# Can Fock space be written as a tensor product? (Cohen Tannoudji Vol. 3 book)

$$\newcommand{\ket}[1]{\left|#1\right\rangle} \newcommand{\tl}[1]{\tag{#1}\label{#1}}$$

Let us define the Fock space for bosons (S) by: $$\mathcal{E}_{Fock}^{S}=\mathcal{E}_{Fock}^{S}(0) \oplus\mathcal{E}_{Fock}^{S}(1) \oplus\mathcal{E}_{Fock}^{S}(2)\oplus \dots \oplus\mathcal{E}^S_{Fock}(N) \oplus \dots$$ where $$\mathcal{E}_{Fock}^{S}(N)$$ is state space of $$N$$ particles after symetrization.

The questions below come from reading the Comments from Vol. 3 from Choen Tannoudji book p. 1595 and 1596.

Is it true or false that

$$\mathcal{E}_{Fock}^{S}$$ can be written as the tensor product of Fock spaces $$\mathcal{E}_{Fock}^{u_i}$$ associated with the individual orthogonal states $$\ket{u_i}$$, each $$\mathcal{E}_{Fock}^{u_i}$$ being spanned by the kets $$\ket{n_i}$$ where $$n_i$$ takes on all integer values (from zero to infinity for bosons, from zero to one for fermions)

($$n_i$$ is the occupation number)

In other words, is the following equality true? $$\mathcal{E}_{Fock}^{S} = \mathcal{E}_{Fock}^{u_1}\otimes \mathcal{E}_{Fock}^{u_2}\otimes\dots\otimes\mathcal{E}_{Fock}^{u_i}\otimes\dots\tl{01}$$

The justification provided in the book is that: The basis states of $$\mathcal{E}_{Fock}^{S}$$ which I know can be written as $$\!\!\!\!\!\!\ket{n_i,n_j,\dots,n_l,\dots} =cS_N \ket{1:u_i;2:u_i;..;n_i:u_i;n_i+1:u_j;..;n_i+n_j:u_j;\dots} \tl{02}$$ (with $$S_N$$ being the symmetrization operator $$S_N=1/N!\sum_\alpha P_\alpha$$ and $$c$$ a normalization constant which is derived in p. 1594, and which I think it is irrelevant)

can also according to the comment of the book be written as the following tensor product:

$$\ket{n_1,n_2,\dots,n_i,\dots}=\ket{n_1}\otimes \ket{n_2}\otimes\dots\otimes\ket{n_i}\otimes\dots \tl{03}$$

I cannot justify and here is a simple example that allow us to compare \eqref{02} and \eqref{03}, from \eqref{02} we have: $$\ket{n_1=2,n_3=1}=c\frac{1}{3!}\sum_{\alpha} P_\alpha \ket{1:u_1;2:u_1;3:u_3}$$ the $$\ket{1:u_1;2:u_1;3:u_3}$$ is what Tannoudji calls the mathematical ket: $$\ket{1:u_1;2:u_1;3:u_3}=\ket{1:u_1}\otimes\ket{2:u_1}\otimes\ket{3:u_3}$$ Explicitly the previous summation yields: $$\ket{n_1=2,n_3=1}=c\frac{2}{3!}(\ket{1:u_1;2:u_1;3:u_3}+\ket{1:u_1;2:u_3;3:u_1}+\ket{1:u_3;2:u_1;3:u_1})$$ On the other hand \eqref{03} tells us that: $$\ket{n_1=2,n_3=1}=\ket{n_1=2}\otimes \ket{n_3=1}$$ and since $$\ket{n_1=2} = \frac{1}{2!}\sum_\alpha P_\alpha\ket{1:u_1;2:u_1}=\ket{1:u_1;2:u_1}$$ and $$\ket{n_3=1} = \frac{1}{1!}\sum_\alpha P_\alpha\ket{3:u_3}=\ket{3:u_3}$$ we find that $$\ket{n_1=2,n_3=1}=\ket{1:u_1;2:u_1}\otimes \ket{3:u_3} = \ket{1:u_1;2:u_1;3:u_3}$$ Clearly \eqref{02} and \eqref{03} are different.

Questions: Where is the mistake and how to make compatible the direct sum point of view on the Fock space and the tensor product point of view? I also would appreciate some other sources that I can read about this subject, so far only in this one I found this tensor product view.

Thank you

• The red coloring is telling you that MathJax doesn’t support \ket. You can use | and \rangle to make a ket. Commented Nov 24, 2022 at 5:27

Yes, $$(2)$$ and $$(3)$$ are formally different kets. The Hilbert spaces $$\mathcal E_{Fock}^S$$ and $$\mathcal E_{Fock}^{u_1} \otimes \mathcal E_{Fock}^{u_2} \otimes \ldots$$ are not equal. They are canonically isomorphic, however, so for all intent and purposes, they can be identified.
The reason for this is that they both have an orthonormal basis of kets $$|n_1,n_2,\ldots\rangle$$ indexed by lists of integers (with finitely many non-zero terms). So there is a natural unitary operator : $$U\left\{\begin{array}{ccc}\mathcal E_{Fock}^{u_1} \otimes \mathcal E_{Fock}^{u_2} \otimes \ldots & \longrightarrow & \mathcal E_{Fock}^S \\ |n_1\rangle\otimes |n_2\rangle\otimes \ldots& \longmapsto & cS_N |1:u_i;2:u_i;..;n_i:u_i;n_i+1:u_j;..;n_i+n_j:u_j;\dots\rangle\end{array}\right.$$
You don't lose much by using $$U$$ to identify both Hilbert spaces and write : $$|n_1\rangle\otimes |n_2\rangle\otimes \ldots= cS_N |1:u_i;2:u_i;..;n_i:u_i;n_i+1:u_j;..;n_i+n_j:u_j;\dots\rangle$$