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The expectation value of a quantum observable is the expected value of its measurement outcome random variable in a specified quantum state. It predicts the mean of outcomes from many identical quantum state preparations. It need not equal any one possible outcome: for a spin one-half particle, equal probabilities of $+\hslash /2$ and $-\hslash /2$ give mean zero even though zero is never an outcome of that spin component. Here $\hslash$ is the reduced Planck constant.

Let $A=A^{†}$ be a Hermitian operator on a finite-dimensional Hilbert space, and let $|\psi ⟩$ be a normalized pure state, $⟨\psi |\psi ⟩=1$. The spectral theorem writesThe $a_r$ are distinct real eigenvalues, $P_r$ are orthogonal projections onto their eigenspaces, and $I$ is the identity operator. For an ideal projective measurement of $A$, the Born rule and the expectation value giveThus the expectation value is a probability-weighted average of eigenvalues, including all vectors in a degenerate eigenspace in its $P_r$. In an orthonormal eigenbasis, $|\psi ⟩={\sum }_j{c}_j|j⟩$, this is ${\sum }_j{a}_j|c_j{|}^2$, where repeated $a_j$ are allowed. The weighted average lies between the smallest and largest eigenvalues; it can lie in a gap between them. The statistical meaning and operator expression are connected in this way in the authored treatment of average energies.^[ref]

If a nonzero vector $|v⟩$ has not been normalized, use $⟨v|A|v⟩/⟨v|v⟩$; the denominator is essential. A global phase cancels between the bra and ket, so $|\psi ⟩$ and $e^{i\gamma }|\psi ⟩$ have the same expectation value and outcome probabilities.

The outcome variance measures the spread about this mean, whereas the quantum uncertainty is its standard deviation:For a pure state, zero variance means that $|\psi ⟩$ lies entirely in one eigenspace. That eigenspace need not be one-dimensional. Knowing only a mean generally does not determine the entire outcome probability distribution.

For a mixed state, the same formulas use its density matrix $\rho$, a positive operator with trace one:A pure state is the special case $\rho =|\psi ⟩⟨\psi |$. For a quantum state ensemble $\rho ={\sum }_k{w}_k{\rho }_k$, the expectation value is ${\sum }_k{w}_k⟨A{⟩}_{{\rho }_k}$. This linearity is useful when preparations vary between trials; different quantum state ensembles can have the same density matrix.

These formulas require no operator-domain qualifications in finite dimension. For an unbounded self-adjoint operator, the spectral theorem instead gives a spectral probability measure; a finite mean requires an integrable absolute first moment, and finite variance requires a finite second moment. Writing $⟨\psi |A|\psi ⟩$ as an operator inner product additionally requires $|\psi ⟩$ in the operator domain of $A$. The quadratic form of a positive quantum Hamiltonian explains the useful wider form-domain version.

For a finite-state homogeneous continuous-time Markov chain with generator matrix $Q$, integrate the indicator of occupying state $j$ over $[0,T]$. Interchanging the bounded time integral and expected value gives the displayed integral of transition probabilities. It counts all visits, not only the first holding time. Summing over states gives $T$. A zero eigenvalue of $Q$ does not cause a problem for this integral, although it prevents casually writing it with an inverse matrix $Q^{-1}$.

For $x^{\prime }-Ax=e^{\lambda t}v$, substituting an exponential particular solution reduces the differential equation to $(\lambda I-A)u=v$. The displayed solution exists whenever $\lambda$ is not an eigenvalue of $A$. Resonance can require polynomial factors in time. The general solution adds $e^{tA}c$, expressed using the matrix exponential.

A real matrix with at least one simple negative eigenvalue cannot be a matrix exponential of a real matrix: a logarithm would commute with it and preserve each one-dimensional real negative-eigenvalue space, where exponentiation can only produce a positive eigenvalue. In particular $\mathrm{diag}(-2,-1/2)$ belongs to the connected special linear group ${\mathrm{SL}}_2(\mathbb{R})$ but has no real logarithm.

An extraordinary subdivision vertex has a valence different from that of the regular refinement lattice. Regular box spline calculations do not alone establish smoothness at this vertex. The finite local subdivision matrix, its eigenvalues, and the associated characteristic map of a subdivision surface determine convergence and tangent behavior.

The density operators on a finite-dimensional Hilbert space form a convex set. Its extreme points are exactly the rank-one projectors, hence exactly the pure states. To prove this, suppose $|\psi ⟩⟨\psi |={\sum }_i{a}_i{\rho }_i$ with $a_i>0$ and ${\rho }_i$ positive semidefinite. For $w\perp \psi$, $0={\sum }_i{a}_i⟨w|{\rho }_i|w⟩$ is a sum of nonnegative numbers, so every term vanishes. Diagonalizing ${\rho }_i$ shows that $⟨w|{\rho }_i|w⟩=0$ implies ${\rho }_{i}w=0$. Thus each ${\rho }_i$ is supported on the line spanned by $\psi$, and its unit trace forces ${\rho }_i=|\psi ⟩⟨\psi |$.

Conversely, a density operator of rank at least two has a spectral decomposition with at least two positive eigenvalues. If $\lambda$ is one of them, $0<\lambda <1$ and $\rho =\lambda |e⟩⟨e|+(1-\lambda )\sigma$, where $\sigma$ is the normalized remainder. These are distinct density operators, proving that $\rho$ is not an extreme point.

