Month: September 2021

(26.1) Consider such that $$[\hat Q_N,\hat \phi_A^\dagger]=\hat \phi_B^\dagger$$ for some generator of a symmetry group such that . Show that .

$$\hat H e^{i\alpha\hat Q_N}|0\rangle=e^{i\alpha\hat Q_N}\hat H|0\rangle=0$$

Thus it must be that $$e^{i\alpha\hat Q_N}|0\rangle=|0\rangle$$ as we assume a complete basis of eigenstates in the Fock space.

Show also that , where and .

$$E_B\hat \phi_B^\dagger|0\rangle=\hat H\hat \phi_B^\dagger|0\rangle\\=\hat H\left[\hat Q_N,\hat\phi_A^\dagger\right]|0\rangle\\=\hat H\hat Q_N\hat \phi_A^\dagger|0\rangle-\hat H\hat\phi_A^\dagger\hat Q_N|0\rangle\\=\hat Q_N\hat H\hat \phi_A^\dagger|0\rangle\\=\hat Q_NE_A\phi_A^\dagger|0\rangle\\=E_A(\hat\phi_B^\dagger+\hat\phi_A^\dagger \hat Q)|0\rangle=E_A\hat\phi^\dagger_B|0\rangle$$

Thus .

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(26.2) Prove the Fabri-Picasso theorem.

$$\langle 0|\hat J(x)\hat Q|0\rangle=\langle 0|e^{i\hat p\cdot a}\hat J(0)e^{-i\hat p\cdot a}\hat Q|0\rangle$$

We know that , because is an internal symmetry. The vacuum has momentum , so

$$\langle 0|\hat J(x)\hat Q|0\rangle=\langle 0|\hat J(0)\hat Q|0\rangle$$

Now considering

$$\langle 0|\hat Q\hat Q|0\rangle=\int d^3x\langle 0|J(x)\hat Q|0\rangle=\int d^3x \langle 0|J(0)\hat Q|0\rangle=\langle 0|J(0)\hat Q|0\rangle\int d^3x$$

This quantity diverges unless , so either has 0 or infinite norm.

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(26.3) Prove Goldstone’s theorem.

Assume there is a continuous symmetry with charge where . Consider a field such that

$$[\hat Q, \hat \phi(y)]=\hat\psi(y)$$

and due to spontaneously broken symmetry.

Show that 

$$\frac{\partial}{\partial x^0} \langle 0|\psi(0)|0\rangle=-\int d\mathbf S\cdot \langle 0|[\hat {\mathbf J}(x),\hat\phi(0)]|0\rangle$$

This is obvious from the continuity equation, since

$$\frac{\partial}{\partial x^0}\hat J^0=-\nabla\cdot \hat J^i$$

The divergence theorem gives the relevant quantity surface integral.