4
$\begingroup$

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)

$\endgroup$
8
  • $\begingroup$ 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

2
$\begingroup$

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

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

$\endgroup$
1
  • $\begingroup$ seems very good to me $\endgroup$
    – Nikos M.
    Commented Oct 22, 2014 at 16:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.