11
This is a really deep question and I urge you to go ahed and read about it in the literature i'll give at the end. I'll try to give a glimpse of what this actually means.
In physics we can construct our theories based solely upon symmetries of a system. When talking about angular momentum and spin in non relativistic quantum mechanics, we are talking about ...
5
There is a priori no relation between observables and irreducible representations: an observable is just an observable, and it can connect states in different representations of a group. An example would be the $\hat z$-component of the dipole moment, which can connect states of different angular momentum. The radial position $r$ is also a observable, and ...
5
The $\otimes$ sign denotes the tensor product. Given two matrices (let’s say $2\times 2$ although they can be $n\times n$ and $m\times m$) $A$ and $B$, then
$A\otimes B$ is the $4\times 4$ matrix
\begin{align}
A\otimes B =\left(
\begin{array}{cc}
A_{11}B&A_{12}B\\
A_{21}B&A_{22}B
\end{array}\right)= \left(\begin{array}{cccc}
A_{11}B_{11}&A_{11}...
4
You are indeed missing some pieces. You can immediately see that there is no way your formula works in general since the full expansion reads
$$
\begin{aligned}
e^{ix\cdot P} K_\mu e^{-ix\cdot P} &= \sum_{n,m=0}^\infty \frac{i^{n-m}}{n!m!} (x\cdot P)^n\,K_\mu\,(x\cdot P)^m\,.
\end{aligned}\tag{1}\label{ini}
$$
The problem is that this is a mess because ...
3
I should concur with the other answers that there is no substitue for reading up in good texts and WP.
You are right that, in a given basis, there is a similarity (equivalence) transformation implied in the equation of your title: it basically means that the tensor product on the l.h.s. is reducible, by an orthogonal basis change to the r.h.s.; that is, ...
2
This is actually the decomposition for the tensor product of irreducible representation of SU(2). We can set your $1/2$ as $j$, which means the (2j+1) dimimension irreducible representation of SU(2). Generally, Clebsch–Gordan series gives:
$$D^{\left(j_{1}\right)} \otimes D^{\left(j_{2}\right)}=\bigoplus_{J=\left|j_{1}-j_{2}\right|}^{j_{1}+j_{2}} D^{(J)}$$
...
2
The example given above of the Clebsch-Gordan $\langle 2,0|2,0,1,0\rangle$ is not an accidental or non-trivial zero! It is a result of the known selection rules. It corresponds to the 3-$j$ case (Mathematica notation)
ThreeJSymbol[{2, 0}, {1, 0}, {2, 0}]
and is zero because the sum of the top 3 three $j$'s, i.e. $2+1+2$, is not even as required (see ...
2
Take the $3\times 3$ rotation matrix $R(\Omega)$ and tensor it with itself: $R(\Omega)\otimes R(\Omega)$ will be a $9\times 9$ matrix acting on the components of the $9\times 1$ $T_{ij}$.
As you have guessed, $R(\Omega)\otimes R(\Omega)$ is reducible. The $5\times 5$ block contains the Wigner D-matrices for $L=2$, i.e the $D^2_{MM'}(\Omega)$ functions. ...
2
Table III of the legendary 1974 paper by Ling-Fong Li, required canonical reading for theory students, details which low-lying Higgs representations break SU(n) groups to what subgroup and why.
The "job" is to SSBreak 12 of the 24 symmetry directions of SU(5) so the remaining 12, so far unbroken at this stage, comprise the 8+3+1=12 of the SU(3)×SU(2)×U(1) ...
2
You are doing quantum mechanics around the clock in a clock of only 4 hours (period 4). Apart from an excessive square root in your normalizations (which I suspect should be just 1/2), your are Discrete Fourier Transforming your "hour" states $| 1\rangle, | 2\rangle,| 3\rangle,| 4\rangle,$ (the discrete analog of position angle eigenstates) to "shift" ...
2
The properly written kinetic term for matrix-valued fields is
$$ \text{Tr}\Big((D_\mu\psi)^\dagger (D^\mu\psi)\Big)$$
If you use this expression it doesn't matter in which order you multiply them, as the trace is cyclic.
2
Indeed, you are doomed. There is no such V.
Suppose there were an equivalence (4,5,6).
Then consider $\gamma_5 = i \gamma_0 \gamma_1 \gamma_2 \gamma_3$, as well as its transform, independently of basis or representation,
$$
V\gamma_5 V^{-1} = i V\gamma_0 \gamma_1 \gamma_2 \gamma_3 V^{-1} \implies \\
\gamma_0 = i \gamma_5 \gamma_1 \gamma_2 \gamma_3 \\ = i \...
1
I don't think this property has a name, but it's not too hard to show. In fact, it pops out automatically if you revisit how the Clebsch-Gordan coefficients are derived in the first place, keeping track of the symmetry.
If you start with spins $s$, then the maximum total $m$ you can get is $2s$, from the state $|m_1 = s, m_2 = s \rangle$. This state must be ...
1
First of all, the equation you write for $\Phi'(x')$ is not correct. The reason is that in principle you are allowed to use the same generator multiple times in the product since the order matters (incidentally, since $\Phi$ is a scalar primary, the action of $D$ and $K$ would vanish). An analogy is the Euler angles where you can parametrize a rotation as a ...
1
To put it simply, to know that the Lie group $SL(2,\mathbb{C})$ of $2\times 2$ complex matrices with unit determinant is (the double cover of) the restricted Lorentz group $SO^+(1,3;\mathbb{R})$ gives a simple and direct way to understand how the restricted Lorentz group can act on a Weyl spinor $\psi\in\mathbb{C}^2$. In contrast, the group action (on a Weyl ...
1
If we're looking at complex representations, then every irreducible representation of $G\times H$ is the tensor product of an irrep of $G$ and an irrep of $H$, see e.g. this math.SE question. So the fundamental representation of $\mathrm{SU}(2)\times\mathrm{U}(1)$ is two-dimensional and $(s,u)\in\mathrm{SU}(2)\times\mathrm{U}(1)$ acts simply with the $s$ ...
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