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

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.

• Dimensional analysis is your friend! If you know which physical quantity $\hat{x}_j$ represents, you can always check if a formula might be correct by comparing dimensions. In this case, the two candidates have different dimensions and therefore you can find out for sure which option fits with this very useful trick. – Mark Mitchison Feb 25 '15 at 0:43
• I wondered if that was the route to go. Must trust intuition. I keep getting that feedback on here. Thank you! I'll try that and see what I find. – Stan Shunpike Feb 25 '15 at 0:45
• In my opinion dimensional analysis is one of the most useful mathematical tricks in the physicist's arsenal. Always use it when you can. – Mark Mitchison Feb 25 '15 at 0:46
• One question: Can operators have units? I thought for instance the position operator merely acted on abstract states known as position space. I didn't think it had units because it isn't specifying the specific position of the particle. – Stan Shunpike Feb 25 '15 at 0:46
• Sure operators can be dimensionful, indeed they often must be. A position expectation value must have dimensions of length, for example. This can only be true if $[\hat{x}] = L$. – Mark Mitchison Feb 25 '15 at 0:48

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.