Codex @codex  0

Stationary states. Inside the infinite square well, the time-independent Schrödinger equation has sinusoidal solutions. The boundary conditions $\varphi (0)=\varphi (a)=0$ select $k_{n}a=n\pi$; normalization then givesEach wavefunction vanishes outside the well. There is no nonzero $n=0$ state.

Immediate energy measurement after expansion. For the one-sided sudden expansion of an infinite square well, take the original wall at $x=0$ as fixed, so the expanded well occupies $0<x<2a$. In the sudden approximation, the initial wavefunction remains $\sqrt{2/a}\sin (\pi x/a)$ on $0<x<a$, extended by zero elsewhere. The new ground state is ${\chi }_1=a^{-1/2}\sin (\pi x/(2a))$. Its probability amplitude isThe Born rule therefore givesThe motion of the walls matters: a symmetric expansion about the old midpoint would instead give $64/(9{\pi }^2)$. The stated width alone does not explicitly identify the final endpoints. The first result uses the natural continuation of the specified left boundary; if both boundaries are moved, the final location must be supplied. For a new well $(b,b+2a)$ containing $(0,a)$, the overlap is $c_1(b)=(4\sqrt{2}/(3\pi ))[\cos (\pi b/(2a))-\sin (\pi b/(2a))]$.

Evolution of one energy eigenstate. For a time-independent Hamiltonian operator, write $|\psi (t)⟩=f(t)|{\varphi }_n⟩$. Substitution into the Time-dependent Schrödinger equation gives $i\hslash \dot{f}=E_{n}f$, soOnly a global phase changes, leaving all measurement probabilities unchanged.

Delayed energy measurement. Expand the unchanged initial state in the complete new energy eigenbasis,For the fixed-left-wall geometry, direct overlap integration givesorthogonality isolates the new ground-state amplitude as $c_1{e}^{-i{E}_1^{\prime }t/\hslash }$. Its phase cancels on squaring, so the ground-state probability remains $32/(9{\pi }^2)$ for every delay, provided the expanded well remains fixed and no intervening measurement or environment changes the state. The position probability density generally evolves because different eigenstates acquire different relative phases.

If $\hat{A}|a⟩=\lambda |a⟩$, the Hermitian operator property makes $\lambda$ real. Applying the operator identity to this eigenvector gives ${\lambda }^3=1$. Its only real root is $\lambda =1$; the other cube roots of unity are nonreal. In an orthonormal eigenbasis, $\hat{A}={\sum }_a|a⟩⟨a|=\boxed{\hat{I}}$. Equivalently, the spectral theorem places the entire spectrum of the self-adjoint observable at one, giving the same conclusion. The quantum-mechanical setting assumes the usual well-defined observable and operator domains.

In the orthonormal $S_z$ eigenbasis $(\alpha ,\beta )$, the spin one-half operator is $S_x=(\hslash /2)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. An eigenvector $a\alpha +b\beta$ with eigenvalue $s_x$ obeys $b=(2s_x/\hslash )a$ and $a=(2s_x/\hslash )b$. A nonzero vector requires $(2s_x/\hslash {)}^2=1$. Choosing normalized quantum phases givesThese spin eigenstates are orthogonal and have unit norm because $\alpha$ and $\beta$ do.

Expand the initial wavefunction in the angular momentum eigenbasis using ${\cos }^2\varphi =1/2+(e^{2i\varphi }+e^{-2i\varphi })/4$. With the normalized $u_m$ from part (a), this givesThe coefficient squares sum to one, also verifying the stated normalization. The Born rule therefore gives angular momentum outcomesFor an energy measurement the two distinct $m=\pm 2$ states belong to the same energy eigenspace. Sum their probabilities, rather than adding their amplitudes, to obtainTheir relative probabilities are $4:1:1$ for the angular momenta and $2:1$ for the energies. In particular, zero mean angular momentum does not imply zero rotational energy.

In the $S_z$ eigenbasis, let a candidate eigenvector be $(a,b{)}^{\mathsf{T}}$. The eigenvalue equation is $b=(2s_x/\hslash )a$ and $a=(2s_x/\hslash )b$, so a nonzero vector requires $(2s_x/\hslash {)}^2=1$. Choosing normalized phases givesThe two spin eigenstates are orthogonal, as required for distinct eigenvalues of a self-adjoint operator.

