Unitarily Inequivalent Representations The definition of unitarily equivalent representations I am using is the one given here: https://en.wikipedia.org/wiki/Haag%27s_theorem.
Now in this text http://www.sa.infn.it/Massimo.Blasone/documents/cantrans.pdf we are given two sets of operators on some Hilbert space that both satisfy the canonical commutation relations from quantum field theory; $|0\rangle$ is the vacuum for one set of operators, and $|0(\theta)\rangle$ is the vacuum from the other. On page five of the author makes the claim that because $\langle 0|0(\theta)\rangle =0\,,$ we have two inequivalent representations. I am a little confused by why this is true. 
I am also a little confused by why the author is claiming that acting by $U(\theta)$ on the vacuum leads out of the original Hilbert space when it seems everything there belongs to the same Hilbert space, the one originally defined, but maybe this is just a semantics issue.
 A: The claim is that $|0(\theta)\rangle$ lies outside the Hilbert space built on the original vacuum $|0\rangle$.
To check that this true, consider the overlap of the new vacuum $|0(\theta)\rangle$ and the (unnormalized) basis states $(a_k^\dagger)^n|0\rangle$ generated from $|0\rangle$, taking into account that $a_k(\theta) = a_k + \theta_k$, $a_k = a_k(\theta) - \theta_k$, and $a_k(\theta)\;|0(\theta)\rangle = 0$. We have successively:
\begin{eqnarray}
\langle 0|a_k^n|0(\theta)\rangle &=& \langle0|a_k^{n-1}[a_k(\theta)-\theta_k]|0(\theta)\rangle\\
&=& \langle 0|a_k^{n-1}a_k(\theta)|0(\theta)\rangle - \theta_k \langle 0|a_k^{n-1}|0(\theta)\rangle \\
&=& -\theta_k \langle 0|a_k^{n-1}|0(\theta)\rangle \\
&=& (-\theta_k)^2 \langle 0|a_k^{n-2}|0(\theta)\rangle \\
&=& \ldots\\
&=& (-\theta_k)^n \langle 0|0(\theta) \rangle\\
&=& 0
\end{eqnarray}
A similar result holds for basis states involving multiple modes $k$. In other words, the new vacuum $|0(\theta) \rangle$ is orthogonal not only on the original vacuum, but also on each of the canonical basis states generated from it. Hence it must lie outside the corresponding Hilbert space, despite being generated using a unitary $U(\theta)$ that is formulated in terms of $a_k$-s, $a_k^\dagger$-s.
A: Let H be the prehilbertian vector space of the possible states of One particle. And F its associated Fock space. F has an orthonormal Hilbert occupancy number basis of vectors (each one is a sequence of entire numbers all equal to 0 except for a finite number of them). F is generated by finite linear combinations of these vectors AND by the converging  $\Sigma_{n_1,n_2,n_3,.}c(n_1,n_2,n_3,.) |n_1,n_2,n_3,..\rangle$ where $\Sigma_{n_1,n_2,n_3,.} |c(n_1,n_2,n_3,.)|^2$ is finite.
If dim(H) = 1 we have only one mode and the occupancy basis is (0) (1) (2) ...
If dim(H) = 2 the sequences are (0,0) (0,1) (0,2) .... (1,0) (1,1) (1,2).... 
A CCR representation is given by annihilation and création opérations $a_k$ and  $a^\dagger_k$ (one for each mode).
Another representation is given by $a_k + \theta_k$ and $a^\dagger_k + \theta^*_k$. (here $\theta$ is a complex number. They are both canonical because $$[a_j,a^\dagger_k] = [a_j + \theta_j,a^\dagger_k + \theta_k^*]= \delta_{jk}$$ 
If these two representations are unitarily equivalent there is a unitary map U such that $$  a_k U = U a_k + U \theta_k = U a_k + \theta_k U$$ for each mode. If 
$|x\rangle = U |0\rangle$ then $$\langle x|x \rangle = 1 $$  and
      $$ a_k |x\rangle = U (a_k + \theta_k) |0\rangle =  \theta_k |x\rangle$$ 
this equality gives recurrence relations between the coordinates of x in the occupatin basis. One can easily find that if dim H = 1 (one mode)
$$ |x\rangle = c_0 |0\rangle + c_1 |1\rangle + .... = c_0 (1 + \theta_1 + \theta_1^2/2! ...)|0\rangle $$ so
$$\langle x | x\rangle = |c_0|^2 exp (|\theta_1|^2)$$
there is no problem to find  $c_0$ so that x is unitary.
If dim H = 2 we find
$$\langle x | x\rangle = |c_{00}|^2 exp (|\theta_1|^2 + |\theta_2|^2)$$
if the set of modes is enumerable we have
$$\langle x | x\rangle = |c_{00...}|^2 exp (\Sigma_k |\theta_k|^2)$$
this shows that when the sum does not converge, we cannot find $c_{00...}$ so
thar $\langle x | x\rangle = 1$ and it is the case when all the $\theta_k$ are equal. In this case we get a contradiction with the existence of unitary equivalentrepresentations.
There is no need to consider that U | 0> is outside the Fock space (and orthogonal to it) because the unitary map U does not exist.
