Modular Spectrum and Spectral Measures

Finite density matrices turn modular spectra into ratios of probabilities. A continuum algebra need not retain a countable list of levels: the useful object may instead be a projection-valued measure, a vector spectral measure, or a state-independent factor invariant. These objects must be kept distinct, and continuity or type-III$_1$ behavior must be proved from a limiting construction or a structural theorem rather than inferred from a dense finite plot.

Required background. Modular Hamiltonian definitions and domains supply the support compression, logarithmic functional calculus, and unbounded-operator domains used here.

Helpful background. Tomita–Takesaki flow supplies the standard pair, modular conjugation, and invariant vector used to derive reciprocal spectral symmetry.

The chapter’s route from standard pairs to geometric flow identifies which spectra are intrinsic. The representative-flow table separates exact continuum objects from regulated representatives, and the claim-licensing checklist tests any proposed continuum inference.

Spectral calculus for a standard pair

Let $(\mathcal A,\Omega)$ be standard and let $\Delta$ be its modular operator. Then $\Delta$ is positive, self-adjoint, and injective. Injectivity means $E_\Delta(\{0\})=0$; it does not prevent $0$ from belonging to the continuous spectrum. The spectral theorem gives

$$\Delta=\int_{[0,\infty)}\lambda\,E_\Delta(d\lambda), \qquad K=-\log\Delta =\int_{\mathbb R}k\,E_K(dk),$$

where $k=-\log\lambda$ on $\lambda>0$. For a vector $\xi$,

$$\mu_\xi^K(B)=\langle\xi,E_K(B)\xi\rangle$$

is a finite positive measure, and it is a probability measure only when $\lVert\xi\rVert=1$. The exact domain test for a Borel function is

$$\xi\in\mathcal D(f(K)) \quad\Longleftrightarrow\quad \int_{\mathbb R}\lvert f(k)\rvert^2d\mu_\xi^K(k)<\infty.$$

Cross measures

$$\mu_{\xi,\eta}^K(B)=\langle\xi,E_K(B)\eta\rangle$$

are generally complex. They encode modular correlators, but they are not probability distributions. The invariant vector satisfies $K\Omega=0$; this gives one zero eigenvector without making zero isolated or supplying an eigenbasis for the rest of the Hilbert space.

If a normal state is not faithful, first compress to its support projection $p=s(\omega)$ and work with $p\mathcal A p$. In a type-I model this means discarding the zero eigenspace before taking $\log\rho$. Writing an infinite “entanglement energy” for every vector in the discarded kernel is regulator notation, not the spectral calculus of a standard faithful pair.

Reciprocal symmetry and finite ratio spectra

Tomita–Takesaki theory gives

$$J\Delta J=\Delta^{-1}, \qquad JKJ=-K.$$

Hence

$$0<\lambda\in\operatorname{spec}(\Delta) \Longleftrightarrow \lambda^{-1}\in\operatorname{spec}(\Delta), \qquad k\in\operatorname{spec}(K) \Longleftrightarrow -k\in\operatorname{spec}(K),$$

and, for every Borel set $B\subset\mathbb R$,

$$J E_K(B)J=E_K(-B), \qquad \mu_{J\xi}^K(B)=\mu_\xi^K(-B).$$

The measure of a generic $\xi$ need not be even: the reflected measure belongs to $J\xi$.

For a faithful finite density matrix $\rho=\sum_i p_i\lvert i\rangle\langle i\rvert$, use the Hilbert–Schmidt standard representation. Matrix units obey

$$\Delta(\lvert i\rangle\langle j\rvert) =\frac{p_i}{p_j}\lvert i\rangle\langle j\rvert, \qquad K(\lvert i\rangle\langle j\rvert) =-\log\!\frac{p_i}{p_j}\lvert i\rangle\langle j\rvert.$$

The standard modular spectrum is therefore a ratio set. The one-sided entanglement levels $-\log p_i$ are different data; only their pairwise differences occur as eigenvalues of $K$. Multiplicity in the standard representation also differs from multiplicity in the one-sided density matrix.

One state’s spectrum versus the Connes invariant

The Connes $S$-invariant of a von Neumann algebra is not the spectrum of one chosen modular operator. In its weight formulation,

$$S(\mathcal M) =\bigcap_{\varphi} \operatorname{spec}(\Delta_\varphi),$$

where the intersection is over faithful normal semifinite weights in the standard construction. Connes gives this definition in Connes 1973, Definition 3.1.1, printed p. 188. For type-III factors, the resulting classification is

factor	$S(\mathcal M)$
$\mathrm{III}_0$	$\{0,1\}$
$\mathrm{III}_\lambda$, $0<\lambda<1$	$\{0\}\cup\{\lambda^n:n\in\mathbb Z\}$
$\mathrm{III}_1$	$[0,\infty)$

These cases appear at the opening of Connes 1973, Part IV, printed p. 214. The zero in $S(\mathcal M)$ is a spectral boundary point; injective modular operators still have no zero eigenvector.

