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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: $$ \begin{align*} \int\mathcal{D}\phi\,\phi(x_1)\phi^*(x_2)e^{iS[\phi]} &= \int\mathcal{D}\phi^\prime\,\phi^\prime(x_1)\phi^{*\prime}(x_2)e^{iS[\phi^\prime]}\\ &\overset{(1)}{=}\int\mathcal{D}\phi\,\phi^\prime(x_1)\phi^{*\prime}(x_2)e^{iS[\phi^\prime]}\\ &\overset{(2)}{=} \int\mathcal{D}\phi\,\phi(x_1)\phi^{*}(x_2)e^{iS[\phi]}\\ &+ \int \mathcal{D} \phi\Bigg[i \alpha(x_{1}) \phi(x_{1}) \phi^{*}(x_{2})-i \alpha(x_{2}) \phi(x_{1}) \phi^{*}(x_{2})\\ &+\phi(x_{1}) \phi^{*}(x_{2}) \int d^x x \frac{i \delta S}{\delta \alpha} \alpha(x)\Bigg] e^{i S[\phi]}. \end{align*} $$

Questions

  • 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...
Qmechanic
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Sito
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1 Answers1

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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!

Cream
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