# Annihilation and creation operator on fock space

I try to understand why the equality/inequality you can see below holds.

Let $$\mathfrak{h}$$ be a separable Hilbert space and define the Fock space $$\mathcal{F}:= \oplus_{N=0}^\infty\otimes^N\mathfrak{h}$$. For $$\varphi\in\mathfrak{h}$$ let $$(a(\varphi), D(\sqrt{\mathcal{N}}))$$ and $$(a^\dagger(\varphi), D(\sqrt{\mathcal{N}}))$$ be the annihilation and creation operator with $$\mathcal{N}$$ the number operator. Then for all $$\Psi \in D(\sqrt{\mathcal{N}}))$$ is holds that

\begin{align} \vert\vert a^\dagger(\varphi)\Psi\vert\vert_\mathcal F &= \vert\vert\varphi\vert\vert_\mathfrak h \cdot \vert\vert \sqrt{\mathcal N + 1}\Psi \vert\vert_\mathcal F\\ \vert\vert a(\varphi)\Psi\vert\vert_\mathcal F &\leq \vert\vert\varphi\vert\vert_\mathfrak h \cdot \vert\vert\sqrt{\mathcal N + 1}\Psi\vert\vert_\mathcal F \end{align}

For the first i tried to use the definition of the operator \begin{align} \vert\vert a^\dagger(\varphi)\Psi\vert\vert_\mathcal F = \vert\vert (0,\varphi\cdot \psi_0, \sqrt{2}\varphi\otimes\psi_1, \sqrt{3}\varphi\otimes\psi_2,\dots)\vert\vert \end{align} but i am note sure how to proceed and how to deal with the tensor product inside the norm.

When you write $$\mathcal F = \bigoplus_{n=0}^\infty \otimes^n \mathfrak h$$, the $$\oplus$$ and $$\otimes$$ are understood as direct sum and tensor product of Hilbert spaces. They carry a hermitian product and a norm built from that of $$\mathfrak h$$.
This is the definition of the norm in the case of binary $$\oplus$$ and $$\otimes$$. They generalize easily to infinite operations.
For the direct sum, the different summand are taken to be orthogonal : if $$x\in \mathfrak h,y \in \mathfrak f$$, then $$(x,y)\in \mathfrak h\oplus \mathfrak f$$ has norm : $$\|(x,y)\|_{\mathfrak h\oplus \mathfrak f}^2 = \|x\|_{\mathfrak{h}}^2 + \| y\|_{\mathfrak f}^2$$
For the tensor product, the hermitian product is defined by , if $$x,x' \in \mathfrak h$$, $$y,y'\in \mathfrak f$$ : $$(x\otimes y , x'\otimes y')_{\mathfrak h\otimes \mathfrak f} = (x,x')_{\mathfrak h} (y,y')_{\mathfrak f}$$ In particular, $$\|x\otimes y\|_{\mathfrak{f}\otimes \mathfrak f} = \|x \|_{\mathfrak h} \| y\|_{\mathfrak f}$$