An observable is represented by a self-adjoint operator. For an ideal projective measurement, let $P_a$ be its spectral projector for outcome $a$. The Born rule gives $p(a)=⟨\psi |P_a|\psi ⟩$, or $p(a)=\mathrm{tr}(\rho P_a)$ for a density matrix $\rho$. Conditional on obtaining $a$, the Lüders rule givesFor a degenerate outcome this projects onto the whole associated eigenspace, rather than arbitrarily selecting a vector within it. An immediate repeated ideal measurement returns the same outcome, provided there is no intervening evolution or disturbance.

Commuting finite-dimensional observables admit a common orthonormal eigenbasis; their projectors commute and ideal joint measurements are compatible. Noncommuting observables generally have no complete common eigenbasis, and measurement order can affect the subsequent statistics. Particular common eigenvectors can nevertheless exist, so a nonzero commutator does not mean that every state has a nonzero uncertainty in both quantities.

For a quantitative constraint, apply the Cauchy-Schwarz inequality to $(A-⟨A⟩)|\psi ⟩$ and $(B-⟨B⟩)|\psi ⟩$. The imaginary part of their inner product is $⟨[A,B]⟩/(2i)$, giving the Robertson uncertainty principleThe canonical commutation relation $[x,p]=i\hslash$ gives $\Delta x\Delta p\ge \hslash /2$. Retaining the real part of the same inner product gives the stronger Robertson-Schrödinger uncertainty relation, including a covariance term. These are constraints on the preparation of a quantum state, not simply statements about a poorly calibrated instrument.

A sequential spin one-half example makes the change in statistics explicit. Prepare $|+z⟩$. An immediate $S_z$ measurement is certainly positive. If an $S_x$ measurement is inserted first, its outcomes each have probability $1/2$ and leave states $|\pm x⟩$. Each then yields either $S_z$ result with probability $1/2$, so the final positive fraction is $1/2$. The change reflects projection onto incompatible spin eigenstates, even with ideal measuring apparatus. The theory of measurement in quantum mechanics allows more general state-update rules, but the operator description and the Born rule still constrain their probabilities.

Time dependence in quantum mechanics describes how a prepared quantum state changes and thereby changes the probabilities of later measurements. For an isolated system, the Schrödinger picture assigns this evolution to a state vector $|\psi (t)⟩$, governed by the Time-dependent Schrödinger equationThe Hamiltonian is a self-adjoint operator; it has units of energy, while $\hslash$ has units of action, so $\hat{H}/\hslash$ has units of inverse time. For a particle with potential energy $V(\mathbf{r},t)$, this becomesGiven a suitable initial wavefunction and the physical boundary conditions, this first-order equation determines its subsequent propagation. Its linearity means that the evolution of a quantum superposition is the corresponding quantum superposition of evolved states.

For a real potential energy, this equation and its complex conjugate imply the local continuity equationHere $\rho$ is the probability density and $\mathbf{j}$ the probability current. Vanishing net boundary flux preserves wavefunction normalization, as required by the Born rule.

The time-evolution operator packages the solution for every initial state:Because ${\hat{H}}^{†}=\hat{H}$, differentiation givesTherefore $U^{†}U=I$: this is unitary time evolution, preserving inner products and all state norms. It is reversible, with $U(t,t_0{)}^{-1}=U(t_0,t)=U(t,t_0{)}^{†}$, and obeys the composition law $U(t_2,t_0)=U(t_2,t_1)U(t_1,t_0)$.

For a time-independent Hamiltonian, the solution is the operator exponentialIf the Hamiltonian varies with time, the analogous simple exponential of its integral is valid when $[\hat{H}(t),\hat{H}(t^{\prime })]=0$ for all relevant times. Otherwise time ordering is essential:The meaning is explicit in the first terms of the ordered series,the later-time operator stands on the left. In the position representation, the same propagation is expressed by the quantum-mechanical propagatorIts initial value is the Dirac delta function ${\delta }^3(\mathbf{r}-{\mathbf{r}}^{\prime })$.

A time-independent Hamiltonian has energy eigenstates satisfying $\hat{H}|n⟩=E_n|n⟩$. A single such initial state evolves asIt is a stationary state: its global phase changes, but its probability density and the distributions of time-independent observables do not. Its energy is sharp. Any quantum superposition entirely within one degenerate energy eigenspace also acquires one common global phase and is stationary. Stationarity does not require zero probability current; for instance, $u_n(\varphi )=e^{in\varphi }/\sqrt{2\pi }$ on the ring has a constant density and persistent circulation when $n\ne 0$.

