Unitary Operators as Inner-Product Isometries

Derivation

Unitary Operators as Inner-Product Isometries

D-083 Home PU-104 Threads symmetry Depends on Existence and Uniqueness of the Adjoint, Cauchy–Schwarz and the Triangle Inequality

Statement

Let \(U\) be a bounded linear operator on a complex Hilbert space \(\mathcal{H}\) with a well-defined adjoint \(U^\dagger\), and take the inner product antilinear in its first slot, linear in its second. Then \(U\) preserves the inner product, \(\langle Ux, Uy\rangle = \langle x, y\rangle\) for all \(x,y\in\mathcal{H}\), if and only if \(U^\dagger U = I\). On a finite-dimensional space this is equivalent to full unitarity \(U^\dagger U = U U^\dagger = I\), and every eigenvalue \(\lambda\) of \(U\) then satisfies \(\lvert\lambda\rvert = 1\), i.e. \(\lambda = e^{i\theta}\) for some real \(\theta\).

Why it matters

Unitarity is the mathematical backbone of quantum dynamics: closed-system time evolution \(U(t)=e^{-iHt/\hbar}\) is unitary precisely because it must conserve total probability \(\langle\psi\,|\,\psi\rangle\). Inner-product preservation is the same statement as "probabilities add up to one at all times," so the abstract identity \(U^\dagger U = I\) is what forbids amplitude from being created or destroyed under evolution.

The eigenvalue result \(\lvert\lambda\rvert=1\) is why symmetry generators exponentiate to pure phases rather than growing or decaying modes. It underlies the reality of measurable frequencies, the stability of stationary states, and the classification of representations of symmetry groups — one of the central threads linking symmetry to conservation laws.

Assumptions

The scalar field is \(\mathbb{C}\) and the inner product is positive-definite.Over \(\mathbb{R}\) the eigenvalue conclusion weakens: real orthogonal operators can have no real eigenvalues at all (a rotation), so the phase interpretation \(\lambda=e^{i\theta}\) genuinely needs complex spectra. Without positive-definiteness (an indefinite or Krein metric) "isometries" become pseudo-unitary and eigenvalues can leave the unit circle. \(U\) is linear and everywhere-defined with an adjoint \(U^\dagger\) satisfying \(\langle Ux,y\rangle=\langle x,U^\dagger y\rangle\).Antilinear operators (e.g. time reversal) can also preserve \(\lvert\langle\cdot,\cdot\rangle\rvert\) but are antiunitary, not unitary; the adjoint identity used in every step below would fail. In infinite dimensions, \(U\) is also surjective (bounded below with dense range).\(U^\dagger U=I\) alone makes \(U\) only an isometry (injective, norm-preserving) but possibly not onto — e.g. the right shift on \(\ell^2\). Full unitarity \(UU^\dagger=I\) additionally requires surjectivity; in finite dimensions this is automatic, but not otherwise.

Derivation

\[ \langle Ux, Uy\rangle = \langle x, y\rangle \quad \text{for all } x,y\in\mathcal{H} \]

Definition of inner-product preservation — the hypothesis we translate into operator form. A

\[ \langle Ux, Uy\rangle = \langle x, U^\dagger U y\rangle \]

Apply the adjoint identity \(\langle Ux,z\rangle=\langle x,U^\dagger z\rangle\) with \(z=Uy\); uses adjoint existence and uniqueness (prior result). A

\[ \langle x, U^\dagger U y\rangle = \langle x, y\rangle \;\Longrightarrow\; \big\langle x,\, (U^\dagger U - I)\,y\big\rangle = 0 \]

Equate Steps 1 and 2, then use linearity of the inner product in its second slot to collect terms. A

\[ \big\langle x,\,(U^\dagger U - I)y\big\rangle = 0\ \ \forall x \;\Longrightarrow\; (U^\dagger U - I)y = 0 \]

A vector orthogonal to every \(x\) is the zero vector (non-degeneracy of the inner product). Taking \(x=(U^\dagger U-I)y\) forces \(\lVert(U^\dagger U-I)y\rVert^2=0\). B

\[ (U^\dagger U - I)y = 0 \ \ \forall y \;\Longrightarrow\; \boxed{\,U^\dagger U = I\,} \]

