One-mode quartic ground-energy increment bound

Write $a_{\lambda }=q_{\lambda }+2\mathrm{\parallel }\cdot {\mathrm{\parallel }}_2^2$ for the shifted form inner product of the $\lambda$ completion. On $C_c^{\mathrm{\infty }}$, $V_{\lambda }\ge V_0$ gives $a_{\lambda }(u,u)\ge a_0(u,u)$, so a $Q_{\lambda }$-Cauchy sequence is $Q_0$-Cauchy and the identity extends continuously to $Q_{\lambda }\to Q_0$ with the same $L^2$ representative. The unshifted forms $q_{\lambda }=a_{\lambda }-2\mathrm{\parallel }\cdot {\mathrm{\parallel }}_2^2$ and $q_0=a_0-2\mathrm{\parallel }\cdot {\mathrm{\parallel }}_2^2$ are continuous on their respective form spaces. For any $u\in Q_{\lambda }$, approximate it in $Q_{\lambda }$ by $u_r\in C_c^{\mathrm{\infty }}$; then $q_{\lambda }(u_r)\ge q_0(u_r)$ for every $r$, and continuity of both forms gives $q_{\lambda }(u)\ge q_0(u)$ in the limit. For normalized $u\in Q_{\lambda }$, [F3] then gives $q_{\lambda }(u)\ge q_0(u)\ge E_g(0)$. Taking the infimum yields $E_g(\lambda )\ge E_g(0)$.

The Gaussian ${\varphi }_b$ from [F2] and its derivative are polynomial times a Gaussian. For cutoffs ${\chi }_R$ as in the cutoff construction of Anisotropic Hermite oscillator and quadratic form basis, ${\chi }_R{\varphi }_b\in C_c^{\mathrm{\infty }}$ converges to ${\varphi }_b$ in the $\lambda$ form norm: the derivative error is $(1-{\chi }_R){\varphi }_b^{\mathrm{\prime }}-({\chi }_R^{\mathrm{\prime }}){\varphi }_b$, whose first term has vanishing Gaussian tail and whose second has norm at most $C\mathrm{\parallel }{\varphi }_b{\mathrm{\parallel }}_2\mathrm{/}R$; the weighted error has squared magnitude bounded by $(m^2{q}^2\mathrm{/}2+\lambda q^4+2)\mathrm{\mid }{\varphi }_b{\mathrm{\mid }}^2$, an integrable polynomial-Gaussian tail. Thus ${\varphi }_b\in Q_{\lambda }$.

