Fock space is defined as
$$\mathcal{F}_s(\mathcal{H})=\bigoplus_{k=0}^\infty \overset{k}{\bigotimes}_s\mathcal{H},$$
and we can write it as
$$\mathcal{F}_s(\mathcal{H})=H_0\oplus H_1 \oplus H_2 \oplus \dots,$$
where $H_i$ means the Hilbert space of $i$ particles.
I understand the spaces with nonzero number of particles. But what is the Hilbert space of zero number of particles, $H_0$? It seems like the only state should be $\vert 0 \rangle$. Is that still a legitimate Hilbert space (that follows the axioms)?
Best Answer
The complex numbers $\mathbb{C}$ form a one-dimensional Hilbert space. The axioms of a Hilbert space are that it be a complete metric space with a metric $d(x,y)=\sqrt{\langle x-y,x-y\rangle}$ that is defined in terms of an inner product $\langle x,y\rangle$. The complex numbers are endowed with a sesquilinear inner product $\langle x,y\rangle=x^{*}y$, which generates the usual distance measure $d(x,y)=|x-y|^{2}$, under which $\mathbb{C}$ is, of course, complete.
The one-dimensional Hilbert space $H_{0}$ with the bare vacuum $|0\rangle$ as its only basis state is just isomorphic to $\mathbb{C}$. You may be getting confused by the fact that the physical state of a quantum-mechanical system is not specified by an element of a Hilbert space, but by a ray in Hilbert space (or, equivalently, an element of a projective Hilbert space). The norm of a state $|0\rangle$, versus say $2|0\rangle$, is not physically meaningful. They both lie along the same ray (as does every element of $H_{0}$), but they are different elements of the Hilbert space.