How do you compute the partial trace in the oblique coordinates $a=u+\frac{1}{2}x, b=x$? For my master's thesis I am reading the paper https://arxiv.org/abs/1204.5627 about quantum reference frame changes to center of mass coordinates. On Page 4, there is a calculation i just can't make sense of. Given is a quantum wave function $|\psi(a,b)\rangle $ living in $\mathbb{H}_A \otimes \mathbb{H}_B$ of which i want to calculate the partial trace 
$$
\rho_B=Tr_{A}(|\psi(a,b)\rangle \langle\psi(a,b)|).
$$
This would be
$$
\int da \langle a| \big[\int \psi(a',b') |a'\rangle |b'\rangle da'db' \int \psi^*(a'',b'') \langle a''|\langle b''| da''db'' \big] |a\rangle =\\ 
\int \psi(a,b)\psi^*(a,b') |b\rangle \langle b'| da\; db\; db'
$$
So far so good. The problem is that a coordinate transformation to oblique coordinates $a=u+\frac{1}{2}x, b=x$ is performed "for convenience". The outcome is never stated, only that the matrix elements are given by
$$
\langle x|\rho_B|x+\delta\rangle = \int du e^{u\partial_x}[\psi(u,x)\psi^*(u-\frac{\delta}{2}, x-\frac{\delta}{2})]
$$
I know that $\int du e^{u\partial_x}$ has something to do with a translation over all of u, or a. But how exactly does this operator come into play? And how is the volume element $da\;db$ transformed? My calculation leads to $da\;db = (du + \frac{1}{2}dx)dx$. How do you perform an integration with a term $dx^2$? 
I am sorry if this is not much information to help solve the problem, but thats all that I got myself. I would be forever grateful to whoever helps me understand this derivation.
 A: First of all: Substitution for integrals over several variables works slightly different than one might expect. One has to use something called the Jacobi determinant, see e.g.:
https://en.wikipedia.org/wiki/Integration_by_substitution#Substitution_for_multiple_variables
I had a look at the paper. I will follow the notation of the paper, page 4, because I have the impression that there might be a mistake with $\psi$ instead of $\tilde{\psi}$ in your question.
So we start from the expression:
\begin{equation}
|\psi \rangle = \int \mathrm d x_0 \mathrm d x_1 \psi(x_0,x_1) | x_0\rangle_0 |x_1\rangle_1
\end{equation}
Now the paper introduces center-of-mass and relative coordinates as 
\begin{align}
x_{cm} = \frac{m_0 x_0 + m_1 x_1}{M}\\
x_{r_1} = x_1 - x_0
\end{align}
and furthermore, they define new states as
\begin{align}
|x_0\rangle_0 |x_1\rangle_1 =: |x_{cm}\rangle_{cm} |x_{r_1}\rangle_{r_1}
\end{align}
Now we want to express $|\psi\rangle$ in the new coordinates, and this is done using the Jacobi determinant $\det{J}$:
\begin{equation}
\mathrm{d} x_{cm} \mathrm{d} x_{r_1} = |\det{J}| \mathrm{d} x_0 \mathrm{d} x_1
\end{equation}
The Jacobian is defined as
\begin{align}
J =
\begin{pmatrix}
\frac{\partial x_{cm}}{\partial x_0} & \frac{\partial x_{cm}}{\partial x_1} \\ \frac{\partial x_{r_1}}{\partial x_0} & \frac{\partial x_{r_1}}{\partial x_1}
\end{pmatrix}
= 
\begin{pmatrix}
\frac{m_0}{M} & \frac{m_1}{M}\\ -1 & 1
\end{pmatrix}
\end{align}
so we find 
\begin{equation}
|\det{J}| = \left|\frac{m_0}{M}\cdot 1 - \frac{m_1}{M}\cdot (-1)\right| = 1
