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Assume the given wavefunction is zero outside $[0,L]$. Wavefunction normalization requiresAn arbitrary overall quantum phase is unobservable, so choose the positive real representative $N=\sqrt{3}/L^{3/2}$.

The linear ramp state in an infinite square well has a boundary subtlety: the limiting value at $L$ is $\sqrt{3/L}$, rather than zero. The function is still a normalized element of the well's Hilbert space $L^2(0,L)$, so position measurement and energy-projection probabilities are meaningful. It is not in the domain of the Dirichlet Hamiltonian, nor is it a finite-energy box state. Altering the value at the single endpoint does not cure its nonzero boundary trace; a physical finite-energy preparation must smooth the function to zero near the wall. The following probabilities refer to the supplied idealized state.

For a fixed propagation direction and spatial mode, a photon has two independent transverse polarization amplitudes. Orthogonal horizontal and vertical states span its two-dimensional Hilbert space: $\alpha |H⟩+\beta |V⟩$, with $|\alpha {|}^2+|\beta {|}^2=1$. Polarization analyzers measure bases in this space, and phase plates implement transformations of its relative amplitudes. Additional spatial or frequency modes enlarge the space and must not be silently discarded.

Take an orthonormal basis $\{|j⟩\}$ and insert its completeness relation:No normalization or orthogonality of $u,v$ is required. This is the trace of a rank-one operator; in a Hilbert space the sum converges by the Cauchy-Schwarz inequality.

An ensemble with normalized quantum states $|{\psi }_a⟩$ and probabilities $p_a\ge 0$, ${\sum }_a{p}_a=1$, has density matrixEach rank-one projector is Hermitian, so ${\rho }^{†}=\rho$. Taking the trace in any orthonormal basis givesIt is also positive: $⟨v|\rho |v⟩={\sum }_a{p}_a|⟨{\psi }_a|v⟩{|}^2\ge 0$. Its eigenvalues ${\lambda }_a$ are therefore nonnegative and sum to one. Hencewith equality exactly when one eigenvalue is one and the others are zero. Thus the purity of a density operator distinguishesEquivalently, a pure state has ${\rho }^2=\rho$ and rank one. In a $d$-dimensional Hilbert space, purity is at least $1/d$, attained by the maximally mixed state $I/d$.

The Holstein–Primakoff transformation replaces each finite-dimensional spin algebra by a bosonic creation operator and annihilation operator, while restricting the local occupation number to $0\le n_j\le 2S$. The number $n_j=a_j^{†}{a}_j$ counts deviations from the fully polarized reference quantum state. Its zero-boson Fock vacuum has ${\tilde{S}}^z=S$.

The representation is exact on this physical subspace: the ordered square root gives ${\tilde{S}}^{+}|n⟩=\sqrt{n(2S-n+1)}|n-1⟩$ and ${\tilde{S}}^{-}|n⟩=\sqrt{(n+1)(2S-n)}|n+1⟩$, reproducing the spin commutation relations. For $n_j\ll S$,so the leading Hamiltonian is quadratic in bosons and describes spin waves.

The square root generates boson interactions beyond leading order and its operator ordering matters. Truncation also enlarges the Hilbert space to unphysical occupations unless the low-density approximation remains self-consistent. Large fluctuations, particularly near a gap closing or in low dimensions, can invalidate the linear spin-wave approximation even when its algebraic manipulation is straightforward.

The Holstein–Primakoff transformation maps the finite spin basis $|S,m⟩$ to bosonic Fock states $|n⟩$ with $n=S-m$. Thus $n=0$ is the fully polarized reference state, and creating a boson lowers $S^z$ by one. The physical subspace has the Holstein–Primakoff occupation constraint $0\le n\le 2S$, not the whole infinite oscillator Hilbert space.

The specified ordering of the square root givesThese are the exact spin ladder operator matrix elements; in particular, the lowering amplitude vanishes at $n=2S$. The underlying oscillator obeys its usual canonical commutation relation on the enlarged space, while the spin operators preserve the constrained physical subspace. Projecting the oscillator operators themselves to a finite space would change their commutator at its upper endpoint.

Expanding $\sqrt{2S-n}=\sqrt{2S}(1-n/(4S)+\cdots )$ converts an ordered spin problem to approximately free magnons, with interactions in higher orders of $1/S$. Two complications are the nonpolynomial, order-sensitive square roots, which generate interacting bosons, and loss of the occupation bound when a truncated expansion uses the full bosonic Fock space. Large depletion, high temperature, or soft modes near a phase transition can invalidate the assumption $⟨n⟩\ll S$ even when $S$ is large.

A nondegenerate quantum code is perfect at correction radius $t$ if its distinct correctable Pauli error images exhaust the physical Hilbert space. Each image has dimension $2^k$, and their orthogonal direct sum is the whole $2^n$-dimensional space precisely when the quantum Hamming bound is an equality. This perfectness concerns unknown-location errors of weight at most $t$; it does not imply correction of every higher-weight error.

A basis of apparatus records robust under the relevant environmental interaction. Its meaning is dynamical and depends on the coupling and the time scale, rather than on an arbitrary coordinate choice for the apparatus Hilbert space.

