There seems to be something very strange about the relationship between quantum field theory and quantum mechanics. It is bothering me; perhaps somebody can help.
I'll consider a free Klein-Gordon field. In standard treatments (e.g. Peskin & Schroeder and Schwartz) the one-particle momentum eigenstates $| \vec{k} \rangle$ are normalized so that
$$ \langle \vec{p} | \vec{k} \rangle = 2 \omega_{\vec{p}} (2\pi)^3 \delta^{(3)}(\vec{p}-\vec{k}), \qquad 1 = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 \omega_{\vec{p}}}| \vec{p} \rangle \langle \vec{p} |. $$ Now, assuming $\langle \vec{x}' | \vec{x} \rangle = \delta^{(3)}(x-x')$ as usual, it follows that $$ \langle \vec{x} | \vec{p} \rangle = \sqrt{2 \omega_{\vec{p}}} e^{i \vec{p} \cdot \vec{x}}. $$ Now, one can compute (here in the Schrodinger picture; see Schwartz 2.76 or P&S 2.42) that $$ \langle 0 | \phi(\vec{x}) | \vec{p} \rangle = e^{i \vec{p} \cdot \vec{x}}. $$ This is supposed to mean that $\phi$ creates a particle localized at position $\vec{x}$. P&S are a little cautious about the details, but Schwartz claims that the calculation implies $$ \phi(\vec{x}) |0 \rangle = | \vec{x} \rangle. $$ But this is false because $\langle \vec{x} | \vec{p} \rangle \neq e^{i \vec{p} \cdot \vec{x}}$ with the normalization conventions used. I suppose it could be true with some weird normalization of $| \vec{x} \rangle$, but I can't see what that might be (and at the very least this is not spelled out on the text).
Even if this works out, it seems extremely strange for there to be a relative normalization between the one-particle states of field theory and the states of one-particle relativistic quantum mechanics. One ought to be able to redo the correspondence to make the normalization work out, but I don't see how. (Note that the normalizations can be easily made to agree in the non-relativistic limit $\omega \approx m$, but that's besides the point. Even if fully relativistic quantum mechanics is inconsistent [as some texts claim without reference], at the very least the perturbative corrections for $v \ll 1$ should be recoverable from field theory.)
[Edit: This seems to go beyond normalization. We can get a feel for what kind of state $\phi(\vec{x})|0\rangle$ is by computing its wavefunction as a function of $\vec{x}'$, $$ \langle \vec{x}' | \phi(\vec{x}) |0 \rangle = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2 \omega_{\vec{p}}}} e^{i\vec{p} \cdot (\vec{x}-\vec{x}')}. $$ This wavefunction is peaked (I think divergent) at $\vec{x}'=\vec{x}$, so in some sense the particle is centered at $\vec{x}$, but it seems to be quite a stretch to say that it is at $\vec{x}$ (as the books do). I would go so far as to say the claim is incorrect, since in quantum mechanics saying the particle is at particular position means the wave function is a delta-function there. I guess the same language is used in the Heisenberg picture, when two-point functions are called amplitudes for particles to propagate from one spacetime point to another. This similarly seems false by the conventional meaning of amplitude as the overlap between two localized states. Words of wisdom would be appreciated.]