An operator sending every \(y\) to \(0\) is the zero operator; this establishes the forward direction. A

\[ U^\dagger U = I \;\Longrightarrow\; \langle Ux, Uy\rangle = \langle x, U^\dagger U y\rangle = \langle x, I y\rangle = \langle x,y\rangle \]

Converse: substitute \(U^\dagger U=I\) back through the adjoint identity. The equivalence now holds both ways. A

\[ U v = \lambda v,\quad v\neq 0 \;\Longrightarrow\; \langle v,v\rangle = \langle Uv, Uv\rangle = \langle \lambda v, \lambda v\rangle = \lvert\lambda\rvert^2\,\langle v,v\rangle \]

Feed an eigenvector into the preservation identity; antilinearity in the first slot pulls out \(\overline{\lambda}\), giving \(\overline{\lambda}\lambda=\lvert\lambda\rvert^2\). B

\[ \langle v,v\rangle > 0 \;\Longrightarrow\; \lvert\lambda\rvert^2 = 1 \;\Longrightarrow\; \boxed{\,\lvert\lambda\rvert = 1,\quad \lambda = e^{i\theta}\,} \]

Divide by \(\langle v,v\rangle\neq0\) (positive-definiteness). Every eigenvalue lies on the unit circle. A

\[ \dim\mathcal{H}<\infty:\quad \det(U^\dagger U)=\lvert\det U\rvert^2 = \det I = 1 \;\Longrightarrow\; U^{-1}=U^\dagger \;\Longrightarrow\; UU^\dagger = I \]

In finite dimensions a left inverse is a two-sided inverse, so \(U^\dagger U=I\) upgrades to full unitarity \(UU^\dagger=I\). C

Result

\[ \langle Ux, Uy\rangle = \langle x,y\rangle\ \ \forall x,y \;\Longleftrightarrow\; U^\dagger U = I, \qquad Uv=\lambda v \Rightarrow \lvert\lambda\rvert = 1 \]

Reading. Preserving all inner products is exactly the algebraic condition \(U^\dagger U=I\) — geometry (angles and lengths are unchanged) and algebra (the adjoint is the inverse) are the same statement. Because the length of an eigenvector cannot change, an eigenvalue can only rotate a vector by a phase, never rescale it, so it must lie on the unit circle.

Units check. In the standard quantum convention state vectors are dimensionless and normalized, so \(\langle x,y\rangle\) is dimensionless; \(U\) maps states to states and is dimensionless; \(U^\dagger U=I\) equates dimensionless operators. Eigenvalues \(\lambda=e^{i\theta}\) are pure phases with dimensionless argument \(\theta\) (radians), consistent with \(\lvert\lambda\rvert=1\) being a dimensionless statement.

Limiting cases

\(U=I\): trivially inner-product preserving, \(I^\dagger I=I\), single eigenvalue \(\lambda=1=e^{i0}\).
Real orthogonal \(U\) (\(U^\top U=I\)): the special case over \(\mathbb{R}\); lengths preserved, but eigenvalues come in conjugate pairs \(e^{\pm i\theta}\) and may be non-real.
Diagonal phase operator \(U=\mathrm{diag}(e^{i\theta_1},\dots,e^{i\theta_n})\): the general finite unitary written in its eigenbasis; each diagonal entry sits on the unit circle by construction.
\(U(t)=e^{-iHt/\hbar}\) with \(H^\dagger=H\): unitary for all \(t\); the \(t\to0\) limit returns \(U\to I\) and recovers Step 1 trivially.

Breaks when

Isometry without surjectivity (infinite dimensions). The right shift \(S(x_1,x_2,\dots)=(0,x_1,x_2,\dots)\) on \(\ell^2\) satisfies \(S^\dagger S=I\) and preserves inner products, yet \(SS^\dagger\neq I\): it is not unitary and has no eigenvalues at all. Step 9 relies on finite-dimensional invertibility and fails here.
Antiunitary operators. Time reversal \(T\) obeys \(\langle Tx,Ty\rangle=\overline{\langle x,y\rangle}=\langle y,x\rangle\), preserving \(\lvert\langle\cdot,\cdot\rangle\rvert\) but not the inner product itself. \(T\) is antilinear, so Step 2's adjoint identity and Step 7's factoring break; \(T\) is not captured by a linear \(T^\dagger T=I\).
Indefinite (Krein / Minkowski) metric. If \(\langle v,v\rangle\) can be zero or negative, Step 8's division is invalid; "pseudo-unitary" operators preserving an indefinite form (e.g. Lorentz boosts preserving \(\eta\)) have real eigenvalues \(e^{\pm\eta}\) off the unit circle.
Non-normal near-unitary numerics. A finite-precision \(U\) with \(U^\dagger U = I + E\), \(\lVert E\rVert\neq0\), no longer preserves inner products exactly; eigenvalues drift off \(\lvert\lambda\rvert=1\) by \(O(\lVert E\rVert)\), which accumulates in long unitary time-stepping.

