# Question regarding translation operators in momentum and position

Given the translation operators (and/or boost) on position and momentum as ($\hbar$=1) $$U_x=e^{iap}\rightarrow U^\dagger_xxU_x=x+a$$ $$U_p=e^{-abx}\rightarrow U^\dagger_ppU_p=p+b$$ And am asked to calculate $$\langle x|U_p^\dagger U_x^\dagger |\psi\rangle$$

My approach (notation is a little messy in terms of missing daggers when acting left or right, etc):

$$U_p|x\rangle=|x\rangle=\langle x|U_p^\dagger$$ $$U_x|x\rangle = \langle x|U_x^\dagger=|x+a\rangle$$ So we operate $U_p$ left then $U_x$ left to get $$\langle x|U_p^\dagger U_x^\dagger |\psi\rangle=\langle x+a|\psi\rangle = \psi(x+a)$$ I feel like I am missing something about $U_p|x\rangle$ , but I can't think of what

• You are equating bras to kets which is completely wrong. – jinawee Oct 1 '15 at 21:35

You are certainly missing something out. $$U_p |x_0 \rangle = \exp(-i b \hat{x}) |x_0 \rangle =\exp(-i b x_0) |x_0 \rangle$$ Your equations are clearly wrong which you can see from $$\langle x_0|U_p^\dagger p U_p |x_0\rangle =\langle x_0|p |x_0\rangle+b$$ using your second equation. Using your third equation one would get $$\langle x|U_p^\dagger p U_p |x\rangle =\langle x|p |x\rangle$$ which is wrong.
So $$\langle x|U_p^\dagger U_x^\dagger |\psi \rangle =\exp(i b x) \psi(x+a)$$