Contradiction in TISE of two-particle system $V(x_1-x_2)$ For this Hamiltonian, we have the classical Poisson bracket $\{H,p_1+p_2\}=0$. Quantising gives $$[H,P_1+P_2]=0.$$
Eigenvectors of $P_1+P_2$ are the functions $e^{i(p_1x_1+p_2x_2)}$. I say these are the eigenvectors because $$(-i\frac{d}{dx_1}-i\frac{d}{dx_2})e^{i(x_1p_1+x_2p_2)}=(p_1+p_2)e^{i(x_1p_1+x_2p_2)}.$$
These must also be the eigenvectors of $H$ because of the commutation relation.
But, all these eigenvectors are states of definite momentum (as they are also eigenvectors of $P_1$ ans $P_2$ individually). How can the eigenvectors of the Hamiltonian have definite momentum when a potential is present?
 A: The answer to this question is that while the two operators $\hat{H}$ and $\hat{P}= \hat{P}_1+\hat{P}_2$ share a simultaneous eigenbasis, it doesn't have to be the same as the eigenbasis of states that look like $e^{i(p_1x_1+p_2x_2)}$.  This is because there is an infinite degeneracy for each subspace corresponding to total momentum eigenvalue $p$.  To see this, we note that
$$
e^{i(p_1x_1+p_2x_2)}
=
e^{i(p_1-p_2)r/2}e^{ipR}
=
e^{i(2p_1-p)r/2}e^{ipR}\,,\,,
$$
where $r=x_1-x_2$ it the relative coordinate, $R=(x_1+x_2)/2$ is the center-of-mass coordinate, and $p=p_1+p_2$.  Note that as long as $p=p_1+p_2$ is fixed, this is an eigenstate of $\hat{P}$, but we can integrate over, say $p_1$ and still have an eigenstate of $\hat{P}$, i.e.,
$$
\int_{-\infty}^{\infty}dp_1\,c(p_1)e^{i(2p_1-p)r/2}e^{ipR}
=
e^{ipR}\int_{-\infty}^{\infty}dp_1\,c(p_1)e^{i(2p_1-p)r/2}\,,
$$
is an eigenstate of $\hat{P}$ while being an eigenstate of neither $\hat{P}_1$ nor $\hat{P}_2$, generally speaking.  The eigenstates of the Hamiltonian will be of this general form.

This is very general situation.  You have three operators $\hat{X}$, $\hat{Y}$, and $\hat{Z}$.  They don't all mutually commute, because $[\hat{X},\hat{Z}]\neq0$, but $[\hat{X},\hat{Y}]=[\hat{Y},\hat{Z}]= 0$.  This means that it is impossible to find a simultaneous eigenbasis for all of them at once.  However, because $\hat{Y}$ commutes with $\hat{X}$, there exists a simultaneous eigenbasis that we might label as $\lvert x,y\rangle$.  Then, because $\hat{Y}$ also commutes with $\hat{Z}$, there must be basis of eigenstates of both of these operators. Eigenstates of $\hat{Z}$ can then be written as linear combinations of these states with fixed $y$, i.e.,
$$
\lvert y,z\rangle = \sum_x c_{x,z}\lvert x,z\rangle\,.
$$
