Why is there $1/2\pi$ in $\int\frac{dp}{2\pi}|p\rangle\langle p|$?

I'm reading Richard MacKenzie's lectures on path integrals and on page 7 he derives the propagator for the free particle as follows:

\begin{align} K &= \langle q'|e^{-iHT}|q\rangle \\ &= \langle q'|e^{-iT\hat p^2/2m} \int\frac{dp}{2\pi}|p\rangle\langle p|q\rangle \\ &= \int \frac{dp}{2\pi} e^{-iT p^2/2m} \langle q'|p\rangle\langle p|q\rangle \\ &= \int \frac{dp}{2\pi} e^{-T(p^2/2m)+i(q'-q)p} \\ &= \boxed{\left(\frac{m}{2\pi iT}\right)^{1/2} e^{im(q'-q)^2/2T}} \end{align}

I'm comfortable with all the steps except the first. What I want to know is: where does the factor of $1/2\pi$ come from? I'm familiar with the idea of expanding a state function with respect to a basis, but $\int dp |p\rangle \langle p| = 1,$ not $2\pi$. Is there an implicit Fourier transform when he switches to the momentum basis? Does it have something to do with the fact that he's using units where $\hbar = 1$?

The factor of $1/2\pi$ is an artifact of the normalization convention being used for the momentum eigenstates.

To begin to see how this is so, let us note that the choice of normalization of a Dirac-orthogonal continuous basis completely determines the form of the resolution of the identity. Writing an arbitrary state $|\psi\rangle$ in a given continuous" basis gives \begin{align} |\psi\rangle = \int da'\,\psi(a')|a'\rangle, \end{align} and applying $\langle a'|$ to both sides gives \begin{align} \langle a|\psi\rangle = \int da' \,\psi(a')\langle a|a'\rangle. \end{align} Now if this basis is Dirac orthogonal with normalization $K(a)$; \begin{align} \langle a |a'\rangle = K(a)\,\delta(a-a') \end{align} then we find $\langle a|\psi\rangle = K(a)\psi(a)$ which gives \begin{align} |\psi\rangle = \int \frac{da'}{K(a)}\,\langle a'|\psi\rangle|a'\rangle \end{align} which implies \begin{align} \int \frac{da'}{K(a)} |a'\rangle\langle a'| = I. \end{align} Now specializing to quantum mechanics, recall a standard convention is to set \begin{align} \langle q|q'\rangle = \delta (q-q'), \qquad \langle p|p'\rangle = 2\pi\delta(p-p') \tag{DefCon1} \end{align} and therefore by the remarks above, this implies the following resolutions of the identity: \begin{align} \int dq'\, |q'\rangle\langle q'| = I, \qquad \int \frac{dp'}{2\pi} \,|p'\rangle\langle p'| = I. \end{align}

Motivation for normalization conventions

To try to motivate the conventional normalizations of the position and momentum eigenstates written above, suppose we were to adopt the following general normalizations: \begin{align} \langle q|q'\rangle = C(q)\,\delta(q-q'), \qquad \langle p|p'\rangle = K(p)\,\delta(p-p'), \end{align} for functions $C(q)$ and $K(p)$. Let $Q$ be the position operator, and let $P$ be the momentum operator. Recall that the canonical commutation relation $[Q,P] = iI$ can be used to show that $e^{-iaP}$ is the generator of translations, namely that $e^{iap}|q\rangle = |q+a\rangle$. It follows that $\langle p|q\rangle = \langle p|e^{-iqP}|0\rangle = e^{iqp}\langle p|0\rangle$. If we let $\langle p|0\rangle=\phi(p)$ for some undetermined function $\phi$, then in summary we have so far shown that \begin{align} \langle p|q\rangle= e^{iqp}\phi(p) \end{align} If we now consider the quantity $\langle q|\psi\rangle$, and if we insert one resolution of the identity integrating over $p$ and the other integrating over $q'$, then we get \begin{align} \langle q|\psi\rangle = \int dq'\left[\int \frac{dp}{2\pi}e^{-i(q-q')p}\frac{2\pi|\phi(p)|^2}{C(q')K(p)}\right] \langle q'|\psi\rangle \end{align} Now, in order for this equation to be true, the term in brackets must equal $\delta(q-q')$, which means that \begin{align} \frac{2\pi|\phi(p)|^2}{C(q')K(p)} = 1 \end{align} Another way of saying this is that position and momentum representations must be related by Fourier transform. As you can see, we can arbitrarily fix $C$ and $K$ to satisfy this condition because we have complete freedom to choose $\phi(p)$. The conventional choice leading to $(\text{DefCon1})$ is basically as simple as you can get because two of the degrees of freedom, namely $C(q)$, and $\phi(p)$ are set to 1, and this fixes $K(p)$ to be $2\pi$.

Addendum. Other conventions in the literature

As Qmechanic points out in his comment below, it's useful to stress that there is at least one other common convention in the literature that is just as simple as $(\text{DefCon1})$, and that is to choose \begin{align} C(q) = 1, \qquad K(p) = 1, \qquad \phi(p) = \frac{1}{\sqrt{2\pi}} \end{align} which would yield \begin{align} \langle q|q'\rangle = \delta (q-q'), \qquad \langle p|p'\rangle = \delta(p-p') , \qquad \langle p|q\rangle = \frac{1}{\sqrt{2\pi}}e^{-ipq}\tag{DefCon2} \end{align} This convention is used by Cohen-Tannoudji and Sakurai in their quantum texts. Of course, in practice, as long as one sticks to a single convention, no one is stopping her from choosing any convention she chooses, except for possibly a research advisor or physics gods.

• Thank you for spelling this out so precisely! I had gotten hung up on Dirac orthogonality before and ultimately came to the conclusion that the normalization coefficient was a matter of convention—only I always assumed the convention was to set K=1 in both cases. Now this begs the question: what motivates this particular convention? – user3736508 Oct 17 '14 at 4:51
• @user3736508 Sure thing. I added an extra section attempting to motivate the convention, but seeing as how that convention isn't universal, it's clear that it is, in fact, just a convention, so in the end, it's really just an aesthetic choice as far as I can tell. – joshphysics Oct 17 '14 at 7:11
• That's a long way to say that the $\pi$ makes the Fourier transforms work :) – DanielSank Oct 17 '14 at 7:15
• @joshphysics: Great follow up! This actually helps me a lot, and it made me think of one reason normalization conventions might be more than aesthetics: QFT is built on the canonical commutation relations, one of which says $[a(\vec k), a^\dagger(\vec k)] = \langle k|k' \rangle = (2\pi)^3 2\omega \delta^3(\mathbf k - \mathbf k')$ where $\omega = \sqrt{\mathbf k^2 +m^2}$ (Srednicki p. 49). Without this choice of normalization, the theory wouldn't be Lorentz-invariant. – user3736508 Oct 17 '14 at 13:02
• @user3736508: Conventional factors of $2\pi$ do not affect the status of Lorentz-invariance. – Qmechanic Oct 17 '14 at 15:51