Semibounded Hamiltonians
A lower-bounded Hamiltonian has a finite floor to its energy spectrum. That floor need not be attained by a normalizable ground state, and it does not by itself make the Gibbs operator trace class. This page develops those distinctions and the associated variational and imaginary-time constructions. Here “semibounded” means bounded below; a bound from above alone is a different condition.
Helpful background. The Unbounded Spectral Theorem supplies spectral integration. Quadratic Forms extends energy expectations to the form domain. Trace-Class Operators explains when a Gibbs operator can define a density operator.
Lower bounds and the spectral bottom
Section titled “Lower bounds and the spectral bottom”Let be self-adjoint on a nonzero Hilbert space. The bound means
It is equivalent to . One direction follows by integrating against the spectral measure. For the other, if there were spectrum below , a nonzero spectral projection onto a bounded interval strictly below would supply a normalized vector in with expectation below .
Set . The associated form is conveniently written
on . Shifting by another lower bound gives the same form domain with an equivalent form norm.
Self-adjointness and lower boundedness answer different questions. Self-adjointness constructs autonomous unitary evolution. Lower boundedness constrains the energy and makes the form and imaginary-time constructions below available. Neither property should be inferred from the formal appearance of a kinetic-plus-potential expression.
Variational energy and attainment
Section titled “Variational energy and attainment”The spectral bottom has the variational characterization
The lower inequality is the spectral bound. For the reverse inequality, the projection is nonzero for every . A normalized vector in its range has bounded spectral support and expectation less than . This proves both infimum statements without assuming an eigenbasis.
The infimum is attained precisely when is an eigenvalue. If a normalized form-domain vector satisfies , then
The integrand is nonnegative, so the measure is supported at . Thus and . Conversely every normalized ground eigenvector attains the bound.
For the free Hamiltonian on , the bottom is zero but no normalized ground vector exists. If and is smooth with compact support, the dilated state
has kinetic energy . Attaining zero would require a weakly constant function, which must vanish. The minimizing sequence spreads out rather than converging to a normalized minimizer.
Imaginary time suppresses higher spectral energies
Section titled “Imaginary time suppresses higher spectral energies”For with units of inverse energy, define
This is a strongly continuous contraction semigroup. Spectral dominated convergence gives
If the remainder of the spectrum lies in with , then
Without a gap, strong convergence still holds, but a uniform exponential estimate need not. When , normalized imaginary-time evolution converges to . If the ground space is degenerate, this retains the initial projection within it; it does not select a preferred basis vector.
If no ground eigenvector exists, the unnormalized evolution tends strongly to zero. Renormalizing it does not manufacture a Hilbert-space ground state. Numerical imaginary-time methods consequently need a separate analysis of finite-volume confinement, spectral gaps, and the limit of their normalized states; see Imaginary-Time Projection.
Lower boundedness is not a partition-function theorem
Section titled “Lower boundedness is not a partition-function theorem”For , the operator is bounded because its spectral multiplier is bounded on . Trace-class membership is stronger:
must be established separately before writing a Gibbs density operator. For example, on an infinite-dimensional Hilbert space is self-adjoint and nonnegative, but has infinite trace.
Even a discrete spectrum tending to infinity does not guarantee finite at every temperature. For , , the resolvent is compact, while
is finite exactly when . Growth of energy levels and degeneracy both matter.
Exercises
Section titled “Exercises”- Why does minimizing the energy over a finite-dimensional subspace of give an upper bound on , and why can a poorly chosen subspace miss the ground state?
Solution
The restricted admissible set is smaller, so its infimum cannot be below the unrestricted one. A subspace orthogonal to the ground eigenspace may only sample excited energies. Obtaining convergent variational estimates requires approximation in the relevant form norm, not merely increasing the matrix size without control of the trial spaces.
- Suppose is not an eigenvalue. Show that for every finite , even though for every fixed vector.
Solution
The spectrum approaches , so the supremum of over the spectrum is one. Hence the operator norm is one. The strong limit is zero because the spectral projection at vanishes. Vectors concentrated closer and closer to the spectral bottom obstruct uniform convergence.
References
Section titled “References”- K. Schmüdgen, Unbounded Self-adjoint Operators on Hilbert Space, Graduate Texts in Mathematics 265, Springer, 2012, Chapter 10, publisher edition — semibounded operators and forms.
- G. Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, Sections 2.3 and 4.4, author’s text — lower bounds and variational spectral characterizations.