I know that there several questions that deal with this question but I‘ve found no satisfactory answer. In QFT we want that a scalar field is invariant under Poincare transformations $\mathcal{P}$ which means: $$\mathcal{P}(\phi(x))=\phi^{\prime}(x^{\prime})=\phi(x).$$ Intuitive this means if we e.g. rotate our coordinates and our field in the same way the field value stays the same . But since the transformation is passive I see no difference to an arbitrary change of coordinates $\mathcal{T}$ for which the following should also apply: $$\mathcal{T}(\phi(x))=\phi^{\prime}(x^{\prime})=\phi(x).$$ I know that the difference between a Poincare transformation and an arbitrary diffeomorphism is that $\mathcal{P}^*g=g$ should hold for a Poincare transformation where $g$ is the metric but I don’t see what this changes if we look at fields.

  • $\begingroup$ I lack knowledge in this area, so maybe what I write is rubbish. I am not sure what you mean by arbitrary change of coordinates $\mathcal{T}$. Aren't Poincare transformations ones that take you from one inertial frame to another: $x^{\mu}\rightarrow x'^{\mu} = {\Lambda^{\mu}}_{\nu}x^{\nu}+a^{\mu}$. Do you mean instead of e.g. the Lorentz transformation $\Lambda$ you use something different? In that case, don't you break Lorentz invariance? $\endgroup$
    – Vangi
    Mar 31, 2023 at 13:26
  • $\begingroup$ @Vangi an arbitrary change of coordinates could be for example something non linear. This shouldn‘t break Lorentz invariance because it isn‘t an active transformation and we are just choosing different coordinates to describe the same physics. $\endgroup$
    – Silas
    Mar 31, 2023 at 16:41
  • $\begingroup$ Poincare invariance in QFT means the Lagrangian is invariant to such a transformation. For classical field theory, you can write your Lagrangians in terms of a general metric and replace partial derivatives with covariant derivatives. That makes them invariant to a general coordinate transform. I am not sure to what extent you can do this in QFT. $\endgroup$
    – Virgo
    Apr 1, 2023 at 0:46
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    $\begingroup$ If you declare the metric to be fixed, then a diffeomorphism given by something other than a Killing vector of that metric is not just relabelling coordinates. If you let the diffeomorphism act on the metric (and covariant derivatives) too, then you recover invariance up to anomalies as Virgo said. $\endgroup$ Apr 2, 2023 at 12:25
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    $\begingroup$ Whilst the answer below seems fine, I think it is the previous comment which actually answers your question... $\endgroup$ Apr 2, 2023 at 20:35

1 Answer 1


First, there is something important to clarify here, QFT has nothing to do with this. This is a statement that arises in classical field theory before quantization ever comes into the picture. Typically classical field theory is not taught as its own independent course and thus, since it is necessary for the study and understanding of QFT, many aspects of it are covered at the beginning of QFT courses and textbooks, unfortunately leading people to conflate the two.

Now, lets say we want to construct a classical field theory; in other words, find an action $S=\int d^4x\mathcal{L}$ on Minkowski space $\mathbb{R}^{1,3}$. This can then be used to derive equations of motion for the degrees of freedom in the theory via the least action principle. But what are the theory's "degrees of freedom"? Well, for a local theory, the simplest answer is that you construct your lagrangian density $\mathcal{L}$ out of local fields which take values in Minkowski space, i.e. smooth functions $\varphi:\mathbb{R}^{1,3}\rightarrow \mathcal{T}$. Here $\mathcal{T}$ is known as the target space, in general it could be a wide variety of mathematical objects but for our purposes we will restrict it to a linear space such as a vector space. These will be our degrees of freedom.

Now note that Minkowski space doesn't have a natural coordinate system: you can preform a Poincare transformation on it and obtain the same space back. This is because as you said, the Poincare group is the group of metric preserving diffeomorphisms, or more compactly, the isometry group. Thus, if we are to construct an action on Minkowski space using local fields, we don't want it to depend on the basis we might be working in, as we know it is not unique but related to others via Poincare transformations. Ultimately this is an axiom of physics but it's certainly a reasonable one!

So, if we want the action to be independent of our coordinate system, the lagrangian density must be built out of objects that are independent of Poincare transformations. The simplest choice would be the one you have offered, which is a scalar field $\phi$ where $\mathcal{T}$ is either $\mathbb{R}$ or $\mathbb{C}$. In which case we could add all sorts of terms like $\phi^n$ to $\mathcal{L}$, and they would all be not only Poincare invariant but also, as you pointed out, generically coordinate invariant. But this would be quite a dull theory; a simple application of the Euler-Lagrange equations would demonstrate that the only allowed field configurations would be $\phi=Constant$, no matter how many terms of the form $\phi^n$ you add. How do we remedy this? (I will be using common relativistic notation in what follows)