Failure modes

Confusing \(U^\dagger U=I\) with \(UU^\dagger=I\). Students assume they are automatically the same; they coincide only in finite dimensions. Quoting "left inverse = right inverse" without the finiteness hypothesis is the classic slip.
Dropping the conjugate in Step 7. Writing \(\langle\lambda v,\lambda v\rangle=\lambda^2\langle v,v\rangle\) instead of \(\lvert\lambda\rvert^2\langle v,v\rangle\). This wrongly concludes \(\lambda=\pm1\) rather than \(\lambda=e^{i\theta}\).
Assuming unitary means Hermitian. Unitary (\(U^\dagger=U^{-1}\)) and Hermitian (\(A^\dagger=A\)) coincide only when \(U=U^{-1}\), i.e. eigenvalues \(\pm1\); in general they are different classes of operator.
Believing norm preservation is weaker than inner-product preservation. Over \(\mathbb{C}\) the polarization identity makes \(\lVert Ux\rVert=\lVert x\rVert\) and \(\langle Ux,Uy\rangle=\langle x,y\rangle\) equivalent; students often prove only the norm version and think the full statement needs more.
Reading \(\lvert\lambda\rvert=1\) as \(\lambda=1\). Forgetting the whole unit circle is available; e.g. a phase gate has \(\lambda=e^{i\pi/4}\neq1\).

Discussion

The theorem is really a bridge between three languages for one object. Geometrically, \(U\) is a rigid motion of Hilbert space fixing the origin: it preserves lengths and angles, hence maps orthonormal bases to orthonormal bases. Algebraically, that rigidity is captured by the single identity \(U^\dagger U=I\), i.e. the adjoint is the inverse. Spectrally, the constraint radiates outward to the eigenvalues, pinning them to the unit circle. Each viewpoint is nearly trivial to state once you hold the others, but the equivalence is the content.

Physically, the unit-circle spectrum is why continuous symmetries generate pure phases. Writing \(U=e^{-iG}\) with \(G^\dagger=G\) Hermitian, the eigenvalues of \(U\) are \(e^{-ig}\) with \(g\in\mathbb{R}\) real — the observable \(G\) (energy, momentum, angular momentum) has a real spectrum precisely because its exponential is unitary. Probability conservation and the reality of measured quantities are thus two faces of one geometric fact.

The finite/infinite-dimensional gap is not a technicality but a doorway to operator theory. An isometry \(V\) with \(V^\dagger V=I\) but \(VV^\dagger=P\neq I\) is characterized by its defect: \(P\) projects onto the part of the space \(V\) reaches, and \(I-P\) measures what it misses. Wold's decomposition splits any isometry into a unitary piece plus copies of the unilateral shift, and the deficiency indices of a symmetric operator decide whether a Hermitian \(H\) even has a unitary \(e^{-iHt/\hbar}\) — the self-adjointness question that ordinary "Hermitian" glosses over. Stone's theorem then certifies that strongly continuous one-parameter unitary groups are exactly the exponentials of self-adjoint generators.

Common misconceptions. "Unitary = Hermitian" (false; they intersect only where \(\lambda^2=1\)). "\(U^\dagger U=I\) always gives a unitary" (false in infinite dimensions — you also need onto). "Eigenvalues of a unitary are real" (false — they are unimodular, generally complex). "Antiunitary operators are a kind of unitary" (false — they are antilinear and preserve only the magnitude of the inner product).

