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I have a question specific to Kibble's proof of Goldstone's theorem, as found in: http://www.scholarpedia.org/article/Englert-Brout-Higgs-Guralnik-Hagen-Kibble_mechanism#Proof_of_the_theorem

In this proof, he is only considering a complex scalar field with global $U(1)$ symmetry.

I have trouble understanding the very last line of the proof: "If we insert a complete set of intermediate states into (16), we see that this implies that there must be states that couple to the vacuum through $\phi$ for which $k_0\rightarrow0$ as $\mathbf{k}\rightarrow\mathbf{0}$, i.e., massless particle states."

It would be great if someone could help me through the steps that he is outlining here. I've tried to insert intermediate states but I don't know how to get to the final conclusion.

Here's how far I got: $$f^{0}(k^0,\mathbf{k})=-i\int d^4xe^{ik\cdot x}\langle 0\rvert [j^{0}(x),\phi(0)]\lvert 0\rangle \\ = -i\int d^4xe^{ik\cdot x}\int\frac{d^3\mathbf{k'}}{(2\pi)^32 {k^{0}}'}(\langle0\rvert j^0(x)\lvert k'\rangle\langle k'\rvert\phi(0)\lvert 0\rangle-\langle0\rvert \phi(0)\lvert k'\rangle\langle k'\rvert j^0(x)\lvert 0\rangle)$$, where I have inserted intermediate states using relativistic normalisation. We know that $f^0(k^0,\mathbf{0})\propto\delta(k^0)$ so the contribution above centres on $k^0=0$ as $\mathbf{k}\rightarrow\mathbf{0}$. But I expect there to be a $\delta^{(3)}(\mathbf{k}-\mathbf{k'})$ of some sort to get rid of the integral over k primed so that we can come to some conclusion about k instead of k primed. Perhaps $j^0(x)$ needs to be expanded in terms of operators?

Thank you very much for your help!

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  • $\begingroup$ It's probably easier if you never write the 2nd line, but, instead, set k =0. The remaining exponential exp(itk<sub>0</sub>) will give you the $\delta(k^0)$ only if there is some invertible state with 0-energy dependence, that is that does not depend on t , so the rest of the integrand is a constant. Indeed, you are seeking the t-independent mode of $j^0$. $\endgroup$ – Cosmas Zachos Feb 20 '18 at 0:57
  • $\begingroup$ I'm sorry to have to make you connect the dots here, but how does the time independence of $\langle 0\rvert [j^0(x),\phi (0)]\lvert 0\rangle$ imply the existence of some massless state? $\endgroup$ – Ziruo Zhang Feb 20 '18 at 11:22
  • $\begingroup$ I meant "intermediate state" up there. You ignore the positive energy states and seek some zero energy state s.t. $\langle 0|j^0(0) \exp(-it k_0')|k'\rangle$ that is time independent, so $k_0'=0$. It won't cancel with its following c.c., and so will yield a constant, leading to the δ-fctn in $k^0$ sought in the integral. I don't want to fuss with the relativistic normalization, which is why I didn't venture into your second line. $\endgroup$ – Cosmas Zachos Feb 20 '18 at 14:26
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Following what Cosmas Zachos has set out in his comments,

$$f^0(k^0,\mathbf{0})=-i\int d^4xe^{ik^0t}\langle 0\rvert [j^0(x),\phi(0)]\lvert 0\rangle = g\delta(k^0),$$ for some non-zero constant g. For this to hold, we need $\langle 0\rvert [j^0(x),\phi(0)]\lvert 0\rangle$ to be time-independent. Inserting intermediate states, $$\langle 0\rvert [j^0(x),\phi(0)]\lvert 0\rangle = \sum_{k'}(\langle 0\rvert j^0(x)\lvert k'\rangle\langle k'\rvert\phi(0)\lvert 0\rangle - c.c.)$$ Now $$\langle 0\rvert j^0(x)\lvert k'\rangle=\langle 0\rvert e^{i\hat{P}\cdot x}j^0(0)e^{-i\hat{P}\cdot x}\lvert k'\rangle=\langle 0\rvert j^0(0)e^{-ik'\cdot x}\lvert k'\rangle,$$ where $\sum_{k'}$ is a shorthand for $\int\frac{d^3\mathbf{k'}}{(2\pi)^3 2k'^{0}}$ and $\hat{P}=(\hat{H},\hat{\mathbf{p}})$ ($\hat{H}$ is the Hamiltonian and $\hat{\mathbf{p}}$ is the total momentum operator). For this to be time-independent, the states contributing to the matrix element must have $k'^0=0$, i.e. they are massless. There must be such states since $g\neq 0$.

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