# Which definition of a quantum field is right?

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?

• excitation of a quantum field is a particle. excitation of a phonon field, which is actually just the array of atoms, is a phonon, also viewed as a particle because of course, its (meaning atom array) vibrations can be quantized. So, one field is this fundamentl quantum field of, for example, electrons and the othr is just an array of atoms. – Žarko Tomičić Oct 25 '15 at 8:21