Worked examples

\[ U(\theta)=\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix},\qquad \theta = 30^\circ = \frac{\pi}{6} \]

Real \(2\times2\) rotation (orthogonal): test \(U^\top U=I\) and find its spectrum. A

\[ U^\top U = \begin{pmatrix}\cos\theta & \sin\theta\\ -\sin\theta & \cos\theta\end{pmatrix}\!\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix} = \begin{pmatrix}\cos^2\theta+\sin^2\theta & 0\\ 0 & \cos^2\theta+\sin^2\theta\end{pmatrix} = I \]

Off-diagonals cancel; diagonals use \(\cos^2+\sin^2=1\). Inner-product preservation confirmed. A

\[ \det(U-\lambda I)=(\cos\theta-\lambda)^2+\sin^2\theta=0 \;\Rightarrow\; \lambda = \cos\theta \pm i\sin\theta = e^{\pm i\theta} \]

Characteristic polynomial; the eigenvalues are the rotation phase and its conjugate. B

\[ \theta=\frac{\pi}{6}:\quad \lambda = \cos 30^\circ \pm i\sin 30^\circ = \frac{\sqrt3}{2} \pm \frac{i}{2},\qquad \lvert\lambda\rvert=\sqrt{\tfrac34+\tfrac14}=1 \]

Insert the number; verify the modulus. A

\[ \lambda = \tfrac{\sqrt3}{2}\pm\tfrac{i}{2}=e^{\pm i\pi/6},\qquad \lvert\lambda\rvert = 1 \]

Reading. A \(30^\circ\) rotation preserves all lengths and angles in the plane; both eigenvalues sit on the unit circle at \(\pm30^\circ\). No real eigenvector exists — a real rotation turns every vector — illustrating why the complex field is needed for the phase picture.

Units check. Entries of \(U\) are dimensionless (cosines and sines of a radian angle); the eigenvalues are dimensionless unit-modulus complex numbers.

\[ H = \frac{\hbar\omega}{2}\,\sigma_z = \frac{\hbar\omega}{2}\begin{pmatrix}1&0\\0&-1\end{pmatrix},\qquad U(t)=e^{-iHt/\hbar} \]

Two-level system (qubit) evolving under a Hermitian Hamiltonian; build its time-evolution operator. A

\[ U(t)=\begin{pmatrix}e^{-i\omega t/2} & 0\\ 0 & e^{+i\omega t/2}\end{pmatrix},\qquad U^\dagger(t)=\begin{pmatrix}e^{+i\omega t/2} & 0\\ 0 & e^{-i\omega t/2}\end{pmatrix} \]

Exponentiate the diagonal Hamiltonian; the adjoint conjugates the phases. B

\[ U^\dagger U = \begin{pmatrix}e^{+i\omega t/2}e^{-i\omega t/2} & 0\\ 0 & e^{-i\omega t/2}e^{+i\omega t/2}\end{pmatrix} = \begin{pmatrix}1&0\\0&1\end{pmatrix}=I \]

Phases cancel entry by entry — unitarity holds for every \(t\), so probability is conserved throughout the evolution. A

\[ \omega = 1.0\times10^{15}\ \mathrm{rad\,s^{-1}},\quad t = 1.0\times10^{-15}\ \mathrm{s} \;\Rightarrow\; \frac{\omega t}{2}=0.50\ \mathrm{rad} \]

Insert numbers; the phase angle is dimensionless (rad s\(^{-1}\) times s). A

\[ \lambda_\pm = e^{\mp i(0.50)} = \cos 0.50 \mp i\sin 0.50 = 0.8776 \mp 0.4794\,i,\quad \lvert\lambda_\pm\rvert=\sqrt{0.8776^2+0.4794^2}=1.000 \]

Evaluate the eigenvalue phases and check the modulus. A

\[ \lambda_\pm = e^{\mp i\,0.50}=0.878\mp0.479\,i,\qquad \lvert\lambda_\pm\rvert = 1.00 \]

Reading. Over \(1\ \mathrm{fs}\) the two energy eigenstates acquire opposite phases \(\mp0.50\ \mathrm{rad}\); neither amplitude changes magnitude, so \(\lvert\psi\rvert^2\) is unchanged. The eigenvalues are pure phases on the unit circle exactly because \(H\) is Hermitian, making \(U(t)\) unitary.

Units check. \(\omega t\) is (rad s\(^{-1}\))(s) = rad, dimensionless; \(Ht/\hbar\) is (J)(s)/(J s) = dimensionless; every eigenvalue is a unit-modulus complex number.

