In introductory quantum field theory, I was taught that, given a single-particle Hilbert space $\mathcal H$, the quantum field operator for that type of particle was a mapping $\varphi(x)$ from $k$-tensors of single-particle states to (k+1)-tensors of single-particle states given by creating a position eigenstate at $x$, namely,
$$\varphi(x)~\cdot ~|\psi_1\rangle\otimes \cdots \otimes|\psi_k\rangle:=|x\rangle\otimes |\psi_1\rangle\otimes \cdots \otimes|\psi_k\rangle.$$
However, in my solid state physics class, it's a whole other story. For phonons, the quantum field operator $u(x)$ is the observable corresponding to the displacement of the atom at site $x$ in the lattice. For example, for a 1D lattice $X=\{a,2a,\cdots Na\}\subset \mathbb S^1$,
$$u(na)~(|\psi(x)\rangle_a\otimes \cdots \otimes|\psi(x)\rangle_{Na}):=|\psi(x)\rangle_a\otimes \cdots \otimes |x\psi(x)\rangle_{na}\otimes\cdots \otimes|\psi(x)\rangle_{Na}$$
So one operator adds a position eigenstate at site $x$, and the other simply is the position operator at site $x$. Where's the equivalence?