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I am reading P&S, particularly Chapter 5.5. The authors are trying to derive an expression for the Ward identity (not formally, but still). They claim that the amplitude describing a photon emission (stripped off from the polarization vector of the respective photon) in any QED process can be written as $$\mathcal{M}^{\mu}(k)=\int d^4x e^{ik\cdot x}\langle f|j^{\mu}(x)|i\rangle$$ and then they dot product $\mathcal{M}^{\mu}(k)$ with the photon momentum $k_{\mu}$ to demonstrate that the Ward identity holds, $$k_{\mu}\mathcal{M}^{\mu}(k)=0$$ exploitting the fact that the current $j^{\mu}(x)$ is conserved (provided that classical equations of motion hold at the quantum level). My questions are the following:

  1. Is $\mathcal{M}^{\mu}(k)$ a Feynman amplitude (i.e. the scattering amplitude with the momentum conserving $\delta$ function being stripped off) or a scattering amplitude? (My guess is that $\mathcal{M}^{\mu}(k)$ is a scattering amplitude! )

  2. In deriving the Ward identity, we basically had to substitute $k$ with $-i\partial$ and then to exploit the fact that the four-divergence of the expression $e^{ik\cdot x} \langle f|j^{\mu}(x)|i\rangle$ vanishes inside the spacetime integral (otherwise we can not reduce $k_{\mu}\mathcal{M}^{\mu}(k)$ to a single integral $\int d^4x e^{ik\cdot x}\langle f|\partial_{\mu}j^{\mu}(x)|i\rangle$, that vanishes due to the divergence of the current being zero. Does this imply that the spinor fields vanish at the asymptotic regions in space? Or does the vanishing of the four-divergence occur because of something entirely different?

Any help will be appreciated.

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    $\begingroup$ What is a "Feynman amplitude"? And how do you think it differs from a scattering amplitude? $\endgroup$
    – hft
    Jun 29 at 20:26
  • $\begingroup$ The scattering amplitude $S_{\beta\alpha}$ is related to the Feynman amplitude according to the relation $S_{\beta\alpha}=(2\pi)^4\delta^4(P_{\beta}-P_{\alpha})\ i\mathcal{M}_{\beta\alpha}$, where $|\beta\rangle$ and $|\alpha\rangle$ represent the final and initial state of the scattering, and $P_{\beta}$, $P_{\alpha}$ the final and initial total four-momentum. $\mathcal{M}_{\beta\alpha} is the corresponding Feynman amplitude $\endgroup$
    – schris38
    Jun 30 at 6:40

1 Answer 1

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  1. Using translation operators the position space amplitude can be written $$\mathcal{M}^\mu(x)\equiv \langle f|j^{\mu}(x)|i\rangle=\langle f|j^{\mu}(0)|i\rangle e^{i(\sum p_f-p_i)x}$$ After Fourier transforming the exponential you indeed get a delta function ensuring $k$ is compatible with momentum conservation. This is not a problem because the delta function is not setting $k=0$, it is setting $k$ equal to the difference between the initial and final on-shell momentum.

  2. The vanishing of the four-divergence is a manifestation of charge conservation. Classically $\partial_\mu j^\mu = 0$ too. This is the whole point of the Ward identity, it is the analogue of classical conservation equations but for quantum mechanical correlation functions.


Here is the Fourier transform in (2) in more detail

$$G(k) \equiv k_\mu \langle f|j^{\mu}(k)|i\rangle$$ Fourier transform $G(k)$, $$G(x)= \int \frac{d^dk}{(2\pi)^d} e^{-ikx} k_\mu \langle f|j^{\mu}(k)|i\rangle$$ $$=i\partial_\mu\int \frac{d^dk}{(2\pi)^d} e^{-ikx} \langle f|j^{\mu}(k)|i\rangle=i\partial_\mu \langle f|j^{\mu}(x)|i\rangle$$ Now we say that the final equality vanishes since $\partial_\mu j^\mu=0$ as a classical conservation law (this is the non-trivial part). So $G(x)=0$ and thus its Fourier transform $G(k)=0$.

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    $\begingroup$ Hi @octonion and thanks for the reply. I will have to think about (1), as I can not yet see how translational invariance helps me deduce whether or not $\mathcal{M}^{\mu}(k)$ is a Feynman amplitude or a scattering ($S-$matrix) amplitude. For (1), are you suggesting that the Feynman amplitude is the Fourier transform of the scattering amplitude? If so, I haven't thought about it like that so far, but most probably everything you say on (1) makes sense in that way. $\endgroup$
    – schris38
    Jun 30 at 6:48
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    $\begingroup$ As far as (2) is concerned, I was asking about the boundary behavior of spinor fields actually. Because in identifying $k$ as $-i\partial$ and to proceed to proving that $k_{\mu}\mathcal{M}^{\mu}$, we have to assume that the spinor fields vanish at the asymptotic regions of space, according to my understanding. Is that true? This is what I have been meaning to ask (sorry if I didn't convey it very well) $\endgroup$
    – schris38
    Jun 30 at 6:51
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    $\begingroup$ For (1), consider the unitary operators $e^{i\hat{p}}$ which translate position of any field $e^{i\hat{p}y}\phi(x)e^{-i\hat{p}y}=\phi(x+y)$. You can use this to shift the argument of $j$ to zero, but the in and out states are momentum eigenstates so the operators just become an ordinary exponential and after Fourier transforming an exponential it becomes a delta function. $\endgroup$
    – octonion
    Jun 30 at 15:47
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    $\begingroup$ For (2), $\partial_\mu\langle f|j^\mu(x)|i\rangle$ vanishes on its own, not just within an integral. You don't need to use an argument like the divergence theorem which relies on the asymptotic behavior of $\langle f|j^\mu(x)|i\rangle$. We know the exact x-dependence of this correlation function, by the way, it is the complex exponential we derived in part (1). So no it doesn't asymptotically vanish. $\endgroup$
    – octonion
    Jun 30 at 15:54
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    $\begingroup$ Hi. Thanks again for answering. I am okay with (1), I understand. But for (2), when we dot product the momentum vector on the F.T. of the amplitude, i.e. $k_{\mu}\mathcal{M}^{\mu}(k)$ and then substitute the latter with $\int d^4x e^{ik\cdot x} \mathcal{M}^{mu}(x)$, then don't we need to use the fact that the surface terms vanish asymptotically? What I get is this: $\int d^4x k_{\mu}e^{ik\cdot x}\mathcal{M}^{\mu}(x)=\int d^4x (-i\partial_{\mu})e^{ik\cdot x}\mathcal{M}^{\mu}(x)$ and then use the product rule, arriving thus in a total derivative and the term $\sim \partial_{\mu}j^{\mu}(x)=0$ $\endgroup$
    – schris38
    Jun 30 at 16:02

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