In three dimensions let $f\in L^{6/5}({\mathbb{R}}^3)$ have compact support. Let $v_R\in H_0^1(B_R)$ solve $-\Delta v_R=f$ and extend it by zero. There is a unique $v\in {\dot{H}}^1({\mathbb{R}}^3)$ with $\int \nabla v\cdot \nabla \varphi =\int f\varphi$ for all $\varphi \in {\dot{H}}^1$. Indeed Holder inequality and the homogeneous Sobolev inequality make the right side a continuous linear functional in the gradient norm, and the Riesz representation theorem applies to this Hilbert space.

The weak formulation says that $v_R$ is the orthogonal projection of $v$ onto the closed subspace consisting of zero extensions of $H_0^1(B_R)$. These subspaces increase with $R$ and their union contains every test function, which is dense by the definition of the homogeneous Sobolev space. The best-approximation property of orthogonal projection therefore gives $\Vert \nabla (v_R-v){\Vert }_2\to 0$. The homogeneous Sobolev inequality also gives $\Vert v_R-v{\Vert }_6\to 0$. This convergence does not generally extend to the inhomogeneous Sobolev space $H^1$, as the spherical mean of a Poisson solution outside its source shows for a source of nonzero total mass.

An orthogonal projection $P$ splits a one-particle Hilbert space into empty and filled modes. Take completed exterior Fock spaces of empty particles and of holes in the filled modes, with the graded tensor product enforcing the canonical anticommutation relations between both factors. The vacuum is killed by particle annihilation in $PH$ and by particle creation in $(I-P)H$. Only finite particle-hole excitations are needed for a dense basis. Normal ordering subtracts the filled-vacuum expectation from bilinear observables.

A symmetric operator $L$ on a real Hilbert space is strictly positive definite when $⟨Lv,v⟩>0$ for every nonzero $v$ in its operator domain. Strict positivity proves uniqueness of a solution of $Lu=f$, but in infinite dimension it need not prove existence for every $f$. A uniformly positive definite symmetric operator additionally has a positive lower bound independent of $v$. Symmetry must be stated separately when positivity is defined only by a real quadratic expression: adding a skew-symmetric matrix changes the operator without changing that expression.

A bounded Hermitian operator $T$ on a Hilbert space is positive, also called a positive semidefinite operator, whenThe value must be real as well as nonnegative. This permits zero values on nonzero vectors: the zero linear operator is positive. For a bounded Hermitian operator, positivity is equivalent to its spectrum being contained in $[0,\infty )$; in finite dimension, this is the condition that all eigenvalues are nonnegative.^[ref]

The Hermitian operator diagram separates three cases. $D=\mathrm{diag}(1,-1)$ is self-adjoint but not positive, because $q_D((0,1{)}^{\mathrm{T}})=-1$. The orthogonal projection $P=\mathrm{diag}(1,0)$ is positive semidefinite, with $q_P(x)=|x_1{|}^2$, but is not strictly positive on every nonzero vector. The complex Hermitian matrix $H=\left(\begin{array}{cc}2& i\\ -i& 2\end{array}\right)$ is positive definite because $q_H(a{e}_1+b{e}_3)=|a{|}^2+3|b{|}^2>0$ whenever $(a,b)\ne (0,0)$. For real symmetric matrices, MIT's lecture 27 relates these tests to the geometry of quadratic forms and their minima.^[ref]

A fundamental source of positive operators is $T^{\ast }T$ for a bounded linear operator $T:H\to K$:For the complex matrix $T=\left(\begin{array}{cc}1& i\\ 2& 1-i\end{array}\right)$ used above,Its eigenvalues are $4-\sqrt{6}$ and $4+\sqrt{6}$, both positive. Positivity organizes the squared error minimized by a least-squares solution: the same $T^{\ast }T$ occurs in the normal equation for a linear inverse problem. The positive square root of an operator gives the converse squared-length description, $q_T(x)=\Vert T^{1/2}x{\Vert }^2$ for every bounded positive operator $T$.

On a real Hilbert space, nonnegativity of $⟨Tx,x⟩$ alone does not imply self-adjointness, so the Hermitian operator hypothesis is part of the definition. On a complex Hilbert space, real-valuedness of all these expressions already forces self-adjointness by the polarization identity.

Strict positivity and a uniform positive lower bound also differ in infinite dimension. On the l2 sequence space, the bounded Hermitian operator $(Tx{)}_n=x_n/n$ satisfiesConsequently no $\gamma >0$ satisfies $q_T(x)\ge \gamma \Vert x{\Vert }^2$ for every $x$. Compare the positive definite symmetric operator and uniformly positive definite symmetric operator conditions.

A positive operator-valued measure is a family of positive operators $E_i$ satisfying ${\sum }_i{E}_i=I$. Each $E_i$ is called an effect and determines a probability $p_i=\mathrm{Tr}(E_i\rho )$ for input density operator $\rho$. The effects need not be orthogonal projections or commute. This permits measurements with more outcomes than the dimension of the Hilbert space, such as the trine qubit POVM. Coupling to a quantum ancilla and measuring the larger system supplies a physical route to generalized measurements. ^[ref].

