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A densely defined operator $T$ on a Hilbert space is symmetric whenEquivalently, $T$ agrees with its adjoint operator on $D(T)$, while $D(T^{\ast })$ may be larger. A self-adjoint operator additionally requires equality of these operator domains.

For a bounded operator on a real Hilbert space, $⟨Lv,v⟩=⟨Sv,v⟩$ with $S=(L+L^{\ast })/2$. Consequently the Euler-Lagrange equation of $⟨Lv,v⟩-2⟨f,v⟩$ is $Sv=f$. It is $Lv=f$ only when $L$ is self-adjoint or when the relevant solution also annihilates the skew part. For example $L=\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$ has positive quadratic form $\Vert v{\Vert }^2$, but that form contains no information about its skew part.

A continuous symmetric coercive bilinear form $a$ on a real Hilbert space and a continuous linear functional $\ell$ give a unique minimum of $I$. The parallelogram identity makes any minimizing sequence Cauchy in the energy norm; coercivity and completeness give its limit. Differentiation gives $a(u,w)=\ell (w)$, and $I(v)-I(u)=a(v-u,v-u)$ proves uniqueness. A nonsymmetric operator instead differentiates to its symmetric part. Mere pointwise positivity without a lower bound in the chosen norm need not yield a minimizing vector.

For a bounded linear operator between Hilbert spaces, the spectral regularization method for quadratic Tikhonov regularization has scalar coefficient $s/(s^2+\alpha )$. Since $(s-\sqrt{\alpha }{)}^2\ge 0$, this coefficient is at most $1/(2\sqrt{\alpha })$. The spectral theorem for normal operators transfers this scalar bound to the operator norm, providing the noise term in the noise-bias decomposition for linear regularization.

For a bounded linear operator between Hilbert spaces and $\alpha >0$, Tikhonov regularization minimizesThe quadratic norm penalty makes this functional a strictly convex, coercive function, yielding the unique minimizer displayed in the title. Here $A^{\ast }$ is the adjoint operator and $I$ the identity operator. Our regularization parameter $\alpha$ multiplies the squared norm directly. Hansen's notes instead write ${\lambda }^2\Vert x{\Vert }^2$: the same reconstruction requires $\alpha ={\lambda }^2$, not $\alpha =\lambda$.^[ref]

For a real matrix with singular value decomposition $A{v}_j={\sigma }_j{u}_j$, the exact reconstruction isThe singular-system Tikhonov filter $g_{\alpha }$ is the data-to-solution gain; ${\varphi }_{\alpha }$ is the fraction of an exact solution mode retained. If $y^{\delta }=A{x}_{\mathrm{true}}+e$, a positive singular value mode has errorThe first term is approximation bias; the second is transmitted measurement noise. Suppressing a weak mode suppresses its true signal too.

For the two-pixel matrix in the inverse problem example and $\alpha =0.01$ ($\lambda =0.1$), the mean and difference attenuation factors are $100/101$ and $1/2$. Since $y^{\delta }=q_{+}/\sqrt{2}+0.12q_{-}/\sqrt{2}$, the reconstruction isIts noise contribution is $(0.05,-0.05{)}^T$, half the direct inverse's, but its exact-data reconstruction is $(0.74505,0.24505{)}^T$. Consequently its total error Euclidean norm, about $0.28293$, exceeds the direct inverse's $0.14142$: this parameter sacrifices too much contrast. Taking $\alpha =0.001$ instead gives $(1.04496,-0.04595{)}^T$ and error about $0.06429$. Regularization parameter choice matters; attenuation alone does not guarantee accuracy.

