query: Does the canonical commutation relation give a unique solution for the momentum operator?
positive_passages.docid: 26125
positive_passages.text: Does the canonical commutation relation fix the form of the momentum operator?
negative_passages.docid: 112187
negative_passages.text: Is there a way to see that $ \nabla_\mu g_{\nu \rho} = 0 $ without explicit computation, where $\nabla_\mu$ refers to the covariant derivative?
negative_passages.docid: 26896
negative_passages.text: Why can't $ i\hbar\frac{\partial}{\partial t}$ be considered the Hamiltonian operator?
negative_passages.docid: 52935
negative_passages.text: Do eigenvectors of quantum operators span the whole Hilbert Space?
negative_passages.docid: 113610
negative_passages.text: Spinor field normalisation from poles in the propagator
negative_passages.docid: 121802
negative_passages.text: Prove that every self-adjoint unitary linear operator can be expressed in the form $U\alpha = \beta - \gamma$
negative_passages.docid: 128276
negative_passages.text: Quantization of orbital angular momentum
negative_passages.docid: 112216
negative_passages.text: Geometric description of canonical commutation relation
negative_passages.docid: 55231
negative_passages.text: Is there a simple way of finding the eigenstates of the creation and annihilation operator in QM?
negative_passages.docid: 94717
negative_passages.text: Systems in quantum mechanics solved by ladder operator method
negative_passages.docid: 26417
negative_passages.text: Why do we use Hermitian operators in QM?
negative_passages.docid: 116427
negative_passages.text: $(c^{\mu}\partial_{\mu}-m)\phi(x)=0$ not Lorentz-invariant?
negative_passages.docid: 44239
negative_passages.text: Why the extra term $\frac{1}{2}(\partial_{\rho}A^{\rho})^2$ in the photon Lagrangian?
negative_passages.docid: 70880
negative_passages.text: Invariance of canonical Hamiltonian equation when adding the total time derivative of a function of $q_i$ and $t$ to the Lagrangian
negative_passages.docid: 136184
negative_passages.text: Why we dont have "direct" velocity operator just as $p$? ( as use $p$ space not $v$? ) in quantum mechanics?
negative_passages.docid: 62693
negative_passages.text: Eigenspaces of angular momentum operator and its square (Casimir operator)
negative_passages.docid: 103579
negative_passages.text: Why are the zeroth order terms in degenerate perturbation theory the eigenstates of the perturbing Hamiltonian?