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bio website winterhunters.blogspot.com
location Germany
age 35
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postdoc in Technische Universität München.


Dec
23
revised Curved spacetime point particle Lagrangian density
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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) $.
Jul
30
answered Hermite polynomials for expected value of harmonic oscillator
Mar
8
revised What if the binding energy becomes larger than the rest mass?
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Mar
8
comment What if the binding energy becomes larger than the rest mass?
The example of the numerical calculation of the complete proton structure you can find in G. S. Adkins, C.R. Nappi, Nucl.Phys. B233 (1984) 109. Although the real dynamics is still unknown, I would like to stress here that neither baryons nor mesons are the bound states of any number of quarks and gluons (in the conventional sense), they are nontrivial low-temperature collective excitations of the quark and gluon fields.
Mar
8
comment What if the binding energy becomes larger than the rest mass?
For example, pions can be considered as pseudo-Goldstone bosons of the chiral-flavor symmetry breaking. There are plenty of phenomenological models which consider baryons (like proton) as a nontrivial soliton of some effective chiral fields (see, e.g., Skyrme model, Nambu—Jona-Lasinio model etc).
Mar
8
comment What if the binding energy becomes larger than the rest mass?
For example, the QCD coupling constant strongly depends on the distance, thus it is believed that there is the second order phase transition, e.g., during the cooling down of quark-gluon plasma, so that a non-trivial non-perturbative QCD vacuum appears and it is filled with the quark condensate. The resulting excitations (hadrons) are absolutely non-trivial collective excitations of this vacuum.