Is true that $F(H)\otimes F(H)\cong F(H)$? where $F(H)$ is the full fock space of Hilbert space $H$

Question

Let $H$ be a Hilbert space. Define the full fock space $F(H)=\bigoplus\limits_{n=0}^\infty H^{\otimes n}$ where $H^{\otimes 0}=\Bbb{C}\Omega$, $\Omega$ is an element of $H$ of unit norm. Let us define the map $\phi:F(H)\otimes F(H)\to F(H)$ by $$\phi((a\Omega\oplus x_1\oplus x_2\oplus\cdots)\otimes(b\Omega\oplus y_1\oplus y_2\oplus\cdots))=ab\Omega\oplus (a y_1+b x_1)\oplus (a y_2+x_1\otimes y_1+b x_2)\oplus\cdots$$

I think this $\phi$ defines the unitary operator between $F(H)\otimes F(H)$ and $F(H)$. I can verify that $\phi$ preserves inner product and it is onto since $\phi((a\Omega\oplus x_1\oplus x_2\oplus\cdots)\otimes(1\Omega\oplus 0\oplus 0\oplus\cdots))=a\Omega\oplus x_1\oplus x_2\oplus\cdots$. Therefore, it unitary. But I want to compute what is the adjoint of $\phi$. Initially, I thought it would be $\phi^*(\xi)=\xi\otimes (1\Omega\oplus 0\oplus 0\cdots)$ for $\xi\in F(H)$. But with this formula $\langle \phi^*(\xi),\eta\otimes\zeta)=0$ if the vaccum of $\zeta$ is $0$ i.e. the first component of $\zeta$ is $0$ for any $\xi$. That should not happen. There is something that I'm missing, but couldn't figure out. There is one doubt: is $\phi$ well-defined?

Can anyone help me in this regard? Thanks for your help in advance.

Answer 1

(Disclaimer: I want to warn against a common mistake when dealing with tensor products, which is to forget that the tensor product is the (maybe closed) linear span of the elementary tensors. For some things one can comfortably work with elementary tensors, but for other things, like calculating norms, sums of elementary tensors cannot be avoided. Mentioning this as someone who made this mistake several times)

Your $\phi$ is well-defined, as it is a linear extension of $$ (a\Omega\oplus x_1\oplus x_2\oplus\cdots)\otimes (0\oplus\cdots\oplus 0\oplus y_n\oplus0\oplus\cdots)\longmapsto (\overbrace{0\oplus\cdots\oplus 0}^n\oplus ay_n\oplus x_1\otimes y_n\oplus\cdots) $$

There is an issue, though, which is how you see $F(H)\otimes F(H)$ as a Hilbert space. What I mean is that in general the inner product on the tensor product is defined as $$ \langle x\otimes y,z\otimes w\rangle=\langle x,z\rangle\langle y,w\rangle.$$ With this definition, your $\phi$ does not preserve the inner product. Indeed, $$ \langle \phi((\Omega\otimes(0\oplus y)),\phi((0\oplus z)\otimes\Omega)\rangle=\langle 0\oplus y,0\oplus z\rangle=\langle y,z\rangle, $$ while $$ \langle (\Omega\otimes(0\oplus y),(0\oplus z)\otimes\Omega\rangle=\langle \Omega\oplus0, 0\oplus z\rangle\langle 0\oplus y,\Omega\oplus0\rangle=0. $$

But what you seem to be doing is to consider $F(H)$ as having an intrinsic tensor product, by which you have $F(H)\otimes F(H)=F(H)$ precisely via your $\phi$. The problem is that $\phi$ is not injective (and this is the mistake I mentioned at the beginning): for example $$ \phi((0\oplus x_1)\otimes (0\oplus0\oplus x_1\otimes x_1))=0\oplus0\oplus0\oplus (x_1\otimes x_1\otimes x_1)=\phi((0\oplus0\oplus x_1\otimes x_1)\otimes (0\oplus x_1)). $$ So the element $(0\oplus x_1)\otimes(0\oplus0\oplus x_1\otimes x_1)-(0\oplus0\oplus x_1\otimes x_1)\otimes (0\oplus x_1)$, which is not an elementary tensor, is in the kernel of $\phi$.

And this shows that problem with using $\phi$ to define the inner product on $F(H)\otimes F(H)$: what you get is not an inner product. If this is not the inner product you are using on $F(H)\otimes F(H)$, you need to tell us what you use.