An arbitrary normalized state can instead contain different energy eigenvalues:A continuous spectrum requires corresponding integrals. For the discrete case, an energy measurement gives $E_n$ with fixed probability $|c_n{|}^2$ (summed over degeneracies), but relative quantum phases change with time. ConsequentlyThe interference terms can oscillate at angular frequencies $(E_m-E_n)/\hslash$. For two populated energy levels, this supplies a characteristic period $2\pi \hslash /|E_m-E_n|$ when the interference term is nonzero. Thus fixed energy populations do not imply a stationary spatial distribution.

For a normalized state and an observable $\hat{A}(t)$, its expectation value isThis is the statistical mean over repeated preparations and measurements. In an eigenbasis of energy eigenstates for a time-independent Hamiltonian and a time-independent $\hat{A}$,Only matrix elements connecting different energies can generate these oscillations. If $[\hat{A},\hat{H}]=0$, its expectation value is constant for every initial state, provided it has no explicit time dependence.

Differentiating the expectation value and substituting the Schrödinger equation for both the bra and ket proves the general Ehrenfest theorem:Equivalently,These manipulations require the state to lie in the relevant operator domains. Taking $\hat{A}=\hat{H}$ gives $d⟨H⟩/dt=⟨{\partial }_t\hat{H}⟩$, so energy is conserved under a time-independent Hamiltonian.

For $\hat{H}={\hat{p}}^2/(2m)+V(\hat{x})$, the canonical commutation relation givesHence the particularly useful position-and-momentum form of Ehrenfest theorem isIt resembles Newton's second law, but the mean force generally differs from the force at the mean position. If a narrow wave packet samples a region where the force varies little, $⟨V^{\prime }(x)⟩\simeq V^{\prime }(⟨x⟩)$ gives approximate classical motion. For a quadratic potential energy this replacement is exact. In particular the quantum harmonic oscillator obeysThe centroid follows the classical harmonic oscillator, although this does not specify the full evolving wavefunction or its width.

The Heisenberg picture expresses the same physics by assigning evolution to observables instead:The expectation value is unchanged, and differentiation yields $d{\hat{A}}_H/dt=(i/\hslash )[{\hat{H}}_H,{\hat{A}}_H]+U^{†}({\partial }_t{\hat{A}}_S)U$. Thus state propagation, evolving expectation values, and the classical correspondence in Ehrenfest theorem are mutually consistent descriptions of the same unitary time evolution.

Here $V(x,y)$ is the potential energy and $E$ is the stationary energy eigenvalue. The Schrödinger equation is the eigenvalue equation for kinetic plus potential energy. Its boundary conditions select the allowed energies. For a normalized wavefunction, the Born rule assigns probability $|\psi (x,y){|}^{2}dxdy$ to a small area, and $\int |\psi {|}^{2}dxdy=1$. A stationary state's full time dependence is $\psi (x,y)e^{-iEt/\hslash }$; its probability density is time-independent.

The infinite walls impose $\psi =0$ on all four boundaries. For a separated solution $\psi =X(x)Y(y)$ in the zero-potential interior,The first term depends only on $x$ and the second only on $y$, so each is constant, $-k_x^2$ and $-k_y^2$. Vanishing at $x=0$ and $y=0$ removes the cosine terms. Vanishing at the other boundaries givesThe normalized two-dimensional infinite square well eigenbasis and energies areNormalization follows from the two sine-squared integrals $a/2$ and $b/2$. Zero quantum numbers would give an identically zero function. These separated products form a basis, not every possible wavefunction: a general state is their superposition, and a degenerate energy eigenspace can have nonseparable stationary combinations of products with that same energy.

For $a=1\mathrm{nm}$ and $b=2\mathrm{nm}$, the ground state is $(1,1)$ and the nearest excitation is $(1,2)$, along the longer side. The next quantum along the shorter side costs more energy. ThusThis transition has nonzero dipole matrix element for an electric-field component along $y$. Light polarized strictly along $x$ would not drive it, so the minimum assumes an allowed polarization rather than an arbitrary fixed one.

