# Tag Info

## New answers tagged lie-algebra

0

Apart from the issues of $SU(2)$ vs. $SL(2,\mathbb{C})$, complexification, etc, which is covered e.g. in this Phys.SE post and links therein, it seems OP's core question (v2) is essentially the following question. How does the Lie algebra generator $J_3=J^L_3+J^R_3$ act on a vector space $V_L\otimes V_R$ for a tensor representation? The short answer ...

2

Firstly, $m$ does not have to be an integer, it is entirely possible for $m$ to be 1/2 for instance. Your points ,1-3 are fine. There are is a maximal and a minimal value of $m$. Call the maximal value $M$ (we have to call it something). Now we can apply the lower operator any number of times, each time it lowers the value of $m$ by a full integer ...

1

Any linear transformation wrought on the Lie algebra of a Lie group yields a valid Lie algebra as I think you understand (the Gell Mann matrices are actually $i$ times the skew-symmetric Lie algebra members), and your proposed $\lambda_3$ is a linear combination of the Gell Mann matrices. The basis comprising $i$ times the Gell Mann matrices does indeed span ...

1

Magnetic moment, in classical physics, is related to current in a loop, which in turn can be connected to angular momentum of a charged particle. Thus, in classical physics, magnetic moment and angular momentum are connected. In fact, they are proportional with the constant of proportionality being the gyromagnetic ratio. Moving to quantum mechanics, some ...

6

That spin follows the angular momentum algebra is no accident - like angular momentum, it is part of the conserved quantity - the Noether charge - associated to rotations. The reason why the $\mathfrak{so}(3)$ transformations of spin should be indeed those associated to the $\mathfrak{so}(3)$ of spatial rotations is not answerable in QM alone - you have to ...

4

You appear confused by how spin is introduced in ordinary QM. It is rather ad hoc: Given a Hilbert space without spin degrees of freedom of a particle $\mathcal{H}_0$, and the spin $s$ of the particle, we take the total space of states of the particle to be $\mathcal{H}_0\otimes \mathcal{S}_s$, where $\mathcal{S}_s$ is a $2s+1$-dimensional complex Hilbert ...

2

Comment to the question (v1): No, such decomposition is in general not unique. E.g. the unit element ${\bf 1}_{2\times 2}\in SU(2)$ can be written with parameters $b\in 4\pi\mathbb{Z}$ and $a=0=c$.

10

You may indeed identify the generators in the way you did. However, the Lie algebras and Lie groups are different because – as quickly said by Qmechanic – you must use different reality conditions for the coefficients. A general matrix in the $SU(2)$ group is written as $$M = \exp[ i( \alpha J_+ + \bar\alpha J_- + \gamma J_0 )]$$ where $\alpha\in ... 7 The ladder operators do belong to the real Lie algebra $$\quad su(1,1)~\cong~ so(2,1)~\cong~sl(2,\mathbb{R}),$$ but they do not belong to the real Lie algebra $$su(2)~\cong~ so(3).$$ All the above real Lie algebras have complexifications isomorphic to$sl(2,\mathbb{C})$. 1 You should have two boost generators. You have constructed one for boost in the$x$direction, but there is also one for boost in$y$. 0 Clebsch-Gordan coefficients let you treat n spins (or generaly - any n particles with arbitrary angular momentum) as a single composite system. The coefficients are simply the matrix element of basis transformation from seperated to composite system. For 2 particles with total angular momentum eigenvalues$ l_1,l_2 $, such that for example$ ...

3

Every Hermitian traceless matrix $H$ is in $\mathfrak{su}(N)$ since $\mathrm{Tr}(H) = 0$ and so $$\exp(\mathrm{tr}(\mathrm{i}H)) = \det(\exp(\mathrm{i}H)) = 1$$ so $\exp(\mathrm{i}H)$ is unitary with determinant $1$, hence in $\mathrm{SU}(N)$. The gauge field is always in the Lie algebra of the gauge group since it is introduced to cancel terms that are ...