\end{equation}
Furthermore, we have to rewrite $x_0,x_1$ using the new coordinates:
\begin{align}
x_0 = x_{cm} - \frac{m_1}{M}x_{r_1}\\
x_1 = x_{cm} + \frac{m_0}{M}x_{r_1}
\end{align}
This gives us 
\begin{equation}
|\psi \rangle = \int \mathrm d x_{cm} \mathrm d x_{r_1} \ \psi\left(x_{cm} - \frac{m_1}{M}x_{r_1}, x_{cm} + \frac{m_0}{M}x_{r_1} \right) | x_{cm}\rangle_{cm} |x_{r_1}\rangle_{r_1}
\end{equation}
They define a new function for the integrand as
\begin{equation}
\tilde{\psi}(x_{cm},x_{r_1}) := \psi\left(x_{cm} - \frac{m_1}{M}x_{r_1}, x_{cm} + \frac{m_0}{M}x_{r_1} \right)
\end{equation} 
Therefore:
\begin{equation}
|\psi \rangle = \int \mathrm d x_{cm} \mathrm d x_{r_1} \ \tilde{\psi}(x_{cm},x_{r_1}) | x_{cm}\rangle_{cm} |x_{r_1}\rangle_{r_1}
\end{equation}
Now they define the reduced state as
\begin{equation}
\rho_{r_1} = \mathrm{Tr}_{cm}[|\psi\rangle \langle \psi |]
\end{equation}
Here: 
\begin{align}
|\psi\rangle \langle \psi | = \int \mathrm{d} x_{cm} \mathrm{d} x'_{cm} \mathrm{d} x_{r_1} \mathrm{d} x'_{r_1} \ \tilde{\psi}(x_{cm},x_{r_1}) \tilde{\psi^*}(x'_{cm},x'_{r_1}) \ |x_{cm}\rangle_{cm} |x_{r_1}\rangle_{r_1} \langle x'_{cm} |_{cm} \langle x'_{r_1}|_{r_1}
\end{align}
This means
\begin{align}
&\rho_{r_1} =  \mathrm{Tr}_{cm}[|\psi\rangle \langle \psi |] = \int dy \ \langle y |\psi\rangle \langle \psi | y \rangle_{cm} \\ =&\int \mathrm d y 
 \int \mathrm{d} x_{cm} \mathrm{d} x'_{cm} \mathrm{d} x_{r_1} \mathrm{d} x'_{r_1} \ \tilde{\psi}(x_{cm},x_{r_1}) \tilde{\psi^*}(x'_{cm},x'_{r_1})
  \langle y|x_{cm}\rangle \langle x'_{cm} |y\rangle |x_{r_1} \rangle_{r_1}  \langle x'_{r_1}|_{r_1}
\end{align}
and thus
\begin{align}
\rho_{r_1} = & \int \mathrm d y 
 \int \mathrm{d} x_{cm} \mathrm{d} x'_{cm} \mathrm{d} x_{r_1} \mathrm{d} x'_{r_1} \ \tilde{\psi}(x_{cm},x_{r_1}) \tilde{\psi^*}(x'_{cm},x'_{r_1}) \ \delta(x_{cm} - y) \delta(x'_{cm} -y) \\ 
&|x_{r_1}\rangle_{r_1}  \langle x'_{r_1}|_{r_1}
\end{align}
We choose to evaluate the delta distributions such that $y:=x'_{cm} := x_{cm}$, i.e. we keep the integral over $x_{cm}$:
\begin{align}
\rho_{r_1} =
\int \mathrm{d} x_{cm} \mathrm{d} x_{r_1} \mathrm{d} x'_{r_1} \ \tilde{\psi}(x_{cm},x_{r_1}) \tilde{\psi^*}(x_{cm},x'_{r_1}) \ |x_{r_1}\rangle_{r_1} \langle x'_{r_1}|_{r_1}
\end{align}
Now we can evaluate the matrix elements:
\begin{align}
\langle \chi | \rho_{r_1} | \chi + \delta \rangle =& \int \mathrm{d} x_{cm} \mathrm{d} x_{r_1} \mathrm{d} x'_{r_1} \ \tilde{\psi}(x_{cm},x_{r_1}) \tilde{\psi^*}(x_{cm},x'_{r_1}) \ \langle \chi|x_{r_1}\rangle \langle x'_{r_1}| \chi + \delta \rangle \\
=& \int \mathrm{d} x_{cm} \mathrm{d} x_{r_1} \mathrm{d} x'_{r_1} \ \tilde{\psi}(x_{cm},x_{r_1}) \tilde{\psi^*}(x_{cm},x'_{r_1}) \delta(x_{r_1} - \chi) \delta(x'_{r_1} - \chi - \delta) \\
=& \int \mathrm{d} x_{cm} \ \tilde{\psi}(x_{cm},\chi) \tilde{\psi^*}(x_{cm},\chi + \delta)
\end{align}
Now, following the paper, we introduce the new parametrization $x_{cm} =: u+\frac{m_1}{M} \chi$. One can directly see that $\mathrm d u = \mathrm d x_{cm}$, noting that $\chi$ is a constant that we fix on the left hand side.
