Volume element under change of variables

Question

Consider the following function: $$f(q, p)=a\left[(b \cdot q)^{2}+(c \cdot p)^{2}-1\right] e^{-(b \cdot q)^{2}-(c \cdot p)^{2}}$$ with $a,b,c\in \mathbb{R}$. I want to express it in terms of $x\equiv b \cdot q,\ y\equiv c \cdot p$. How do I take into account the volume element under change of variables?

Answer 1

In order to simplify the given expression $(q,p)\mapsto f(q,p)$ you introduced new variables $x$, $y$ via $$x:=b\,q,\quad y:=c\,p\ .\tag{1}$$ This means that you in fact parametrize the $(q,p)$-plane by an auxiliary map $$\psi:\quad(x,y)\mapsto\left\{\eqalign{q&=x/b\cr p&=y/c \cr}\right.\quad\ .$$ (Your momentary troubles stem from the fact that the given formulas $(1)$ do not describe $\psi$, but its inverse $\psi^{-1}$.)

You now have the pullback $$\hat f:=f\circ\psi:\quad(x,y)\mapsto \hat f(x,y)= f\left({x\over b},{y\over c}\right)\ ,$$ which computes to $$\hat f(x,y)=a(x^2+y^2-1)e^{-x^2-y^2}\ ,$$ as originally desired. So far no "volume correction" is necessary.

Now it can be that a domain $B$ in the $(q,p)$-plane is given, and you want to compute the integral $$\int_B f(q,p)\>{\rm d}(q,p)\ .$$ It is possible to convert this integral to an integral concerning the pullback $\hat f$. To this end we need the description of the domain $\hat B:=\psi^{-1}(B)$ in the $(x,y)$-plane, that is transformed by $\psi$ into the given $B$, and we need the Jacobian $J_\psi(x,y)={1\over bc}$. The general transformation formula then says that $$\int_B f(q,p)\>{\rm d}(q,p)=\int_{\hat B}\hat f(x,y)|J_\psi(x,y)|\>{\rm d}(x,y)={1\over|bc|}\int_{\hat B}a(x^2+y^2-1)e^{-x^2-y^2}\>{\rm d}(x,y)\ .$$