Qm Math Fock And Equilibrium — Examples

[Divergence of the one-particle weights] By the definition of the mode list, $e^{-\beta {\epsilon }_j}=e^{-\beta {\epsilon }_0}>0$ for every $j$, a constant strictly positive sequence that does not converge to $0$; by [F3] the series ${\sum }_j{e}^{-\beta {\epsilon }_j}$ diverges, and since all its terms are positive its value is $+\mathrm{\infty }$. This is (a).

[Divergent partial products and the finite-$N$ identity] By [F1] applied to the constant list, for every $N\ge 1$, ${\sum }_{I\subseteq \{1,\dots ,N\}}{e}^{-\beta {\lambda }_I}={\prod }_{j=1}^N(1-e^{-\beta {\epsilon }_0}{)}^{-1}=(1-e^{-\beta {\epsilon }_0}{)}^{-N}$. Since $0<1-e^{-\beta {\epsilon }_0}<1$, these finite powers strictly increase. Put $x:=e^{-\beta {\epsilon }_0}\in (0,1)$. By [F4] and the logarithm product and reciprocal laws, $\log ((1-x{)}^{-N})=-N\log (1-x)\ge Nx$, so monotonicity of $\log$ gives $(1-x{)}^{-N}\ge e^{Nx}$. For any $M>0$, choose $r:=\log M\mathrm{/}x$; if $N>r$, then $Nx>\log M$, hence $e^{Nx}>M$ because the exponential is the increasing inverse of $\log$. Thus the partial products tend to $+\mathrm{\infty }$, proving the quantitative bound in (b).

[Divergence of the grand-canonical weights, no trace] By step 1.1 the series ${\sum }_j{e}^{-\beta {\epsilon }_j}$ is not finite, so the equivalence in [F1] gives ${\sum }_{I\in \mathcal{M}}{e}^{-\beta {\lambda }_I}=+\mathrm{\infty }$; by step 1.2 its finite partial sums are finite and increasing, consistently. By [F2], $D_{\beta }$ is not trace class and $\mathrm{tr}{D}_{\beta }$ does not exist as a finite number. This is (c).

[The single-mode contrast] For $H=\mathbb{C}$ with $\omega ={\epsilon }_0$, [F5] gives $Z(\beta )=1\mathrm{/}(1-e^{-\beta \omega })<\mathrm{\infty }$ and a trace-class weight operator with trace $Z(\beta )$; the same closed-form expression appears in step 1.2 with each of the infinitely many factors equal to it, and step 1.2 proves that these constant-factor products diverge by the quantitative bound $(1-e^{-\beta \omega }{)}^{-N}\ge e^{N{e}^{-\beta \omega }}\to +\mathrm{\infty }$. Thus the infinite multiplicity, not the value of the positive frequency, produces the divergence. This is (d).

[Conclusion and scope] Steps 1.1-1.2 prove (a) and (b), step 2.1 proves (c), and step 2.2 proves (d). The scope paragraph of the example records that no thermodynamic interpretation of $\beta$, $D_{\beta }$ or the divergence is asserted. This completes the verification.

[$U(t)$ on eigenstates] For $n\in \mathbb{N}$ the vector $e_n$ satisfies $N{e}_n=n{e}_n$ by [F1], hence $A{e}_n=\omega n{e}_n$; by [F3] with $T=A$, $a=\omega n$ and $x=e_n$, $\int (\lambda -\omega n{)}^{2}d{E}^{e_n}=0$ with a nonnegative integrand, so by [F4] the measure $E^{e_n}$ gives full mass to $\{\omega n\}$. Applying [F3] to the bounded Borel function $h_n(\lambda ):=e^{-it\lambda \mathrm{/}\mathrm{\hslash }}$ and ${\lambda }_0=\omega n$ gives $\mathrm{\parallel }{h}_n(A)e_n-e^{-it\omega n\mathrm{/}\mathrm{\hslash }}{e}_n{\mathrm{\parallel }}^2=\int \mathrm{\mid }{h}_n-h_n(\omega n){\mathrm{\mid }}^{2}d{E}^{e_n}=0$; hence $U(t)e_n=h_n(A)e_n=e^{-it\omega n\mathrm{/}\mathrm{\hslash }}{e}_n$. For $n\in \{0,1\}$ this gives $U(t)e_0=e_0$ and $U(t)e_1=e^{-i\theta }{e}_1$, so by linearity $\psi (t)=(e_0+e^{-i\theta }{e}_1)\mathrm{/}\sqrt{2}$, which is (a); also $\mathrm{\parallel }{\psi }_0{\mathrm{\parallel }}^2=(\mathrm{\parallel }{e}_0{\mathrm{\parallel }}^2+\mathrm{\parallel }{e}_1{\mathrm{\parallel }}^2)\mathrm{/}2=1$.

[The expectation] By [F2], $Y{e}_0=A_{+}(1)e_0=e_1$ and $Y{e}_1=A_{-}(1)e_1+A_{+}(1)e_1=e_0+\sqrt{2}{e}_2$, so $Y$ maps $\mathrm{span}\{e_0,e_1\}$ into $\mathrm{span}\{e_0,e_1,e_2\}$; since $e_0,e_1\in D(A_{+}(1))\cap D(A_{-}(1))$ by the operator equalities in [F2] and both domains are linear subspaces, both $\psi (t)$ and $Y\psi (t)$ lie in the finite-particle space, and $⟨\psi (t),Y\psi (t)⟩$ is well-defined. With $Y\psi (t)=\frac{1}{\sqrt{2}}(e_1+e^{-i\theta }{e}_0+\sqrt{2}{e}^{-i\theta }{e}_2)$, orthonormality of $(e_0,e_1,e_2)$ and linearity in the first argument give $⟨\psi (t),Y\psi (t)⟩=\frac{1}{2}(⟨e_0,e_1⟩+e^{i\theta }⟨e_0,e_0⟩+e^{-i\theta }⟨e_1,e_1⟩)=\frac{1}{2}(e^{i\theta }+e^{-i\theta })=\cos \theta$.

