My question is: how does the path integral functional measure transform under the following field redefinitions (where $c$ is an arbitrary constant and $\phi$ is a scalar field): \begin{align} \phi(x)&=\theta(x)+c \,\theta^3(x) \tag{1}\\ \phi(x)&=c\,\theta^3(x) \tag{2}\\ \phi(x)&=\sinh\big(\theta(x)\big)\tag{3} \end{align} My naive guess for the transformation in Eq.(3) is \begin{align} \mathcal{D}\phi&=\mathcal{D}\theta\,\,\text{Det}\Bigg[\frac{\delta \phi(x)}{\delta\theta(x')}\Bigg]=\mathcal{D}\theta \,\,\text{Det}\bigg[\cosh(\theta(x))\delta(x-x')\bigg]\\ &=\mathcal{D}\theta\exp\bigg[\text{Tr}\,\Big(\log\big(\cosh(\theta(x))\delta(x-x')\big)\Big)\bigg]\\ &=\mathcal{D}\theta\exp\bigg[\int dx\,\,\log\Big(\cosh(\theta(x))\delta(x-x')\Big)\bigg] \end{align} But that seems very wrong.

  1. It seems natural to generalize OP's setting to several fields $\phi^{\alpha}$ in $d$ spacetime dimensions. Under ultra-local field redefinitions$^1$ $$ \phi^{\prime\alpha}(x)~=~F^{\alpha}(\phi(x),x) ~=~\phi^{\alpha}(x)-f^{\alpha}(\phi(x),x), $$ the Jacobian functional determinant in the path/functional integral is formally given as $$J~=~{\rm Det} (\mathbb{M})~=~\exp {\rm Tr}\ln (\mathbb{M}) ~=~ \exp\left(-\sum_{j=1}^{\infty} \frac{1}{j}{\rm Tr} (\mathbb{m}^j)\right) $$ $$~=~ \exp\left(\delta^d(0) \int\! d^dx ~{\rm tr} (\ln M(x))\right), $$ where we have defined $$ \mathbb{M}~\equiv~\mathbb{1}-\mathbb{m},$$ $$ \mathbb{M}^{\beta}{}_{\alpha}(x^{\prime},x) ~:=~\frac{\delta F^{\beta}(x^{\prime})}{\delta\phi^{\alpha}(x)} ~=~ M^{\beta}{}_{\alpha}(x^{\prime})\delta^d(x^{\prime}\!-\!x),\qquad M^{\beta}{}_{\alpha}(x)~:=~ \frac{\partial F^{\beta}(x)}{\partial\phi^{\alpha}(x)}~=~\delta^{\beta}_{\alpha}-m^{\beta}{}_{\alpha}(x),$$ $$ \mathbb{m}^{\beta}{}_{\alpha}(x^{\prime},x) ~:=~\frac{\delta f^{\beta}(x^{\prime})}{\delta\phi^{\alpha}(x)} ~=~ m^{\beta}{}_{\alpha}(x^{\prime})\delta^d(x^{\prime}\!-\!x),\qquad m^{\beta}{}_{\alpha}(x)~:=~ \frac{\partial f^{\beta}(x)}{\partial\phi^{\alpha}(x)}.$$

  2. If we discretize spacetime, then the Jacobian becomes a product of ordinary determinants $$ J~=~\prod_i \det (M(x_i)), $$ where the index $i$ labels lattice points $x_i$ of spacetime. The Dirac delta at zero $\delta^d(0)$ is here replaced by a reciprocal volume of a unit cell of the spacetime lattice, which can viewed as a UV regulator, cf. e.g. my Phys.SE answer here.

  3. In dimensional regularization (DR), the Dirac delta at zero $\delta^d(0)$ vanishes, cf. Refs. 1 - 3. Heuristically, DR only picks up residues of various finite parameters of the physical system, while contributions from infinite parameters are regularized to zero. As a consequence, in DR the Jacobian $J=1$ becomes one under local field redefinition (if there are no anomalies present).


