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3

I'm not altogether sure what you are asking, but I suspect the following may help. To represent rotations, spins and vectors in $SU(2)$ we work as follows. Rotations live in $SU(2)$. Vectors (in the physicist's sense) live in the algebra $\mathfrak{su}(2)$. The position vector $(x,\,y,z)$ is: $$X =x\,\hat{s}_x+y\,\hat{s}_y+z\,\hat{s}_z = ...


-1

I'm not knowledgeable in some aspects of the question, but I will provide an answer unrelated to others. An object can rotate so fast that some representations of angular velocity cannot be valid. For example, when an object rotates more than 180-degrees or 360-degrees (pi or 2*pi radians) per unit time, the representation must be able to represent such ...


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 ...


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 $sl(2,\mathbb{C})$.


0

Yes, your intuition is correct: two different boosts do contain one rotation, and precisely two boosts along two orthogonal axes contain one rotation around the third orthogonal axis --- the most direct way to see that is by considering that the commutator of two different boosts is one rotation, and more completely the Lorentz algebra of rotations $R_{a}$ ...


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 $y$.


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 ...


1

Can these two pictures be connected in some way? Yes, that's why the Wikipedia spinor article features a picture of a Möbius strip: GNUFDL image by Slawekb, see Wikipedia The Mobius strip also features in the Mathspages Dirac's belt article where you can read that it's "reminiscent of spin-1/2 particles in quantum mechanics, since such particles must be ...


1

I) Perhaps it is helpful to point out that even if the physical system $S$ has no rotational symmetry (e.g. if the system $S$ is a 3D an-isotropic harmonic oscillator), then the Lie group $G=SO(3)$ of rotations still has a group action $G \times S \to S$ on the system. See also e.g. this Phys.SE post. In particular the Hilbert space ${\cal H}$ of the system ...


0

My understanding of this limited, but this might help (too long for a comment): The state space is spanned by the set of simultaneous eigenstates of the Hamiltonian, $ \hat L^2$, and $ L_z $. In fact, they form an orthonormal basis of a Hilbert space $ H $ which is the state space. Out of convenience, we denote the eigenstates by the quantum numbers, ...


2

For orbital angular momentum, indeed, $L = x\times p$ even as a quantum operator, see this question. When writing a ket $\lvert l,m \rangle$, this is meant to live in the $2l+1$-dimensional space $\mathcal{H}_l = \mathbb{C}^{2l+1}$ on which the representation of the angular momentum algebra labelled by $l$ exists ($m$ is the eigenvalue of the ket for ...


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 ...


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 $X\in\mathfrak{su}(N)$, ...


1

I think the strong CP problem is still a pretty important problem. Although some people don't worry too much about these fine-tuning problems. Just like some people didn't consider the flatness problem a big deal before the inflation theory explained it. So like every problem, it's as big as you make it. It depends on how much you value naturalness... ...


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 ...



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