Commutator of Position Operator and Generator of Translations

Question

Problem:

Let $|\psi' \rangle = \hat T(\delta x) | \psi \rangle$ for infinitesimal $\delta x$, show that $\langle x \rangle'= \delta x +\langle x \rangle$

Attempt at Solution:

Using $\hat T(\delta x) = 1-\frac{i}{\hbar} \hat P_x \delta x$ and $[\hat x, \hat P_x] = i \hbar$, I calculate $[\hat x, \hat T(\delta x)]=\delta x$

Therefore,

$$\hat x \hat T(\delta x)=\delta x + \hat T(\delta x) \hat x$$

Plug this into $\langle \psi | \hat T^{\dagger}(\delta x) \ \hat x \ \hat T(\delta x) | \psi \rangle$

$$\langle \psi | \hat T^{\dagger}(\delta x) (\delta x + \hat T(\delta x) \hat x) | \psi \rangle$$

$$\langle \psi | \hat T^{\dagger}(\delta x) \delta x | \psi \rangle + \langle \psi | \hat T^{\dagger}(\delta x) \hat T(\delta x) \hat x | \psi \rangle$$

Since $\hat T$ is unitary, this almost simplifies to the result:

$$\langle \psi | \hat T^{\dagger}(\delta x) \delta x | \psi \rangle + \langle \hat x \rangle$$

Question:

What am I doing wrong here? I've seen on wikipedia that $[\hat x, \hat T(\delta x)]=\hat T(\delta x) \delta x$, which would solve the problem since

$$\langle \psi | \hat T^{\dagger}(\delta x) \ \delta x \ \hat T(\delta x) | \psi \rangle$$

$$\delta x \langle \psi | \hat T^{\dagger}(\delta x) \hat T(\delta x)| \psi \rangle$$

$$\delta x $$

but I'm really not seeing how to get there with the way Townsend has defined things.

Answer 1

You're almost there. Note that

$$T(\delta x) \delta x = \delta x + \mathcal O(\delta x^2)$$

so if you stick to first order (i.e. consider infinitesimal $\delta x$), your result and the more general result from wikipedia agree.