Well, we could add terms of the form $\partial_{\mu}\phi$, ($\partial_{\mu}\equiv\frac{\partial}{\partial x^{\mu}}$), to give our theory some dynamics, but note that these no longer transform as simply as the previous terms. These transform covariantly under an active Poincare transformation $x^{'\mu}=\Lambda^{\mu}_{\ \ \nu} x^{\nu} + a^{\mu}$ as

$$ \partial_{\mu}\phi(x)\rightarrow\partial^{'}_{\mu}\phi'(x')=\frac{\partial x^{\nu}}{\partial x^{'\mu}}\partial_{\nu}\phi(x)=(\Lambda^{-1})^{\nu}_{\ \ \mu}\partial_{\nu}\phi(x). $$

Thus, in order to add this to the lagrangian density we must do something about that extra index, this is where the Minkowski metric, and thus the necessity of Poincare symmetry comes in. If we look at the term $$ \eta^{\mu\nu}\partial_{\mu}\phi(x)\partial_{\nu}\phi(x) $$ we can see immediately that this is Poincare invariant, since we've already computed how the $\partial\phi$'s transform and we know that for any Lorentz transformation $\Lambda_{\ \ \mu}^{\nu}\eta^{\mu\rho}\Lambda_{\ \ \rho}^{\sigma}=\eta^{\rho\sigma}$ by definition. But this term is not generally coordinate invariant! This is obvious from the fact that for a general coordinate transformation $\frac{\partial x^{\mu}}{\partial x^{'\nu}}\neq(\Lambda^{-1})^{\mu}_{\ \ \nu}$ and thus the term would transform as $$ \eta^{\mu\nu}\partial_{\mu}\phi(x)\partial_{\nu}\phi(x)\rightarrow\frac{\partial x^{\rho}}{\partial x^{'\mu}}\frac{\partial x^{\sigma}}{\partial x^{'\nu}}\eta^{\mu\nu}\partial_{\rho}\phi(x)\partial_{\sigma}\phi(x). $$

Now, I have argued that for theories on $\mathbb{R}^{1,3}$, Poincare invariance is necessary so that the theory doesn't depend on the basis you've chosen to define Minkowski space. Further, we have shown that there are terms that one would very reasonably like to have in a lagrangian density which are Poincare invariant but not generally coordinate invariant. I believe this answers your original question, but in the comments of your post you have further stated that "In my opinion demanding that our theory is invariant under Poincare transformations has no physical content because we are then just saying that we can relabel our coordinates in a special way". I will now explain why this is not the correct way to view things.

Thus far I have only talked about scalar fields, but we know from elementary physics courses that this is not the only game in town. The electric and magnetic fields are both 3d-vector valued, that is, $\mathcal{T}=\mathbb{R}^3$, and perhaps we would like to build a lagrangian field theory to describe these. The problem however, is that they are not Lorentz covariant like $\partial\phi$ was, so if we change our coordinates they do not simply get acted on by $\Lambda$, making it quite difficult to construct a Poincare invariant action by simply contracting with $\eta$. In order to build more complicated field theories of vector, tensor, or even spinor valued fields, one must study the representation theory of the Poincare group, something 20th century physicists and mathematicians did very thoroughly. This is the way in which the Poincare group gives physical content: it tells us what kind of fields we are allowed to have in our theory and what kinds of terms we can construct out of them to keep the action invariant.

Now I believe I have made my point, but there is one last remark I would like to make. It is very natural to ask why we would care so much about preserving $\eta$. Surely we should allow for more general coordinate transformations: after all, the philosophy that led us here was not wanting our action to depend on a basis for Minkowski space. Shouldn't we say that any coordinates are manmade and thus the theory shouldn't care about them? This is a fundamental point and one that leads down the road toward the general theory of relativity, as general coordinate frames correspond to accelerating observers. However, even there, Minkowski space is our model space, i.e. the general manifold $\mathcal{M}$ we work on must look locally like $\mathbb{R}^{1,3}$, and here the Lorentz group appears again, now as the structure group of the tangent bundle. Thus, even in GR, the fields need to be valued in vector spaces forming representations of the Lorentz group.

I could keep going on about the usefulness of conserved charges and the importance of symmetry groups in quantum theory, but I believe that this response is quite long enough. I hope this helps.

  • $\begingroup$ Thank you! So to sum it up the problem is that we are building scalars from derivatives and they are living in a Minkowski vectorspace where the structure preserving maps are Lorentz transformations and not general diffeomorphisms. Is that correct? $\endgroup$
    – Silas
    Apr 2, 2023 at 19:47
  • $\begingroup$ Yes, it is the fact that we are fixing the spacetime as Minkowski space with metric $\eta$. Scalar derivatives are not the only concern: you may also want to have non-scalar fields (like the electromagnetic 4-vector potential) as I mentioned and those would transform non-trivially even without derivatives. Thus their action terms would also require contractions with the metric, making them not generally coordinate invariant. As I remarked, all of this can be relaxed to full coordinate invariance but it would require the machinery of GR. $\endgroup$
    – Wintermute
    Apr 2, 2023 at 20:09

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