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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.

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    $\begingroup$ What do you mean by unphysical? In your example the Lagrangian is clearly symmetrical in the two fields, so for sure you cannot that one is unphysical without further information. $\endgroup$ – fqq Mar 29 '17 at 9:00
  • $\begingroup$ @fqq actually I have to agree with you but that is the point of my question: I have an exercise which ask me to explain why one of the two fields with that Lagrangian density is unphysical. And I have also to rewrite the Lagrangian density in such a way that it is clear that reason. I do not have much more about that example, because I do not have the solution. What I would like to know is exactly what unphysical fields are and maybe I will be able to understand why the exercise ask me to prove that one of the two field is not physical. You're comment is maybe the whole reason why I asked this $\endgroup$ – Alessandro Mininno Mar 29 '17 at 9:07
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    $\begingroup$ You have kinetic mixing, and you should write down all operators allowed, including conventional ones, then make field redefinitions st the kinetic term is canonical $\endgroup$ – innisfree Mar 29 '17 at 10:31
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    $\begingroup$ What is your definition of "unphysical" here? To me, a field would be unphysical if it does not occur in the Lagrangian at all, or if it is an auxiliary field whose equation of motion is trivial or merely a constraint for another field $\endgroup$ – ACuriousMind Mar 29 '17 at 10:46
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    $\begingroup$ @innisfree I think that I have undestood what you mean: if I redefine the fields $\phi_1=(\tilde{\phi}_1+\tilde{\phi}_2)/4$ and $\phi_2=(\tilde{\phi}_1-\tilde{\phi}_2)/4$, I get a Lagrangian density with $1/2(\partial_\mu \tilde{\phi}_1)^2 - 1/2(\partial_\mu \tilde{\phi}_2)^2$, and $\tilde{\phi}_2$ has the wrong kinetic sign. But, can I do it? What does it mean? If I try to do the same in, for instance, Klein gordon field with two fields, do I get something different and maybe meaningful? However it seems that is $\tilde{\phi}_2$ to be "unphysical" and not $\phi_2$. What does it mean? $\endgroup$ – Alessandro Mininno Mar 29 '17 at 10:50
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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 :)

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  • $\begingroup$ FP ghosts have negative norm as well, so your second category is contained in the first one. Moreover, your two categories are not exhaustive: there are many other kind of fields that are, in a sense, unphysical: Pauli-Villars ghosts (which also happen to have negative norm), the longitudinal fields that are eaten out by gauge fields in the Higgs mechanism, auxiliary fields, acausal solutions in higher spin theories, etc. $\endgroup$ – AccidentalFourierTransform Mar 29 '17 at 14:08
  • $\begingroup$ I didn't know this, I'll edit my post accordingly thank you for your input ! And indeed there are many other examples but I felt like OP was most interested in the negative norm case. $\endgroup$ – gertian Mar 29 '17 at 14:10
  • $\begingroup$ You have answered what I asked and you have also given to me the right direction to look into if in the future I will have to do something similar! Thank you! $\endgroup$ – Alessandro Mininno Mar 29 '17 at 14:23
  • $\begingroup$ You are very welcome. Glad I could help ! $\endgroup$ – gertian Mar 29 '17 at 14:36

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