4

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)

Community
  • 1
jak
  • 9,937
  • 4
  • 35
  • 106
  • 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 – Nikos M. Oct 22 '14 at 14:12
  • Thanks for your comment. I changed it in the question to avoid confusion. In the notes they are considering a global transformation – jak Oct 22 '14 at 14:15
  • ok, still $T$ represents the infinitesimal Lie group generators,while $Q$ represents the overall total charge – Nikos M. Oct 22 '14 at 14:16
  • 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 – Nikos M. Oct 22 '14 at 14:20
  • 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...?! – jak Oct 22 '14 at 14:26
  • not necessarily, since the representations are representations on the same space. One can use different representations of the same space and all these are related through the Lie algebra. Sth like representing the same matrix on a coordinate transformation. – Nikos M. Oct 22 '14 at 14:34
  • I've never heard of that. If, for example, $\phi$ lives in the tangent space at the identity of the group (=the Lie algebra), the group acts on elements of this vectors space as $g \phi g^{-1}$ (an the generators as $[T_a,\Phi]$, which is called the adjoint representation and what corresponds to transformation $2)$ from above. How can the group act on this vector space in any other way? – jak Oct 22 '14 at 14:47
  • hmm, i'm not sure. both $T$ and $Q$ should be the same. Dont see a difference there. Plus there is no acting on the space in other way, acting is done througnthe Lie algebra, the problem is the representations of the operators in your question. – Nikos M. Oct 22 '14 at 15:15

1 Answers1

2

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

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

Qmechanic
  • 201,751