Good evening, I am learning about path integrals in QFT and I was wondering, can you simplify the path integral by shifting the fields? To make it more clear I will give you an example. Suppose that I have the following path integral \begin{equation} \int \mathcal{D} \phi \mathcal{D} A \mathcal{D} B \ \text{exp}[\int d^d x \ \phi(x)(\partial^2 +m^2+A(x)+B(x))\phi(x)+ B(x)(\partial^2 +m^2)B(x)] \end{equation} where $\phi,A,B$ are scalar fields and the rest are constants. Can I change variable $A(x) \rightarrow A(x)- B(x)$ in order to get \begin{equation} \int \mathcal{D} \phi \mathcal{D} A \mathcal{D} B \ \text{exp}[\int d^d x \ \phi(x)(\partial^2 +m^2+A(x))\phi(x)+ B(x)(\partial^2 +m^2)B(x)] \end{equation} where $\phi$ and $B$ are decoupled?
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2$\begingroup$ The short answer is that it depends on the space of fields over which you are integrating (think limits of integration). If there are no "restrictions", kind of like integrating from $-\infty$ to $\infty$, then yes you can. Moreover, just like in regular calculus for more complicated transformations you need a Jacobian. $\endgroup$– AaronFeb 17 at 15:46
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$\begingroup$ @Aaron thank you for the explanation! $\endgroup$– AndreasFeb 25 at 15:36
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You can think of the integration measure as $$ \mathcal{D}A \sim \prod_x dA(x), $$ i.e., for each $x$ you have a variable called $A(x)$ to be integrated over. Your question then reduces to whether the following is valid $$ dA(x) dB(x) = dA'(x) dB(x) $$ with $A'(x) = A(x) - B(x)$, and the answer is yes since the Jacobian is $$ \begin{pmatrix} 1 & 0\\ -1 & 1 \end{pmatrix}, $$ whose determinant is $1$.