Notation for a Magnetic Translation of a State as a Similarity Transformation

Question

Equation 180 of Aravos' notes gives: $$t(\mathbf{d})\psi(\mathbf{r})t^\dagger(\mathbf{d})=\psi(\mathbf{r}+\mathbf{d})$$

Where $t(\mathbf{d})$ is the magnetic translation operator. Why does this operator act on both sides of the state? It makes sense if it was an operator instead of a state because then if it acted on a translated state, it would be untranslated, acted on by the operator in the untranslated basis, then translated back.

Or is this just an overuse of notation where $\psi$ doesn't represent a state $\left|\psi\right>$ with the $\mathbf{r}$ showing that it is untranslated and not in a basis?

Answer 1

Didn’t look at the notes, but this is standard notation for second quantization. $\psi$ is an annihilation operator so that: $$ \psi(r)|r\rangle =|0\rangle $$ The action on the kets: $$ t(d)|r\rangle =|r+d\rangle $$ is equivalent to the action the operators: $$ t(d)\psi(r)t(d)^{-1}=\psi(r+d) $$ The correspondence between the two is spatial analogue of the equivalence between the Schrödinger picture and the Heisenberg picture for time.

Hope this helps.