Interacting fields and Hilbert space "Two fields with different mass parameters lies in two different Hilbert space."
How am I supposed to prove this using the concept that a field consist of a collection of harmonic oscillators? Tom banks in his book on qft gives a suggestion of using overlap of ground state to arrive at a conclusion. But I am missing something in between. I am not able to figure out how to start.
For reference, one could look on to problem 2.4 of Banks's book. 
 A: The question is essentially the first step of Haag's Theorem. 
In his original paper "On Quantum Field Theories"  [Dan. Mat. Fys. Medd. 29, n12 (1955) pp.1-37]
Haag actually works out this problem.  Anyone looking for a more detailed answer to this question should start there.  Here's the link:
http://cds.cern.ch/record/212242/files/
it starts on page 17. 
There is also a decent discussion of this problem in Duncan's "Conceptual Framework of QFT", chapter 10.5.
The general gist of the proof is that for two finite collections of harmonic oscillators, there exists a unitary operator connecting their Hilbert spaces, 
but for an infinite number of oscillators no such operator can exist.
Here's a brief, incomplete sketch of the proof.
Suppose we have two finte colletions of creation and annihilation operators.
$a_1(n),a_1^\dagger(n)$ for mass $m_1$ and
$a_2(n),a_2^\dagger(n)$ for mass $m_2$.  (both with $n=1,2,3,...,N$) 
Then the Hilbert spaces we're interested in, $\mathcal{H}_1$ and $\mathcal{H}_2$,  can be generated by applying finite polynomials of the creation operators to the respective ground states, $|0_1\rangle$ and $|0_2\rangle$
If we assume the two sets can be related by the equations
$$a_1(n)=\cosh(\epsilon(n))a_2(n)+\sinh(\epsilon(n))a^\dagger(n)$$
$$a_1^\dagger(n)=\sinh(\epsilon(n))a_2(n)+\cosh(\epsilon(n))a^\dagger(n)$$
for some function $\epsilon(n)$.
Then we can construct a unitary operator, 
$U=e^{iT}$  with  
$$T=\sum_{n=1}^N\frac{i\epsilon(n)}{2}
\left(a_2^\dagger(n)a_2^\dagger(n)-a_2(n)a_2(n)\right),$$
that realizes the above transformation for the creation and annihilation operators.  
$$a_1(n)=Ua_2(n)U^\dagger$$
$$a_1^\dagger(n)=Ua_2^\dagger(n)U^\dagger$$
Note that as long as $N$ is finite, $T$ is a perfectly fine self adjoint operator.
Assuming that the ground states are unique vectors satisfying, 
$$a_j(n)|0_j\rangle=0\  for\  all\  n\ \ (j=1,2),$$
leads to $|0_1\rangle=U|0_2\rangle$, and a similar expression for every vector in $\mathcal{H}_1$.  Meaning that every vector in $\mathcal{H}_1$ can be written as a finite combination of vectors in $\mathcal{H}_2$
For the creation and annihilation operators associated with free fields there are an infinite number of creation and annihilation operators.  By playing around with the usual expressions for $\phi$ and $\pi$ it's possible to get the relationship between creation and annihilation operators into the form we assumed earlier, but this time with the continuous parameter $\bar p$ replacing $n$.
$$a_1(\bar p)=\cosh(\epsilon(\bar p))a_2(\bar p)+\sinh(\epsilon(\bar p))a^\dagger_2(\bar p)$$
$$a_1^\dagger(\bar p)=\sinh(\epsilon(\bar p))a_2(\bar p)+\cosh(\epsilon(\bar p))a^\dagger_2(\bar p)$$
Here $\epsilon(\bar p)$ is equal to something like $\frac{1}{2}\ln\left(\frac{\bar p^2+(m_1)^2}{\bar p^2+(m_2)^2}\right)$, depending on the conventions you are using.
(note that to get this expression you'll have to assume that the fields and the conjugate momenta coincide at some fixed time, $\phi_{m_1}(\bar x,0)=\phi_{m_2}(\bar x,0)$, etc)
In this case we find that the generator of this transformation is given by,
$$T'=\int d\bar p\frac{i\epsilon(\bar p)}{2}
\left(a_2^\dagger(\bar p)a_2^\dagger(\bar p)-a_2(\bar p)a_2(\bar p)\right)$$
So if $|0\rangle_1$ is in $\mathcal{H}_2$ it would have to have the form 
$$|0_1\rangle=e^{iT'}|0_2\rangle$$.
But if you try to calculate the norm of this vector you run into some trouble,
Namely that $$\|T|0_2\rangle\|=\infty.$$
Meaning that $e^{iT'}$ is not a proper operator, and $|0_1\rangle$ cannot be a normalizable vector in $\mathcal{H}_2$.  A similar argument can be made for every vector in $\mathcal{H}_1$, since an arbitrary vector can be expressed in terms of creation operators applied to $|0_1\rangle$.  (or at least sufficiently approximated by such vectors)
For anyone looking for more information about the conceptual implications of this problem, I would recommend Earman and Fraser's paper (Erkenntnis 64 (2006) 305) found here http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.476.8781
, and the references they cite.
A: A thorough statement, proof, and discussion is given in Theorem 3.1 (Haag's theorem for free fields) of the recent Ph.D. thesis http://edoc.hu-berlin.de/dissertationen/klaczynski-lutz-2015-11-06/PDF/klaczynski.pdf by Lutz Klaczynski.
''different Hilbert space'' must be viewed with a grain of salt as all separable Hilbert spaces in infinite dimensions are isomorphic. What is meant is Hilbert spaces with inequivalent unitary representations of the Poincare group. so that no unitary map respecting the Poincare commutation rules exists.