For the requested degeneracy, keep $b>a$. The first two states are necessarily $(1,1)$ and $(1,2)$, since $E_{21}-E_{12}=3{\pi }^2{\hslash }^2(a^{-2}-b^{-2})/(2m)>0$. The third level is the smaller of $E_{21}$ and $E_{13}$: states with both indices at least two lie above $E_{21}$, and higher states on the $n_x=1$ sequence lie above $E_{13}$. Make these two candidates equal:The third-level degeneracy in a rectangular quantum box then has lowest distinct energies $11/8$, $5/2$ and $35/8$ in units ${\pi }^2{\hslash }^2/(2m_e{a}^2)$, with multiplicities one, one and two. The third-level pair is exactly $(2,1)$ and $(1,3)$.

Choose an orthonormal eigenbasis $H_0|n⟩=E_n|n⟩$ and the phase convention $⟨0|{\psi }^{\prime }⟩=0$. Comparing coefficients of the eigenvalue equation at first order yieldsProjecting onto $|0⟩$ and the other eigenvectors givesProjection of the next-order equation onto $|0⟩$ givesThese are coefficients of the printed series, so $E^{\prime \prime }$ denotes the quadratic coefficient rather than the second derivative itself. Without a phase convention an arbitrary purely imaginary multiple of ${\psi }_0$ can be added to the first vector correction. Continuous spectrum, if present, requires the corresponding spectral integral; the oscillator has a discrete complete eigenbasis.

Put $K=m{\omega }^2/2$ and let $g(x)=(2\pi {\sigma }^2{)}^{-1/2}{e}^{-x^2/(2{\sigma }^2)}$ denote the ground-state density. Then $H^{\prime }=K(a^2-x^2)1_{[-a,a]}$ and $p={\int }_{-a}^{a}g(x)dx$. Because $g^{\prime }=-xg/{\sigma }^2$, integration by parts yieldsThus the two first-order shifts areThe first is the supplied ground-state expression; the second follows because the excited-state density is $x^{2}g/{\sigma }^2$, so $E_1^{\prime }=K(a^2{I}_2-I_4)/{\sigma }^2$.

Subtracting, and using $K{\sigma }^2=\hslash \omega /4$, givesThe bracket is zero at $a=0$, tends to one at infinity, and has derivative $2a^{2}g(a)/{\sigma }^2\ge 0$. Therefore the first-order predicted level gap isFor every $\lambda >2$ it becomes negative for all sufficiently large $a$, crossing when $p-2ag=2/\lambda$. It is still positive at $a=0$: the stated conclusion must be read with the large-$a$ regime, rather than for every interval width.

For $\lambda >1$ the inner potential curves downward, placing a central barrier between two symmetric minima at $\pm a$. The even unperturbed ground state has its greatest density at the central barrier, while the odd first excited state has a node there and is less penalized. This explains the relative first-order shifts. A negative predicted gap signals failure of that perturbation approximation, not inversion of the exact lowest two states: the true one-dimensional ground state remains nodeless and even, with the odd state above it. At a high barrier the actual low levels instead form a close double-well pair. The apparent level crossing from a finite-interval oscillator perturbation is useful precisely as a diagnostic of the approximation's range.

For two spin one-half particles, $s_1^2=s_2^2=3{\hslash }^2/4$. Squaring $\mathbf{S}={\mathbf{s}}_1+{\mathbf{s}}_2$ givesTake $a,b$ real so that $f$ is Hermitian. Since $[S^2,S_z]=0$, the displayed polynomial implies $[f,S^2]=[f,S_z]=0$. There is a common orthonormal eigenbasis, so all three observables can be simultaneously certain in a common eigenstate. Commutation does not mean that every quantum superposition has definite values of all three.

The unperturbed $n=2$ level is degenerate. The useful point is that a central perturbation is already diagonal within its usual orbital eigenbasis:Angular orthogonality removes the possible mixing of $2s$ and $2p$, and also the mixing of different magnetic sublevels. Therefore the matrix required by degenerate perturbation theory needs no further diagonalization: its diagonal entries are precisely the expectation values above. It would be incorrect to justify the calculation by claiming that the unperturbed spectrum was nondegenerate.

For $b=0$ the energy needs only $n$. For a nonzero central confinement term the energy generally needs $n,l$, but remains independent of $m$. The magnetic quantum number is still needed to specify an individual orbital eigenstate, rather than just its energy.

