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0

What Qmechanic said in comments is pretty solid, "Lagrangian (2) is not bounded from below because the kinetic term of $A_0$ field has the wrong sign, and hence the theory is not physical in the first place", but I think your Question needs a change of emphasis. Your Lagrangian allows us to construct four equations of motion for four non-interacting fields. ...


0

Field $\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}}$ with a given spin and mass (i.e. field which transforms under irrep of the Poincare group) must satisfy some determined conditions called irreducibility conditions: $$ \tag 1 \hat{W}^{2}\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}} = -m^{2}\frac{n + m}{2}\left(\frac{n + m}{2} + ...


0

They do not lie in $\mathfrak{so}(3)$ but they lie in its complexification, which would be $A_1$ in the usual mathematical classification. Much of Lie representation theory is set up this way: you work at the level of the complexification then go back to the real form. For compact groups it's not a big deal; for non-compact groups extra care is needed. So ...


2

I guess what you are missing is the following: given a representation $\rho(g)$ of $g\in$SU(2) acting on some vector space $V$. We define the representation $\rho_\otimes$ of SU(2) (not of SU(2)$\times$SU(2)) on $V\otimes V$ as $$\rho_\otimes(g) (v_1 \otimes v_2) = \rho(g) v_1 \otimes \rho(g) v_2.$$ So in fact we are defining the tensor product of two ...


0

To find the spin eigenstates corresponding to a multi-particle state, all one needs to do is build the appropriate multi-particle spin operator using the direct product, e.g. $$J_z^{(2)}=J_z \otimes 1 + 1\otimes J_z \\ J_z^{(3)}=J_z \otimes 1 \otimes 1 + 1\otimes J_z \otimes 1 + 1\otimes 1\otimes J_z $$ and then diagonalize the resulting operator to find ...


0

So this explanation (my first post on stackexchange!) is based on H. Georgi's "Lie algebras in particle physics", chapter 4. Since $Q_{ij}$ is symmetric, real and traceless, it has 5 independent degrees of freedom. So it's possible to express $Q_{ij}$ into a 'spherical' basis $Q^s_l$, where $s=2$ in this case and $l$ takes on values -2, -1, 0, 1, 2. (For ...


2

None of this is coincidental. But there is also not much mysterious going on: The energy of any state is the expectation value of the Hamiltonian in that state. The Hamiltonian generates indeed a one-parameter group, the time translations, but it is not $\mathrm{U}(1)$, but rather $(\mathbb{R},+)$ - it is abelian, but not compact (since otherwise we could ...


1

I assume to deal with an autonomous, first order (at least $C^1$ or smooth) system of ordinary differential equations and that the hypotheses sufficient for existence and uniqueness of maximal solutions are satisfied. You may always reduce to the case of a first order system by adding auxiliary variables, $\dot{x}$, to the initial system of differential ...


2

Be careful. It may be the case that $\mathfrak{su}(2)=\mathfrak{so}(3)$, but it is not the case that $SU(2)=SO(3)$. $SU(2)=\mathrm{Spin}(3)$ and $\rho :SU(2)\rightarrow SO(3)$ is the two-sheeted universal cover of $SO(3)$. It thus turns out that only the integer spin representations of $SU(2)$ factor through $\rho$ to give well-defined representations of ...


0

When I studied at first course and investigated the special theory of relativity the lecturer said about old interpretation of relativity. In this approach instead pseudo-euclidean metric and four-vectors $(t,\bf x) $ people use euclidean metric and four-vectors $(it,\bf{x})$. But it does not mean that we use SO(4) group! We use also SO(3,1) group but we do ...


2

In quantum mechanics, the canonical commutation relations $[q,p]=i\mathbb{1}$, $[q,\mathbb{1}]=[p.\mathbb{1}]=0$ (I am taking $\hbar=1$) between position ( $q$ ), momentum ( $p$ ) and identity ($\mathbb{1}$) operators form an algebra, that is the Lie algebra of the Heisenberg group. There are problems of domains because the operators are unbounded, so ...


1

Explicit symmetry breaking, i.e. adding a mass term for a particle simply implies that this particle is massive. In the case of a vector boson, there will be three degrees of freedom. The "additional" degree of freedom does not have to come from somewhere else by a certain mechanism, it is simply there from the beginning. An example would be Proca theory, ...


2

The trace is just the inner product for the Lie algebra. The field strengths are Lie algebra valued, i.e., $\mathbf{F}_{\mu\nu}$ is an element of the Lie algebra, and can be written as a linear combination of generators: $\mathbf{F}_{\mu\nu} = \sum_a F^a_{\mu\nu} t^a$. One usually normalizes the generators such that $\left \langle t^a, t^b\right\rangle = ...


6

The Poincare group has two Casimir Invariants - namely $p^2$ and $W^2$ where $$ W_\mu = \frac{1}{2} \epsilon_{\mu\nu\rho\sigma} J^{\nu\rho} p^\sigma $$ is the Pauli-Lubanski pseudo-vector. Thus, representations of the Lorentz group are labelled by the eigenvalues of both $p^2$ and $W^2$. When $p^2 = -m^2$, we have the property $W^2 = -m^2 {\bf J}^2$. ...


8

The point is that the symmetries in QM (bijective operations sending states to states preserving the transition probability) can be represented by either unitary or antiunitary operators. This is the statement of a famous theorem due to Wigner. It is possible to prove also that, if the Hamiltonian of a system is bounded below, time reversal must be ...


1

I expand my comment into an answer. The idea is to fix $\alpha, \beta \in \mathbb R$ in order that, if $P:= U(\alpha, \beta)$ (which is automatically unitary), we have (i) $PP=e^{ik}I$ for some $k\in \mathbb R$, (ii) $P\hat{x}P^\dagger = -\hat{x}$, (iii) $P\hat{p}P^\dagger = -\hat{p}$ Since $\hat{x}$ and $\hat{p}$ has to be treated symmetrically, we ...


0

If I know the explicit form of the states I would first apply the SU(2) generators to them. For example apply the Cartan generator $L_3$ and see whether you get the eigenvalues $-5/2, -3/2,...,+5/2$.


3

I) First recall the fact that $SL(2,\mathbb{R})\times SL(2,\mathbb{R})$ is (the double cover of) the identity component $SO^{+}(2,2;\mathbb{R})$ of the split orthogonal group $O(2,2;\mathbb{R})$. This follows partly because: There is a bijective isometry from the split real space $(\mathbb{R}^{2,2},||\cdot||^2)$ to the space of $2\times2 $ real ...



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