Spectral Theory Cheat Sheet

The spectral theorem replaces an observable by a projection-valued measure, but its formulas have different domain requirements. Probabilities exist for every normal state; an unbounded mean or variance may not. This sheet collects the distinctions for self-adjoint operators and links the complete arguments. It does not extend the same formulas to arbitrary nonnormal operators.

Self-adjoint spectral formulas with their conditions

Let $A=A^\dagger$ on a Hilbert space, let $E_A$ be its PVM, and set $\mu_\psi(B)=\langle\psi,E_A(B)\psi\rangle$. For a density operator $\rho$, write $\mu_\rho(B)=\operatorname{Tr}[\rho E_A(B)]$.

Object	Formula or test	Conditions and explanation
Spectral reconstruction	$A=\int\lambda\,E_A(d\lambda)$	Unbounded integral with its domain; unbounded spectral theorem
Function domain	$D(f(A))=\{\psi:\int\lvert f\rvert^2\,d\mu_\psi<\infty\}$	Borel $f$, finite almost everywhere for the spectral measure; functional calculus
Outcome probability	$\Pr(A\in B)=\operatorname{Tr}[\rho E_A(B)]$	Borel $B$, every density operator; Born rule
Eigenvalue test	$E_A(\{\lambda\})\ne0$	Its range is $\ker(A-\lambda)$; spectral projections
Resolvent	$R_A(z)=(A-zI)^{-1}$	$z\notin\sigma(A)$; resolvent operators
Resolvent norm	$\lVert R_A(z)\rVert=1/\operatorname{dist}(z,\sigma(A))$	Self-adjoint $A$; the same distance equality is not universal for nonnormal operators
Unitary evolution	$U_t=\int e^{-it\lambda/\hbar}E_A(d\lambda)$	Defined on every vector; the derivative requires $D(A)$; Stone’s theorem

For a bounded Borel $f$, the norm of $f(A)$ is its essential supremum relative to the PVM, not necessarily its pointwise supremum over all real numbers. Changing $f$ on an $E_A$-null set does not change the operator. For continuous $f$ on the spectrum, the familiar spectral supremum formula applies.

Means, variances, and spectrum types are separate data

A finite probabilistic mean requires $\int|\lambda|\,d\mu_\rho<\infty$. A finite second moment requires $\int\lambda^2\,d\mu_\rho<\infty$. For a vector state, the second condition is exactly $\psi\in D(A)$, while the first is $\psi\in D(|A|^{1/2})$. The form-defined mean can therefore exist when the vector expression $\langle\psi,A\psi\rangle$ is not defined. See Entropy and Trace for a related integrability warning and Expectation Functionals for observable expectations.

The point/continuous/residual classification concerns the invertibility and range of $A-\lambda I$. For self-adjoint $A$, residual spectrum is absent. The pure-point/absolutely-continuous/singular-continuous decomposition instead concerns spectral measures and reducing subspaces. These are related classifications, not identical uses of the word “continuous.” For the two classifications, see Point, Continuous, and Residual Spectrum and Continuous Spectrum Without Eigenvectors.

Worked example: a continuum plus an isolated level

On $\mathcal H=L^2([-1,1],dx)\oplus\mathbb C$, define

$$A(f,c)=(xf(x),3c).$$

It is bounded self-adjoint, with spectrum $[-1,1]\cup\{3\}$. The isolated level $3$ is an eigenvalue. Multiplication by $x$ has no eigenvectors in the first summand, since a function supported on a single point is zero in $L^2$.

Choose the normalized state $\psi=(\sqrt{p/2},\sqrt{1-p})$ with $0\le p\le1$, where the first entry is a constant function. Its scalar spectral measure is

$$d\mu_\psi(\lambda) =\frac p2\,\mathbf1_{[-1,1]}(\lambda)\,d\lambda +(1-p)\delta_3(d\lambda).$$

Thus

$$\Pr(A\in[0,4])=1-\frac p2,\qquad \langle A\rangle=3(1-p),\qquad \langle A^2\rangle=\frac p3+9(1-p).$$

The operator spectrum is independent of $p$, but the observed distribution can suppress an entire spectral sector. A state’s measured support need not reveal every part of the observable’s spectrum.

Exercises

Resolvent at an imaginary argument. For the worked operator and $z=i\eta$, $\eta>0$, compute the resolvent norm and its action.

Solution

$R_A(i\eta)(f,c)=((x-i\eta)^{-1}f,(3-i\eta)^{-1}c)$. The distance from $i\eta$ to $[-1,1]$ is $\eta$, while its distance to $3$ is $\sqrt{9+\eta^2}$. Therefore $\|R_A(i\eta)\|=1/\eta$. The continuous sector controls the norm even when the chosen state has $p=0$ and only occupies the isolated level.

An exact outcome and a narrow bin. In the same model with $p=1$, compare $\Pr(A=0)$ with $\Pr(|A|\le\varepsilon)$ for $0<\varepsilon<1$.

Solution

The exact point has probability zero because the measure has no atom there. The bin has probability $\varepsilon$, obtained from a uniform density $1/2$ over an interval of length $2\varepsilon$. This is the distinction used by continuous-variable measurement models: finite detector resolution concerns bins, not a normalized eigenvector at an exact continuous outcome.

References

• Hall, Brian C. Quantum Theory for Mathematicians. Graduate Texts in Mathematics 267, Springer, 2013. doi:10.1007/978-1-4614-7116-5. Spectral measures and observable probabilities.
• Teschl, Gerald. Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed. Graduate Studies in Mathematics 157, American Mathematical Society, 2014. Author’s edition. Spectral calculus, resolvents, and spectral types.