  1. M. Henneaux & C. Teitelboim, Quantization of Gauge Systems, 1994; Subsection 18.2.4.

  2. G. Leibbrandt, Introduction to the technique of dimensional regularization, Rev. Mod. Phys. 47 (1975) 849; Subsection IV.B.3 p. 864.

  3. A.V. Manohar, Introduction to Effective Field Theories, arXiv:1804.05863; p. 33-34 & p. 51.


$^1$ Much of this can be generalized to local field redefinitions $$ \begin{align}\phi^{\prime\alpha}(x)&~=~F^{\alpha}(\phi(x),\partial\phi(x),\partial^2\phi(x), \ldots ,\partial^N\phi(x) ,x)\cr &~=~\phi^{\alpha}(x)-f^{\alpha}(\phi(x),\partial\phi(x),\partial^2\phi(x), \ldots ,\partial^N\phi(x) ,x),\end{align}$$ and derivatives $\partial^j\delta^d(0)$ of the Dirac delta at zero. Local field redefinitions correspond to insertion of IR-irrelevant vertices in the action.

  • $\begingroup$ I have provided a tentative answer, but my answer conflicts with your statement that the Jacobian is unity in dimensional regularization. Could you please elaborate on this point? $\endgroup$ – Luke Jan 22 '19 at 15:44
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Jan 24 '19 at 17:55

All three cases, (1)-(3), are local redefinitions, meaning that the value of $\phi(x)$ for any given $x$ is determined only by the value of $\theta(x)$ at that same value of $x$ (and conversely, assuming it's invertible).

Conceptually, the parameter $x$ is just a continuous index labeling different integration variables. In fact, the most generally-applicable way we have for defining a functional integral (at least in QFT) is to replace this continuous parameter with a discrete index. Then you have an ordinary multi-variable integral, and the rule for changing integration variables is the usual one. So the cases (1)-(3) just describe changes-of-variable in a bunch of single-variable integrals.

Thinking about things this way (with $x$ discretized) should help track down what's really going on with the $\delta(x-x')$ factor.

  • $\begingroup$ Can one perform a field redefinition which is non-invertible such as $\phi(x)=\tanh(\theta(x))$? In the case of 1-dimensional integrals one cannot do this, but I would just like to make sure that the same applies in the path integral case too. $\endgroup$ – Luke Jan 22 '19 at 15:49
  • 1
    $\begingroup$ @Luke When $x$ is a discrete index, the same rules should apply. For example, $\phi(x)=\theta^2(x)$ won't work, because for each $x$, the integral over $\phi(x)$ goes from $-\infty$ to $+\infty$, but $\theta^2(x)$ can't be $<0$. And $\phi(x)=\tanh(\theta(x))$ won't work because $\tanh$ is bounded and so can't go from $-\infty$ to $+\infty$. But you can go the other way: If $\phi(x)$ is the original variable from $-\infty$ to $+\infty$, you can define $\theta(x)=\tanh(\phi(x))$. The new interval is $-1$ to $+1$, and that's fine, because it represents all the original values of $\phi(x)$. $\endgroup$ – Chiral Anomaly Jan 23 '19 at 0:57

