# 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.

## 1 Answer

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.

• So since $|p\rangle$ is non-normalizable, we cannot compute $\langle p | X^n|p\rangle$. Is this correct? – zahbaz Oct 20 '17 at 4:50
• Essentially yes. Your expectation value does not have the proper dimensions because the state is not normalized. – lcv Oct 20 '17 at 11:03