= Solution
Write $d\gamma_1(x)=(2\pi)^{-1/2}e^{-x^2/2}\,dx$. On the dense <polynomial> algebra in $L^2(\gamma_1)$, the <Gaussian creation and annihilation operators> are
$$
\boxed{a^-f=f',\qquad a^+f=xf-f'.}
$$
<Gaussian integration by parts> gives $\langle a^-f,g\rangle=\langle f,a^+g\rangle$. Their closed realizations are adjoints. More explicitly, for the <orthonormal basis> $h_n=\operatorname{He}_n/\sqrt{n!}$ of normalized <Probabilists' Hermite polynomials>,
$$
a^-h_n=\sqrt n\,h_{n-1},\qquad a^+h_n=\sqrt{n+1}\,h_{n+1}.
$$
Thus the closed annihilation operator has domain $\{\sum c_nh_n:\sum n|c_n|^2<\infty\}$, and the closed creation operator has domain $\{\sum c_nh_n:\sum(n+1)|c_n|^2<\infty\}$. These describe the same set of $L^2$ functions, the <Gaussian Sobolev space>, though the two operators differ. Directly on <polynomials>, $[a^-,a^+]=I$, and
$$
\boxed{Lf=-a^+a^-f=f''-xf'.}
$$
Use the standard <carré du champ operator> convention $\Gamma(f,g)=\tfrac12(L(fg)-fLg-gLf)$. The product rule for this <semigroup generator> gives
$$
L(fg)-fLg-gLf=2f'g',
$$
so the requested squared-gradient and <joint energy> formulas, for real functions in the standard algebra, are
$$
\boxed{\Gamma(f,g)=f'g',\qquad
\mathcal E_{\gamma_1}(f,g)=-\int fLg\,d\gamma_1=\int f'g'\,d\gamma_1.}
$$
For complex functions the <joint energy> is sesquilinear, with $\overline{g'}$ in the second factor. If “squared gradient” is defined without the factor $1/2$ in the product-rule expression, its value is instead $2f'g'$ and the <joint energy> is one half of its integral; the <joint energy> and logarithmic Sobolev constant used here remain as displayed.
For the <Gaussian logarithmic Sobolev inequality>, first take smooth bounded $f$ with bounded first and second <derivatives>, bounded away from zero. The <Mehler formula for the Ornstein-Uhlenbeck semigroup> is
$$
P_tf(x)=\int_{\mathbb R}f\left(e^{-t}x+\sqrt{1-e^{-2t}}\,y\right)\,d\gamma_1(y).
$$
It preserves $\gamma_1$ and differentiating in $x$ gives the <Ornstein-Uhlenbeck gradient commutation identity>
$$
(P_tf)'=e^{-t}P_t(f').
$$
Put $u_t=P_tf$. Differentiate its <entropy functional>. Since $\int u_t\,d\gamma_1=\int f\,d\gamma_1$ is constant, and <Gaussian integration by parts> applies,
$$
\frac d{dt}\operatorname{Ent}_{\gamma_1}(u_t)
=\int (1+\log u_t)Lu_t\,d\gamma_1
=-\int\frac{|u_t'|^2}{u_t}\,d\gamma_1.
$$
For such data, Mehler's formula and dominated convergence give $P_tf(x)\to\int f\,d\gamma_1$ as $t\to\infty$. The uniform positive upper and lower bounds justify integrating this limit inside the entropy, whose limit is zero. Hence
$$
\operatorname{Ent}_{\gamma_1}(f)
=\int_0^\infty\int\frac{|(P_tf)'|^2}{P_tf}\,d\gamma_1\,dt.
$$
Apply the allowed weighted <Cauchy-Schwarz inequality> with $g=f'$:
$$
\frac{|(P_tf)'|^2}{P_tf}
=e^{-2t}\frac{(P_tf')^2}{P_tf}
\leq e^{-2t}P_t\left(\frac{|f'|^2}{f}\right).
$$
After integration, invariance of $\gamma_1$ gives
$$
\int\frac{|(P_tf)'|^2}{P_tf}\,d\gamma_1
\leq e^{-2t}\int\frac{|f'|^2}{f}\,d\gamma_1.
$$
Integrating in $t$ yields
$$
\operatorname{Ent}_{\gamma_1}(f)\leq\frac12\int\frac{|f'|^2}{f}\,d\gamma_1.
$$
For a real smooth compactly supported $g$, take $f=g^2+\epsilon$. Then
$$
\frac{|f'|^2}{f}
=\frac{4g^2|g'|^2}{g^2+\epsilon}\leq4|g'|^2.
$$
Letting $\epsilon\downarrow0$ proves
$$
\boxed{\operatorname{Ent}_{\gamma_1}(g^2)\leq
2\int|g'|^2\,d\gamma_1=2\mathcal E_{\gamma_1}(g).}
$$
Smooth approximation and cutoff in the Gaussian Sobolev <norm> extend this to the full form domain. The entropy is lower semicontinuous under the resulting $L^2$ convergence: the square densities converge in $L^1$, their masses converge, and, along an almost-everywhere convergent subsequence, Fatou's lemma applies to $h\log h$ after adding the bound $1/e$. The same inequality for complex $g$ follows by applying it to $|g|$ and using the weak-derivative bound $|(|g|)'|\leq|g'|$. Thus the constant is at most $2$.
For <sharpness of the Gaussian logarithmic Sobolev constant>, take $g_s(x)=e^{sx/2}$ with real $s\ne0$. These functions belong to the <Gaussian Sobolev space>. The <Gaussian moment-generating function> gives
$$
\int g_s^2\,d\gamma_1=e^{s^2/2},\qquad
\int x e^{sx}\,d\gamma_1=s e^{s^2/2}.
$$
Therefore
$$
\operatorname{Ent}_{\gamma_1}(g_s^2)
=s^2e^{s^2/2}-e^{s^2/2}\frac{s^2}{2}
=\frac{s^2}{2}e^{s^2/2},
\qquad
\mathcal E_{\gamma_1}(g_s)=\frac{s^2}{4}e^{s^2/2}.
$$
The entropy-to-energy ratio is exactly $2$, so no smaller constant can work. Combining this lower bound with the proved inequality gives
$$
\boxed{c_{\mathrm{LS}}(\gamma_1)=2.}
$$
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