# Algebra, commutators and test functions

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I am trying to make sense out of the algebra of the generators of the conformal group and I am running into some issues regarding how to calculate commutators.

For instance, for translations of a "test operator" $\phi(x)$, we know that $\phi(x+a) = \phi(x) + a^\mu \partial_\mu \phi(x) + \cdots$ if we simply Taylor expand. Moreover, in the Heisenberg picture, we can also write

$$\exp(i a^\mu P_\mu) \; \phi(x) \; \exp(-i a^\mu P_\mu) = \phi(x) + i a^\mu [ P_\mu, \phi(x)] + \cdots$$

from which we conclude that $[P_\mu, \phi(x)] = -i \partial_\mu \phi(x)$. Why then do we say that $P_\mu = -i \partial_\mu$ is the generator of translation? From the above definition, the equality does not hold directly since we used a commutator to define $P_\mu$'s action on $\phi(x)$, but introductory quantum mechanics teaches us that $P_\mu = -i \partial_\mu$. My problem seems to be that I can't reconcile this with rigorous group theoretical concepts.

What made this confusion arise was the calculation of commutators of algebra generators for the conformal group. For example, taking $D$ to be the generator of dilatations, to prove that $[D, P_\mu] = i P_\mu$, one needs to apply the commutator $[D, P_\mu]$ onto a test function $\phi(x)$ outside of the commutator after replacing the $D$ and $P_\mu$ by their differential definition. So how do we go from $[P_\mu, \phi(x)] = -i \partial_\mu \phi(x)$ to $P_\mu = - i\partial_\mu$ where the equality makes sense?

This post imported from StackExchange Physics at 2014-06-21 21:34 (UCT), posted by SE-user physguy
Maybe you are looking for this? en.wikipedia.org/wiki/Stone%E2%80%93von_Neumann_theorem

This post imported from StackExchange Physics at 2014-06-21 21:34 (UCT), posted by SE-user Robin Ekman

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The issue here is that there two distinct actions of the translation group on fields present in your computations.

Definitions of the group actions.

The two group actions to which I'm referring are as follows. For conceptual simplicity, let $\phi$ denote an operator-valued field defined on $\mathbb R$; generalization to higher dimensions is straightforward. Let $\mathcal H$ denote the Hilbert space of the theory, then we assume (at least in theories for which we want to talk about translation-invariance) that there is a unitary representation $\hat U$ of the translation group of $\mathbb R$ acting on the Hilbert space.

This unitary representation then induces an action $\rho_1$ of the translation group acting on fields as follows: \begin{align} (\rho_1(a)\hat \phi)(x) = \hat U(a)\hat\phi(x)\hat U(a)^{-1}. \tag{1} \end{align} On the other hand, we can define a second action of the translation group acting on fields as follows: \begin{align} (\rho_2(a)\hat \phi)(x) = \hat \phi(x-a) \tag{2} \end{align}

Infinitesimal generators.

Each of the group actions above possesses an infinitesimal generator.

To determine what that is for $\rho_1$, we write $\hat U(a) = e^{-ia\hat P}$, so that $\hat P$ is the infinitesimal generator of $\hat U$, and we notice that if we expand the right hand side of $(1)$ in $a$ we have \begin{align} \hat U(a)\hat\phi(x)\hat U(a)^{-1} &= (\hat I - ia\hat P)\hat \phi(x) (\hat I + ia\hat P) + O(a^2) \\ &= \hat \phi(x) + ia\hat\phi(x)\hat P - ia\hat\phi(x) \hat P + O(a^2) \\ &=\hat \phi(x) -ia\Big(\hat P\hat \phi(x)-\hat \phi(x) \hat P\Big) + O(a^2) \\ &= \hat\phi(x) -ia[\hat P,\hat \phi(x)] + O(a^2) \end{align} inspecting the term that is first order in $a$, we see immediately that the operator \begin{align} \hat \phi(x) \mapsto [\hat P,\hat \phi(x)] \end{align} is the infinitesimal generator of the first group action $\rho_1$. It turns out by the way that this operator has a special name: the adjoint operator, and it is often denoted $\mathrm{ad}_{\hat P}$. So all in all, we see that $\mathrm{ad}_{\hat P}$ is the infinitesimal generator of $\rho_1$ since we have shown that \begin{align} (\rho_1(a)\hat \phi)(x) = \big(\hat I -ia \,\mathrm{ad}_{\hat P} + O(a^2)\big)\hat\phi(x) \end{align} As an aside, this is all intimiately related to the so-called Hadamard Lemma for the Baker-Campbell-Hausdorff formula.

To determine the infinitesimal generator for $\rho_2$, we expand the right hand side of $(2)$ in $a$ using Taylor's formula to obtain \begin{align} \hat\phi(x+a) &= \hat \phi(x) - a \partial\phi(x) + O(a^2) \\ &= \left(1-ia(-i\partial) + O(a^2)\right)\hat\phi(x) \\ \end{align} so that the infinitesimal generator of the group action $\rho_2$ is $-i\partial$. We can summarize these results as follows. Let's call the infinitesimal generator of $\rho_1$ $P_1$ and the infinitesimal generator of $\rho_2$ $P_2$, then we have shown that \begin{align} P_1 = \mathrm{ad}_{\hat P}, \qquad P_2 = -i\partial \end{align} Notice, in particular, that these are not the same mathematical objects.

Fields that transform in special ways under translations.

Although the group actions $\rho_1$ and $\rho_2$ are distinct and have distinct generators, it is sometimes the case in field theory that one considers fields $\hat \phi$ that transform as follows: \begin{align} \hat U(a)\hat\phi(x) \hat U(a)^{-1} = \hat\phi(x-a). \end{align} Notice that the left hand side of this is just the action of $\rho_1$ on $\hat\phi$, and the right hand side is the action of $\rho_2$ on $\hat\phi$, so for this special class of fields, the two group actions agree! In this special case, it also follows that the infinitesimal generators of $\rho_1$ and $\rho_2$ agree on these special fields; \begin{align} P_1 \hat\phi(x) = P_2\hat\phi(x), \end{align} or more explicitly \begin{align} \mathrm{ad}_{\hat P}\hat\phi(x) = -i\partial\hat\phi(x). \end{align}

This post imported from StackExchange Physics at 2014-06-21 21:34 (UCT), posted by SE-user joshphysics
answered Jun 21, 2014 by (835 points)
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I don't have time to give a complete answer, but the gist of it is, $P_\mu = -i \partial_\mu$ is not true for the quantum fields you are talking about. This paper might be of help: http://arxiv.org/pdf/hep-th/0206008.pdf

This post imported from StackExchange Physics at 2014-06-21 21:34 (UCT), posted by SE-user DrEntropy
answered Jun 20, 2014 by (0 points)

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