# Commutator of Position Operator and Generator of Translations

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.

• Which reference by Townsend? – Qmechanic Nov 10 '19 at 6:23
• I'm suspicious of using $\hat T(\delta x) = 1-\frac{i}{\hbar} \hat P_x \delta x$ to solve this problem – s0ggyj0hns0n Nov 10 '19 at 6:26

$$T(\delta x) \delta x = \delta x + \mathcal O(\delta x^2)$$
so if you stick to first order (i.e. consider infinitesimal $$\delta x$$), your result and the more general result from wikipedia agree.