Unphysical fields in quantum (and classical) field theory Good morning, my question is very specific and at the same time very general. How can I understand from a lagrangian density if a field is physical or unphysical. How can I make it evident from the expression of the Lagrangian? 
Let me make an example. Let $\phi_1$ and $\phi_2$ be two real scalar fields, with Lagrangian density:
\begin{equation}
\mathcal{L}=2\partial_\mu\phi_1\partial^\mu\phi_2
\end{equation}
How can I say that one of the two fields is unphysical?
My thinking is that if I complete the square I would have a term like
\begin{equation}
\mathcal{L}=\partial_\mu(\phi_1+\phi_2)\partial^\mu(\phi_1+\phi_2)-(\partial_\mu\phi_1)^2-(\partial_\mu\phi_2)^2
\end{equation}
but how may this Lagrangian make me to conclude that one of the two field is unphysical?
I would like an answer yes, specific, but not on the example, in general, I would like to understand the way to determine if a field is physical since I know that in the example one of them is unphysical I want to know how to prove it.
 A: The kinetic term for a field should always have a positive sign as $L = T -V$. If some field has a negative sign for its kinetic term than it is unphysical.
Let met clarify this with an example from q.e.d. after gauge fixing we obtain:
$$\mathcal{L} = -\frac{1}{2}\partial_\mu A_\nu \partial^\mu A^\nu$$ 
If we use the metric $g = tr(- + + +)$ this becomes:
$$\mathcal{L} = \frac{1}{2}\partial_\mu A_0 \partial^\mu A_0 - \sum_i\frac{1}{2}\partial_\mu A_i \partial^\mu A_i$$
$$\mathcal{L} = \underbrace{-\frac{1}{2}\partial_tA_0\partial_tA_0}_\text{Wrong sign} + \underbrace{\sum_i\frac{1}{2}\partial_tA_i \partial_t A_i}_\text{correct sign} + \underbrace{\text{terms with spatial derivatives}}_{irrelevant}$$
We see that the first term resembles a kinetic term with the wrong sign such that it must be unphysical. The second term contains the kinetic terms for the three $A_i$ fields that are physical. The last term is irrelevant for this discussion.
If you were to correctly quantize such a field you would find that $<0|A_0|0>\  <\  0$ such that is has a negative norm ! This is even more proof that the field is unphysical.

As accidental Fourier Transform pointed out there are many other cases of unphysical fields such as the Fadeev-Popov ghosts that arise in non-abelian gauge theories. Pauli-Villars ghosts that are used in regularisation of certain propagators. Longitudinal photon polarizations that are "killed" by the gauge constraint. And Goldstone bosons that are used to give mass to other particles can all be considered unphysical in some sense. 
I will not go into detail on them because your question seemed to be focused on negative norm states and not more general ghosts.

As for the answer to your specific Lagrangian, completing the squares will not work and to be fairly honest, I think that it is plain wrong as you are no longer studying the original Lagrangian.
All you can do is to redefine your fields as linear combinations of others (say A,B) such that you get recognizable kinetic terms.
For instance: $\phi_1 = \frac{1}{2}(A+B)$ and $\phi_2 = \frac{1}{2}(A-B)$. Plug this into your Lagrangian and you should find:
$$\mathcal{L}_{example} = -\frac{1}{2}\partial_\mu A\partial^\mu A + \frac{1}{2}\partial_\mu B \partial^\mu B$$
From this you can extract the kinetic terms (use the metric of your lecturer not the one I gave above !) and find which one has the wrong sign, this is your unphysical field!
Good luck, and I hope this helped :)