This distinction also limits QFT claims. Fredenhagen proves full positive modular spectra, and hence type-III$_1$ central components, for local algebras under an asymptotic scale-invariance hypothesis Fredenhagen 1985, abstract and pp. 79–89. That result does not license the sentence “every continuum local algebra is type III$_1$” without the model’s locality, standardness, scaling, and factoriality hypotheses. Nor does type III$_1$ imply that every chosen vector measure is absolutely continuous.

A finite Gaussian sequence for a free wedge

The free wedge supplies a concrete continuous target. In rapidity variables, longitudinal boosts act by translations, so the one-particle boost generator has continuous spectrum on $\mathbb R$. The Bisognano–Wichmann theorem fixes the wedge modular generator to $2\pi$ times that boost generator, with orientation set by convention Bisognano and Wichmann 1976, §III, pp. 306–310.

Consider a free-fermion spectral Galerkin model. Choose a cutoff $L$, spacing $\delta$, and midpoint grid

$$k_j=-L+\left(j+\frac12\right)\delta, \qquad j=0,\ldots,N-1, \qquad N=\frac{2L}{\delta}.$$

Define the one-particle Gaussian modular matrix and correlation matrix by

$$h_{\delta,L}=\operatorname{diag}(k_j), \qquad C_{\delta,L}=\left(I+e^{h_{\delta,L}}\right)^{-1}.$$

Thus every eigenvalue of $C$ lies strictly between zero and one and

$$h_{\delta,L} =\log\!\left[(I-C_{\delta,L})C_{\delta,L}^{-1}\right].$$

This is the number-conserving Gaussian reconstruction derived in Peschel 2003, Eqs. (5)–(12), printed pp. L205–L206; a transpose appears in Peschel’s site-index convention and is immaterial in this diagonal basis.

Let $H_{\delta,L}=\sum_j k_jc_j^\dagger c_j$ and $\rho_{\delta,L}=Z^{-1}e^{-H_{\delta,L}}$. In Hilbert–Schmidt standard form,

$$K_{\delta,L}(X)=H_{\delta,L}X-XH_{\delta,L}.$$

If $\lvert0\rangle$ is the Fock vacuum and $\lvert j\rangle=c_j^\dagger\lvert0\rangle$, then $X_j=\lvert j\rangle\langle0\rvert$ has modular eigenvalue $k_j$. Choose

$$\xi_{\delta,L}=\sum_j\sqrt{w_j}\,X_j, \qquad w_j= \frac{\delta e^{-k_j^2/2}} {\sum_\ell\delta e^{-k_\ell^2/2}}.$$

Its exact vector spectral measure and characteristic function are

$$\mu_{\delta,L}^K=\sum_jw_j\delta_{k_j}, \qquad \chi_{\delta,L}(s)=\sum_jw_je^{-isk_j}.$$

In the joint limit $\delta\downarrow0$, $L\uparrow\infty$, these Riemann sums converge weakly to the selected continuum boost-frequency wavepacket

$$d\mu(k)=\frac{e^{-k^2/2}}{\sqrt{2\pi}}\,dk, \qquad \chi(s)=e^{-s^2/2}.$$

This is a vector-measure convergence test for a one-particle free-wedge sector. It is not a proof that a particular spatial lattice recovers the full wedge algebra, its spectral multiplicity, or its Connes invariant.

For reproduction, use midpoint sums in double precision and the sample set $\mathcal S=\{0.5,1,2\}$. Separate grid and full-target errors:

$$\begin{aligned} E_{\mathrm{grid}}&= \max_{s\in\mathcal S} \lvert\chi_{\delta,L}(s)-\chi_{\delta/2,L}(s)\rvert,\\ E_{\mathrm{full}}&= \max_{s\in\mathcal S} \lvert\chi_{\delta,L}(s)-e^{-s^2/2}\rvert. \end{aligned}$$

$L$	$\delta$	$N$	$E_{\mathrm{grid}}$	$E_{\mathrm{full}}$
$4$	$0.500$	$16$	$1.05\times10^{-5}$	$7.61\times10^{-5}$
$4$	$0.250$	$32$	$2.92\times10^{-6}$	$8.51\times10^{-5}$
$4$	$0.125$	$64$	$7.51\times10^{-7}$	$8.75\times10^{-5}$
$2$	$0.125$	$32$	$1.72\times10^{-4}$	$6.09\times10^{-2}$
$5$	$0.125$	$80$	$1.17\times10^{-8}$	$9.78\times10^{-7}$

At fixed $L=4$, refinement reduces the grid error while exposing the nonzero window bias; increasing $L$ then removes that bias. The displayed digits are rounded from the stated sums, so the last quoted digit carries ordinary floating-point and rounding uncertainty.