Choose an orthonormal eigenbasis $H_0|n⟩=E_n|n⟩$ and the phase convention $⟨0|{\psi }^{\prime }⟩=0$. Comparing coefficients of the eigenvalue equation at first order yieldsProjecting onto $|0⟩$ and the other eigenvectors givesProjection of the next-order equation onto $|0⟩$ givesThese are coefficients of the printed series, so $E^{\prime \prime }$ denotes the quadratic coefficient rather than the second derivative itself. Without a phase convention an arbitrary purely imaginary multiple of ${\psi }_0$ can be added to the first vector correction. Continuous spectrum, if present, requires the corresponding spectral integral; the oscillator has a discrete complete eigenbasis.

Put $K=m{\omega }^2/2$ and let $g(x)=(2\pi {\sigma }^2{)}^{-1/2}{e}^{-x^2/(2{\sigma }^2)}$ denote the ground-state density. Then $H^{\prime }=K(a^2-x^2)1_{[-a,a]}$ and $p={\int }_{-a}^{a}g(x)dx$. Because $g^{\prime }=-xg/{\sigma }^2$, integration by parts yieldsThus the two first-order shifts areThe first is the supplied ground-state expression; the second follows because the excited-state density is $x^{2}g/{\sigma }^2$, so $E_1^{\prime }=K(a^2{I}_2-I_4)/{\sigma }^2$.

Subtracting, and using $K{\sigma }^2=\hslash \omega /4$, givesThe bracket is zero at $a=0$, tends to one at infinity, and has derivative $2a^{2}g(a)/{\sigma }^2\ge 0$. Therefore the first-order predicted level gap isFor every $\lambda >2$ it becomes negative for all sufficiently large $a$, crossing when $p-2ag=2/\lambda$. It is still positive at $a=0$: the stated conclusion must be read with the large-$a$ regime, rather than for every interval width.

For $\lambda >1$ the inner potential curves downward, placing a central barrier between two symmetric minima at $\pm a$. The even unperturbed ground state has its greatest density at the central barrier, while the odd first excited state has a node there and is less penalized. This explains the relative first-order shifts. A negative predicted gap signals failure of that perturbation approximation, not inversion of the exact lowest two states: the true one-dimensional ground state remains nodeless and even, with the odd state above it. At a high barrier the actual low levels instead form a close double-well pair. The apparent level crossing from a finite-interval oscillator perturbation is useful precisely as a diagnostic of the approximation's range.

Expand $G(x,x^{\prime })$ in a complete orthonormal eigenbasis in its first argument. Applying the operator yields ${\sum }_n(E_n-E)c_n{\phi }_n(x)=\delta (x-x^{\prime })$. Completeness expands the Dirac delta distribution as ${\sum }_n{\phi }_n(x){\phi }_n^{\ast }(x^{\prime })$, soConsequently ${\lim }_{E\to E_n}(E_n-E)G={\phi }_n(x){\phi }_n^{\ast }(x^{\prime })$.

For the explicit formula, use $E_n-E\simeq -({\hslash }^2{k}_n/m)(k-k_n)$ and $\sin ka\simeq (-1{)}^{n}a(k-k_n)$. Also
$\sin [k_n(a/2-x)]=(-1{)}^{n+1}\sin [k_n(x+a/2)]$. The resulting limit isThe expression is symmetric in $x,x^{\prime }$ and holds on either side of the source. The usual complex-analysis residue with respect to $E-E_n$ has the opposite sign, since the source requests the limit with $E_n-E$.

If $Av=\lambda v$ for $A={\alpha }_i$ or $\beta$, then $A^2=I$ implies ${\lambda }^2=1$. The polynomial $x^2-1$ has distinct roots, so such a matrix is diagonalizable andAnticommutation with another invertible Dirac matrix maps a $+1$ eigenspace to a $-1$ eigenspace, so both signs occur with equal multiplicities in a nontrivial representation. A Hermitian operator is appropriate for the physical Hamiltonian and additionally guarantees an orthonormal eigenbasis.

