Grisha Kirilin
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 Nov 10 awarded Yearling Apr 24 awarded Nice Answer Mar 18 awarded Good Answer Feb 12 awarded Nice Question Nov 10 awarded Yearling Dec 23 revised Curved spacetime point particle Lagrangian density deleted 37 characters in body Dec 23 revised Curved spacetime point particle Lagrangian density deleted 7 characters in body Dec 23 answered Curved spacetime point particle Lagrangian density Nov 10 awarded Yearling Sep 27 awarded Nice Answer Aug 7 comment Calculating an expression for the trace of generators of two Lie algebra Generally, you can include $SU(N)$ as a subgroup in some noncompact group $G$, and then you can find an Abelian noncompact subgroup of $G$, so that $SU(N)$ would be a normalizer for this subgroup en.wikipedia.org/wiki/Centralizer_and_normalizer Probably Jackiw kept something like this in his mind. Aug 7 comment Calculating an expression for the trace of generators of two Lie algebra But you should keep in mind that it is only about finite matrices. For example, the components of the angular momentum $J^{i}$ are generators for $SU(2)$, and the commutator relation for any "vector" operator $V^{j}$ is $[J^{i},V^{j}]=i\epsilon^{ijk}V^{k}$. Hence if you put $V^{j}=p^{j}$, i.e., the operator of momentum, so that $[p^{i},p^{j}]=\delta^{i,j}$, then you find $[J^{i},p^{j}]=i\epsilon^{ijk}p^{k}$, i.e. the algebra you asked for. But you should remember that translations are not compact (group) thus you cannot define the notion of "trace" for them. Aug 7 comment Calculating an expression for the trace of generators of two Lie algebra It is difficult to comment Jackiw's paper without reading it carefully. You question was about the trace thus my proof is valid only for finite matrices. As I noticed it is a consequence of that fact that $SU(N)$, algebraically, is a simple Lie group, i.e., its Lie algebra is simple. Thus if you require that the set of matrices $R^{a}$ has as many components as $Q^{a}$, then it immediately follows that $R^{a}\equiv Q^{a}$. Aug 7 comment Calculating an expression for the trace of generators of two Lie algebra I added alternative way to show the same. Aug 7 revised Calculating an expression for the trace of generators of two Lie algebra alternative proof is added Aug 5 comment Calculating an expression for the trace of generators of two Lie algebra Probably you know that $\left( C^{a}\right) _{bc}=-if^{abc}$ are the generators of adjoined representation. Since for finite matrices the trace of commutator vanishes, thus $\mathrm{tr}\left[ Q^{a},R^{b}\right] =0=if^{abc}\mathrm{tr}R^{c}=-\left( C^{c}\right)_{ab}\mathrm{tr}R^{c}$ implies a linear relation for $\hat{C}^{a}$ which is only possible for all $\mathrm{tr}R^{c}=0$. You can also use the identity $f^{acd}f^{acd^{\prime}}=C_{A}\delta^{dd^{\prime}}$, hence $R^{c}=-iC_{A} ^{-1}f^{abc}\left[ Q^{a},R^{b}\right]$. Aug 2 revised Calculating an expression for the trace of generators of two Lie algebra fixed grammar Aug 2 answered Calculating an expression for the trace of generators of two Lie algebra Jul 30 comment Hermite polynomials for expected value of harmonic oscillator So why do you not accept my answer? Jul 30 comment Hermite polynomials for expected value of harmonic oscillator I used dimensionless integration variable $\alpha\, x\to x$ and omitted the overall factor $k/(2\alpha^2)$. Hence the expectation value of the potential energy is $I_{n} k/(2\alpha^2)$.