The exponential of the derivative translates a function 𝑓(𝑥) by an amount εpic.twitter.com/mwQ1aLpOHt
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I tried to interpret this intuitively but am struggling because it relies on the spooky action at a distance enjoyed by analytic functions and is false for general infinite-dimensional functions. Pretty cool though.
that's true .. I think my Physicist is showing .. we damn near always assume analyticity and function/finite number or arguments
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There's something I find very weird about assuming analyticity in physics. It basically means you are assuming perfect knowledge of what's going on outside of your lightcone, right? Which seems to me... odd.
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Sure but that's why we impose locality of operators
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No analyticity is needed. As any mathematical physicist knows, there's a self-adjoint operator i d/dx on L^2(R) and a theory of exponentiating self-adjoint operators that makes exp(t d/dx)(f) = f(x+t) true for all functions f in L^2(R). It's not just wishful thinking.
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I’m quite familiar with the momentum operator id/dx but how does that factor of i save us from what
@wtgowers was saying? is it the hermiticity? - 7 more replies
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