For bounded linear operators between Hilbert spaces with $\Vert Bu\Vert \ge \beta \Vert u\Vert$ and $\beta >0$, the operator $K^{\ast }K+\alpha B^{\ast }B$ is coercive and invertible for $\alpha >0$. The reconstruction uniquely minimizes $\Vert Ku-f{\Vert }^2/2+\alpha \Vert Bu{\Vert }^2/2$. Testing its normal equation for coercive Tikhonov regularization by $u$ gives $\Vert R_{\alpha }\Vert \le 1/(2\beta \sqrt{\alpha })$. If $K$ is injective, exact-data consistency and $\alpha \to 0$, $\delta /\sqrt{\alpha }\to 0$ give norm convergence to $K^{†}f$.

For an absolutely one-homogeneous functional $J=\mathrm{TV}$ on a real Hilbert space, a subgradient $q$ at $u$ is characterized by $⟨q,v⟩\le J(v)$ for every $v$ and $⟨q,u⟩=J(u)$. A bounded vector field with $|z|\le 1$ and $q=\mathrm{div}z$ proves the inequality through the dual definition, provided cutoff approximation is valid in the ambient space. Equality is called a calibration. It certifies a global minimizer of total variation denoising, rather than only a minimizer within a chosen ansatz.

An internal qubit uses two distinguishable electronic or hyperfine structure quantum states, denoted $|g⟩,|e⟩$. The ion's centre-of-mass motion has a different Hilbert space. In one approximately harmonic direction, a number state $|n⟩$ has energy $E_n=\hslash \nu (n+1/2)$, where $\nu$ denotes a motional angular frequency, not ordinary frequency. The tensor product of quantum systems gives $|g,n⟩$. Cooling towards $n=0$ prepares motion; it does not automatically prepare the internal qubit. Nor does the ground state mean a stationary point particle: zero-point energy and position spread remain.

For several ions, Coulomb forces couple their displacements into collective normal modes. Each selected normal mode has its own quantum harmonic oscillator number states, while each ion retains internal qubit states. A pulse addressing one ion can change a shared mode. This supplies a route to correlations between distant internal quantum states; it also makes residual motional excitation and unwanted excitation of other normal modes relevant. A single-ion pulse calculation alone establishes no experimental multi-ion quantum gate performance.

A densely defined linear operator on a complex Hilbert space is self-adjoint when its adjoint operator has exactly the same domain and agrees there. Symmetry only requires $A\subseteq A^{\ast }$ and does not guarantee self-adjointness. A self-adjoint operator is a closed operator with real spectrum. Domains matter for differential operators and unbounded multiplication operators.

A positive definite symmetric operator is uniformly positive definite if $⟨Lv,v⟩\ge \gamma \Vert v{\Vert }^2$ for some $\gamma >0$. For a bounded linear operator defined on a whole Hilbert space, this is a symmetric coercive operator, and the Lax-Milgram theorem supplies a unique solution to $Lu=f$ for every $f$. For a differential operator, the corresponding bounded coercive form is used on its form space.

An associative algebra with a two-sided multiplicative identity. For a subalgebra of bounded operators, the term unital here means that it contains the identity of the represented Hilbert space.

Two operators on Hilbert spaces are unitarily equivalent if a unitary operator maps their operator domains onto one another and intertwines their actions. Their spectra and eigenvalue multiplicities then agree. For nonnegative self-adjoint operators it suffices that the unitary maps their closed quadratic form domains onto one another and preserves the forms, since the associated operator is determined by the form.

A linear isometry from a subspace of a finite-dimensional Hilbert space into the same ambient space extends to a unitary operator. Complete an orthonormal basis of the input subspace and its isometric image to full orthonormal bases, then map the first to the second. The equal complement dimensions make this possible. Orthogonality of the specified image columns, not normalization alone, is essential. In an infinite-dimensional space arbitrary isometries need not be surjective; the finite-dimensional same-space hypothesis cannot be dropped.

A time-independent Hamiltonian evolves a state by the unitary operator $U(t)=e^{-iHt/\hslash }$.

