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

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  • Thread starter fxdung
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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>?
 

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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.
 

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