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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?

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  • $\begingroup$ 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. $\endgroup$ – Žarko Tomičić Oct 25 '15 at 8:21
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None of these is the action definition of a quantum field.

They are apparently used in several approximations to quantum field theory, but they are not valid definitions in usual relativistic QFT. The first one doesn't work because there is no position operator in QFT, and hence no notion of "position eigenstate". The second one doesn't work because QFT is not, in general, working with lattices of atomes, and thus has no notion of "displacement".

Also, the two do evidently not agree, they are not equivalent. They may be useful/valid in dealing with the special cases they are meant for, but you should think of neither of those as a generic quantum field, since, in general, a quantum field is an operator-valued distribution on spacetime, and nothing more. It has, in full generality, neither necessarily a relation to particle states nor to displacements or any other specific setup.

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  • $\begingroup$ Suppose I were only interested in non-relativistic lattice QFT - then are these equivalent? $\endgroup$ – Dave Oct 25 '15 at 16:23
  • $\begingroup$ @Dave: Uh, how could they? One is an operator that preserves particle number, the other doesn't. $\endgroup$ – ACuriousMind Oct 25 '15 at 16:25
  • $\begingroup$ Exactly. that's the problem. So there are two notions of "field" in this setting. $\endgroup$ – Dave Oct 25 '15 at 17:03

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