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I A question about assumptions made in derivation of LSZ formula

  1. Feb 11, 2017 #1
    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##?!
     
    Last edited: Feb 11, 2017
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  3. Feb 12, 2017 #2

    A. Neumaier

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    The first assumption simply amounts to the assumption that the vacuum expectation value has been moved to zero. This must be assumed, as it is not automatic.

    For the second assumption, note that ##\langle k|\phi(0)|\Omega\rangle## is a priori not (as you claim) a Lorentz scalar!

    In both cases it is far more natural form a formal point of view to assume the two covariant statements rather than to derive them from two special noncovariant cases (which must be assumed instead) and an additional argument.
     
  4. Feb 12, 2017 #3
    Why though do they relate ##\phi(x)## to ##\phi(0)## (via the 4-momentum operator, ##\phi(x)=e^{iP\cdot x}\phi(0)e^{-iP\cdot x}##) to end up with $$\langle\Omega\vert\phi(x)\vert\Omega\rangle=\langle\Omega\vert\phi(0)\vert\Omega\rangle$$ I get that ##\langle\Omega\vert\phi(x)\vert\Omega\rangle\neq 0## in general. Is the point of relating the expectation value at two spacetime points so that we can rewrite the above as $$\langle\Omega\vert\phi(x)\vert\Omega\rangle-\langle\Omega\vert\phi(0)\vert\Omega\rangle=0$$ such that we can shift ##\phi(x)## to ensure that its vacuum expectation value vanishes?!
     
  5. Feb 12, 2017 #4

    A. Neumaier

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    What they did is still Lorentz invariant,. They could have as well related it to ##\phi(y)## at any other ##y## by a similar formula and then be fully Poincare invariant. Since I haven't seen the text itself that you refer to it is hard to say why they did what they did. There is generally a lot of freedom in making a sound exposition, and not every detail can be justified, except by a sense of taste that differs among people.
     
  6. Feb 12, 2017 #5
    Here are the notes I've been looking at: http://www2.ph.ed.ac.uk/~egardi/MQFT_2013/MQFT_2013_lecture_2.pdf (pages 2-3). What I'm unsure about is why do this in the first place? Is it just to elucidate how one can shift the field ##\phi## to ensure that its vev vanishes?
     
  7. Feb 12, 2017 #6

    A. Neumaier

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    It is not assumptions. Instead, conclusions are drawn from the setting in Section 2.1, and the goal of the discussion is to establish the equivalence of the approach of Section 2.1 with the earlier one in the part you didn't link to.
     
  8. Feb 12, 2017 #7
    Apologies, I didn't have a link to that part of the notes, but I've since found a link: http://www2.ph.ed.ac.uk/~egardi/MQFT_2013/MQFT_2013_lecture_1.pdf
    What I don't understand is, why ##\phi(0)##? ##\phi(x)## also annihilates the vacuum in the free theory. When the author refers to the previous notes (in particular eq. (5)) they simply point the reader to the Fourier decompostion of the free scalar field and the fact that ##\langle 0\vert a^{\dagger}_{\mathbf{k}}=0## and ##a_{\mathbf{k}}\vert 0\rangle=0##. Surely, it is automatically true that ##\langle 0\vert\phi(x)\vert 0\rangle=0##, why do we even need to consider ##\langle 0\vert\phi(0)\vert 0\rangle## and then translate it to ##\langle 0\vert\phi(x)\vert 0\rangle##?! (Sorry if I'm being stupid here, it just doesn't seem obvious to me, but maybe I'm missing something?!)
     
  9. Feb 12, 2017 #8

    A. Neumaier

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    ##\phi(x)## does not annihilates the vacuum, not even for ##x=0##, in neither version of the theory. Only the expectation vanishes.

    The lecture notes are obviously sloppy; they also state wrongly that the matrix element with ##k### is Lorentz invariant. Try to understand their results, and not their details!
     
  10. Feb 12, 2017 #9
    Sorry, I didn't think about that point. What I meant was that ##\langle 0\vert\phi(x)\vert 0\rangle=0##, so why consider ##\langle 0\vert\phi(0)\vert 0\rangle=0## and then evolve it to ##\langle 0\vert\phi(x)\vert 0\rangle=0##?! This seems to be an important point in how one can satisfy the required condition ##\langle \Omega\vert\phi(x)\vert\Omega\rangle=0##, it's also mentioned in Peskin & Schroeder and Srednicki.

    I think I understand the results of the LSZ reduction formula, but I also want to understand the assumptions made and how one can ensure that they are satisfied.
     
    Last edited: Feb 12, 2017
  11. Feb 12, 2017 #10
    Is the point that ##\langle\Omega\vert\phi(0)\vert\Omega\rangle## is a Lorentz invariant number whereas ##\langle\Omega\vert\phi(x)\vert\Omega\rangle## is a (Lorentz invariant) function (at least, in principle it is a function, however, it is shown that it is trivially related to ##\langle\Omega\vert\phi(0)\vert\Omega\rangle##). As such, in general, ##\langle\Omega\vert\phi(0)\vert\Omega\rangle=\text{const.}\equiv v\neq 0##, and we can use this fact to redefine the field ##\phi(x)\rightarrow\phi(x)-v## such that ##\langle\Omega\vert\phi(x)\vert\Omega\rangle =0## ##\forall\;x^{\mu}##?!
     
  12. Feb 13, 2017 #11

    A. Neumaier

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    There is no point in the details, only in the results.
     
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