Normal density states and spectral moments — Examples

Under the retained choice hypotheses [A1], the imported pure-state criterion applies. Suppose $\mathrm{\mid }\mathrm{\Phi }⟩=a\otimes b$, where $a=a_0\mathrm{\mid }0⟩+a_1\mathrm{\mid }1⟩$ and $b=b_0\mathrm{\mid }0⟩+b_1\mathrm{\mid }1⟩$. Comparing coefficients with $(\mathrm{\mid }00⟩+\mathrm{\mid }11⟩)\mathrm{/}\sqrt{2}$ gives $a_0{b}_0=a_1{b}_1=1\mathrm{/}\sqrt{2}$ but $a_0{b}_1=a_1{b}_0=0$. The first pair makes all four factors nonzero, contradicting the second pair. Thus $\mathrm{\mid }\mathrm{\Phi }⟩$ is not a product vector, so [F2] makes ${\rho }_{\mathrm{\Phi }}$ entangled under the exact definition [F1].

In the product basis, the partial-trace rule [F3] removes $\mathrm{\mid }00⟩⟨11\mathrm{\mid }$ and $\mathrm{\mid }11⟩⟨00\mathrm{\mid }$, since the traced factor basis indices differ, and retains the two diagonal terms. Therefore both reductions of ${\rho }_{\mathrm{\Phi }}$ are $I_2\mathrm{/}2$; positivity and trace preservation in [F3] make them density matrices. The mixture already has only diagonal terms, and tracing either factor likewise gives $I_2\mathrm{/}2$. The spectrum of ${\rho }_{\mathrm{\Phi }}$ is $(1,0,0,0)$, whereas that of ${\rho }_{\mathrm{m}\mathrm{i}\mathrm{x}}$ is $(1\mathrm{/}2,1\mathrm{/}2,0,0)$; each marginal has spectrum $(1\mathrm{/}2,1\mathrm{/}2)$. Substituting these eigenvalues into the entropy definition [F4] gives the displayed three entropy values. In particular both pairs obey the finite subadditivity inequality in [F4].

The mixture is the convex combination with weights $1\mathrm{/}2,1\mathrm{/}2$ of the two pure product densities $\mathrm{\mid }00⟩⟨00\mathrm{\mid }$ and $\mathrm{\mid }11⟩⟨11\mathrm{\mid }$; hence it lies in the finite convex hull in [F1] and is separable. Put $\mathrm{\mid }{\mathrm{\Phi }}_{\pm }⟩=(\mathrm{\mid }00⟩\pm \mathrm{\mid }11⟩)\mathrm{/}\sqrt{2}$; then $W\mathrm{\mid }{\mathrm{\Phi }}_{\pm }⟩=\pm \mathrm{\mid }{\mathrm{\Phi }}_{\pm }⟩$, and $W$ vanishes on the orthogonal span of $\mathrm{\mid }01⟩,\mathrm{\mid }10⟩$. The Bell state is $\mathrm{\mid }{\mathrm{\Phi }}_{+}⟩$, so [F5] assigns probability one to $+1$. Each of $\mathrm{\mid }00⟩,\mathrm{\mid }11⟩$ has squared overlap $1\mathrm{/}2$ with each of $\mathrm{\mid }{\mathrm{\Phi }}_{+}⟩,\mathrm{\mid }{\mathrm{\Phi }}_{-}⟩$, so the equal mixture gives probability $1\mathrm{/}2$ to each of $+1,-1$ and zero to $0$.

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.

Write $r(h)=\sqrt{(ϵ\mathrm{/}2{)}^2+h^2}$. Direct matrix multiplication gives $H(h{)}^2=r(h{)}^{2}I$ and $\mathrm{Tr}H(h)=0$. Splitting the exponential power series into even and odd powers therefore gives $e^{-\beta H(h)}=\cosh (\beta r(h))I-(\sinh (\beta r(h))\mathrm{/}r(h))H(h)$, hence $Z(h)=2\cosh (\beta r(h))$. Also $\mathrm{Tr}(H(h){\sigma }_x)=-2h$, so $⟨{\sigma }_x{⟩}_h=h\tanh (\beta r(h))\mathrm{/}r(h)$. Dividing by $h$ and taking $h\to 0$ yields $\chi =\tanh (\beta ϵ\mathrm{/}2)\mathrm{/}(ϵ\mathrm{/}2)=\beta \tanh (x)\mathrm{/}x$.

At $h=0$, the Gibbs density is diagonal in the displayed basis, so $\mathrm{Tr}({\rho }_0{\sigma }_x)=0$. By the ordinary variance definition in [F1], ${\mathrm{Var}}_{{\rho }_0}({\sigma }_x)=\mathrm{Tr}({\rho }_0{\sigma }_x^2)-\mathrm{Tr}({\rho }_0{\sigma }_x{)}^2=1$. The commutator of $\mathrm{diag}(-ϵ\mathrm{/}2,ϵ\mathrm{/}2)$ with ${\sigma }_x$ has nonzero off-diagonal entries for $ϵ>0$, so the commuting equality case in [F1] does not apply. Finally, for $x>0$, $x-\tanh x={\int }_0^x(1-{\mathrm{sech}}^{2}t)dt>0,$ because the integrand is strictly positive for $t>0$. Thus $0<\chi <\beta$. At the zero-gap boundary, ${\lim }_{x\downarrow 0}\tanh (x)\mathrm{/}x=1$, so the response tends to the commuting value $\beta$.