How Does the Operator e^{i\ell q_0/\~R} Affect Quantum State Windings?

[Jimmy Snyder]

Homework Statement

Verify that acting on a state with winding m, the operator
$$e^{i\ell q_0/\~R}$$
gives a state with winding $m + \ell$

Homework Equations

Equation (17.83)
$$e^{-i\ell q_0/\~R}we^{i\ell q_0/\~R} = w + \frac{\ell}{\~R}$$

The Attempt at a Solution

It seems easy: given |m> where w|m> = m|m> we have
$$w|e^{i\ell q_0/\~R}m> = e^{i\ell q_0/\~R}e^{-i\ell q_0/\~R}we^{i\ell q_0/\~R}|m> = e^{i\ell q_0/\~R}(w + \frac{\ell}{\~R})|m> = (m + \frac{\ell}{\~R})|e^{i\ell q_0/\~R}m>$$
So is there a typo in the book?

 

[Jhero]

No, there is not a typo in the book. Your solution is correct. The operator e^{i\ell q_0/\~R} does indeed give a state with winding m + \ell when acting on a state with winding m. This can be seen from the equation (17.83) provided in the homework, which shows that the operator e^{-i\ell q_0/\~R}we^{i\ell q_0/\~R} = w + \frac{\ell}{\~R}. This equation directly corresponds to the solution you have provided. Therefore, there is no need to doubt the correctness of the book.