I Understanding the proof of Goldstone theorem

[pines-demon]

I was reviewing the proof of Goldstone theorem in Aitchinson-Hay book, vol 2, and I do not understand it. It starts by saying that there is some continues symmetry
$$Q=\int \mathrm d^3 x\, J_0 (x)$$
where $J_\mu$ is the Noether current. Then one considers the expectation value $\langle 0|[Q,\phi(0)]|0\rangle\neq 0$. Using the conservation of the current $\partial^\mu J_\mu =0$ and integrating one can show that

$$
\sum_n (2\pi)^3 \delta(\mathbf p_n)\left(\langle 0 |J_0(0)|n\rangle \langle n|\phi(0)|0\rangle e^{-\mathrm i p_{n0}x_0} -\langle 0 |\phi(0)|n\rangle \langle n|J_0(\mathbf x)|0\rangle e^{\mathrm i p_{n0} x_0}\right)\neq0\tag{17.67}
$$
where the sum is over eigenstates $|n\rangle$ and this calculation is shown to be also independent of $x_0$. Then the book says:

But this expression is independent of $x_0$. Massive states $|n\rangle$ will produce explicit $x_0$-dependent factors $e^{i M_n x_0}$ ($p_{n0}\to M_n$, as the $\delta$-function constrains $\mathbf p_n=0$); hence, the matrix elements of $J_0$ between $|0\rangle$ and such a massive state must vanish, and such states contribute zero to (17.67). Equally, if we take $|n\rangle=|0\rangle$, (17.67) vanishes identically.

I do not get the $|n=0\rangle$ case, if it is massive we have the same issue as with the massive $|n\rangle$ , so it has to vanish. But the expression only vanishes identically if it is not massive. I don't get it.

But it has been assumed to be not zero. Hence, some state or states must exist among $|n\rangle$ such that $\langle 0|J_0|n\rangle\neq 0$ and yet (17.67) is independent of $x_0$. The only possibility is states whose energy $p_{n0}$ goes to zero as their 3-momentum does (from $\delta(\mathbf p_{n})$). Such states are, of course, massless: they are called generically Goldstone modes. Thus, the existence of a non-vanishing vacuum expectation value for a field, in a theory with a continuous symmetry, appears to lead inevitably to the necessity of having a massless particle, or particles, in the theory.

Can somebody help me understand what is going on? What is zero and what is not? What is so special of the $|0\rangle$ case?

 

[Demystifier]

The "mass" is a property of an excitation (particle or quasiparticle) around the vacuum. The vacuum $|n=0\rangle$ itself is neither massive nor massless, that's what makes $n=0$ special. In fact, the energy of the vacuum is defined as zero, $p_{00}=0$. (If it was not defined so, then the vacuum state would be time dependent in the Schrodinger picture, so the vacuum would not be time-translation invariant.) When we talk of a massive or massless state, we always assume a state with $n\neq 0$.

Since $p_{00}=0$, the $n=0$ does not contribute to the sum, so we can take the sum to be taken over states with $n\neq 0$. How can this sum be time-independent and yet non-zero? It's only possible if $p_{n0}=0$ for some $n\neq 0$. Since ${\bf p}_n=0$ (due to the $\delta$-function) we have $p_{n0}=M_n$, so there must be some excited states $n\neq 0$ with $M_n=0$. That's the Goldstone theorem.

 

[pines-demon]

The "mass" is a property of an excitation (particle or quasiparticle) around the vacuum. The vacuum $|n=0\rangle$ itself is neither massive nor massless, that's what makes $n=0$ special. In fact, the energy of the vacuum is defined as zero, $p_{00}=0$.

Thank you I was missing that.

(If it was not defined so, then the vacuum state would be time dependent in the Schrodinger picture, so the vacuum would not be time-translation invariant.) When we talk of a massive or massless state, we always assume a state with $n\neq 0$.

Since $p_{00}=0$, the $n=0$ does not contribute to the sum, so we can take the sum to be taken over states with $n\neq 0$. How can this sum be time-independent and yet non-zero? It's only possible if $p_{n0}=0$ for some $n\neq 0$. Since ${\bf p}_n=0$ (due to the $\delta$-function) we have $p_{n0}=M_n$, so there must be some excited states $n\neq 0$ with $M_n=0$. That's the Goldstone theorem.

But what is this condition of that $$\langle n | J |0\rangle\neq 0$$ why is it important in this derivation?

 

[Demystifier]

But what is this condition of that $$\langle n | J |0\rangle\neq 0$$ why is it important in this derivation?

If it was zero for all $n$, then the entire sum would be zero, so the symmetry would not be (spontaneously) broken. The Goldstone theorem says that massless excitations appear when the symmetry is broken.

 

[pines-demon]

If it was zero for all $n$, then the entire sum would be zero, so the symmetry would not be (spontaneously) broken. The Goldstone theorem says that massless excitations appear when the symmetry is broken.

Gotcha, I think I was looking at it as if it was more mysterious than what it is. Thanks