[proofplan]
We prove the operator identity first on the algebraic Fock space, namely the linear span of the vacuum vector and all elementary tensors. The computation has two cases: the vacuum vector and a tensor of positive length. In each case, creation by $g$ places $g$ in the first tensor position, and annihilation by $f$ removes that first position while contributing the scalar $(g,f)_H$. Since both sides are bounded operators and the algebraic Fock space is dense in $\mathcal{F}(H)$, equality extends to all of $\mathcal{F}(H)$.
[/proofplan]
[step:Define the dense subspace on which the computation will be made]
For each $n \in \mathbb{N}$, let $H^{\otimes n}$ denote the $n$-fold Hilbert [tensor product](/page/Tensor%20Product) of $H$, and set $H^{\otimes 0} := \mathbb{C}\Omega$. Define the algebraic Fock space
\begin{align*}
\mathcal{F}_{\mathrm{alg}}(H) := \operatorname{span}\{\Omega\} + \operatorname{span}\{h_1 \otimes \cdots \otimes h_n : n \in \mathbb{N},\ h_1,\dots,h_n \in H\}.
\end{align*}
By construction of the Hilbert direct sum $\mathcal{F}(H)$, the subspace $\mathcal{F}_{\mathrm{alg}}(H)$ is dense in $\mathcal{F}(H)$.
For $f \in H$, the adjoint $\ell(f)^*: \mathcal{F}(H) \to \mathcal{F}(H)$ satisfies the annihilation formula
\begin{align*}
\ell(f)^*\Omega = 0
\end{align*}
and, for every $n \in \mathbb{N}$ and $h_1,\dots,h_n \in H$,
\begin{align*}
\ell(f)^*(h_1 \otimes \cdots \otimes h_n) = (h_1,f)_H h_2 \otimes \cdots \otimes h_n,
\end{align*}
with the convention that the remaining tensor is $\Omega$ when $n=1$. This formula is exactly the adjoint formula corresponding to the convention that $(\cdot,\cdot)_H$ is linear in the first argument.
[/step]
[step:Compute the composite on the vacuum vector]
Using the definition of the left creation operator, we have $\ell(g)\Omega = g$. Applying the annihilation formula with the one-tensor $g \in H^{\otimes 1}$ gives
\begin{align*}
\ell(f)^*\ell(g)\Omega = \ell(f)^*g = (g,f)_H \Omega.
\end{align*}
Since $I_{\mathcal{F}(H)}\Omega = \Omega$, this is
\begin{align*}
\ell(f)^*\ell(g)\Omega = (g,f)_H I_{\mathcal{F}(H)}\Omega.
\end{align*}
[guided]
The vacuum case checks the length-zero sector of the Fock space. The creation operator $\ell(g)$ sends the vacuum vector $\Omega$ to the one-particle vector $g \in H$. The annihilation operator $\ell(f)^*$ then removes this one-particle tensor and contributes the [inner product](/page/Inner%20Product) of the removed tensor with $f$. Therefore
\begin{align*}
\ell(f)^*\ell(g)\Omega = \ell(f)^*g = (g,f)_H \Omega.
\end{align*}
The right-hand side of the claimed operator identity sends $\Omega$ to
\begin{align*}
(g,f)_H I_{\mathcal{F}(H)}\Omega = (g,f)_H \Omega.
\end{align*}
Thus the two operators agree on the vacuum vector.
[/guided]
[/step]
[step:Compute the composite on every elementary tensor of positive length]
Let $n \in \mathbb{N}$ and let $h_1,\dots,h_n \in H$. By the definition of $\ell(g)$,
\begin{align*}
\ell(g)(h_1 \otimes \cdots \otimes h_n) = g \otimes h_1 \otimes \cdots \otimes h_n.
\end{align*}
Applying the annihilation formula for $\ell(f)^*$ to this $(n+1)$-tensor gives
\begin{align*}
\ell(f)^*\ell(g)(h_1 \otimes \cdots \otimes h_n) = (g,f)_H h_1 \otimes \cdots \otimes h_n.
\end{align*}
Since
\begin{align*}
(g,f)_H I_{\mathcal{F}(H)}(h_1 \otimes \cdots \otimes h_n) = (g,f)_H h_1 \otimes \cdots \otimes h_n,
\end{align*}
the two operators agree on every elementary tensor of positive length.
[/step]
[step:Extend the equality from the algebraic Fock space to the full Fock space]
The preceding two steps show that
\begin{align*}
\ell(f)^*\ell(g)\xi = (g,f)_H I_{\mathcal{F}(H)}\xi
\end{align*}
for every $\xi \in \mathcal{F}_{\mathrm{alg}}(H)$, because both sides are linear maps and $\mathcal{F}_{\mathrm{alg}}(H)$ is spanned by $\Omega$ and elementary tensors.
Both $\ell(f)^*\ell(g)$ and $(g,f)_H I_{\mathcal{F}(H)}$ are bounded linear operators on $\mathcal{F}(H)$. Since they agree on the dense subspace $\mathcal{F}_{\mathrm{alg}}(H)$, they agree on all of $\mathcal{F}(H)$. Hence
\begin{align*}
\ell(f)^*\ell(g) = (g,f)_H I_{\mathcal{F}(H)}.
\end{align*}
This proves the creation-annihilation relation.
[/step]
