In several of the undergraduate-level treatments of quantum mechanics I've seen, including really good ones like Allan Adams and Leonard Susskind, there seems to be a subtle contradiction that is glossed over which I would like to understand better.
For a discrete superposition state, let's take spin, it is clear to see how orthogonal spin bases are complementary to one another and what is meant when it is said that "An eigenstate in one basis is a maximally uncertain state in the complementary basis." For example:
\begin{equation} |z_{+}\rangle=\frac{1}{\sqrt{2}}(|x_{+}\rangle+|x_{-}\rangle)\tag{1} \end{equation}
But when this is carried over to something like the double slit experiment, it seems that the fact that we've moved from a discrete basis (spin) to a continuous basis (position, momentum) isn't really treated in much detail.
What I mean is this, what we undergraduates tend to be taught with respect to the double slit experiment is "When the which-path information isn't known, the position of the particle is uncertain, therefore the particle has a well-defined momentum. If the particle has a well-defined momentum, it has a well-defined wave number and so behaves like a wave, displaying interference. However, if the position is known, the momentum becomes maximally uncertain, and the wave behavior vanishes because there is no well-defined wave number."
The question I'm getting around to is this: How is a double-slit superposition state uncertain enough to correspond to a well-defined momentum state? It feels to me that in the double slit experiment, we know the position of the particle pretty well regardless. We know that it is somewhere in the laboratory, somewhere between our source and detector, and -- with respect to the particles that made it to the interference screen -- we know that it was in exactly one of the two possible slits:
\begin{equation} |\psi\rangle=\frac{1}{\sqrt{2}}|\psi_{L}\rangle+\frac{1}{\sqrt{2}}|\psi_{R}\rangle\tag{2} \end{equation}
The implication I'm speaking of seems to be that:
\begin{equation} |\psi\rangle=\frac{1}{\sqrt{2}}|\psi_{L}\rangle+\frac{1}{\sqrt{2}}|\psi_{R}\rangle=|p\rangle\tag{3} \end{equation}
Where $|p\rangle$ is a momentum eigenstate. Even acknowledging that there is obviously some simplification going on, I don't see how equation (3) is even close to being true.
How do we quantify the uncertainty "spread" in complementary variables given a discrete superposition state in a continuous basis?
I think I can at least see that if the slits have width $L$, we can quantify the "continuous" uncertainty in $x$ by:
\begin{equation} \Delta{x}=2L\rightarrow\Delta{p}\approx\frac{\hbar}{4L}\tag{4} \end{equation}
Where if $L$ is comparatively large, even on the scale of nanometers, $\Delta{p}$ is still very very small because $\hbar$ is very, very small. And so we take that small uncertainty in $p$ to be arbitrarily close to being a well-defined $p$. Is that all there is to it?
The problem is, this construction would suggest that when we know the WPI of the particle, $\Delta{x}$ simply goes to $L$ instead of $2L$, which shouldn't change the momentum uncertainty all that much.
