# How does the functional measure transform under a field redefinition?

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).

References:

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.

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$$^1$$ 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.

• 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? – Luke Jan 22 at 15:44
• I updated the answer. – Qmechanic Jan 24 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.

• 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. – Luke Jan 22 at 15:49
• @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 $\phi(x)$. – Chiral Anomaly Jan 23 at 0:57

Consider the more general case of a (not necessarily local) field redefintion of the form: $$$$\phi(x)=\int d y \,\,f(x,y)\,g\big(\theta(y)\big)\tag{1}$$$$ For example, in the question above we have $$$$\phi(x)=\int d y \,\,\delta(x-y)\,\sinh\big(\theta(y)\big)$$$$ which is an example of a local field redefinition. In the discrete case where we think of the path integral as taking place on a lattice, Eq.(1) takes the form, where $$\phi_i$$ is shorthand for the discrete variable $$\phi(x_i)$$: $$$$\phi_i=\sum_j F_{ij}\, g(\theta_j) \tag{2}$$$$ where $$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)$$ we encounter the highly singular term $$$$Tr(\log(\delta^d(x-y)))=\int d^dx\,\,\log\bigg(\delta^{d}(x-x)\bigg)=\int d^dx\,\,\log\bigg(\delta^{d}(0)\bigg)$$$$ So to answer the original question, I think the measure transforms as: \begin{align} \mathcal{D}\phi=\mathcal{D}\theta\,\,\exp\bigg(\int dx\,\log\big(\delta(0)\big)+\log\big(1+3c\theta^2)\big)\bigg)\tag{1}\\ \mathcal{D}\phi=\mathcal{D}\theta\,\,\exp\bigg(\int dx\,\log\big(\delta(0)\big)+\log\big(3c\theta^2\big)\bigg)\tag{2}\\ \mathcal{D}\phi=\mathcal{D}\theta\,\,\exp\bigg(\int dx\,\log\big(\delta(0)\big)+\log\big(\cosh(\theta(x))\big)\bigg)\tag{3} \end{align} Apparently one can ignore the $$\delta^d(0)$$ when working in dimensional regularization as we can interpret $$\delta^d(0)$$ as the volume of spacetime, which in $$d-\epsilon$$ dimensions is $$$$\delta^{d}(0)=\frac{2\pi^{d/2}}{\Gamma(d/2)}\frac{\Gamma(-d)}{\Gamma(1-d)}$$$$ which is $$\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.