For an elementary unsharp $Z$ example on a qubit, takeOn the eigenstate $|0⟩$, $p_{\pm }=(1\pm \eta )/2$; at $\eta =1/2$ they are $3/4$ and $1/4$. At $\eta =0$ the outcome is a fair coin independent of the input, and at $\eta =1$ these effects specify a sharp $Z$ projective measurement. Choosing $K_{\pm }=\sqrt{E_{\pm }}$ specifies one quantum instrument, but $U_{\pm }\sqrt{E_{\pm }}$ for outcome-dependent unitary operators has the same effects and different state updates. Thus POVM does not determine the post-measurement state.

A Hermitian form is positive semidefinite when $h(v,v)\ge 0$ for every vector. It need not define an inner product, because a nonzero vector may have zero squared norm. The Cauchy-Schwarz inequality still holds: applying nonnegativity to $v+zw$ and minimizing the quadratic expression in $z$ gives $|h(v,w){|}^2\le h(v,v)h(w,w)$ when $h(w,w)>0$; when $h(w,w)=0$, varying $z$ forces $h(v,w)=0$. Thus its zero-norm vectors are exactly the radical of a Hermitian form. Quotienting this radical of a Hermitian form gives a positive inner product; taking its Hilbert space completion then gives a Hilbert space. This is the final positivity step in the Gupta-Bleuler null-state quotient.

A bounded linear operator $T:L^1(\Omega )\to L^1(\Omega )$ preserves positivity when it takes nonnegative functions to nonnegative functions. By linearity it also preserves the pointwise order: $u\le v$ implies $Tu\le Tv$. For example, integration against a nonnegative kernel preserves positivity, as does the identity operator. This is order positivity, which differs from a positive quadratic form on a Hilbert space.

For an isolated quantum system, pure states are normalized rays in a complex Hilbert space, observables are self-adjoint operators, and their probabilities obey the Born rule. A spectral projector $P_a$ has outcome probability $⟨\psi |P_a|\psi ⟩$; an ideal repeatable projective measurement has conditional state $P_a\psi /\sqrt{⟨\psi |P_a|\psi ⟩}$. The Schrödinger equation specifies continuous unitary evolution between such ideal measurements, given a Hamiltonian operator. General measurements need not have the projective state-update rule.

Composite systems use the tensor product of their state spaces. Identical particles additionally obey the bosonic or fermionic exchange symmetry appropriate to their species. The photoelectric effect, Davisson–Germer experiment and double-slit experiment motivate energy quanta, matter-wave phases and probability interference, respectively; these experiments constrain the framework but do not separately prove every abstract postulate or specify every system's Hamiltonian.

If every allowed input density operator has support of a positive operator inside a subspace with orthogonal projector $P$, completeness need only equal $P$ on that subspace. To define a measurement on the full Hilbert space, one can append $M_{\perp }=I-P$ when the original operators act only on the subspace. The added outcome has zero probability on supported inputs. The effects $M_r^{†}{M}_r$ form a POVM; the operators themselves additionally specify state updates.

For a real Hilbert space, $E(x)=\alpha J(cx-y)+⟨x,z⟩$, $\alpha >0$ and $c\ne 0$, identify $z$ with its vector by the Riesz representation theorem. Completing the square in $\Vert x-q{\Vert }^2/2+⟨x,z⟩$ shifts the input to $q-z$. Substitute $v=cx-y$ and multiply by $c^2$ to obtain the displayed formula, valid also for negative $c$. If $c=0$ and $J(-y)<\infty$, the answer is $q-z$; if $J(-y)=\infty$, the objective is identically infinite and no proper proximal problem remains.

The Euclidean norm distance between normalized pure state vector representatives in a common Hilbert space. Its square is $2-2\mathrm{Re}⟨\psi |\varphi ⟩$. If vectors differing by a global phase are identified, minimize over that phase to obtain $\sqrt{2-2|⟨\psi |\varphi ⟩|}$. This is a different distance convention from trace distance or one minus squared quantum fidelity; stating the convention is necessary when defining $\epsilon$-closeness.

Let $V$ be a closed linear subspace of a real Hilbert space, and let $Q(v)=a(v,v)$ for a symmetric bounded bilinear form that is nonnegative on the ambient space and coercive on $V$. Then $Q$ has a unique minimizer on every closed affine subspace $g+V$. If $Q(v_j)\to m={\inf }_{g+V}Q$, the midpoint belongs to $g+V$ and the parallelogram law gives $Q(v_j-v_k)\le 2Q(v_j)+2Q(v_k)-4m\to 0$. Coercivity on $V$ makes the minimizing sequence Cauchy; completeness and continuity give a minimizer. The vanishing first variation is $a(u,v)=0$ for all $v\in V$. Two minimizers differ by $w\in V$ with $Q(w)=0$, so coercivity gives uniqueness. This proves the symmetric existence theorem directly, without invoking the Lax-Milgram theorem.