Units and the momentum eigenstate in position basis Consider $\psi(x)=\langle x|\psi\rangle$. Such a wavefunction must have dimensions of inverse square root length to satisfy the normalization condition. Why then does $\langle x | p \rangle = \frac{1}{\sqrt{2\pi\hbar}}e^{-ikx}$ have units of inverse square root action? (We can flip this question around and ask why $\langle p | x\rangle$ doesn't have units of inverse root momentum too.)
This came up when I considered the expectation value of operator $X^n$ for a momentum eigenket. What's wrong with this calculation? The units make no sense!
\begin{align}
\langle X^n\rangle &=\langle p|X^n|p\rangle 
\\
&= \langle p |\int dx |x\rangle \langle x|X^n|p\rangle
\\
&= \int dx\ x^n \langle p |x\rangle \langle x|p\rangle
\\
&= \int \frac{dx}{2\pi\hbar} x^n
\\
&= \lim_{a\to\infty}\frac{x^{n+1}}{(n+1)2\pi\hbar} \bigg|_{-a}^a
\end{align}
such that $\langle X^n\rangle$ seems to vanishes for $n$ odd and diverge for $n$ even. That's strange enough, but prior to evaluation, the units of the formal integral are those of $[x^n]/[p]$ - not what I expected... I feel the issue may be rooted in the first part of my question, and perhaps is linked to the assumptions in QM behind defining position and momentum eigenkets.
 A: In quantum mechanics, we often want to decompose our state into a basis.  There are two qualitatively different types of bases, discrete and continuous.  In the discrete case, we can simply write
$$
| \psi \rangle = \sum_n c_n | n \rangle \ ,
$$
where $| \psi \rangle$, $| n \rangle$, and the $c_n$ are all dimensionless.
In the continuum case, we need an integral.  For example, for the questioner's momentum states, we could write
$$
| \psi \rangle = \int dp \, | p \rangle \langle p | \psi \rangle \ . 
$$
The wave packet $|\psi \rangle$ is dimensionless and assumed to be properly normalized $\langle \psi | \psi \rangle = 1$.
But now the $dp$ in the measure has units of momentum.  It follows that $| p \rangle$ and $\langle p |$ should have units of the square root of the inverse of momentum.  We could run a similar argument for the position eigenstates $|x \rangle$ from which we could conclude that $|x \rangle$ and $\langle x |$ have units of the inverse of the square root of length.  Furthermore, $\langle x | p \rangle$ has units of the square root of the inverse of momentum times length, or the same units as $1 / \sqrt{\hbar}$, as the questioner has observed.
When we take an expectation value of an operator like $\hat x^n$, we should take it with respect to a wave-packet $|\psi \rangle$ that is properly normalized.  With respect to the wave packet, we find 
$$
\langle \hat x^n \rangle = \langle \psi | \hat x^n | \psi \rangle .
$$
We can then further decompose $| \psi \rangle$ into momentum eigenstates or position eigenstates, but it should be clear at this point that the expectation value will have the correct units.  The measure factors $dp$ or $dx$ will cancel the extra dimensional factors from the $|p\rangle$'s, $\langle p |$'s, $|x \rangle$'s, and $\langle x|$'s.