Problems

Show the phase operator \(U=\mathrm{diag}(e^{i\alpha},e^{i\beta})\) is unitary and give its eigenvalues. Evaluate for \(\alpha=\pi/3,\ \beta=-\pi/4\).
Solution
\(U^\dagger=\mathrm{diag}(e^{-i\alpha},e^{-i\beta})\), so \(U^\dagger U=\mathrm{diag}(e^{-i\alpha}e^{i\alpha},e^{-i\beta}e^{i\beta})=\mathrm{diag}(1,1)=I\). The matrix is diagonal, so its eigenvalues are the diagonal entries \(\lambda_1=e^{i\alpha},\ \lambda_2=e^{i\beta}\), each with \(\lvert\lambda\rvert=1\). Numerically \(\lambda_1=e^{i\pi/3}=0.500+0.866\,i\) and \(\lambda_2=e^{-i\pi/4}=0.707-0.707\,i\); both have modulus \(1.000\).
Prove that the product of two unitaries \(U,V\) is unitary, and that \(U^\dagger\) is unitary.
Solution
Using \((UV)^\dagger=V^\dagger U^\dagger\): \((UV)^\dagger(UV)=V^\dagger U^\dagger U V=V^\dagger I V=V^\dagger V=I\), and likewise \((UV)(UV)^\dagger=I\), so \(UV\) is unitary. For the adjoint, \((U^\dagger)^\dagger(U^\dagger)=U U^\dagger=I\) and \((U^\dagger)(U^\dagger)^\dagger=U^\dagger U=I\); hence \(U^\dagger=U^{-1}\) is unitary. Together with associativity these are the group axioms of the unitary group \(U(n)\).
Determine whether \(U=\dfrac{1}{\sqrt2}\begin{pmatrix}1 & 1\\ i & -i\end{pmatrix}\) is unitary, and find \(\lvert\det U\rvert\).
Solution
\(U^\dagger=\dfrac{1}{\sqrt2}\begin{pmatrix}1 & -i\\ 1 & i\end{pmatrix}\) (transpose then conjugate). Then \(U^\dagger U=\dfrac12\begin{pmatrix}1 & -i\\ 1 & i\end{pmatrix}\begin{pmatrix}1 & 1\\ i & -i\end{pmatrix}=\dfrac12\begin{pmatrix}1+1 & 1-1\\ 1-1 & 1+1\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}=I.\) So \(U\) is unitary. \(\det U=\dfrac12\big(1\cdot(-i)-1\cdot i\big)=\dfrac12(-2i)=-i\), and \(\lvert\det U\rvert=1\), as required for any unitary.
Show that if \(U\) preserves norms, \(\lVert Ux\rVert=\lVert x\rVert\) for all \(x\), then over \(\mathbb{C}\) it preserves the full inner product.
Solution
Use the complex polarization identity \(\langle x,y\rangle=\dfrac14\sum_{k=0}^{3} i^{k}\,\lVert x + i^{k} y\rVert^2\). Apply \(U\) inside each norm: \(\lVert Ux+i^kUy\rVert^2=\lVert U(x+i^ky)\rVert^2=\lVert x+i^ky\rVert^2\) by linearity of \(U\) and the norm hypothesis. Summing, \(\langle Ux,Uy\rangle=\dfrac14\sum_k i^k\lVert x+i^ky\rVert^2=\langle x,y\rangle\). Hence norm preservation and inner-product preservation are equivalent over \(\mathbb{C}\), and by the main theorem both equal \(U^\dagger U=I\).
Let \(U\) be unitary with \(Uv_1=\lambda_1 v_1\), \(Uv_2=\lambda_2 v_2\) and \(\lambda_1\neq\lambda_2\). Show \(v_1\perp v_2\).
Solution
By inner-product preservation, \(\langle v_1,v_2\rangle=\langle Uv_1,Uv_2\rangle=\langle\lambda_1 v_1,\lambda_2 v_2\rangle=\overline{\lambda_1}\lambda_2\langle v_1,v_2\rangle\) (antilinearity in the first slot). Thus \((\overline{\lambda_1}\lambda_2-1)\langle v_1,v_2\rangle=0\). Since \(\lvert\lambda_1\rvert=1\) we have \(\overline{\lambda_1}=1/\lambda_1\), so \(\overline{\lambda_1}\lambda_2=\lambda_2/\lambda_1\neq1\) because \(\lambda_1\neq\lambda_2\). The prefactor is nonzero, forcing \(\langle v_1,v_2\rangle=0\). Eigenvectors of a unitary for distinct eigenvalues are orthogonal — the basis for diagonalizing it by an orthonormal eigenbasis (spectral theorem for normal operators).