New answers tagged

2

Supersymmetry generators are not always Hermitian. If you impose SUSY, and then compute de corresponding Noether's currents, and then you calculate the conserved charge, i.e., the fermionic Lorentz generators, you will get two non-Hermitian conserved currents. (By the way, the relation $Q^\dagger=\bar Q$ is only valid in Lorentzian signature, in Euclidean ...


2

The Noether charge is the generator of the symmetry it belongs to, see e.g. this answer by Qmechanic. This relationship is also preserved in the quantum theory, see this question, in the sense that the quantum Noether charge $Q$ must commute with the Hamiltonian $H$, at least in the absence of anomalies and if we do not run into "quantization issues" when ...


9

Translational symmetry in the sense of the standard formulation of Noether theorems means that the Lagrangian is invariant under the action of the group of spatial translations. This is not the case in your example because $U$ does not admit such invariance. However there is another, more physical, version of the idea of translational invariance for a ...


3

I) Let the Lagrangian be $$\tag{1} L~=~\frac{m}{2}v^2-U(x), \qquad v~:=~\dot{x}.$$ Let the force $$\tag{2} F~=~-U'(x) $$ be a constant. II) Infinitesimal translations $$\tag{3} \delta x~=~\varepsilon $$ is a quasi-symmetry $$\tag{4} \delta L ~=~\varepsilon \frac{df}{dt}, \qquad f~:=~Ft $$ of the Lagrangian (1). Here $\varepsilon$ is an ...



Top 50 recent answers are included