Why can't we interprete /x> in relativistic QFT as position eigenfunc?

[fxdung]

Why can't we interprete /x> in relativistic QFT as position eigenstate?And by the way what is the difference between /x> and /1x>=Phi(field operator)(x)/0>?

 

[Demystifier]

In the meantime I made a short answer here: https://www.physicsforums.com/threa...f-notion-of-wave-function.994115/post-6400897

 

[Demystifier]

And by the way what is the difference between /x> and /1x>=Phi(field operator)(x)/0>?

In non-relativistic QFT it's the same. In relativistic QFT one changes the normalization of $\phi$ so that it becomes Lorentz invariant, the consequence of which is that $|1x\rangle$ differs from $|x\rangle$. See e.g. my
http://de.arxiv.org/abs/hep-th/0202204
Eqs. (55)-(60). In particular, the function (59) is Lorentz invariant, but is not a $\delta$-function.