[The stationary thermal weight] By [F6], $G_{\beta }$ is the direct sum of the scalars $e^{-\beta \omega n}$ on the one-dimensional sectors of $F(\mathbb{C})$; since the sector $n$ is spanned by $e_n$, this says $G_{\beta }{e}_n=e^{-\beta \omega n}{e}_n$, so $G_{\beta }$ is diagonal in the complete orthonormal family $(e_n)$. By step 1.1, $U(t)G_{\beta }U(t{)}^{\ast }{e}_n=U(t)G_{\beta }{e}^{i\theta n}{e}_n=e^{i\theta n}{e}^{-\beta \omega n}{e}^{-i\theta n}{e}_n=e^{-\beta \omega n}{e}_n=G_{\beta }{e}_n$ for every $n$, $\theta n=\omega nt\mathrm{/}\mathrm{\hslash }$. Both $U(t)G_{\beta }U(t{)}^{\ast }$ and $G_{\beta }$ are bounded and agree on the complete orthonormal family, hence on its dense span, hence everywhere; dividing by the finite positive scalar $Z(\beta )$ gives $U(t){\rho }_{\beta }U(t{)}^{\ast }={\rho }_{\beta }$. This is (d).

[Failure of convergence of the expectation and of the state] By step 2.1, $⟨\psi (t),Y\psi (t)⟩=\cos (\omega t\mathrm{/}\mathrm{\hslash })$ equals $1$ at $t_n=2\pi \mathrm{\hslash }n\mathrm{/}\omega$ and $-1$ at $s_n=(2n+1)\pi \mathrm{\hslash }\mathrm{/}\omega$ for every $n$; if a limit existed along $t\to +\mathrm{\infty }$, [F7] would give $1=-1$, a contradiction. This is (b). For two times $t,s$, [F5] gives $\mathrm{\parallel }\psi (t)-\psi (s){\mathrm{\parallel }}^2=\frac{1}{2}\mathrm{\mid }{e}^{-i\omega t\mathrm{/}\mathrm{\hslash }}-e^{-i\omega s\mathrm{/}\mathrm{\hslash }}{\mathrm{\mid }}^2=1-\cos (\omega (t-s)\mathrm{/}\mathrm{\hslash })$, which equals $2$ for $t=t_n$, $s=s_n$; hence the family $(\psi (t))$ is not Cauchy at $+\mathrm{\infty }$ and has no limit in norm. This is (c).

[Conclusion and scope] Steps 1.1 and 2.1 establish the setting and the expectation $\cos (\omega t\mathrm{/}\mathrm{\hslash })$, step 3.1 proves (b) and (c), and step 2.2 proves (d). The scope paragraph of the example records that no physical interpretation of the state, the coupling or the weights is asserted and that relaxation to equilibrium is not derived. This completes the verification.

[The two-particle exclusion] For $u\in H$ the swap identity $U_{\tau }(u\otimes u)=u\otimes u$ of [F1] gives $P_{-}^{(2)}(u\otimes u)=\frac{1}{2}(u\otimes u-U_{\tau }(u\otimes u))=0$. This is (a).

[Slater vector with repeated orbital] By [F2] with $u_1=u_2=u$, $S(u,u)=\frac{1}{\sqrt{2}}(u\otimes u-u\otimes u)=0$; equivalently $S(u,u)=\sqrt{2}{P}_{-}^{(2)}(u\otimes u)=0$ by step 1.1. For orthonormal $u_1,u_2$, [F2] gives $\mathrm{\parallel }S(u_1,u_2)\mathrm{\parallel }=1$, so the hypothesis of pairwise orthonormal orbitals in [F2] cannot be dropped. This is (b).

[Superposition check] Let $u=e_1+e_2\in {\mathbb{C}}^2$. Bilinearity of the tensor product gives $u\otimes u=e_1\otimes e_1+e_1\otimes e_2+e_2\otimes e_1+e_2\otimes e_2$. By [F1], $U_{\tau }$ fixes $e_1\otimes e_1$ and $e_2\otimes e_2$ and interchanges $e_1\otimes e_2$ with $e_2\otimes e_1$; hence $P_{-}^{(2)}(u\otimes u)=\frac{1}{2}((e_1\otimes e_2+e_2\otimes e_1)-(e_2\otimes e_1+e_1\otimes e_2))=0$, in agreement with step 1.1. This is (c).

[Consequence in the antisymmetric sector] If $u\mathrm{\ne }0$ then $u\otimes u\mathrm{\ne }0$ (the tensor product of nonzero vectors has norm $\mathrm{\parallel }u{\mathrm{\parallel }}^2>0$), while step 1.1 gives $P_{-}^{(2)}(u\otimes u)=0$; so no nonzero two-particle antisymmetric vector arises by antisymmetrising $u\otimes u$. By [F3] this is a statement about the sector $\mathrm{ran}{P}_{-}^{(2)}$ of $H^{\otimes 2}$ ; the zero projection belongs to that linear sector, but no nonzero vector of that form exists. The bosonic $F(H)$ has the distinct symmetric sector of [F3]. This is (d).

[Conclusion and scope] Steps 1.1 and 2.1 prove (a) and (b), step 2.2 proves (c), and step 2.3 proves (d). The scope paragraph of the example records that the symmetric/antisymmetric sector assignment is a postulate and that no spin-statistics statement is proved. This completes the verification.