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I am going through Weinberg's second proof of Goldstone's theorem. I am able to follow it up to Eqn. (19.2.34), where he asserts that the following relation holds by Lorentz invariance

$$\langle \Omega| J^{\lambda}(x)|B\rangle = i \frac{F p_{B}^{\lambda} e^{ip_{B}\cdot x}}{(2\pi)^{3/2}\sqrt{2p_{B}^{0}}},$$

such that $|B\rangle$ is a single-particle spin-zero state of momentum $p_{B}$, $|\Omega\rangle$ is a vacuum state, $J^{\lambda}(x)$ is the Noether current of some global symmetry that has been spontaneously broken.

Via explicit evaluation, we have through Lorentz invariance

$$\langle \Omega| J^{\lambda}(x)|B\rangle=\langle \Omega| J^{\lambda}(x) \ a_{B}(p_{B})^\dagger|\Omega\rangle=\langle \Omega|\Lambda^{\lambda}{}_{\lambda'} J^{\lambda'} \big(\Lambda^{-1}x\big) \ U(\Lambda)^{-1}a_{B}\big(\Lambda^{-1} p_{B}\big)^\dagger U(\Lambda)|\Omega\rangle$$ $$=\sqrt{\frac{(\Lambda p)^{0}}{p^{0}}}\Lambda^{\lambda}{}_{\lambda'}\langle \Omega| J^{\lambda'}\big(\Lambda^{-1}x\big) \ a_{B}\big(\Lambda^{-1} p_{B}\big)^\dagger |\Omega\rangle.$$

From here, I can imagine tucking away some factors into $F$ and arguing that $p^{\lambda}$ is really the only sensible quantity that makes the vacuum expectation value in the last equality covariant. But I'm not sure how one obtains the complex exponential from the vacuum expectation value. If we were instead doing this for the free scalar field (as opposed to $J^{\lambda}(x)$), then one would immediately obtain the complex exponential by expanding the field operator in Fourier space, to which the complex exponential would fall out immediately up to a factor.

There must be a better way to obtain this expression. Perhaps there is something in earlier chapters that I did not appropriately grasp.

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When you see a plane wave like that you should immediately think about translations. The basic point here is to first and foremost observe that we can obtain $J^\lambda(x)$ by translating $J^\lambda(0)$. Then also using that the vacuum is translation invariant because it is an eigenstate of the four-momenta $P^\mu$ with eigenvalue $0$ and that $B$ is an eigenstate of the four-momenta with eigenvalue $p_B^\mu$ we get $$\langle\Omega|J^\lambda(x)|B\rangle=\langle \Omega|e^{-ix\cdot P}J^\lambda(0)e^{ix\cdot P}|B\rangle=\langle \Omega|J^\lambda(0)e^{ix\cdot p_B}|B\rangle=e^{ix\cdot p_B}\langle \Omega|J^\lambda(0)|B\rangle\tag{1}$$

As for why the dependence in $p_B^\lambda$, the argument you give is quite standard. The state $|B\rangle$ carries as its only label the momentum $p_B$. Therefore this amplitude is, in fact, a function of $p_B$. Moreover, since $J^\lambda(0)$, as an operator, transforms as a four vector, this amplitude also transforms as a four-vector. So it is a function only of $p_B$ and transforms as a four-vector. The only way to satisfy these two constraints is to have something proportional to $p_B^\lambda$. Therefore you should be able to write $$\langle \Omega|J^\lambda(0)|B\rangle = Cp_B^\lambda\tag{2}.$$

Now take $\langle \Omega|J^\lambda(0)|B\rangle$ and insert $U(\Lambda)$. We have $$\langle \Omega|J^\lambda(0)|B\rangle=\langle \Omega|U(\lambda)^\dagger U(\Lambda)J^\lambda(0)U(\Lambda)^\dagger U(\Lambda)|B\rangle=\sqrt{\dfrac{(\Lambda p_B)^0}{p_B^0}}(\Lambda^{-1})^\lambda_{\phantom\lambda\sigma}\langle \Omega|J^\sigma(0)|\Lambda B\rangle\tag{3}$$

Using (2) twice, once in the LHS with $p_B$ itself and once in the RHS with $\Lambda p_B$ we find

$$C p_B^\lambda = \sqrt{\dfrac{(\Lambda p_B)^0}{p_B^0}}(\Lambda^{-1})^\lambda_{\phantom\lambda\sigma}C'\Lambda^\sigma_{\phantom\sigma\kappa} p_B^\kappa=\sqrt{\dfrac{(\Lambda p_B)^0}{p_B^0}}C' p_B^\lambda\tag{4}$$

Finally this tells us that

$$C=\sqrt{\dfrac{(\Lambda p_B)^0}{p_B^0}}C'\tag{5}.$$

So Lorentz invariance only fixes how $C$ changes between two frames. In particular $C = \alpha\frac{1}{\sqrt{2p_B^0}}$ is compatible with (5). My opinion is that you can also reach this conclusion by going to a frame in which $p_B$ assumes the standard massless momentum form $(1,0,0,1)$, setting $C$ there and then using (5) to define $C$ in all the other frames.

The way Weinberg fixes $\alpha$ is then probably for later convenience. Combining it all in (1) you get the form Weinberg proposes, out of Poincare symmetry only.

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  • $\begingroup$ This is perfect - this clears everything up. I very much appreciate it! $\endgroup$
    – user107053
    Feb 17, 2021 at 20:59

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