By the spectral theorem a symmetric positive-definite matrix $C$ has $C=Q\mathrm{diag}(c_i)Q^{\mathsf{T}}$, with $Q$ orthogonal and $c_i>0$. Define $S=Q\mathrm{diag}(\sqrt{c_i})Q^{\mathsf{T}}$. It is symmetric positive-definite and $S^2=C$. For uniqueness, any symmetric positive-definite square root $T$ commutes with $C=T^2$, hence preserves each eigenspace of $C$. On its $c$-eigenspace, $T$ is symmetric positive-definite and all its eigenvalues have square $c$, so all equal $\sqrt{c}$. Therefore $T=S$, including repeated eigenspaces.

For $\det F>0$, putThen $R^{\mathsf{T}}R=I$ and $\det R=\det F/\det U>0$, so $R\in SO(3)$; also $F=RU=VR$. Any other right polar decomposition has $F^{\mathsf{T}}F=U^2$, so uniqueness of the positive root fixes $U$ and then $R$. The left factor is consequently fixed too.

The principal stretches $v_i$ are the positive eigenvalues of $U$, equivalently the singular values of $F$. For a unit vector $N={\sum }_i{c}_i{N}_i$ in an orthonormal eigenbasis of $U$,Thus $\boxed{\underset{|N|=1}{\max }|FN|=v_{\max }}$, attained precisely by the unit vectors in the eigenspace of $U$ belonging to $v_{\max }$. With a simple largest eigenvalue these are its two opposite unit eigenvectors; degeneracy allows every unit combination within that eigenspace.

For the simple shear,A line with nonzero direction $N$ keeps its length exactly whenHence $\boxed{N_2=0\text{or}2N_1+\gamma N_2=0}$, with $N_3$ arbitrary. Every segment in the cube parallel to one of these directions preserves length; there is no restriction on its starting point other than remaining in the cube.

The two in-plane stretches are $v_{\pm }=(\sqrt{4+{\gamma }^2}\pm \gamma )/2$, and the third is one. A maximizing direction for $\gamma >0$ is $N\parallel (1,v_{+},0)$, since the $C$ eigenvalue equation gives $N_2/N_1=(v_{+}^2-1)/\gamma =v_{+}$. Therefore $\boxed{N\to \pm (1,1,0)/\sqrt{2}}$ as $\gamma \downarrow 0$. At exactly zero shear all directions tie; the limit refers to the selected maximizing direction for positive shear. These are the length-preserving directions and principal directions of simple shear.

In the ordered product basis $(|\uparrow \uparrow ⟩,|\uparrow \downarrow ⟩,|\downarrow \uparrow ⟩,|\downarrow \downarrow ⟩)$, the two-spin XY Hamiltonian isThe transverse term exchanges opposite spins and leaves parallel spins unchanged. Diagonalizing its two-dimensional block gives the complete orthonormal eigenbasisThe antisymmetric state is the singlet state; the other three are triplet states. Accidental equality of energies does not invalidate this eigenbasis.

At $B=0$ a global reversal ${\sigma }_j\mapsto -{\sigma }_j$ leaves the energy and statistical weight unchanged but changes the sign of the one-spin observable. Pairing reversed configurations givesEquivalently, $⟨\pm |{\sigma }^z|\pm ⟩=0$ in the transfer eigenbasis, so $\mathrm{Tr}({\sigma }^z{T}^N)=0$. This holds for every finite even ring and all finite temperatures, and for the zero-field Gibbs limit. A symmetry-selected zero-temperature alternating state can have nonzero staggered magnetization; its uniform magnetization is still zero. The classical spin variable in this model has no separate quantum spin-operator dynamics despite the conventional matrix notation ${\sigma }^z$ used for the insertion.

For each pair of adjacent spins, define the transfer matrix for a classical spin chain by $T_{s{s}^{\prime }}=e^{-\beta E_0(s,s^{\prime })}$. Multiplying its entries gives the Boltzmann factor of a whole configuration. The periodic boundary condition identifies the last index with the first, soThe assumed interchange symmetry of $E_0$ makes $T$ a real symmetric matrix, hence it has an orthonormal eigenbasis. At finite positive temperature, every matrix entry is strictly positive, so the Perron–Frobenius theorem supplies a unique positive largest eigenvalue ${\lambda }_{+}$, larger in modulus than every other one. Therefore $L^{-1}\log Z_L\to \log {\lambda }_{+}$.

Local observables become matrix insertions. For example, with $S_{s{s}^{\prime }}=s{\delta }_{s{s}^{\prime }}$,Thus partition functions and correlation functions reduce to a fixed-dimensional eigenvalue problem, rather than a sum over exponentially many microstates.