[proofplan]
We prove the formula from the defining adjoint identity by testing against the algebraic Fock space, the dense subspace spanned by the vacuum and elementary tensors. Orthogonality of tensor levels forces nearly all inner products to vanish, and the only nonzero case occurs when the target tensor has degree exactly one larger than the test tensor. In that case, the tensor-product [inner product](/page/Inner%20Product) gives the scalar $(f,h_1)_H$, and conjugate-linearity in the second variable converts this into the coefficient $(h_1,f)_H$ in the adjoint vector.
[/proofplan]
[step:Test the adjoint identity against the vacuum]
Let $\mathcal{F}_{\mathrm{alg}}(H)$ denote the algebraic direct sum of $\mathbb{C}\Omega$ and the finite tensor powers $H^{\otimes n}$, so $\mathcal{F}_{\mathrm{alg}}(H)$ is dense in $\mathcal{F}(H)$. For every elementary tensor $\xi \in \mathcal{F}_{\mathrm{alg}}(H)$, the vector $\ell(f)\xi$ belongs to a positive tensor level. Since the vacuum space $\mathbb{C}\Omega$ is orthogonal to every positive tensor level, we have
\begin{align*}
(\ell(f)\xi,\Omega)_{\mathcal{F}(H)} = 0.
\end{align*}
By the defining property of the adjoint,
\begin{align*}
(\xi,\ell(f)^*\Omega)_{\mathcal{F}(H)} = 0
\end{align*}
for every $\xi \in \mathcal{F}_{\mathrm{alg}}(H)$. Density of $\mathcal{F}_{\mathrm{alg}}(H)$ in $\mathcal{F}(H)$ and nondegeneracy of the Hilbert-space inner product imply $\ell(f)^*\Omega = 0$.
[/step]
[step:Compute the only tensor degree that can contribute]
Fix $n \geq 1$ and $h_1,\dots,h_n \in H$. Define
\begin{align*}
\eta := h_1 \otimes \cdots \otimes h_n \in H^{\otimes n}.
\end{align*}
To determine $\ell(f)^*\eta$, test against an arbitrary elementary tensor
\begin{align*}
\xi := g_1 \otimes \cdots \otimes g_m \in H^{\otimes m},
\end{align*}
where $m \geq 0$, with the convention that $m=0$ means $\xi=\Omega$. Since $\ell(f)\xi$ lies in $H^{\otimes(m+1)}$ and tensor levels are mutually orthogonal, we have
\begin{align*}
(\ell(f)\xi,\eta)_{\mathcal{F}(H)} = 0
\end{align*}
unless $n=m+1$.
[guided]
Fix $n \geq 1$ and choose vectors $h_1,\dots,h_n \in H$. We want to identify the vector $\ell(f)^*(h_1 \otimes \cdots \otimes h_n)$, so we test it against a spanning family of vectors in $\mathcal{F}(H)$. Let
\begin{align*}
\eta := h_1 \otimes \cdots \otimes h_n \in H^{\otimes n}.
\end{align*}
Let also
\begin{align*}
\xi := g_1 \otimes \cdots \otimes g_m \in H^{\otimes m}
\end{align*}
be an elementary tensor, where $m \geq 0$ and $m=0$ means $\xi=\Omega$.
The left creation operator raises tensor degree by one: $\ell(f)\xi$ belongs to $H^{\otimes(m+1)}$. The vector $\eta$ belongs to $H^{\otimes n}$. In the full Fock space, distinct tensor levels are orthogonal. Therefore, if $n \neq m+1$, the two vectors lie in orthogonal summands, and hence
\begin{align*}
(\ell(f)\xi,\eta)_{\mathcal{F}(H)} = 0.
\end{align*}
Thus the adjoint vector can only have a nonzero pairing with test tensors in $H^{\otimes(n-1)}$. This degree bookkeeping is the reason the annihilation operator removes exactly the first tensor factor.
[/guided]
[/step]
[step:Evaluate the matching degree inner product]
Now assume $m=n-1$. If $n=1$, then $\xi=\Omega$ and $\ell(f)\Omega=f \in H$. The tensor-product inner product gives
\begin{align*}
(\ell(f)\Omega,h_1)_{\mathcal{F}(H)} = (f,h_1)_H.
\end{align*}
If $n \geq 2$, then
\begin{align*}
\ell(f)\xi = f \otimes g_1 \otimes \cdots \otimes g_{n-1}.
\end{align*}
Using the tensor-product inner product on $H^{\otimes n}$, we obtain
\begin{align*}
(\ell(f)\xi,\eta)_{\mathcal{F}(H)} = (f,h_1)_H \prod_{j=1}^{n-1} (g_j,h_{j+1})_H.
\end{align*}
Equivalently, if
\begin{align*}
\zeta := h_2 \otimes \cdots \otimes h_n \in H^{\otimes(n-1)}
\end{align*}
for $n \geq 2$, and $\zeta := \Omega$ for $n=1$, then in both cases
\begin{align*}
(\ell(f)\xi,\eta)_{\mathcal{F}(H)} = (f,h_1)_H(\xi,\zeta)_{\mathcal{F}(H)}.
\end{align*}
[/step]
[step:Convert the scalar using the first-variable linear convention]
Since the inner product is linear in the first variable and conjugate-linear in the second variable, scalar multiplication in the second argument satisfies
\begin{align*}
(\xi,(h_1,f)_H\zeta)_{\mathcal{F}(H)} = \overline{(h_1,f)_H}(\xi,\zeta)_{\mathcal{F}(H)}.
\end{align*}
By conjugate symmetry of the Hilbert-space inner product,
\begin{align*}
\overline{(h_1,f)_H} = (f,h_1)_H.
\end{align*}
Therefore
\begin{align*}
(\xi,(h_1,f)_H\zeta)_{\mathcal{F}(H)} = (f,h_1)_H(\xi,\zeta)_{\mathcal{F}(H)}.
\end{align*}
Combining this with the previous step gives
\begin{align*}
(\ell(f)\xi,\eta)_{\mathcal{F}(H)} = (\xi,(h_1,f)_H\zeta)_{\mathcal{F}(H)}
\end{align*}
for every elementary tensor $\xi \in \mathcal{F}_{\mathrm{alg}}(H)$.
[/step]
[step:Use density to identify the adjoint vector]
The left creation operator $\ell(f)$ is bounded on $\mathcal{F}(H)$, so its adjoint is characterized by the identity
\begin{align*}
(\ell(f)\xi,\eta)_{\mathcal{F}(H)} = (\xi,\ell(f)^*\eta)_{\mathcal{F}(H)}
\end{align*}
for every $\xi \in \mathcal{F}(H)$. The previous step proves that the vector $(h_1,f)_H\zeta$ has this same pairing against every $\xi \in \mathcal{F}_{\mathrm{alg}}(H)$. Since $\mathcal{F}_{\mathrm{alg}}(H)$ is dense in $\mathcal{F}(H)$, the two vectors are equal:
\begin{align*}
\ell(f)^*\eta = (h_1,f)_H\zeta.
\end{align*}
Substituting the definition of $\eta$ and $\zeta$ gives
\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 empty tensor interpreted as $\Omega$ when $n=1$. Together with $\ell(f)^*\Omega=0$, this proves the formula.
[/step]