Adversarial window and basis tests

The $L=2$ row is an intentional failure. Shrinking only the energy window increases $E_{\mathrm{full}}$ from $8.75\times10^{-5}$ to $6.09\times10^{-2}$ at the same $\delta=0.125$. The finite measure now has an artificial band edge at $\lvert k\rvert=2$; it cannot establish a gap or compact support in the continuum target.

Grid placement can manufacture an apparent zero mode. The midpoint grid has smallest reported $\lvert k_j\rvert=\delta/2$, whereas a node-centered grid contains $k=0$. At $L=4$ and $\delta=0.125$, the maximum difference of their characteristic functions on $\mathcal S$ is $1.93\times10^{-5}$, even though one finite point set appears “gapped” and the other does not. Both converge to the same continuous measure. Apparent point degeneracies and gaps that move under this shift are discretization artifacts.

A pure basis change provides a sharper control: it must change nothing if every object is transported. Take

$$h=\operatorname{diag}(-1,1), \qquad \psi=\begin{pmatrix}\sqrt3/2\\1/2\end{pmatrix}, \qquad U=\frac1{\sqrt2} \begin{pmatrix}1&1\\1&-1\end{pmatrix}.$$

The original characteristic function is

$$\chi(s)=\frac34e^{is}+\frac14e^{-is}.$$

Transporting both $h'=UhU^*$ and $\psi'=U\psi$ leaves $\chi$ exactly invariant. If one changes the matrix basis but mistakenly keeps the old coordinate vector $\psi$, the new spectral weights are $(2\pm\sqrt3)/4$. At $s=\pi/2$ the wrong answer is $i\sqrt3/2$ instead of $i/2$, an absolute residual

$$R_{\mathrm{basis}}=\frac{\sqrt3-1}{2}=0.3660254\ldots.$$

For genuinely different truncation subspaces there is no exact invariance; report the projection, window, grid placement, vector transport, and convergence of measures or resolvents. A finite eigenvalue list alone cannot decide continuum spectral type.

Reconstruction from modular correlators

For normalized $\xi$,

$$C_\xi(s)=\langle\xi,e^{-isK}\xi\rangle =\int_{\mathbb R}e^{-isk}d\mu_\xi^K(k)$$

is positive definite, and its Fourier transform reconstructs the measure in the distributional sense. A finite time window multiplies $C_\xi$ and therefore convolves the spectral measure with the transform of that window. Report the time range, sampling interval, window function, nominal frequency resolution, positivity residual, and any KMS detailed-balance residual.

Unitary equivalence preserves spectral data only covariantly:

$$K'=UKU^*, \qquad E_{K'}(B)=UE_K(B)U^*, \qquad \mu_{U\xi}^{K'}=\mu_\xi^K.$$

Changing the state or algebra constructs a different modular operator. Restricting to a subalgebra is not “deleting levels,” and a UV regulator is not generally a unitary change of representation.

Common pitfalls

Equating one-sided levels with standard modular eigenvalues. The levels $-\log p_i$ and the differences $-\log(p_i/p_j)$ answer different questions and have different multiplicities.

Reading continuum spectral type from finite eigenvalues. Every finite matrix has pure point spectrum. Establish a joint cutoff limit of measures, correlators, or resolvents before calling a limit continuous or gapped.

Using $S(\mathcal M)$ as another name for $\operatorname{spec}(\Delta_\Omega)$. The former is an intersection over modular data; the latter depends on one algebra–state pair. State which object was computed.

Exercises

1. Ratio spectrum, multiplicity, and support

Let $\rho=\operatorname{diag}(1/2,1/3,1/6)$. Find the spectra of $\Delta$ and $K$ in Hilbert–Schmidt standard form, including the multiplicity of the eigenvalue $1$ of $\Delta$. Compare with the one-sided entanglement levels. What changes if the last probability is set to zero and the remaining support is renormalized?

