Very basic question about QFT at finite density This must be the first question everyone asks when starting to study field theory at finite density and zero temperature. 
To introduce a finite density one adds a Lagrange multiplier which fixes the number of particles in a non-relativistic field theory or the number of particles minus the number of antiparticles in a relativistic field theories. To be concrete, in a non-relativistic field theory one has the Lagrangian density:
$$\mathcal L=i\psi ^*\partial _t \psi -{1\over 2m}\nabla\psi ^*\nabla\psi - V[\psi ^*\psi]$$
and then the same theory at finite density is described by:
$$\mathcal L=i\psi ^*\partial _t \psi -{1\over 2m}\nabla\psi ^*\nabla\psi - V[\psi ^*\psi]+\mu \psi ^*\psi \,\,\,\,\,\,\,\, [1]$$ 
where $\mu$ is the chemical potential and the number of particles is $N=\int \psi ^*\psi$.
What disturbs me is that one may redefine the field:
$$\chi\equiv e^{-i\mu t}\psi$$
with $t$ the time. And after replacing in [1]:
$$\mathcal L=i\chi ^*\partial _t \chi -{1\over 2m}\nabla\chi ^*\nabla\chi - V[\chi ^*\chi]$$
The "finite density term" disappears. Moreover, the path integral measure is invariant under this trivial field redefinition so that the finite density QFT seems equivalent to the zero density one. (Those who are familiarized with the non-relativistic limit of relativistic equation —such as Klein-Gordon or Dirac— will recognize this term and field redefinition.)
Something similar happens in the relativistic case. 
What am I overlooking? 
(Note that the theory is at zero temperature so I am in real time formalism and I don't have to impose periodic boundary conditions (I think). I am interested in time evolution, scattering, etc.)
Reply to Ron's answer:
(I here will overlook Thomas' comments, I will answer him tomorrow.)
Ron makes an interesting point, but I think his answer is not correct. The Hamiltonian density of the theory in the original fields is:$$\mathcal H={1\over 2m}\nabla\psi ^*\nabla\psi + V[\psi ^*\psi]-\mu \psi ^*\psi $$
What he correctly pointed out is that the Hamiltonian in the new fields is not just the previous Hamiltonian written in the new fields. This is right and in classical mechanics it is well-know that one has to add a piece to the Hamiltonian when one makes an explicit time dependent transformation. He pointed out the case of quantum mechanics, and the field theoretic version is:
$$i\partial _t \, \chi =i\partial _t \, (e^{-i\mu t}\psi )=i(-i) \mu  \, e^{-i\mu t}\psi + e^{-i\mu t}[\psi , H]=[e^{-i\mu t}\psi ,\, \mu N + H] $$
And therefore the Hamiltonian for the field $\chi$ is: 
$$H'=\mu N + H=\int {1\over 2m}\nabla\chi ^*\nabla\chi + V[\chi ^*\chi]$$
This is of course the Hamiltonian corresponding to the Lagrangian written in terms  of $\chi$. In this way, one sees how the "standard" theory (without the $\mu$ term) is recovered in the Hamiltonian formalism as it was in path integral formalism. And this makes a lot of sense because the theory with $\mu = -m$ has to be equivalent to the theory with $\mu=0$ because we are just subtracting the rest energy.
Of course, this is true as long as one does not impose periodic boundary conditions. 
 A: This is indeed a little more subtle (You can also see the problem by realizing that the free fermion propagator only depends on $p_0-\mu$, so that integrals over $p_0$ are $\mu$ independent, unless $\mu$ enters into the $i\epsilon$ prescription). There are basically two ways to analyze this more carefully:
1) Start from the path integral representation of the partition function. This means that we do have to go to euclidean space and impose boundary conditions on $\psi$. We find that the path integral contains convergence factors that depend on $\mu$, which turn into $i\epsilon$ prescriptions. See, for example, the text book by Negele and Orland. This can also be seen by carefully discretizing the theory on the lattice. We find that $\mu$ enters as a gauge field, that means as a $\exp(\pm \mu a)$ factor for forward/backward going links with lattice spacing $a$ (see, for example, arXiv:hep-lat/0308016). 2) Use canonical quantization and start from the fact that the ground state of the non-interacting theory is a filled Fermi sphere. This is described in Fetter and Walecka or Abrikosov, Gorkov, Dzyaloshinskii. In the end we get the same $\mu$ dependent $i\epsilon$ prescription. 
A: The transformation you made is well known in quantum mechanics--- it is the time-dependent phase transformation that corresponds to the freedom to add a constant to the Hamiltonian. The freedom to make this redefinition is the statement that the zero point of energy is arbitrary.
In order to do it correctly, you have to shift all energies. The chemical potential term is giving you an energy cost per extra particle, and you need to shift it by a constant in order to make this transformation give the same physics. Once you do this, the system looks exactly the same.
Your change of phase only sort-of works relativistically, you have to pick a frame to do the phase change.