Consider the more general case of a (not necessarily local) field redefintion of the form: \begin{equation} \phi(x)=\int d y \,\,f(x,y)\,g\big(\theta(y)\big)\tag{1} \end{equation} For example, in the question above we have \begin{equation} \phi(x)=\int d y \,\,\delta(x-y)\,\sinh\big(\theta(y)\big) \end{equation} which is an example of a local field redefinition. In the discrete case where we think of the path integral as taking place on a lattice, Eq.(1) takes the form, where $\phi_i$ is shorthand for the discrete variable $\phi(x_i)$: \begin{equation} \phi_i=\sum_j F_{ij}\, g(\theta_j) \tag{2} \end{equation} where $F_{ij}=F(x_i,x_j)$ can be thought of as a matrix and $F_{ij}=\delta_{ij}$ corresponds to the case of a local transformation. Discretizing the path integral we can write the change of variables in Eq.(2) as (here $\wedge$ denotes the wedge product, which is always present for the tensor density $d^dx$ but I make it explicit here only to make the presence of the Jacobian apparent) \begin{align} \int \mathcal{D}\phi&=\int d\phi_1\wedge d\phi_2\wedge...\wedge d\phi_n\\ &= \int \frac{1}{n!}\epsilon_{i_1...i_n}d\phi^{i_1}\wedge...\wedge d\phi^{i_n}\\ &=\int \frac{1}{n!}\epsilon_{i_1...i_n}\,\,\frac{\partial\phi^{i_1}}{\partial\theta^{i_1'}}...\frac{\partial\phi^{i_1}}{\partial\theta^{i_n'}}d\theta^{i_1'}\wedge...\wedge \,d\theta^{i_n'}\\ &=\int \frac{1}{n!}\epsilon_{i_1...i_n}\,\,\bigg(F^{i_1i_1'}\frac{d g(\xi)}{d\xi}\bigg\vert_{\xi=\theta_{i_1'}}\bigg)...\bigg(F^{i_ni_n'}\frac{d g(\xi)}{d\xi}\bigg\vert_{\xi=\theta_{i_n'}}\bigg)d\theta^{i_1'}\wedge...\wedge \,d\theta^{i_n'}\\ &=\int\frac{1}{n!}\epsilon_{i_1'...i_n'}\text{Det}\bigg[F^{ii'}\frac{d g(\xi)}{d\xi}\bigg\vert_{\xi=\theta_{i'}}\bigg]d\theta^{i_1'}\wedge...\wedge \,d\theta^{i_n'}\\ &=\int\text{Det}\bigg[F^{ii'}\frac{d g(\xi)}{d\xi}\bigg\vert_{\xi=\theta_{i'}}\bigg]d\theta^{i_1'}\wedge...\wedge \,d\theta^{i_n'}\\ &=\int d\theta^{i_1'}\wedge...\wedge \,d\theta^{i_n'}\,\,\exp\bigg\lbrace\text{Tr}\bigg(\log\bigg[F^{ii'}\frac{d g(\xi)}{d\xi}\bigg\vert_{\xi=\theta_{i'}}\bigg]\bigg)\bigg\rbrace\tag{3} \end{align} So for example if we have a local field redefintion $F_{ij}=\delta_{ij}$ then we encounter the term $Tr(\log(\delta_{ij}))=\log(n)$, where $n$ is the number of lattice sites. In the continuous case where $F_{ij}\rightarrow f(x,y)=\delta^{d}(x-y)$ we encounter the highly singular term \begin{equation} Tr(\log(\delta^d(x-y)))=\int d^dx\,\,\log\bigg(\delta^{d}(x-x)\bigg)=\int d^dx\,\,\log\bigg(\delta^{d}(0)\bigg) \end{equation} So to answer the original question, I think the measure transforms as: \begin{align} \mathcal{D}\phi=\mathcal{D}\theta\,\,\exp\bigg(\int dx\,\log\big(\delta(0)\big)+\log\big(1+3c\theta^2)\big)\bigg)\tag{1}\\ \mathcal{D}\phi=\mathcal{D}\theta\,\,\exp\bigg(\int dx\,\log\big(\delta(0)\big)+\log\big(3c\theta^2\big)\bigg)\tag{2}\\ \mathcal{D}\phi=\mathcal{D}\theta\,\,\exp\bigg(\int dx\,\log\big(\delta(0)\big)+\log\big(\cosh(\theta(x))\big)\bigg)\tag{3} \end{align} Apparently one can ignore the $\delta^d(0)$ when working in dimensional regularization as we can interpret $\delta^d(0)$ as the volume of spacetime, which in $d-\epsilon$ dimensions is \begin{equation} \delta^{d}(0)=\frac{2\pi^{d/2}}{\Gamma(d/2)}\frac{\Gamma(-d)}{\Gamma(1-d)} \end{equation} which is $\frac{1}{\epsilon}$ dependent and hence we can use $\frac{1}{\epsilon}$ dependent counterterms to get rid of this term. However $\exp(\log(g'))$ terms still remain and it is unclear to me how to show that correlation functions will be unaffected by this term.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.