Well... that's the thing. The majority of renormalization in quantum field theories is in practise done by formally analytically continuing divergent quantities in the number of dimensions, which is taken to be d \in C. The divergences 1/\epsilon can then be dealt with. But 1/2
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they don't usually go away just by doing the analytic continuation, they have to be subtracted by hand (see Dimensional Regularization), hence my surprise. apologies if you obviously know all this. 2/2
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Replying to @Quantensalat @InertialObservr
No I really don't know all of this, just what I pick up talking to physicists. But yeah, my impression was that this extra infinite term was generally subtracted (with only the most superficial understanding). So when I saw this, I thought that perhaps it could be similarly (1/2)
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dealt with through analytic continuation, but the physicists just prefer this method instead. So it's safe to assume that the renormalised infinities in physics don't just go away by analytic continuation, the same as in this -1/12 case, yeh? (2/2)
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Replying to @Quasilocal @InertialObservr
no, not usually, that's what all the so-called subtraction schemes are for. The situation is slightly different though, in high energy physics we usually start at the divergent point d=4 and move away from that, then subtract the divergence and go back to d=4. with the zeta 1/2
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function, we start at a point inside the convergence and work our way to -1, right?
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Replying to @Quantensalat @InertialObservr
Yeah, I guess that's a way to look at it. And also interesting to hear about how the physicists do this stuff. One day I'm going to force myself to spend some time learning this stuff, rather than just asking people stupid questions and hoping I'll eventually learn something
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Please do me a favour: never stop asking "stupid questions". No question is stupid.
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Replying to @aminorex @jacquiavelli and
I'm curious about who this was now -- I was told the same once, and the person who told me got it from *his* analysis professor
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THE analysis professor
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