Finite-Dimensional Systems

Finite-dimensional quantum systems remove domain and convergence difficulties but retain noncommutativity, incompatible measurements, and entanglement. They are exact models when the physical state space is finite, and controlled approximations only when an independent argument bounds truncation errors. This page identifies the mathematical guarantees that finite dimension supplies and the conclusions that do not survive an unchecked infinite-dimensional limit.

Helpful background. Finite-Dimensional Postulates supplies the physical rules. Self-Adjoint Operators explains why domain equality matters in the general theory.

Dense domains fill a finite Hilbert space

Let $\mathcal H=\mathbb C^d$, with its usual Hilbert inner product. Every linear subspace is closed, so a dense linear operator domain must equal $\mathcal H$. Every linear operator on that domain is bounded, and its adjoint is defined everywhere.

Consequently a symmetric densely defined operator is self-adjoint: the algebraic equality $A=A^*$ already includes equality of domains. There is no distinct family of self-adjoint boundary realizations hidden behind one finite matrix.

All norms on the finite-dimensional vector space of matrices give the same topology. In particular, weak operator, strong operator, and operator-norm convergence coincide. The constants comparing norms can depend on $d$, so this statement gives no dimension-independent estimate for a sequence of growing matrix problems.

The spectral theorem gives an orthogonal eigenbasis for each Hermitian matrix. Every density operator is trace class, every state functional is normal, and all moments of each fixed observable are finite. These are consequences of finite dimension, not extra physical postulates.

Eigenvalues are stable under a bounded perturbation

For Hermitian matrices $H$ and $V$, order the eigenvalues increasingly with multiplicity. The min-max principle and

$$-\|V\|I\leq V\leq\|V\|I$$

give

$$|\lambda_j(H+V)-\lambda_j(H)|\leq\|V\|.$$

Indeed every Rayleigh quotient shifts by at most $\|V\|$, and the same bound passes through the infimum and supremum defining each ordered eigenvalue. This proof requires no nondegeneracy assumption.

Eigenvectors are less stable. Near a degeneracy, an arbitrarily small perturbation can choose a very different eigenbasis within the nearly degenerate subspace. A bound on individual eigenvectors requires spectral-gap information; the eigenvalue bound alone does not provide it.

For example, $H=0$ on $\mathbb C^2$ has every vector as an eigenvector. The perturbations $\epsilon\sigma_z$ and $\epsilon\sigma_x$ both tend to zero in norm while selecting different preferred bases. The degenerate eigenspace is well defined, but a preferred basis within it is not.

Dynamics has a uniform perturbation estimate

For time-independent Hermitian $H,K$, differentiating the product of their propagators and integrating gives Duhamel’s identity,

$$e^{-itH/\hbar}-e^{-itK/\hbar} =-\frac{i}{\hbar}\int_0^t e^{-i(t-s)H/\hbar}(H-K)e^{-isK/\hbar}\,ds.$$

Since both propagators are unitary,

$$\|e^{-itH/\hbar}-e^{-itK/\hbar}\| \leq\frac{|t|}{\hbar}\|H-K\|.$$

Thus a norm error in the Hamiltonian controls all normalized initial vectors simultaneously for fixed time. Long times can amplify a small Hamiltonian error; the estimate does not promise uniform accuracy for arbitrarily long evolution.

The formula also holds for suitable bounded perturbations of self-adjoint unbounded operators, but that extension requires the operator-theoretic hypotheses. Finite matrices are the setting where no domain issue interrupts its derivation.

A finite canonical commutator is impossible

If $Q,P$ were finite matrices satisfying $[Q,P]=i\hbar I$, cyclicity of the finite trace would imply

$$0=\operatorname{Tr}[Q,P] =i\hbar d,$$

a contradiction for $d>0$ and $\hbar\ne0$. Hence a finite truncation cannot preserve the exact canonical commutation relation on its entire state space.

This does not prevent useful low-energy approximations. It requires an error statement restricted to the states being approximated. The oscillator model shows the boundary term that appears in truncated ladder operators.

Likewise, diagonalizing larger matrices does not by itself prove convergence to a chosen unbounded operator. One must specify the embeddings, the limiting domain or closed form, and a suitable convergence theorem. A bounded finite spectrum at every cutoff can converge to an unbounded spectrum or fail to approximate the intended problem at all.

Exercise: infer a measurement error from a Hamiltonian error

Suppose $\|H-K\|\leq\delta$ and both models start in the same normalized vector. Bound the difference in the probability of any effect $0\leq E\leq I$ at time $t$.

Solution

For normalized $\psi,\phi$, expanding the difference of quadratic forms gives $|\langle\psi,E\psi\rangle-\langle\phi,E\phi\rangle|\leq2\|\psi-\phi\|$. The propagator estimate therefore gives a probability error at most $\min(1,2|t|\delta/\hbar)$. The bound is independent of the initial vector and the chosen effect. It is useful for a fixed finite model, but says nothing about truncation error unless $\delta$ itself has been justified.

References

• Hall, Brian C., Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, doi:10.1007/978-1-4614-7116-5 — finite-dimensional quantum theory and its infinite-dimensional limitations.
• Teschl, Gerald, Mathematical Methods in Quantum Mechanics, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, author’s text, Sections 4.4 and 6.1 — min-max and perturbation estimates.