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I am trying to understand the topic of Induced representation of the euclidean Group E(2) in A. Zee's Group theory in a Nutshell in Chapter IV.i3.

The Lie algebra of E(2) has three elements $P_1, P_2, J$ with commutation relations

$[P_1, P_2] = 0$ and $[J, P_{1,2}] = \pm i P _{2,1}$

Zee constructs an infinite dimensional representation by identifying the maximally commuting subalgebra $P_1, P_2$ and labeling states by $|\vec{p}\rangle = |p, \phi>$, where $P_{1,2}|\vec{p}\rangle = \vec{p}_{1,2} |\vec{p}\rangle$.

He then states that "evidently under Rotations $R(\theta)|p, \phi\rangle = |p, \phi + \theta \rangle$". While this seems sensible, I have trouble reconciling this with the Lie algebra commutation relations since when considering infinitesimal rotations (and setting $P = (P_1, P_2)^{T}$ )

$$[P, R(\delta \theta) ] |\vec{p}\rangle = P |p, \phi + \delta \theta\rangle - \vec{p}|p, \phi + \delta \theta\rangle \\ = ( \begin{pmatrix} \text{cos}(\delta \theta) & -\text{sin}(\delta \theta) \\ \text{sin}(\delta \theta) & \text{cos}(\delta \theta) \end{pmatrix} - \mathbb{1} ) \ \vec{p} \ | p, \phi + \delta \theta\rangle \\ \approx \quad \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \quad \delta \theta \ \vec{p} \ | p, \phi + \delta \theta\rangle .$$

This should equal $$[P, 1 + i \delta\theta J]|\vec{p}\rangle = i \delta\theta \ [P, J] |\vec{p}\rangle = \delta \theta\quad \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \quad \vec{p} \ | p, \phi \rangle ,$$ which it doesn't. So Zee's "evident" prescription doesn't seem to reproduce the correct commutation relations.

My question is am I misunderstanding something basic or is there more going on, that has simply been swept under the rug?

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    $\begingroup$ Have you bothered to eliminate pieces of $O(\delta \theta ^2)$, as you are supposed to? $\endgroup$ – Cosmas Zachos Sep 4 at 13:40
  • $\begingroup$ I suppose you mean replacing $|p, \phi + \delta \theta \rangle \rightarrow |p, \phi \rangle + \mathcal{O}(\delta \theta) .$ But how can we expand a term like this in theta, when for arbitrarily small $\delta \theta$ the two vectors are orthogonal, i.e. $0 = \langle p, \phi|p, \phi + \delta \theta \rangle$. These perturbations are not small in the usual norms. $\endgroup$ – Zarathustra Sep 4 at 13:49
  • $\begingroup$ I'm not sure what you are saying... The slightly rotated vector is basically the original one plus an infinitesimal orthogonal piece. $\endgroup$ – Cosmas Zachos Sep 4 at 13:54
  • $\begingroup$ Here $ |p, \phi> $ is not the vector $(\text{cos}(\phi), \text{sin}(\phi))^T p$, but a label for a basis element of a vector space. I guess this is analogous to $|y\rangle = \delta(x-y)$ in QM. Here we have $\langle y+\delta y| y\rangle = 0 \ \forall \delta y \neq 0 $. I think these should be orthogonal in any Group representation that is Unitary. Or where the Algebra generators are represented as hermitian operators. $\endgroup$ – Zarathustra Sep 4 at 14:01
  • $\begingroup$ I haven't followed the conventions, but it looks to me like a bland dot product of two 2-vectors in polar coordinates... You are aware this is an aggressively informal and "you know, you know" book... $\endgroup$ – Cosmas Zachos Sep 4 at 14:24
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Well, in standard practice, several things are swept under rugs in QM... The basic relation you may check in standard tasteful texts such as Sakurai & Napolitano, etc, is the translation formula, $$ \bbox [yellow]{|\phi + \delta \theta\rangle= e^{-i\delta \theta J} |\phi\rangle=(1-i\delta\theta J+...)|\phi\rangle }~, $$ so $$ \langle \chi|\phi+ \delta \theta\rangle= \delta (\chi -\phi -\delta \theta) \\ \approx \delta (\chi -\phi) -\delta \theta ~\partial_\chi \delta (\chi -\phi) = \langle \chi|\phi \rangle -i\delta \theta\langle \chi|J|\phi \rangle . $$ Yes, these are infinite-dimensional vectors and one may choose to be careful$^\dagger$; but, by and large, dismissing $O(\delta\theta ^2)$ terms is generally consistent...

The length of the two-vector, p, is a canard, of course.

Your first and second expression agree to lowest order in $\delta \theta$.


$^\dagger$ As a reminder, $\hat p | x\rangle= i\hbar\partial_x |x\rangle$, so that $\exp(-i\hat p a/\hbar)| x\rangle= |x+a\rangle= \exp( a \partial_x )| x\rangle$, that is, somewhat hyper-formally, $\langle y|x+a\rangle=\delta(x+a-y)=\sum_{n=0}^\infty \frac{(a\partial_x)^n}{n!} ~\delta(x-y)$. Consequently, dotting by $\langle \psi|$, observe $\psi^*(x+a)=\sum_{n=0}^\infty \frac{(a\partial_x)^n}{n!} ~ \psi^*(x)= \psi^*(x)+a \psi ~'^*(x)+... $.

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