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meopemuk
meopemuk is offline
#64
Nov18-09, 07:51 PM
P: 1,746
Quote Quote by strangerep View Post
Except that the a/c operators in QFT2 are not bona-fide operators, but are really
operator-valued distributions. Hence fields constructed as linear combinations of them
are also operator-valued distributions. Hence QFT2 suffers exactly the same "ill-defined
equal-point multiplication of distributions" mathematical problem as QFT1.
Could you give an example in which a product of a/c operators or quantum fields is "ill-defined"?