Missing a factor of $\sqrt{\frac{\hbar}{m}}$ in a QFT Practice Problem. Can someone explain why?

Question

I am doing problem 2.3 on page 27 of Quantum Field Theory for the Gifted Amateur.

Use eqns 2.46 and 2.62 to show that \begin{equation} \hat{x}_j = \frac{1}{\sqrt{N}} \left(\frac{\hbar}{m}\right) \sum_k \frac{1}{\sqrt{2\omega_k}} [\hat{a}_k e^{\text{i}kja} +\hat{a}^\dagger_k e^{-\text{i}kja}] \end{equation}

Here's the two equations: \begin{align} \hat{x}_j &= \frac{1}{\sqrt{N}} \sum_k \tilde{x}_k e^{\text{i}kja}\\ \hat{x}_k &= \sqrt{\frac{\hbar}{2m\omega_k}}\left(\hat{a}_k + \hat{a}^\dagger_{-k} \right) \end{align} This seem pretty straightforward. I just substituted (2) into (1), distributed, then adjusted the value of the $e$ exponent by making the indices on the creation operators positive and the exponent negative:

\begin{align} \hat{x}_j &= \frac{1}{\sqrt{N}} \sqrt{\frac{\hbar}{m}}\sum_k\frac{1}{\sqrt{2\omega_k}}\left(\hat{a}_k + \hat{a}^\dagger_{-k} \right) e^{\text{i}kja}\\ \hat{x}_j &= \frac{1}{\sqrt{N}} \sqrt{\frac{\hbar}{m}} \sum_k \frac{1}{\sqrt{2\omega_k}} [\hat{a}_k e^{\text{i}kja} +\hat{a}^\dagger_{-k} e^{\text{i}kja}]\\ \hat{x}_j &= \frac{1}{\sqrt{N}} \sqrt{\frac{\hbar}{m}} \sum_k \frac{1}{\sqrt{2\omega_k}} [\hat{a}_k e^{\text{i}kja} +\hat{a}^\dagger_k e^{-\text{i}kja}]\\ \end{align} But as you can see, this doesn't match what they say I should end up with. I am missing a factor of $\sqrt{\frac{\hbar}{m}}$.

My Question:

Why am I missing this factor of $\sqrt{\frac{\hbar}{m}}$?

Intuitively, the fact that everything else matches up seems like a good sign. But it should work out then and it doesn't. So either I'm missing something or the problem as stated is written wrong.

Answer 1

Dimensional analysis shows the book's answer is wrong. Let's work it out. I should obviously get length because this is a position operator. Since neither $e^{\text{i}jka}$ nor the creation annihilation operators have units, I can ignore those terms.

This reduces to

\begin{equation} \sqrt{\frac{1}{m}s\frac{m \cdot L^2}{s^2}}\sqrt{s} = \sqrt{\frac{L^2}{s}}\sqrt{s} = L \frac{\sqrt{s}}{\sqrt{s}} = L \end{equation}

So the answer I gave was right by the above.

I also realized I could look this up by going to their errata page. Next time I have a problem like this, that's what I'll do first instead of posting to SE. But good to learn!