0

I think it's easier to see this if you start from the matter representations rather than the vector field side. For example, imagine that you have some matter field $\psi$ that transforms under some simple Lie group $G$ representation according $$\psi \to g \psi$$ where $g \in G$. Now, the derivative term is not invariant if $g=g(x)$ as one has ...

3

Commutators are Lie brackets, in this case on the algebra of differential operators. Asking whether $[\dot{},\dot{}]$ is a Lie bracket doesn't really make sense (since, for matrix groups, the Lie bracket is the matrix commutator, anyway). The field strength is not "just a definition", it is the natural curvature associated to the principal connection that ...

0

If the Hamiltonian has $SU(2)$ symmetry then the Hilbert space can be spanned by eigenstates of Cartan subalgebra (one dimensional) operator and Casimir operator. As we know from elementary quantum mechanics \begin{align} \hat{L}^{2}\left|l,m \right\rangle &=l(l+1)\left|l,m \right\rangle,\\[3mm] \hat{L}_{z}\left|l,m \right\rangle &= m\left|l,m ... 2 The gauge potential is an object that, when introduced in the covariant derivative, is intended to cancel the terms that spoil the linear transformation of the field under the gauge group. Every gauge transformation g:\Sigma\to G (on a spacetime \Sigma) connected to the identity may be written as \mathrm{e}^{\mathrm{i}\chi(x)} for some Lie algebra ... -1\begin{align} \left[\hat{A}_{i}, \hat{B}_{j} \right] & = \epsilon_{ijk}\hat{C}_{k}, \\[3mm] \hat{A}_{i}\hat{B}_{j} - \hat{B}_{j}\hat{A}_{i} & = \epsilon_{ijk}\hat{C}_{k}, \\[3mm] \sum\limits_{i=1}^{3}\sum\limits_{j=1}^{3}\epsilon_{ijn}\left( \hat{A}_{i}\hat{B}_{j} - \hat{B}_{j}\hat{A}_{i}\right) & = ...

6

I have read somewhere that commutation relations of the form $$[a_i,b_j]=\epsilon_{ijk} c_k$$ admit a "natural rewriting in terms of cross products", but there weren't any details about this statement. This "natural rewriting" of the canonical commutation relations for angular momenta in term of cross products is: $$... 6 It's not possible to derive the orbital angular momentum L = r \times p from the \mathfrak{so}(3) commutation relations alone, since the spin operator S also fulfills the same commutation relations, but certainly is different from r \times p. 2 Just a guess... The purpose is to reproduce the nice features of SU(2). With that convention, the generators of SU(2) are, in terms of Pauli matrices$$T^i = \frac{1}{2}\sigma^i$$So a transformation with parameters \theta_i is given by$$U=\exp\left(-i\frac{1}{2}\theta_i\sigma^i\right) Things get interesant when you realize that the elements of ...

2

The Lie algebra $\mathfrak{su}(N)$, viewed as a vector space of matrices, can be equipped with the following standard inner product: \begin{align} \langle X,Y\rangle = \mathrm{tr}(X^\dagger Y), \end{align} where $X^\dagger Y$ is the matrix product of $X^\dagger$ and $Y$, and $\mathrm{tr}$ is the trace. Since $X^\dagger = X$ for all $X\in\mathfrak{su}(N)$, ...

4

This can be explained by thinking about the coupling of fermions to the $SU(2)$ weak gauge field. Let's recap what we know Weyl fermions necessarily appear in two complex representations of the Lorentz group $L$ and $R$. Only fermions in the $L$ representation of the Lorentz group couple to the $SU(2)$ gauge field. CPT is a symmetry of the theory. Now ...

4

The color language is not really well-suited to understand why there are eight gluons. Here's why, however: The gluon field transforms in the adjoint representation of the color gauge group $\mathrm{SU}(3)$. The adjoint representation is a representation on the vector space of the generators of $\mathrm{SU}(3)$, the Lie algebra $\mathfrak{su}(3)$. An ...

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