Purely discrete spectrum of a self-adjoint operator

Question

Let $(H,D(H))$ be a self-adjoint operator on the infinite dimensional Hilbert space $\mathcal{H}$ that is bounded below and also assume that $$\text{Tr}_{\mathcal{H}}(e^{-H})<\infty \, .$$

How can I see that $H$ has purely discrete spectrum and that the sequence of eigenvalues $e_0\leq e_1\leq e_2\leq...$ repeated according to multiplicity is unbounded? And moreover why is $\text{Tr}_{\mathcal{H}}\left(e^{-\beta(H-\mu)}\right)<\infty$ for $\beta>0$ and $\mu\in\mathbb{R}$?

Answer 1

Let me try to give you some points, perhaps someone else can post proofs or fill some gaps if you need them (the proofs are not easy).

1. $\text{Tr}_{\mathcal H} \left(e^{-H}\right) := \sum_{n=1}^\infty \left\langle e_n, e^{-H}e_{n}\right\rangle <+\infty $ means that the exponential of the negative Hamiltonian is an operator of finite trace, thus bounded. If you assume the trace is the same for all infinite countable bases in $\mathcal{H}$, then the operator $e^{-H}$ is trace class.
2. A trace class self-adjoint operator has a pure point spectrum. This is Theorem 8.6 of Prugovecki, 2nd. Ed., p.387. This is proven using the spectral theorem.
3. Using the spectral theorem and the functional calculus (which works because $e^{-x}$ is bounded for $x>0$), if $e^{-H}$ is bounded with a pure point spectrum (the content of point 2.), then $H$ has a pure point spectrum.
4. In the proof of theorem 8.7 further below in his text, Prugovecki makes the following argument:

$$ H = \int_{\mathbb{R}} \lambda ~ dE_{\lambda} ^{H} = \sum_{\lambda ~ \in\sigma(H)} \lambda E^H(\{\lambda\}) \tag{1}$$

(the sum is uniformly convergent). Each characteristic subspace $M_\lambda$ which corresponds to a non-zero eigenvalue of $H$ is finite-dimensional.

5. Then we can employ a version of the minimax principle quoted as preposition 3.5.5 from page 64 of David Ruelle's Rigorous Stat. Mech. book. This is a little more general than needed (the Hamiltonian would suffice to be symmetric, but in this case, it is assumed directly self-adjoint).

This proves the fact that the spectral values of the Hamiltonian are increasing. Another proof for the spectral properties of the Hamiltonian would have been from the fact that $e^{-H}$ is completely continuous (from being trace-class), hence its spectrum would also be pure point and decreasing. $e^{H}$ is the inverse of $e^{-H}$, so its spectrum would again be pure point but increasing, and the same goes for the spectrum of $H$.

6. Points 1.-5. show that asking for the Hamiltonian to be bounded and with a pure point spectrum is a consequence of supposing that the quantum canonical partition function $Z$ be well-defined as a trace class operator, which is a very stringent requirement. The converse is not true in general. link math.SE. However, the partition function for the harmonic oscillator is well-defined, see the proof here: Proof that $\exp(-\beta H)$ is a trace-class operator for the harmonic oscillator

7. Your last point assumes the fact that the canonical partition function $Z$ is well-defined and questions why is this also well-defined as a trace class operator in case the system has a shift in ground energy and the rest of its spectrum (at least this is what I make from your deduction of a constant from the Hamiltonian). Well, this is easy because the spectral properties of $H-\mu \hat{1}$ for constant $\mu$ are the same (pure point and increasing). Also, $\beta >0$ to keep the whole operator trace class. Also, $\mu$ must be real, else the sum with the Hamiltonian is not self-adjoint.