Write $p_b(q):=\sqrt{b\mathrm{/}\pi }{e}^{-b{q}^2}$ and ${\mathbb{E}}_b[g]:=\int g(q)p_b(q)dq$. For $g\in \{1,q^2\}$, the real-valued integrand has derivative in $b$ ${\mathrm{\partial }}_b(g(q)p_b(q))=g(q)(\frac{1}{2b}-q^2)p_b(q)\mathrm{.}$ Fix $b_0>0$ and choose a compact interval $K=[b_0-\delta ,b_0+\delta ]\subset (0,\mathrm{\infty })$; put $b_{\ast }:=b_0-\delta$. There is a constant $C_K$ such that, for both choices of $g$ and every $s\in K$, $\mathrm{\mid }{\mathrm{\partial }}_s(g(q)p_s(q))\mathrm{\mid }\le C_K(1+\mathrm{\mid }q{\mathrm{\mid }}^4)e^{-b_{\ast }{q}^2},$ and the right side is integrable on $\mathbb{R}$. For any sequence of nonzero $h_r\to 0$ with $b_0+h_r\in K$, for each fixed $q$ and $g\in \{1,q^2\}$ the function $s\mapsto g(q)p_s(q)$ is smooth on $(0,\mathrm{\infty })$, hence continuous on the closed interval between $b_0$ and $b_0+h_r$ and differentiable in its interior. This interval is contained in $K$ and has distinct endpoints. Thus [F7] bounds the difference quotients $\mid \frac{g(q)p_{b_0+h_r}(q)-g(q)p_{b_0}(q)}{h_r}\mid \le C_K(1+\mathrm{\mid }q{\mathrm{\mid }}^4)e^{-b_{\ast }{q}^2};$ they converge pointwise to ${\mathrm{\partial }}_b(g(q)p_b(q)){\mathrm{\mid }}_{b=b_0}$. Applying [F5] to these quotients gives convergence of their integrals to the integral of that derivative. Since this holds for every such sequence, it proves for each $b_0>0$ that $\frac{d}{db}{\mathbb{E}}_b[g]{\mid }_{b=b_0}={\int }_{\mathbb{R}}{\mathrm{\partial }}_b(g(q)p_b(q)){\mid }_{b=b_0}dq\mathrm{.}$ Thus [F5] is used only after the local difference-quotient convergence and majorant have been established. Since ${\mathbb{E}}_b[1]=1$, differentiating gives $0={\mathbb{E}}_b\text{\hspace{0.17em}⁣}[\frac{1}{2b}-q^2],{\mathbb{E}}_b[q^2]=\frac{1}{2b}\mathrm{.}$ Differentiating this identity for ${\mathbb{E}}_b[q^2]$ gives $-\frac{1}{2b^2}={\mathbb{E}}_b\text{\hspace{0.17em}⁣}\left[q^2(\frac{1}{2b}-q^2)\right]=\frac{1}{4b^2}-{\mathbb{E}}_b[q^4],$ so ${\mathbb{E}}_b[q^4]=3\mathrm{/}(4b^2)$. Since ${\varphi }_b^{\mathrm{\prime }}=-bq{\varphi }_b$, it follows that $\mathrm{\parallel }{\varphi }_b{\mathrm{\parallel }}_2^2=1,\mathrm{\parallel }{\varphi }_b^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2=b\mathrm{/}2,\int q^2\mathrm{\mid }{\varphi }_b{\mathrm{\mid }}^2=1\mathrm{/}(2b),\int q^4\mathrm{\mid }{\varphi }_b{\mathrm{\mid }}^2=3\mathrm{/}(4b^2)\mathrm{.}$

Put ${ϵ}_0:=m\sqrt{a\mathrm{/}2}$. For every $u\in Q_0$, the exact expansion in [F2], together with ${ϵ}_n\ge {ϵ}_0$ and Parseval from [F6] for its complete orthonormal basis, gives $q_0(u)-{ϵ}_0\mathrm{\parallel }u{\mathrm{\parallel }}_2^2={\sum }_{n=0}^{\mathrm{\infty }}({ϵ}_n-{ϵ}_0)\mathrm{\mid }⟨u,{\psi }_n⟩{\mathrm{\mid }}^2\ge 0.$ The normalized ${\varphi }_b={\psi }_0\in Q_0$ has only its ground coefficient nonzero, so [F2] also gives $q_0({\varphi }_b)={ϵ}_0$. In particular the moment calculation of step 1.3 yields $a\mathrm{\parallel }{\varphi }_b^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2+\frac{m^2}{2}\mathrm{\parallel }q{\varphi }_b{\mathrm{\parallel }}_2^2=\frac{ab}{2}+\frac{m^2}{4b}=m\sqrt{\frac{a}{2}}={ϵ}_0,$ where $b=m\mathrm{/}\sqrt{2a}$. Hence the normalized form infimum for $q_0$ is ${ϵ}_0$, and the variational-bound lemma [F3] identifies this infimum with $E_g(0)$; therefore $E_g(0)={ϵ}_0$.

By step 1.2, ${\varphi }_b$ is a normalized vector in $Q_{\lambda }$. Its quartic form contribution is $\lambda \int q^4\mathrm{\mid }{\varphi }_b{\mathrm{\mid }}^{2}dq=\frac{3\lambda }{4b^2}=\frac{3a\lambda }{2m^2}\mathrm{.}$ The variational bound [F3], now for $q_{\lambda }$, gives $E_g(\lambda )\le q_{\lambda }({\varphi }_b)=E_g(0)+\frac{3a\lambda }{2m^2}\mathrm{.}$ Combining this with step 1.1 proves both inequalities, including $\lambda =0$.