Therefore:
\begin{align}
\langle \chi | \rho_{r_1} | \chi + \delta \rangle = \int \mathrm{d} u \ \tilde{\psi}\left(u+\frac{m_1}{M} \chi,\chi\right) \tilde{\psi^*}\left(u+\frac{m_1}{M} \chi,\chi + \delta \right)
\end{align}
Now we replace $\tilde \psi$ with the original $\psi$ again:
\begin{align}
&\langle \chi | \rho_{r_1} | \chi + \delta \rangle \\
=& \int \mathrm{d} u \ \psi\left(u+\frac{m_1}{M} \chi - \frac{m_1}{M} \chi, \ u+\frac{m_1}{M}\chi+ \frac{m_0}{M}\chi \right)\cdot \\
& \qquad \cdot\psi^*\left(u+\frac{m_1}{M} \chi - \frac{m_1}{M} \chi - \frac{m_1}{M}\delta, \ u + \frac{m_1}{M} \chi + \frac{m_0}{M} \chi + \frac{m_0}{M}\delta\right)\\
=& \int \mathrm{d} u \ \psi\left(u, u+\chi \right) \psi^*\left(u- \frac{m_1}{M}\delta, u + \chi + \frac{m_0}{M}\delta\right)
\end{align}
Now the last step is to introduce the translation operator, written as $e^{u \partial_\chi} f(\chi) = f(u+\chi)$. This equation can be verified by expanding the exponential as a series, noting that it gives the Taylor series:
\begin{align}
e^{u \partial_\chi} f(\chi) = \sum_{n=0}^\infty \frac{1}{n!} \frac{\partial^{n} f}{\partial \chi^n} u^n
\end{align}
In the case of the paper, we choose
\begin{equation}
f(\chi) := \psi\left(u, u+\chi \right) \psi^*\left(u- \frac{m_1}{M}\delta, u + \chi + \frac{m_0}{M}\delta\right)
\end{equation}
to find 
\begin{align}
\langle \chi | \rho_{r_1} | \chi + \delta \rangle = \int \mathrm{d} u \ e^{u \partial_\chi}\left[\psi\left(u,\chi \right) \psi^*\left(u- \frac{m_1}{M}\delta, \chi + \frac{m_0}{M}\delta\right) \right]
\end{align}
Now finally, we can introduce $d_0 := \frac{m_1}{M}\delta$ and
$d_1 = \frac{m_0}{M}\delta$ to arrive at their Equation (4):
\begin{align}
\langle \chi | \rho_{r_1} | \chi + \delta \rangle = \int \mathrm{d} u \ e^{u \partial_\chi}\left[\psi\left(u,\chi \right) \psi^*\left(u- d_0, \chi + d_1\right) \right]
\end{align}
This finishes the derivation.
Small additional note:
We implicitly used that the states $|x_{cm}\rangle_{cm}$ and $|x_{r_1} \rangle_{r_1}$ still work like usual position eigenstates when it comes to inner products and orthonormality. That is not obvious, but can be seen as follows:
\begin{align}
\langle x_{cm} | x'_{cm} \rangle_{cm} \langle x_{r_1} | x'_{r_1} \rangle_{r_1} :=& \langle x_0 | x'_0 \rangle_0 \langle x_1 | x'_1 \rangle_1 \\ =& \delta(x_0 - x'_0)\delta(x_1 - x_1')  
= \delta(x_0 - x'_0)\delta(x_1 - x'_1 - (x_0 -x_0')) \\
=& \delta\left(\frac{m_0}{M}(x_0-x'_0) + \frac{m_1}{M}(x_0-x'_0)\right)\delta((x_1 - x'_1) - (x_0 -x_0'))\\
=& \delta\left(\frac{m_0}{M}(x_0-x'_0) + \frac{m_1}{M}(x_1-x'_1)\right)\delta(x_1 - x'_1 - (x_0 -x_0')) = \delta(x_{cm}-x'_{cm})\delta(x_{r_1}-x'_{r_1})
\end{align}
This is consistent with the definition
$\langle x_{cm} | x'_{cm} \rangle_{cm} = \delta(x_{cm} -x'_{cm})$ and $\langle x_{r_1} | x'_{r_1} \rangle_{r_1} = \delta(x_{r_1} -x'_{r_1})$.
