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For a bipartite density operator ${\rho }_{SA}$, the reduced density matrix of $S$ isIt is characterized byfor every observable $M_S$ on $S$. Equivalently, in any orthonormal basis $\{|a_k⟩\}$ of $A$, the partial trace isand the result is independent of that basis.

The compact self-adjoint operator $K$ has an orthonormal basis of eigenvectors $e_n$. Part (c) gives ${\lambda }_n>0$ and ${\sum }_n{\lambda }_n<\infty$, so $K$ is positive and trace class, exactly the required condition for a covariance operator of a Gaussian measure on a Hilbert space.

For independent standard normal variables ${\xi }_n$, define the Hilbert-space Gaussian seriesBecause $\mathbb{E}\Vert U-m{\Vert }^2={\sum }_n{\lambda }_n<\infty$, the series converges in $L^2(\Omega ;X)$ and almost surely. Its law is the Gaussian measure $N(m,K)$.

Write $q=1-p$, $\mu =q-p$, $\sigma =2\sqrt{pq}$, and ${\varphi }_S(x)={\prod }_{j\in S}(x_j-\mu )/\sigma$. The functions ${\varphi }_S$ form the p-biased product measure orthonormal basis, so the Fourier expansion is $f={\sum }_S{\hat{f}}_p(S){\varphi }_S$. The normalized discrete derivative of a Boolean function satisfiesApplying Parseval identity and then exchanging two finite sums gives

The noise operator on the Boolean hypercube acts diagonally on the same basis: $T_{\rho }{\varphi }_S={\rho }^{|S|}{\varphi }_S$. Hence the noise stability isIts derivative isTaking the right-hand value at $\rho =0$ leaves exactly the linear Fourier weight ${\sum }_i{\hat{f}}_p(\{i\}{)}^2$, while taking the left-hand value at $\rho =1$ gives $\mathbf{I}(f)$.

For every $A\subseteq [n]$, define the Walsh characterThese $2^n$ functions form an orthonormal basis. For the lazy walk, which stays put with probability $1/2$ and otherwise flips a uniformly chosen coordinate,Hence the eigenvalue $1-k/n$ has multiplicity $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$, for $0\le k\le n$.

The supplied spectral upper bound for total variation mixing givesAt $t=\frac{1}{2}n\log n+Cn$, the last expression is $e^{e^{-2C}}-1$. Choosing $C=C(\epsilon )$ so that this is at most $4{\epsilon }^2$ proves

Let $e_n(t)=\sqrt{2}\sin (n\pi t)$, let $(g_n)$ be independent standard normal variables, and choose $a_n=e^{-n^2}$. DefineFinite collections of values are limits of centered Gaussian vectors, so this is a centered Gaussian process.

The functions $(e_n)$ are an orthonormal basis of $H=L^2(0,1)$. Hencewhich proves ${\mu }_X(H)=1$. For every integer $k\ge 0$,Thus, almost surely and simultaneously for all $k$, the differentiated series converges uniformly on $(0,1)$. Termwise differentiation gives an almost surely infinitely differentiable version.

Finally, let $F\subset H$ be finite-dimensional. Choose a nonzero $h\in F^{\perp }$. Thenis a centered normal variable of variancebecause every $a_n$ is positive and $(e_n)$ is complete. The event $\{X\in F\}$ is contained in $\{⟨X,h{⟩}_H=0\}$, which has probability zero because a nondegenerate normal distribution has no atoms. Therefore ${\mu }_X(F)=0$ for every finite-dimensional linear subspace $F$.

For a subsystem $A$ of a multipartite system with density operator ${\rho }_{A{E}_1\cdots E_n}$, its reduced density matrix is the unique operatorsuch that $\mathrm{Tr}(M_A{\rho }_A)=\mathrm{Tr}[(M_A\otimes I_E){\rho }_{AE}]$ for every observable $M_A$ on $A$. In an orthonormal basis $\{|e_j⟩\}$ of the other subsystems, the partial trace iswhich is independent of the chosen basis.

A normalized quantum state is a linear combination of orthonormal basis vectors in a Hilbert space. Its density operator is $|\psi ⟩⟨\psi |$. When at least two basis amplitudes are nonzero, it has off-diagonal terms and differs from the mixed state ${\sum }_j|a_j{|}^2|j⟩⟨j|$. Quantum gates act linearly on this superposition and preserve its phase information.

Suppose an array $T_{ij}$ is specified in each right-handed orthonormal basis. If $w_i=T_{ij}{v}_j$ transforms as a vector for every vector $v$, then $T$ is a second-order Cartesian tensor under those basis changes. Indeed $v^{\prime }=Rv$, $w^{\prime }=Rw$ imply $T^{\prime }Rv=RTv$ for every $v$, hence $T^{\prime }=RT{R}^T$. The condition on every test vector is essential; one contraction is insufficient.

