Tag Info

New answers tagged

0

This may not be exactly what you want, but it's something. The definition $$ \vec{L}=\vec{r}\times\vec{p} $$ combined with $\vec{p}=-i\hbar\vec{\nabla}$ implies that the the $z$ projection may be written as $$ L_{z} = -i\hbar\frac{\partial}{\partial \phi}.$$ Where $\phi$ is the azimuthal angle. Since $L_{z}|\ell m\rangle = m\hbar |\ell m\rangle$, the $\phi$ ...


0

Mutually commutative means that every operator in the set commutes with every other one. This implies that, if the operators in question are observables, they can all be measured simultaneously. A complete set of mutually commuting observables is a set of observable, hermitian operators that commute - therefore their eigenvalues can be used to label a ...


4

First, I assume finite dimensional operators: otherwise you need to check certain boundedness conditions on the operators. Because the CBH series is here truncated by the vanishing double commutators, the conditions for linear operators on e.g. $\mathbf{L}^2(\mathbb{R})$ will be mild. You need to practice operations with $\mathrm{Ad}$. Look up the ...


2

I) Before we get to quantization and path integrals there are problems already at the classical level. The Legendre transformation is not well-defined without knowledge of the CCR. For instance if the CCRs for the complex bosonic scalar $\hat{\phi}$ and $\hat{\phi}^{\dagger}$ is zero, this would mean that OP's Hamiltonian density ${\cal H}$ is a pure ...



Top 50 recent answers are included