- #1

Frank Castle

- 580

- 23

I've been reading through a derivation of the LSZ reduction formula and I'm slightly confused about the arguments made about the assumptions: $$\langle\Omega\vert\phi(x)\vert\Omega\rangle =0\\ \langle\mathbf{k}\vert\phi(x)\vert\Omega\rangle =e^{ik\cdot x}$$ For both assumptions the author first relates ##\phi(x)## to ##\phi(0)## by using the 4-momentum operator ##P^{\mu}##, i.e. $$\phi(x)=e^{iP\cdot x}\phi(0)e^{-iP\cdot x}$$ such that, in the case of the first assumption, one has $$\langle\Omega\vert\phi(x)\vert\Omega\rangle =\langle\Omega\vert e^{iP\cdot x}\phi(0)e^{-iP\cdot x}\vert\Omega\rangle =\langle\Omega\vert\phi(0)\vert\Omega\rangle$$ where we have used that the vacuum state satisfies ##P^{\mu}\lvert\Omega\rangle =0##, such that ##e^{-iP\cdot x}\vert\Omega\rangle = \vert\Omega\rangle##.

What I don't understand is, why do we need to do this in the first place? Is it simply so we can use that ##v\equiv\langle\Omega\vert\phi(0)\vert\Omega\rangle## is a Lorentz scalar, to rewrite $$\langle\Omega\vert\phi(x)\vert\Omega\rangle-\langle\Omega\vert\phi(0)\vert\Omega\rangle =\langle\Omega\vert\phi(x)\vert\Omega\rangle-v=\langle\Omega\vert(\phi(x)-v)\vert\Omega\rangle =0$$ and so if ##\langle\Omega\vert\phi(x)\vert\Omega\rangle\neq 0##, then we can shift the field ##\phi(x)\rightarrow\phi(x)-v##, and guarantee that ##\langle\Omega\vert\phi(x)\vert\Omega\rangle=0##?!

Similarly, for the second assumption, if ##\langle\mathbf{k}\vert\phi(x)\vert\Omega\rangle \neq e^{ik\cdot x}##, then we can use that ##P^{\mu}\lvert\mathbf{k}\rangle =k^{\mu}\lvert\mathbf{k}\rangle## such that ##e^{ik\cdot x}\lvert\mathbf{k}\rangle##, and $$\langle\mathbf{k}\vert\phi(x)\vert\Omega\rangle=\langle\mathbf{k}\vert e^{iP\cdot x}\phi(0)e^{-iP\cdot x}\vert\Omega\rangle=e^{-ik\cdot x}\langle\mathbf{k}\vert \phi(0)\vert\Omega\rangle$$ In requiring that ##\langle\mathbf{k}\vert \phi(0)\vert\Omega\rangle =1##, we must (in general) rescale ##\phi(0)##, such that ##\phi(0)\rightarrow\sqrt{Z_{\phi}}\phi(0)##.

Again, is the reason why we relate ##\langle\mathbf{k}\vert\phi(x)\vert\Omega\rangle## to ##\langle\mathbf{k}\vert\phi(0)\vert\Omega\rangle## so that we can "induce" the ##e^{-ik\cdot x}## and then impose a condtion on the Lorentz scalar ##\langle\mathbf{k}\vert \phi(0)\vert\Omega\rangle##?!

What I don't understand is, why do we need to do this in the first place? Is it simply so we can use that ##v\equiv\langle\Omega\vert\phi(0)\vert\Omega\rangle## is a Lorentz scalar, to rewrite $$\langle\Omega\vert\phi(x)\vert\Omega\rangle-\langle\Omega\vert\phi(0)\vert\Omega\rangle =\langle\Omega\vert\phi(x)\vert\Omega\rangle-v=\langle\Omega\vert(\phi(x)-v)\vert\Omega\rangle =0$$ and so if ##\langle\Omega\vert\phi(x)\vert\Omega\rangle\neq 0##, then we can shift the field ##\phi(x)\rightarrow\phi(x)-v##, and guarantee that ##\langle\Omega\vert\phi(x)\vert\Omega\rangle=0##?!

Similarly, for the second assumption, if ##\langle\mathbf{k}\vert\phi(x)\vert\Omega\rangle \neq e^{ik\cdot x}##, then we can use that ##P^{\mu}\lvert\mathbf{k}\rangle =k^{\mu}\lvert\mathbf{k}\rangle## such that ##e^{ik\cdot x}\lvert\mathbf{k}\rangle##, and $$\langle\mathbf{k}\vert\phi(x)\vert\Omega\rangle=\langle\mathbf{k}\vert e^{iP\cdot x}\phi(0)e^{-iP\cdot x}\vert\Omega\rangle=e^{-ik\cdot x}\langle\mathbf{k}\vert \phi(0)\vert\Omega\rangle$$ In requiring that ##\langle\mathbf{k}\vert \phi(0)\vert\Omega\rangle =1##, we must (in general) rescale ##\phi(0)##, such that ##\phi(0)\rightarrow\sqrt{Z_{\phi}}\phi(0)##.

Again, is the reason why we relate ##\langle\mathbf{k}\vert\phi(x)\vert\Omega\rangle## to ##\langle\mathbf{k}\vert\phi(0)\vert\Omega\rangle## so that we can "induce" the ##e^{-ik\cdot x}## and then impose a condtion on the Lorentz scalar ##\langle\mathbf{k}\vert \phi(0)\vert\Omega\rangle##?!

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