Let me start from path integral formulation in quantum mechanics and quantum field theory. In QM, we have U(xb,xa;T)=⟨xb|U(T)|xa⟩=∫DqeiS
In QFT we have U(ϕb,ϕa;T)=⟨ϕb|U(T)|ϕa⟩=∫DϕeiS
By analogy with QM, it is tempting to relate |ϕ⟩↔|x⟩
However, in Peskin and Schroeder's QFT, p24, by computing it is said
⟨0|ϕ(x)|p⟩=eip⋅x
We can interpret this as the position-space representation of the single-particle wavefunction of the state |p⟩, just as in nonrelativistic quantum mechanics ⟨x|p⟩∝eip⋅x is the wavefunction of the state |p⟩.
Based on the quoted statement, seems ˆϕ(x)|0⟩↔|x⟩
If relations (3) and (4) are both correct, I should have ˆϕ(ˆϕ|0⟩)=ϕ(x)(ˆϕ|0⟩)
How to reconcile analogies (3) and (4)?
Answer
No ˆϕ|0⟩ is not an eigenvector of ˆϕ. You can see this, for example, by writing out ˆϕ in terms of creation and annihilation operators, then compare ˆϕ|0⟩ against ˆϕ2|0⟩, and observe that one is not a scalar multiple of the other. So as you suspected, eq. 5 is not correct
To obtain some analogy of |x⟩, you can just take a fourier transform of a†(p) to get a†(x), and a†(x)|0⟩≡|x⟩ is the best analogy of |x⟩ that I can think of
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