Are the creation operators on the fermionic Fock space bounded linear operators (do they have finite operator norm)?

Question

Let $H$ be a Hilbert space, denoting single-particle states, and $\mathfrak{F}$ be the fermionic Fock space. If $f\in H$, then is the creation operator $c^*(f)$ a bounded linear operator on $\mathfrak{F}$? If so, what is its norm?

EDIT: Thanks to @J.Murray for his answer, I think the following should be rigorous.

Let $|n\rangle$ denote the occupational basis where $n = \mathbb{N}^\mathbb{N}$ and $|n|=\sum_0^\infty n_i <\infty$ and $n_i=0,1$. Notice that $c_0^*|n\rangle=0$ if $n_0=1$ and $c_0^*|n\rangle =|n+\delta_0\rangle$ if $n_0=0$. Hence, if $|\psi\rangle$ is in the fermionic Fock space, then $|\psi\rangle = \sum |n\rangle \langle n|\psi \rangle$ and thus $$ ||c_0^* |\psi\rangle ||^2 \le \sum |\langle n|\psi \rangle|^2 =||\psi||^2<\infty $$ Hence, $c_0^*$ is a bounded operator and then it's clear that $||c_0^*||=1$.

(Technically, $c_0^*$ is only defined on a dense linear subspace of the Fock space and we need to use the continuous linear extension so that it is well-defined on the entire Fock space)

Answer 1

Yes, it is bounded with norm $\Vert f \Vert$.

To see this more clearly, note that we are always free to choose an orthonormal basis $\{e_i\}$ of $H$ such that $f = \alpha e_1$ for some $\alpha\in\mathbb C$; from there, we can say $$c^*(f) = \alpha \cdot c^*(e_1) \equiv \alpha \cdot c^*_1$$ $$\Vert c^* (f) \Vert_{op} = |\alpha|\cdot \Vert c^*_1\Vert_{op}$$

From there, let $\psi$ be an arbitrary (normalized) state in the Fock space. The norm of $c^*_1 \psi$ is

$$\Vert c^*_1\psi\Vert = \langle c^*_1\psi ,c^*_1\psi \rangle = \langle \psi,c_1 c^*_1 \psi \rangle = 1-\langle \psi,c^*_1 c_1 \psi \rangle$$

where we've used the anticommutation relations and the fact that $c^*$ and $c$ are adjoints of one another.

If $\psi$ is a Fock state, then this quantity is either zero or one; if it is a superposition of Fock states then it is somewhere in between. In any case, it isn't larger than 1, so

$$\Vert c^*_1 \Vert_{op} = \sup_{\psi \in H} \frac{\Vert c^*_1 \psi\Vert }{\Vert \psi \Vert} = 1$$

and therefore

$$\Vert c^*(f)\Vert_{op} = |\alpha| = \Vert f \Vert$$