In a field of characteristic $p$, take a basis $e_0,\dots ,e_{p-1}$, and let $x{e}_j=e_{j-1}$ with indices modulo $p$, while $y{e}_j=j{e}_j$. Then $[x,y]=x$, so $kx+ky$ is a solvable Lie algebra, but the distinct eigenvalues of $y$ force any common eigenvector to be a coordinate vector, and $x$ cyclically permutes those vectors. There is consequently no common eigenvector, even over an algebraically closed field.

For an integer $0\le k\le n$, the fantope is the convex hull of rank-$k$ orthogonal projection matrices in ${\mathbb{R}}^n$. Equivalently, it consists of real symmetric matrices with eigenvalues in $[0,1]$ and matrix trace $k$. To prove the equivalence, apply the spectral theorem for real symmetric matrices and express the eigenvalue vector as a convex combination of the zero-one extreme points of the capped simplex. The sum of the largest eigenvalues is its support function.

Let $D$ be a finite-dimensional division algebra over an algebraically closed field $k$. Left multiplication by $a\in D$ has an eigenvalue $\lambda \in k$ and a nonzero eigenvector $v$. Thus $(a-\lambda )v=0$; multiplying by $v^{-1}$ gives $a=\lambda$. Every element is scalar, so $D=k$.

If encoding and decoding quantum channels factor through a $k$-dimensional space, their composite Kraus operators all have matrix rank at most $k$. The rank bound for a weighted operator trace and the Kraus formula for entanglement fidelity implyThus high operation fidelity requires the leading $k$ eigenvalues of the source density operator to carry almost all its probability. The bound applies to arbitrary completely positive trace-preserving encoders and decoders.

Every self-adjoint operator on a finite-dimensional real or complex inner product space has an orthonormal eigenbasis, and all its eigenvalues are real. In orthonormal coordinates, a real symmetric matrix has the form $A=Q\Lambda Q^T$, with $Q$ an orthogonal matrix and $\Lambda$ a real diagonal matrix. A complex Hermitian matrix instead has $A=U\Lambda U^{\ast }$, with $U$ a unitary matrix and $U^{\ast }$ its conjugate transpose. These are the self-adjoint cases of the spectral theorem; Axler gives the real and complex finite-dimensional statements in Section 7B.^[ref]

An orthonormal eigenbasis turns a seemingly coupled linear map into independent scalar multiplications while preserving lengths and angles. If $x={\sum }_j{c}_j{e}_j$ and $A{e}_j={\lambda }_j{e}_j$, thenThe last identity removes every cross term from the quadratic form. The eigenvalues tell us whether it is positive, negative, or mixed in sign, and the eigenvectors identify the corresponding directions. This connects positive-definite matrices, quadratic forms, and the Hermitian Rayleigh quotient range.

For a fully computed example, takeDirect multiplication gives $A{e}_{+}=3e_{+}$ and $A{e}_{-}=e_{-}$. Their dot product is zero and each has norm one. ThusWriting $x=c{e}_{+}+d{e}_{-}$ changes $x^{T}Ax=2x_1^2+2x_1{x}_2+2x_2^2$ into $3c^2+d^2$. Its level set $x^{T}Ax=1$ is therefore an ellipse with semiaxis lengths $1/\sqrt{3}$ and $1$, along $e_{+}$ and $e_{-}$. The coordinates change, but the ellipse does not: orthogonal diagonalization reveals its existing geometry.

For distinct eigenvalues, orthogonality follows at once from self-adjointness: $\lambda ⟨u,v⟩=⟨Au,v⟩=⟨u,Av⟩=\mu ⟨u,v⟩$. Within a repeated eigenspace, choose an orthonormal basis; arbitrarily chosen eigenvectors of the same eigenvalue need not be orthogonal. One outline of the full argument finds an eigenvector, observes that its orthogonal complement is invariant, and repeats in that smaller space. Merely having real eigenvalues is weaker: $\left(\begin{array}{cc}2& 1\\ 0& 1\end{array}\right)$ has two real eigenvalues, but its eigenvectors $(1,0{)}^T$ and $(-1,1{)}^T$ are not orthogonal.

Here $\alpha =e^{2\pi i/3}$. The missing-trace matrix has determinant $-\sqrt{3}k\Delta (k)$. Its nonzero zeros are the eigenvalues in the parametrization $Av=i{k}^{3}v$ of $A=-d^3/d{x}^3$ with $v(0)=v(L)=v^{\prime }(L)=0$. Integration by parts gives $\mathrm{Re}⟨v,Av⟩=-|v^{\prime }(0){|}^2/2$. No nonzero zero lies in $\mathrm{Re}(-i{k}^3)\le 0$: on its boundary the determinant equation reduces, after rotation, to an equality between a unit-modulus exponential and $\cosh u-i\sqrt{3}\sinh u$, whose modulus exceeds one unless $u=0$. The zero at the origin is removable in the original boundary problem. There are nevertheless infinitely many genuine negative eigenvalues of $A$, obtained from $e^{-3s/2}=2\cos (\sqrt{3}s/2+\pi /3)$, $s>0$, with $k=-is/L$. Absence of contour poles in the Fokas method does not imply absence of discrete spectrum of the operator.