I found a nice answer to a problem that was bothering me for quite a while in a lecture script (unfortunately in german). The first step of the answer, is what remains unclear to me. The script states that the transformation of a quantum field $\phi$ can be written in two different ways:

$$ 1) \qquad \phi \rightarrow e^{i\alpha_a T_a} \phi $$

with the generators of a SU(N) symmetry group $T_a$ and

$$ 2) \qquad \phi \rightarrow e^{i \alpha_a Q_a} \phi e^{-i \alpha_a Q_a}, $$

where $Q$ denotes the Noether charge. Considering infinitesimal transformations we have

$$ \alpha_a T_a \Phi = [\alpha_a Q_a,\Phi] $$

This can be used to show, that the two criteria for spontaneous symmetry breaking:

I $Q_a |0> \neq 0$

II $<0|\phi |0>\neq 0$

follow from each other.

I means the vacuum is not invariant under this symmetry, because $Q |0> \neq 0 \rightarrow e^{i Q} |0> \neq |0> $. And II means a scalar field, with a non-vanishing Vacuum-Expecation Value exists.

My problem is understanding, why there a two different transformation laws for $\phi$, one using the Noether charge, and one using the generators. I always thought that in QFT we identify those two with each other: $Q \leftrightarrow T$ (For a proof see for example this question: Connection between conserved charge and the generator of a symmetry)

I found a nice answer to a problem that was bothering me for quite a while in a lecture script (unfortunately in german). The first step of the answer, is what remains unclear to me. The script states that the transformation of a quantum field $\phi$ can be written in two different ways:

$$ 1) \qquad \phi \rightarrow e^{i\alpha_a T_a} \phi $$

with the generators of a SU(N) symmetry group $T_a$ and

$$ 2) \qquad \phi \rightarrow e^{i \alpha_a Q_a} \phi e^{-i \alpha_a Q_a}, $$

where $Q$ denotes the Noether charge. Considering infinitesimal transformations we have

$$ \alpha_a T_a \Phi = [\alpha_a Q_a,\Phi] $$

This can be used to show, that the two criteria for spontaneous symmetry breaking:

I $Q_a |0> \neq 0$

II $<0|\phi |0>\neq 0$

follow from each other.

I means the vacuum is not invariant under this symmetry, because $Q |0> \neq 0 \rightarrow e^{i Q} |0> \neq |0> $. And II means a scalar field, with a non-vanishing Vacuum-Expecation Value exists.

My problem is understanding, why there a two different transformation laws for $\phi$, one using the Noether charge, and one using the generators. I always thought that in QFT we identify those two with each other: $Q \leftrightarrow T$ (For a proof see for example this question: Connection between conserved charge and the generator of a symmetry)

I found a nice answer to a problem that was bothering me for quite a while in a lecture script (unfortunately in german). The first step of the answer, is what remains unclear to me. The script states that the transformation of a quantum field $\phi$ can be written in two different ways:

$$ 1) \qquad \phi \rightarrow e^{i\alpha_a(x) T_a} \phi $$

with the generators of a SU(N) symmetry group $T_a$ and

$$ 2) \qquad \phi \rightarrow e^{i Q} \phi e^{-i Q}, $$

where $Q$ denotes the Noether charge. Considering infinitesimal transformations we have

$$ \alpha_a T_a \Phi = [Q,\Phi] $$

This can be used to show, that the two criteria for spontaneous symmetry breaking:

I $Q |0> \neq 0$

II $<0|\phi |0>\neq 0$

follow from each other.

I means the vacuum is not invariant under this symmetry, because $Q |0> \neq 0 \rightarrow e^{i Q} |0> \neq |0> $. And II means a scalar field, with a non-vanishing Vacuum-Expecation Value exists.

My problem is understanding, why there a two different transformation laws for $\phi$, one using the Noether charge, and one using the generators. I always thought that in QFT we identify those two with each other: $Q \leftrightarrow T$ (For a proof see for example this question: Connection between conserved charge and the generator of a symmetry)

I found a nice answer to a problem that was bothering me for quite a while in a lecture script (unfortunately in german). The first step of the answer, is what remains unclear to me. The script states that the transformation of a quantum field $\phi$ can be written in two different ways:

$$ 1) \qquad \phi \rightarrow e^{i\alpha_a T_a} \phi $$

with the generators of a SU(N) symmetry group $T_a$ and

$$ 2) \qquad \phi \rightarrow e^{i \alpha_a Q_a} \phi e^{-i \alpha_a Q_a}, $$

where $Q$ denotes the Noether charge. Considering infinitesimal transformations we have

$$ \alpha_a T_a \Phi = [\alpha_a Q_a,\Phi] $$

This can be used to show, that the two criteria for spontaneous symmetry breaking:

I $Q_a |0> \neq 0$

II $<0|\phi |0>\neq 0$

follow from each other.

I means the vacuum is not invariant under this symmetry, because $Q |0> \neq 0 \rightarrow e^{i Q} |0> \neq |0> $. And II means a scalar field, with a non-vanishing Vacuum-Expecation Value exists.

My problem is understanding, why there a two different transformation laws for $\phi$, one using the Noether charge, and one using the generators. I always thought that in QFT we identify those two with each other: $Q \leftrightarrow T$ (For a proof see for example this question: Connection between conserved charge and the generator of a symmetry)