Translation Operator on two operators On my last HW set, we were asked to show that the operator
$$\hat T = \exp(-ic\hat p /\hbar)$$
acts as a translation operator ($\hat T^\dagger q\hat T=q+c)$.  This was simple to show using commutators and other such things. The same thing can be said with a momentum operator.
Can I have a translation operator that will take some operator
$$\hat A=(\hat p-p_0)(\hat q -q_0)$$
and translate this to eliminate both of the C numbers? I feel like this could be something useful for a multidimensional 'number' type operator that raises some form of state.
 A: Yes:
$$\hat A=(\hat p-p_0)(\hat q -q_0)=  e^{-i(q_0\hat p -p_0 \hat q)/\hbar}  ~\hat p \hat q ~e^{i(q_0\hat p -p_0 \hat q)/\hbar}  .$$
In particular, define 
$$
\hat T ^\dagger \equiv e^{-i~q_0\hat p  /\hbar}, 
$$
so that 
$$
\hat T ^\dagger \hat q   \hat T  =\hat q -q_0, \qquad \hat T ^\dagger \hat p   \hat T=\hat p ,
$$
and similarly
$$
\hat S  \equiv e^{i~p_0\hat q  /\hbar}, 
$$
so that 
$$
\hat S  \hat p   \hat S ^\dagger  =\hat p -p_0, \qquad \hat  S \hat q   \hat S ^\dagger=\hat q .
$$
Moreover,
$$
\hat T \hat S = \hat S \hat T \exp (i(p_0q_0)/\hbar),\\
\hat T \hat S^\dagger = \hat S^\dagger \hat T \exp (-i(p_0q_0)/\hbar)= e^{i(q_0\hat p -p_0 \hat q -p_oq_0/2)/\hbar}.
$$
Consequently,
$$
\hat A=\hat T ^\dagger \hat q  ~ \hat T  \hat S ~ \hat p  ~ \hat S ^\dagger= \hat T ^\dagger \hat S ~\hat q   \hat p  ~ \hat T    \hat S ^\dagger e^{ip_0 q_0/\hbar}= \hat T ^\dagger \hat S ~\hat q   \hat p  ~    \hat S ^\dagger  \hat T , 
$$
whence the compact form given above.
To be sure, this proof is overkill. The above expressions follows directly from the standard combinatorial lemma Ad$_{(e^X)}=e^{\operatorname{ad}_X}$ utilized in proving the CBH expansion--many call it the Hadamard lemma. So, for $X\equiv -i(q_0\hat p-p_0\hat q)/\hbar$, and $[X,\hat p]=-p_0$, $[X,\hat q]=-q_0$,
$$
\operatorname{Ad}_{e^X}  (\hat p \hat q)=  \operatorname{Ad}_{e^X} ( \hat p) \operatorname{Ad}_{e^X}(\hat q) = e^{\operatorname{ad}_X}   (\hat p) ~~e^{\operatorname{ad}_X}   (\hat q),
$$
just like your exercise. Can you see that the order of $\hat p$ and $\hat q$ acted on hardly matters in this operation? Their powers? An arbitrary function $f(\hat q, \hat p)$? To quote Feynman quoting Gibbon v1, ChIV, "The power of instruction is seldom of much efficacy except in those happy dispositions where it is almost superfluous.”
