# Proof of the translation operator identity $\langle x| T(a) | \psi \rangle = \psi(x-a)$

I know the translation operator behaves as follows:

$$T(a) |x \rangle = |x+a \rangle$$

However my notes state that for a general quantum state $$|\psi \rangle$$ we know that

$$\langle x| T(a) | \psi \rangle = \psi(x-a)$$

However I am unsure where this comes from and would like to know how to derive this. I know that we can write that $$\langle x| \psi \rangle = \psi(x)$$ but am unsure what exactly I should do.

We can express $$\vert\psi\rangle$$ as $$\int dy\ \psi(y)\vert y\rangle$$. Thus, we can write \begin{align} \langle x|T(a)|\psi\rangle &= \langle x|T(a)\int dy\ \psi(y)|y\rangle \\ &= \langle x|\int dy\ \psi(y)|y+a\rangle\\ &=\int dy\ \psi(y)\langle x|y+a\rangle \\ &=\int dy\ \psi(y)\delta(x-y-a)\\&=\psi(x-a) \end{align}

Since $$T(a)$$ is unitary, $$T(a)^\dagger=T(a)^{-1}=T(-a)$$, so

$$\langle x|T(a)|\psi\rangle=\Big(T(a)^\dagger|x\rangle\Big)^\dagger|\psi\rangle=\Big(T(-a)|x\rangle\Big)^\dagger|\psi\rangle=\Big(|x-a\rangle\Big)^\dagger|\psi\rangle=\langle x-a|\psi\rangle=\psi(x-a)$$

Operators like $$T(a)$$ can work backwards on the bras too (up to complex conjugate).

$$\langle x| T(a) = \langle x - a|$$

$$\langle x| T(a) | \psi \rangle = \psi(x-a)$$ then becomes

$$\langle x - a | \psi \rangle = \psi(x-a)$$