For a closed quantum system, unitary time evolution changes a quantum state through a unitary operator, preserving norms and inner products. For a time-independent self-adjoint Hamiltonian operator $H$, choose the initial time as zero:Here $\hslash =h/(2\pi )$ is the reduced Planck constant. The same $U(t)$ acts on every initial quantum state; it depends on the Hamiltonian operator and elapsed time, rather than the chosen preparation.^[ref]

The self-adjointness of $H$ makes the exponential a unitary operator. In finite dimension, conjugating its matrix exponential givesThe two exponents commute because both are multiples of $H$. Hence, for any two initial vectors,This short proof explains why normalization survives, why orthogonality of initial vectors survives, and why the transition probability between two states evolved by the same $U$ is unchanged. It does not require that a fixed measurement in quantum mechanics have time-independent probabilities: its measurement basis need not move with the quantum states.

For an unbounded self-adjoint operator, define the exponential through the spectral theorem, rather than assuming that its power series acts on every vector. It is a unitary operator on the entire Hilbert space; the strong Schrödinger equation holds for initial vectors in the operator domain $D(H)$. For a density operator, unitary time evolution reads $\rho (t)=U(t)\rho (0)U(t{)}^{†}$. It preserves eigenvalues and purity of a density operator, but can change its entries in a fixed basis.

The useful distinction is between conserving total probability and conserving each measurement's probabilities. A fixed energy spectrum supplies different ticking rates for different probability amplitudes. Their relative quantum phases can move the Bloch vector and create quantum interference even while every energy eigenstate population stays fixed. Qubit precession under a constant Hamiltonian makes this visible; Ramsey interferometry converts the accumulated relative quantum phase into a population difference that a simple readout can detect.

For a bounded sequence $(u_n)$ in a Hilbert space, put $c_h={\mathrm{limsup}}_{N\to \infty }|N^{-1}{\sum }_{n=0}^{N-h-1}⟨u_{n+h},u_n⟩|$. If $H^{-1}{\sum }_{h=1}^H{c}_h\to 0$, then $N^{-1}{\sum }_{n=0}^{N-1}{u}_n\to 0$ in norm. Since the $c_h$ are bounded, it suffices that $(c_h)$ has convergence in density of a sequence to zero. The proof applies the Cauchy-Schwarz inequality to an average of shifted sums.

A variational problem seeks a stationary point or minimizer of a functional on a specified function space. For a symmetric coercive bilinear form $a$ and bounded linear functional $\ell$ on a real Hilbert space, minimizing $J(v)=a(v,v)/2-\ell (v)$ is equivalent to the weak formulation $a(u,v)=\ell (v)$ for every test vector. The Lax-Milgram theorem gives a unique solution, and $J(u+w)-J(u)=a(w,w)/2$ proves unique minimization. Specifying the admissible space and its boundary constraints is part of the problem.

A unital star-subalgebra of the bounded operators on a Hilbert space that is closed in the weak operator topology, equivalently in the strong operator topology. The Von Neumann double commutant theorem makes these conditions equivalent to equality with the double commutant of an operator algebra. It is norm closed and hence a C-star algebra.

For a unital star-subalgebra $A$ of the bounded operators on a complex Hilbert space, its strong operator topology and weak operator topology closures equal its double commutant of an operator algebra. To approximate $T\in A^{\prime \prime }$ on finitely many vectors, take their tuple in $H^m$. The projection onto its diagonal-$A$ cyclic subspace has matrix entries in $A^{\prime }$, so it commutes with the diagonal action of $T$. Therefore the image tuple under $T$ lies in that cyclic subspace and can be approximated by a single $a\in A$.

For a unitary operator $U$ on a Hilbert space, the Cesàro averagesconverge strongly to the orthogonal projection onto the fixed-point subspace $\ker (U-I)$.

If $x_n\rightharpoonup x$ in a Hilbert space, some subsequence $(x_{n_k})$ has Cesàro means converging in norm to $x$. For $x=0$, choose the subsequence so successive vectors have summably small inner products; expansion of the squared norm of its averages then proves convergence.

Thus Hilbert spaces have the weak form of the Banach-Saks property.