I have some trouble understanding a particular expansion in my QFT lecture. Consider a complex scalar field $\phi$, with the Lagrangian $$\mathcal{L} = \partial_\mu\phi^*\partial^\mu\phi-m^2\phi^*\phi.$$ We now consider a local $U(1)$ transformation of said field, i.e. $$\phi(x)\longmapsto e^{i\alpha(x)}\phi(x) \approx \phi(x)+i\alpha(x)\phi(x).$$ Now consider: $$ \begin{align*} \int\mathcal{D}\phi\,\phi(x_1)\phi^*(x_2)e^{iS[\phi]} &= \int\mathcal{D}\phi^\prime\,\phi^\prime(x_1)\phi^{*\prime}(x_2)e^{iS[\phi^\prime]}\\ &\overset{(1)}{=}\int\mathcal{D}\phi\,\phi^\prime(x_1)\phi^{*\prime}(x_2)e^{iS[\phi^\prime]}\\ &\overset{(2)}{=} \int\mathcal{D}\phi\,\phi(x_1)\phi^{*}(x_2)e^{iS[\phi]}\\ &+ \int \mathcal{D} \phi\Bigg[i \alpha(x_{1}) \phi(x_{1}) \phi^{*}(x_{2})-i \alpha(x_{2}) \phi(x_{1}) \phi^{*}(x_{2})\\ &+\phi(x_{1}) \phi^{*}(x_{2}) \int d^x x \frac{i \delta S}{\delta \alpha} \alpha(x)\Bigg] e^{i S[\phi]}. \end{align*} $$


  • For (1), why is $\mathcal{D}\phi^\prime=\mathcal{D}\phi$. I have a hard time seeing why this transformation should leave the "measure" invariant. What exactly are the restrictions for the measure to be invariant?
  • For (2), direct computation shows that $$\phi^\prime(x_1)\phi^{*\prime}(x_2) \approx \phi(x_1)\phi^{*}(x_2) + i \alpha\left(x_{1}\right) \phi\left(x_{1}\right) \phi^{*}\left(x_{2}\right)-i \alpha\left(x_{2}\right) \phi\left(x_{1}\right) \phi^{*}\left(x_{2}\right).$$ But where exactly does the integral in this step come from? I assume it's from $e^{iS[\phi^\prime]}$, but I can't really figure out how exactly...
  • $\begingroup$ Related: physics.stackexchange.com/q/308929/2451 $\endgroup$
    – Qmechanic
    Dec 10, 2020 at 18:07
  • $\begingroup$ Take at face value the path integral does not have a meaning. You have to give a prescription about how things are computed. In other words you have to specify regularization. In most cases (this is one of them) say dimensional regularization deals easily with Jacobians like that. $\endgroup$
    – nwolijin
    Dec 10, 2020 at 18:39
  • 1
    $\begingroup$ \begin{equation} \mathcal D \phi' =\left | \frac{\mathcal D \phi'}{\mathcal D \phi}\right| \mathcal D \phi= \mathcal D \phi \exp\left( i\int dx \delta (x-y) \alpha(x) \delta(x-y) \right ) = \mathcal D \phi \exp\left[ i \delta (0) \alpha(y) \right ]= \mathcal D \phi, \end{equation} since $\delta(0)=0$ in dim reg. $\endgroup$
    – nwolijin
    Dec 10, 2020 at 18:46
  • 2
    $\begingroup$ @nwolijin Thanks for the comment! Could you elaborate on how to compute $\Big| \frac{\mathcal{D}\phi'}{\mathcal{D}\phi} \Big|$? And is that an absolute value there? If so, should it not be the absolut value of the exponential, which is one even for a finite but non-zero regularization of $\delta(0)$? Okay, this related question (physics.stackexchange.com/q/455711) seems to explain it. $\endgroup$
    – Cream
    Dec 11, 2020 at 8:54
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    $\begingroup$ Do it first for finite dimensional integrals: $\mathcal D\phi = d\phi_1\dots d\phi_n$. Then you have to compute the Jacobian (this is what I denote by $\left | \frac{\mathcal D \phi'}{\mathcal D \phi} \right |$). Clearly for small variations, the transformation matrix becomes $\mathbb 1 + \epsilon$. Its determinant can be computed as $\det (\mathbb 1 + \epsilon) =\exp \log (\mathbb 1 + \epsilon)=\mathrm{Tr} \log (\mathbb 1 + \epsilon) \approx \mathrm{Tr} \epsilon$. Computing trace for the case at hand boils down to multiplying by $\delta(x-y)$ and integrating over $x$. $\endgroup$
    – nwolijin
    Dec 11, 2020 at 9:11

1 Answer 1


Concerning (1): I am not entirely sure. But to draw the comparison with multidimensional integration, the Jacobian determinant would be something like $$J "=" \mathcal{D}\phi' / \mathcal{D} \phi "=" \int d^dx \frac{\delta\phi'(x')}{ \delta \phi(x)} "=" e^{i\alpha(x)}$$ which would mean that $\mathcal{D}\phi' = \mathcal{D} \phi |J| = \mathcal{D} \phi$. But I would be glad to make this (with some help) more rigorous!

