What if the momentum operator didn't commute in regards for the translator operator? This is more of a mathematical/conceptual question. I am reading through Chapter 12 of Shankar's "Principle's Of Quantum Mechanics" 2nd Edition, and in the very first section he describes a "consistency test" for the translation operator in that:
$$T(\mathbf{b})T(\mathbf{a})=T(\mathbf{a+b}) \tag{12.1.6}$$
$$ e^{-i\mathbf{b}\cdot\mathbf{P}/\hbar }e^{-i\mathbf{a}\cdot\mathbf{P}/\hbar }=e^{-i\mathbf{(a+b)}\cdot\mathbf{P}/\hbar } \tag{12.1.7}$$
Should be true and without proof, he says yes because the operators commute:
$$[P_x,P_y]=0 \tag{12.1.8}$$
This is what I don't understand, how would the exponentials combine if $[P_x,P_y]=0$ did not commute? It seems like they would combine in the same exact way. For example, lets suppose we have two translations in the $\vec{x}$ direction so that we have:
\begin{align}
T(\mathbf{b\hat{x}})T(\mathbf{a\hat{x}})&=e^{-i\mathbf{b}\cdot\mathbf{P}/\hbar }e^{-i\mathbf{a}\cdot\mathbf{P}/\hbar }\\
&=e^{-ib{P_x}/\hbar }e^{-ia{P_x}/\hbar  }\\
&=e^{-ib{P_x}/\hbar-ia{P_x}/\hbar}\\
&=e^{-\frac{i}{\hbar} (bP_x+aP_x) }\\
&=e^{-\frac{i}{\hbar} (b+a)P_x }\\
\end{align}
Which still satisfies the "law of combination" for the translation and does not even rely on commutativity. Can some one clear this up for me?
 A: If $A$ and $B$ do not commute, then generally
$$e^A e^B \neq e^{A+B}.$$
Then if $P_x$ and $P_y$ don't commute, we generally have
$$T(a \hat{x}) T(b \hat{y}) \neq T(a \hat{x} + b \hat{y}).$$
You only considered the case of combining translations along the same direction; these will always compose nicely, basically by definition. 

Clarification: let's expand both $e^A e^B$ and $e^{A+B}$ and compare the results. On the left-hand side, 
$$(1+A+A^2/2 + \ldots)(1+B+B^2/2 + \ldots) = 1 + (A + B) + \left(AB + \frac{A^2}{2} + \frac{B^2}{2}\right) + \ldots.$$
On the right-hand side, we have
$$1 + (A+B) + \frac{(A+B)^2}{2} + \ldots.$$
The second order terms are not the same! The cross terms are $AB$ on the left-hand side, and $(AB+BA)/2$ on the right-hand side. So they're only generally equal if $A$ and $B$ commute.
A: In general, for operators $X$ and $Y$ which do not necessarily commute, one has,
$$\ln(\exp X \exp Y) = X + Y + \frac12 [X,Y] + \frac{1}{12} \left([X,[X,Y]]+[Y,[Y,X]] \right) + \dots$$
which is known as the Baker-Campbell-Hausdorff formula. As you can see, if $P_x$ and $P_y$ commute, the higher order terms vanish and indeed the resultant expression is simply $P_x+ P_y$.
As a somewhat ad hoc example, one would have $e^a e^{a^\dagger} = e^{a + a^\dagger + \frac12}$ since $[a,a^\dagger]= 1$, and only the first commutator contributes, since higher terms are nested commutators and obviously constants commute with $a$ and $a^\dagger$.
