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

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)

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

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

• I think this is almost rigorous, except for the fact that the adjoint $c$ of the creation operator $c^*$is not necessarily well-defined unless $c^*$ is a bounded operator, so we have a circular argument. – Andrew Yuan Aug 7 at 7:22
• @AndrewYuan The creation and annihilation operators are typically (initially) defined on the subspace of states with finite particle number, which is dense in the Fock space. On this domain, they can be explicitly shown to be mutually adjoint. Because they are bounded on this dense domain (as shown above), they can be extended to the entire Fock space via topological closure. – J. Murray Aug 7 at 13:08