# 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.

## 1 Answer

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.”