Oscillator trace Gibbs and divergent normalization

Let ${\lambda }_n=\mathrm{\hslash }\omega (n+\frac{1}{2})$. Finite-support vectors lie in $D(H)$, so the domain is dense. Since each ${\lambda }_n$ is real, the diagonal operator is symmetric. If $y\in D(H^{\ast })$, there is $z\in {\mathrm{\ell }}^2$ such that $⟨Hc,y⟩=⟨c,z⟩$ for every $c\in D(H)$. Testing $c=e_n$ gives $z_n={\lambda }_n{y}_n$, so $({\lambda }_n{y}_n)\in {\mathrm{\ell }}^2$, which is exactly $y\in D(H)$. Conversely, if this sequence is square summable, Cauchy–Schwarz gives $⟨Hc,y⟩=⟨c,({\lambda }_n{y}_n)⟩$ for every $c\in D(H)$. Thus $D(H^{\ast })=D(H)$ and $H^{\ast }=H$.

The diagonal exponential has eigenvalues $e^{-x(n+1\mathrm{/}2)}$. The finite geometric sum identity and $r^n\to 0$ give ${\sum }_{n\ge 0}{r}^n=(1-r{)}^{-1}$, so $Z_{\beta }=e^{-x\mathrm{/}2}\mathrm{/}(1-r)<\mathrm{\infty }$. Set $E_{\ast }:=\mathrm{\hslash }\omega >0$ in J. For any compact inverse-energy interval $[b,B]\subset (0,\mathrm{\infty })$, the dimensionless terms $[1+({\lambda }_n\mathrm{/}{E}_{\ast }{)}^2]e^{-\beta {\lambda }_n}$ are bounded by $[1+(n+\frac{1}{2}{)}^2]e^{-b{E}_{\ast }(n+1\mathrm{/}2)}$. The ratio of consecutive terms of this last series tends to $e^{-b{E}_{\ast }}<1$, so it is summable. Since $y\le 1+y^2$ for $y\ge 0$, this gives finite trace and first and second energy moments after multiplying the corresponding bounds by $E_{\ast }$ or $E_{\ast }^2$. Hence [F1] and [F2] make ${\rho }_{\beta }$ a normal density, and [F4] identifies it with the adopted canonical preparation.

The normalized weights are $p_n=(1-r)r^n$. Absolute convergence permits subtracting the two convergent series to get $(1-r){\sum }_{n\ge 0}n{r}^n={\sum }_{n\ge 1}{r}^n=r\mathrm{/}(1-r)$; therefore $\sum n{p}_n=r\mathrm{/}(1-r)$. This gives $U_{\beta }={\sum }_n{p}_n{\lambda }_n=\mathrm{\hslash }\omega (\frac{1}{2}+r\mathrm{/}(1-r))$. Also $\log p_n=\log (1-r)+n\log r$, so summing $-k_B{p}_n\log p_n$ using [F3] yields $S=k_B[-\log (1-r)-(\log r)r\mathrm{/}(1-r)]=k_B[-\log (1-r)+xr\mathrm{/}(1-r)]$. Replacing $r\mathrm{/}(1-r)$ by $1\mathrm{/}(e^x-1)$ gives the displayed formulas. The direct sums verify the moment hypotheses of [F5], whose energy-entropy identity agrees with this calculation.

For $H=0$ on an infinite-dimensional ${\mathrm{\ell }}^2$, every basis vector is an eigenvector of $e^{-\beta H}=I$ with eigenvalue one. Their eigenvalue sum diverges, so [F1] says $I$ is not trace class. Its trace cannot be normalized to a finite partition function, establishing the contrasting failure without assuming a lower-bound-only Hamiltonian is normalizable.