Solution

The ratio set is

$$\operatorname{spec}(\Delta) =\left\{1,\frac32,3,\frac23,2,\frac13,\frac12\right\}.$$

Each diagonal matrix unit $\lvert i\rangle\langle i\rvert$ has ratio one, so $1$ has multiplicity three. All six off-diagonal ordered pairs give the six nonunit ratios above, each once. Therefore

$$\operatorname{spec}(K) =\left\{0,-\log\frac32,-\log3, -\log\frac23,-\log2,-\log\frac13,-\log\frac12\right\},$$

with zero of multiplicity three. The one-sided levels are

$$\{\log2,\log3,\log6\},$$

and are not the same set; their ordered differences produce the eigenvalues of $K$. If the last probability is zero, $\rho^{-1}$ and $\log\rho$ are undefined on that vector. After compression and renormalization the nonzero probabilities are $3/5$ and $2/5$; their common normalization cancels from ratios. The faithful support therefore gives $\{1,3/2,2/3\}$, with the unit ratio of multiplicity two. Infinite symbols attached to the discarded direction do not belong to this faithful standard pair.

2. Build the Gaussian vector measure

For the grid construction, prove that $h=\log[(I-C)C^{-1}]$. Show that $X_j=\lvert j\rangle\langle0\rvert$ is an eigenvector of the standard generator, derive $\chi_{\delta,L}$, and evaluate its joint continuum limit.

Solution

Since $C=(I+e^h)^{-1}$,

$$I-C=e^h(I+e^h)^{-1}, \qquad (I-C)C^{-1}=e^h.$$

Functional calculus therefore gives $\log[(I-C)C^{-1}]=h$. The Fock vacuum has energy zero and $\lvert j\rangle$ has energy $k_j$, so

$$K(X_j)=H\lvert j\rangle\langle0\rvert -\lvert j\rangle\langle0\rvert H =k_jX_j.$$

The $X_j$ are orthonormal in Hilbert–Schmidt norm. Consequently $\xi=\sum_j\sqrt{w_j}X_j$ is normalized and

$$\langle\xi,e^{-isK}\xi\rangle =\sum_jw_je^{-isk_j}=\chi_{\delta,L}(s).$$

At fixed $L$, midpoint Riemann sums converge as $\delta\downarrow0$ to the normalized Gaussian restricted to $[-L,L]$. Sending $L\uparrow\infty$ then gives

$$\frac1{\sqrt{2\pi}} \int_{-\infty}^{\infty}e^{-k^2/2}e^{-isk}dk =e^{-s^2/2}.$$

Pointwise convergence of these characteristic functions, with continuity at $s=0$, is equivalent to weak convergence of the probability measures.

3. Detect a basis and window artifact

For the two-mode basis test, compute $\chi(\pi/2)$ before and after the correctly transported Hadamard rotation. Then keep $\psi$ unrotated, find the wrong spectral weights, and compute the residual. Finally explain why changing $L$ from $4$ to $2$ is not a basis change.

Solution

For $h=\operatorname{diag}(-1,1)$, the weights of $\psi$ are $3/4$ and $1/4$, so

$$\chi(\pi/2) =\frac34 i+\frac14(-i)=\frac i2.$$

If $h'=UhU^*$ and $\psi'=U\psi$, then

$$\langle\psi',e^{-ish'}\psi'\rangle =\langle\psi,e^{-ish}\psi\rangle$$

for every $s$. If $h$ is rotated but the coordinate vector is not, the overlaps with the new eigenvectors give

$$w_-=\frac{2+\sqrt3}{4}, \qquad w_+=\frac{2-\sqrt3}{4}.$$

Thus $\chi_{\rm wrong}(\pi/2)=i(w_--w_+)=i\sqrt3/2$ and

$$\left\lvert\chi_{\rm wrong}(\pi/2)-\chi(\pi/2)\right\rvert =\frac{\sqrt3-1}{2}.$$

This is a bookkeeping failure under a unitary equivalence. By contrast, changing $L$ changes the projected spectral subspace and discards weight; no unitary within the old finite space can restore it. The resulting error must be controlled by increasing the window and checking convergence against the same target measure.

References

• Bisognano, Joseph J., and Eyvind H. Wichmann. “On the Duality Condition for Quantum Fields.” Journal of Mathematical Physics 17 (1976): 303–321. DOI.
• Connes, Alain. “Une classification des facteurs de type III.” Annales Scientifiques de l’École Normale Supérieure 6 (1973): 133–252. DOI; Open PDF.
• Fredenhagen, Klaus. “On the Modular Structure of Local Algebras of Observables.” Communications in Mathematical Physics 97 (1985): 79–89. DOI.
• Peschel, Ingo. “Calculation of Reduced Density Matrices from Correlation Functions.” Journal of Physics A: Mathematical and General 36 (2003): L205–L208. DOI; arXiv PDF.