Edit: It seems to be more complicated than that. A more complete derivation can be found in this answer to this question.

Concerning (2): You do not just want to plug in the infinitesimal transformation for $\phi'$ but compute the infinitesimal transformation of the complete integral. The integral should be invariant, not the field itself.

For the field, you computed the infinitesimal transformation as follows:: $$ \phi(x) \to \Big(1 + \int d^dx' \alpha(x') \frac{\delta}{\delta \alpha(x')}\Big|_{\alpha=0} \Big) \phi'(x) = \phi(x) + i\alpha(x) \phi(x). $$ Now, we want to see how the integral transforms to first order in $\alpha$ by computing: $$ \Big(1 + \int d^dx' \alpha(x') \frac{\delta}{\delta \alpha(x')}\Big|_{\alpha=0} \Big) \int \mathcal{D}\phi'~ \phi'(x_1) \phi'^*(x_2) e^{iS[\phi']} .$$ This will lead to the above result via product rule.

Now, you seem to assume that the path-integral is invariant $U(1)$ transformation, i.e., $$ \int d^dx' \alpha(x') \frac{\delta}{\delta \alpha(x')} \Big|_{\alpha=0} = 0 \int \mathcal{D}\phi'~ \phi'(x_1) \phi'^*(x_2) e^{iS[\phi']} $$ but this is not true for general $S$.

If $S$ is the free action $S[\phi] = \int d^4x \mathcal{L}[\phi] = \int d^4x ( \partial_\mu\phi^* \partial^\mu \phi - m^2 \phi^* \phi )$, it is invariant under $U(1)$ transformation and $\frac{\delta}{\delta\alpha} S = 0$. Then, the path-integral is only invariant when you consider $x_1 = x_2$, basically defining $|\phi|^2$. Otherwise, you could add an interaction Lagrangian to the action which cancels the other terms.

Let me know if I made any mistakes!

  • 1
    $\begingroup$ Thank you for the answer, but I still have a few questions on 2. When I try to calculate $$\Big(1 + \int d^dx' \frac{\delta}{\delta \alpha(x')}\Big) \int \mathcal{D}\phi'~ \phi'(x_1) \phi'^*(x_2) e^{iS[\phi']} \Big|_{\alpha=0}$$ I get the first three terms from my equation above just fine. But in the last term I'm missing the $\alpha(x)$. Could you maybe show how one finds that? Also, do I understand correctly that we should have $$\int\mathcal{D}\phi\,\phi(x_1)\phi^*(x_2)e^{iS[\phi]} \overset{!}{=}\Big(1 + \int d^dx' \frac{\delta}{\delta \alpha(x')}\Big) \int \mathcal{D}\phi'~ \dots\,?$$ $\endgroup$
    – Sito
    Dec 28, 2020 at 15:19
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    $\begingroup$ @Sito The factor $\alpha$ was missing in all terms. It should already appear in $\int d^dx' \alpha(x') \frac{\delta}{\delta\alpha(x')}$ since we are "expanding" in the transformation parameter (in analogy to a Taylor expansion). Regarding the second point: I am not sure why we should have $l.h.s. \overset{!}{=} r.h.s.$. We certainly can have it in special cases but not generally. I added a paragraph on this. $\endgroup$
    – Cream
    Dec 28, 2020 at 19:04
  • $\begingroup$ First of all, thanks for the explanation and the edit. I think I have still two questions: 1.) How exactly did you come to this expansion $$\phi(x) \to \Big(1 + \int d^dx' \alpha(x') \frac{\delta}{\delta \alpha(x')}\Big) \phi'(x)\Big|_{\alpha=0} $$ I have never seen this before and I'm a bit confused since it's coming a bit out of the blue.. 2.) if $l.h.s \overset{!}{=}r.h.s$ is not true in general, then I don't understand what this expansion has to do with the calculation shown in my opening post, or in other words, how does what you have written above agree with the calculation I show? $\endgroup$
    – Sito
    Dec 28, 2020 at 20:37
  • $\begingroup$ @Sito In your question you wrote that $\phi(x) \to e^{i\alpha(x)}\phi(x) \approx \phi(x) + i\alpha(x) \phi(x)$. That means you are considering an infinitesimal transformation with $\alpha \approx 0$. Now, for $\phi(x)$ the result is clear because we all know how to expand $e^{i\alpha(x)}$. When you transform the integral, however, you have to perform an actual taylor expansion to first order. But you are expanding in a function. That means you have to "vary and integrate", see for example the first answer here $\endgroup$
    – Cream
    Dec 29, 2020 at 9:32
  • $\begingroup$ 2.: In your question you stated that $l.h.s. = r.h.s.$, but that is only true in certain cases. So you have to tell me why $l.h.s. = r.h.s.$ should be true in your case. In my answer, I only tried to show how to obtain the $r.h.s.$ for infinitesimal transformations and came to the conclusion that it is not equal to the $l.h.s.$. Since this is from a lecture I can imagine that there is some physical motivation that you want the equality to hold and derive $S$ from it or you are considering the special case $x_1=x_2$. But here I can only guess. $\endgroup$
    – Cream
    Dec 29, 2020 at 9:41

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