In my lecture we defined the Fock space as follows:

Let $\mathfrak{h}_1,\mathfrak{h}_2,\dots $ be a sequence of separable Hilbert spaces. Let $\mathcal{H}_N=\mathfrak{h}_1\otimes \dots \otimes \mathfrak{h}_N$ and $\mathcal{H}_0:= \mathbb{C}$. Then \begin{equation} \mathcal{F}:= \bigoplus_{N=0}^{\infty} \mathcal{H}_N \end{equation} is called Fock space.

I asked myself why we have to set $\mathcal{H}_0:= \mathbb{C}$?

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    $\begingroup$ I don't actually know the answer, but since quantum states are only defined up to a complex number presumably all states in $\mathcal H_0$ are in the same equivalence class and so physically there is only one physically unique state in the zero particle space. $\endgroup$
    – Charlie
    May 2, 2021 at 12:40
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    $\begingroup$ Well, $\mathcal{H}_0$ is a one-dimensional (complex) vector space. Hence, it is isomorphic to $\mathbb{C}$ as a vector space over itself (i.e. over the field of complex numbers). $\endgroup$ May 2, 2021 at 13:19

1 Answer 1


The no-particle vacuum state is one dimensional. Any one-dimensional vector space is isomorphic to ${\mathbb C}$.

  • $\begingroup$ thank you very much! $\endgroup$
    – uzizi_1
    May 2, 2021 at 17:44

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