^\dagger A_{\mathrm{old}}U c_{\mathrm{new}}, \end{aligned}$$

so

$$A_{\mathrm{new}} = U^\dagger A_{\mathrm{old}}U.$$

4. For the unilateral shift $S$ on $\ell^2(\mathbb N)$, compute its adjoint on a sequence and verify that $S^\dagger S=I$ but $SS^\dagger\ne I$.

Solution

The adjoint removes the first component and shifts the rest left:

$$S^\dagger(c_1,c_2,c_3,\ldots) = (c_2,c_3,c_4,\ldots).$$

Therefore

$$S^\dagger S(c_1,c_2,\ldots) = (c_1,c_2,\ldots),$$

but

$$SS^\dagger(c_1,c_2,\ldots) = (0,c_2,c_3,\ldots).$$

Thus

$$SS^\dagger = I-\lvert e_1\rangle\langle e_1\vert.$$

The shift preserves norms but is not onto, so it is not unitary.

References

• S. Axler, Linear Algebra Done Right, 3rd ed., Springer, 2015.
• R. A. Horn and C. R. Johnson, Matrix Analysis, 2nd ed., Cambridge University Press, 2013.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Volume I: Functional Analysis, revised and enlarged ed., Academic Press, 1980.
• J. von Neumann, Mathematical Foundations of Quantum Mechanics, Princeton University Press, 1955.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.