# Tag Info

5

Yes. Both universal covers and central extensions incurred during quantization come from the same fundamental concept: Projective representations If $\mathcal{H}$ is our Hilbert space of states, then distinct physical states are not vectors $\psi\in\mathcal{H}$, but rays, since multiplication by a complex number does not change the expectation values given ...

5

Recall that the Faraday tensor in this form is a linear mapping that maps a charged particle's contravariant four-velocity to the latter's rate of change, wrt proper time (modulo scaling by invariant rest mass $m$ and invariant charge $q$): $$m\,\frac{\mathrm{d} v^\mu}{\mathrm{d}\tau} = q\, F^\mu{}_\nu\,v^\nu\tag{1}$$ Now let's think of a particle's ...

0

We assume that OP asks apart from the facts that: Dirac representations by definition are complex; It is much easier to work with an algebraically closed field; Any real representation can be extended to a (possibly reducible) complex representation, so one is not missing anything by going complex. In other words, OP is interested in why certain real ...

2

Comments to the question (v2): The idea to consider the planar large $N_c\to \infty$ limit in $SU(N_c)$ QCD goes back to Ref. 1. In light-cone membrane theory, pioneered in Ref. 2, the group $SU(\infty)$ is naturally identified with area-preserving diffeomorphisms ${\rm SDiff}_0(T^2)$ on the torus $T^2$ connected to the identity. Concretely, OP's proposal ...

3

There are apparently several thousand references to "SU(\infty)" on arxiv.org, and some of them are definitely talking about gauge fields or Yang-Mills. I suspect that some of the time, this will just be a way of talking about the large N limit of SU(N), i.e., not referring to a literal SU(∞) field theory, but rather the N→∞ limit of some quantity in SU(N) ...

0

When I was asking this question, I didn't understand the relation between the commutativity of two operators and their eigenspaces: If an operator $A$ commutates with another operator $B$, then $A$ leaves the eigenspaces of $B$ invariant: $$B\psi = \epsilon\psi \implies BA\psi = AB\psi = \epsilon A\psi$$ But this does not imply that $\psi$ is an ...

6

OP is essentially pondering if commutativity is a transitive relation, ie. if three normal operators$^{1}$ $A$, $B$, and $C$ satisfies $$[A,B]~=~0\quad \wedge\quad [B,C]~=~0 \quad\stackrel{?}{\Rightarrow} \quad [A,C]~=~0 .\tag{T}$$ The answer is No, but OP argues via the existence of a common basis of eigenvectors for two commuting normal operators that ...

2

The $L_i$ has many eigenspaces corresponding to many eigenvalues. Each of those eigenspaces is also an eigenspace of the Casimir operator. So they share common eigenspaces in the sense that there are eigenspaces that are eigen to both. But they don't share them in the sense that they are the same. Look at the hydrogen atom. There are energy eigenspaces and ...

4

Two operators may be simultaneously diagonalized if and only if they commute. As you can see, $L_z$ commutes neither with $L_x$ nor with $L_y$ – and not with any other linear combinations different from a multiple of $L_z$ – so there's no way to diagonalize two different components of $L_i$ at all. However, $L_z$ (and similarly other components) commutes ...

3

The eigenspaces of the quadratic Casimir $L^2 = L_x^2+L_y^2+L_z^2$ of the Lie algebra of infinitesimal rotations $\mathfrak{so}(3)$ are precisely the irreducible representations of $\mathfrak{so}(3)$ - we usually label a representation by its highest weight $l$, which is in this case just a number telling you what the largest possible value for any of the ...

4

The Lorentz group is the group of matrices that conserve the quadratic form: $$\mathscr{Q}(X,\,Y) = X^T\,\eta\,Y\tag{1}$$ where here $X$ and $Y$ are $1\times 4$ column vectors, the $4\times 4$ group member matrices act on these from the left and $\eta$ is the Minkowski (pseuso) metric. Therefore, $\Lambda\in O(1,\,3)$ if and only if: ...

0

$SO(1,\,2)$, with two spatial dimensions, only has rotations in one plane - that of all space. So if our co-ordinates are $(t,\,x,\,y)$ (spatial co-ords last), its unique (up to a real scale factor of course) rotational Lie algebra generator must be: $$R=\left(\begin{array}{ccc}0&0&0\\0&0&-1\\0&1&0\end{array}\right)$$ where we've ...

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