Translation operator

The translation operator is

\begin{align*} \hat T_{x_0} = e^{-i x_p \hat p / \hbar}. \end{align*}

This means that the momentum operator is the generator of position translation, just as the Hamiltonian operator is the generator of time evolution.

Let’s see how the translation operator affects the expectation of position. Consider a position eigenstate $\ket{x_1}$ and a state $\ket \phi = \hat T_{x_0} \ket{x_1}$ . First consider in $p$ space.

\begin{align*} \braket{p|T_{x_0}|x_1} &= \braket{p|e^{-ix_0p/\hbar}|x_1} = e^{-ix_0p/\hbar} \braket{p|x} \\ &= e^{-ix_0p/\hbar} e^{-ix_1p/\hbar} \frac{1}{\sqrt{2\pi\hbar}} \\ &= \frac{1}{\sqrt{2\pi\hbar}} e^{-i(x_1+x_0)} p /\hbar = \braket{p|x_0+x_1}. \end{align*}

By the completeness of $p$ eigenstates and by linearity, conclude $\hat T_{x_0} \ket{x_1} = \ket{x_1 - x_0}$ . It follows that $T_{x_0}$ translates an arbitary state $\ket \psi$ .

\begin{align*} \psi(x) &= \braket{x|\psi} \\ \braket{x|T_{x_0}|\psi} &= \braket{x-x_0|\psi} = \psi(x-x_0). \end{align*}

We can show that

\begin{align*} [T_{x_0},T_{x_1}] = 0 \quad\implies\quad T_{x_0+x_1} = T_{x_0} T_{x_1}. \end{align*}

So translation forms a commutative group. Translation is also obviously unitary so $T_{x_0}\adj = T_{x_0}\inv$ .

A generalization of the translation operator is the displacement, which allows a complex $x_0$ .