Purely discrete spectrum of a self-adjoint operator 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 $\mu\in\mathbb{R}$?
 A: 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).

*

*$\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.

*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.

*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.

*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.


*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 $H$.


*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


*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.
