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)

  • $\begingroup$ if i am not mistaken, 1) is local transformation, while 2) is global (at least there is this difference in the 2 examples you give), also $T$ are the Lie infinitesinal group generators, while $Q$ is the overall charge $\endgroup$
    – Nikos M.
    Commented Oct 22, 2014 at 14:12
  • $\begingroup$ Thanks for your comment. I changed it in the question to avoid confusion. In the notes they are considering a global transformation $\endgroup$
    – jak
    Commented Oct 22, 2014 at 14:15
  • $\begingroup$ ok, still $T$ represents the infinitesimal Lie group generators,while $Q$ represents the overall total charge $\endgroup$
    – Nikos M.
    Commented Oct 22, 2014 at 14:16
  • $\begingroup$ after the last edit, i dont see a difference, both $T$ and $Q$ are just different representations of the same Lie algebra of group generators $\endgroup$
    – Nikos M.
    Commented Oct 22, 2014 at 14:20
  • $\begingroup$ It looks like this, but isn't a representation a map (homomorphism) to the space of linear operators over a vector space: $Lin(V)$. Therefore, once we specify the vector space $\phi$ lives in, we know which representation we must act on it. Two transformation laws, would mean $\phi$ lives in two vectors spaces at the same time...?! $\endgroup$
    – jak
    Commented Oct 22, 2014 at 14:26

1 Answer 1


I) It is difficult to comment without seeing the textbook, but one interpretation is that it is essentially just a matter of assigning appropriate representations as follows. Let $G$ be a Lie group with the corresponding Lie algebra $L$. Let $\exp: L\to G$ be the exponential map. Let $t_a\in L$ be a Lie algebra generator. Let $A$ be an algebra with a set $A^{\times}$ of invertible elements.

II) Let

$$r: G ~\to~ A^{\times}$$

be a Lie group homomorphism. The Lie group homomorphism induces a corresponding Lie algebra homomorphism

$$r:L~\to~ A,$$

which we also call $r$. Let

$$ Q_a~:=~r(t_a)~\in~ A.$$

III) Consider a Lie group/algebra representation

$$R:G~\to~ GL(A,\mathbb{C}), \qquad R:L~\to~ gl(A,\mathbb{C}),$$

defined as

$$R(g)\phi~:=~ r(g)\phi r(g)^{-1}, \quad g\in G, \quad \phi\in A,$$ $$R(t)\phi~:=~ [r(t),\phi], \quad t\in L, \quad \phi\in A,$$

respectively. Define

$$T_a ~:=~R(t_a)~=~[r(t_a), \cdot]~=~[Q_a, \cdot]~\in~gl(A,\mathbb{C}) .$$

  • $\begingroup$ seems very good to me $\endgroup$
    – Nikos M.
    Commented Oct 22, 2014 at 16:30

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