In a well $-a<x<a$, the normalized state proportional to $x$ on $0<x<a$ and zero elsewhere has stationary-state coefficientsThey follow from the orthonormal basis inner products. Their squared magnitudes give energy-measurement probabilities by the Born rule; splitting even and odd indices in the Parseval identity gives convergent integer-coefficient series identities. The expansion is interpreted in the square-integrable state norm.

In an orthonormal basis, rank-one dephasing is the quantum channel $\Delta (\rho )={\sum }_i|{\psi }_i⟩⟨{\psi }_i|\rho |{\psi }_i⟩⟨{\psi }_i|$. It removes off-diagonal entries while preserving the basis probabilities. Its output Von Neumann entropy is their Shannon entropy, and the relative-entropy identity for rank-one dephasing quantifies the entropy increase.

For $Y\sim N(X\beta ,V_{\theta })$ and a full-column-rank design matrix $X\in {\mathbb{R}}^{n\times p}$, take $A\in {\mathbb{R}}^{n\times (n-p)}$ with $A^{T}A=I$ and $A^{T}X=0$. Then $A^{T}Y\sim N(0,A^T{V}_{\theta }A)$, so restricted maximum likelihood maximizesChanging the orthonormal basis by an orthogonal matrix leaves this expression unchanged.

If an array specified in each orthonormal basis has invariant Frobenius inner product with every Cartesian second-rank tensor, it transforms as such a tensor. For a component change $R$, invariance says $(R^{T}A\prime R-A):B=0$ for every matrix $B$. Nondegeneracy of the Frobenius inner product forces the difference to vanish. Testing every tensor, rather than a single selected tensor, is essential.

A scaling function generates the approximation space $V_0$ of a multiresolution analysis by its integer translates. In an orthonormal construction those translates form an orthonormal basis, and ${\phi }_{j,k}(x)=2^{j/2}\phi (2^{j}x-k)$ generate $V_j$. Integrable orthonormal scaling functions have $|\hat{\phi }(0)|=1$, as follows from density and the MRA projection Fourier identity.

Nesting in an orthonormal multiresolution analysis expresses a scaling function in the finer-scale orthonormal basis. The associated low-pass symbol is $m(\xi )=2^{-1/2}{\sum }_k{h}_k{e}^{-ik\xi }$, and $\hat{\phi }(2\xi )=m(\xi )\hat{\phi }(\xi )$. If $\phi$ has compact support, only finitely many coefficients $h_k$ are nonzero. With $\hat{\phi }(0)=1$, iteration gives $\hat{\phi }(\xi )={\prod }_{j\ge 1}m(\xi /2^j)$ when the limit exists.

Let $(w_m)$ be the orthonormal basis of Dirichlet eigenfunctions for a strictly positive Dirichlet realization of an elliptic operator, with eigenvalues ${\lambda }_m$. ThenFor integers $k$, these domains are the Sobolev domains of powers of an elliptic Dirichlet operator. At $k=0$ the condition is just Parseval identity and imposes no boundary condition. At $k=1,2$ it describes $H_0^1$ and $H^2\cap H_0^1$, respectively. Higher $k$ impose traces of powers of $L$, not only the trace of $u$. For $L=-d^2/d{x}^2$ on $(0,\pi )$, the smooth Dirichlet function $u=x(\pi -x)$ has sine coefficients proportional to $m^{-3}$ for odd $m$, so the weighted sum diverges at $k=3$ despite its ordinary $H^3$ regularity.

Every real symmetric matrix has an orthonormal basis of real eigenvectors. Equivalently, it admits an orthogonal diagonalization $A=O^{T}DO$ with $D$ real diagonal and $O$ an orthogonal matrix.

If two pure quantum states overlap by $(1/\sqrt{2}{)}^{1/n}$, their $n$-copy states have overlap $1/\sqrt{2}$. Each block spans an effective qubit, with orthonormal basis $e_0={\Psi }_1$ and $e_1=\sqrt{2}{\Psi }_2-{\Psi }_1$. Applying the two-block PBR exclusion measurement for zero and plus uses $2n$ independently prepared copies. A common single-copy ontic overlap of positive mass remains positive under this finite tensor power.

For a positive quadratic bending operator $K$ with a real orthonormal basis of eigenfunctions, the equipartition theorem gives $C(x,y)=k_{B}T{\sum }_n{W}_n(x)W_n(y)/{\mu }_n$, where $K{W}_n={\mu }_n{W}_n$. This is the inverse-operator Green function multiplied by Boltzmann constant and temperature. Unconstrained zero-energy filament modes prevent a normalizable canonical ensemble, and negative modes signal an unstable quadratic model.

For both ends clamped and $K=A{\partial }_x^4$, the diagonal is $C(x,x)=k_{B}T{x}^3(L-x{)}^3/(3A{L}^3)$. This differs from a clamped-free tip variance. The center variance is $k_{B}T{L}^3/(192A)$.

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.

Wavefunction normalization makes the total Born rule probability equal to one. In an orthonormal basis of energy eigenstates, $\psi ={\sum }_n{c}_n{\varphi }_n$ is normalized precisely when ${\sum }_n|c_n{|}^2=1$.