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There is a well-known isomorphism between the linear space ${\mathcal M}_{m, n}$ of $m\times n$ matrices and typical (vectorial) linear spaces ${\mathcal L}_{m\times n}$ of dimension $\text{dim}({\mathcal L}_{m\times n}) = m\times n$. Everything that is valid in ${\mathcal L}_{m\times n}$ has an equivalent in ${\mathcal M}_{m, n}$ and conversely. For this ...


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That the Hamiltonian is zero is completely correct. The system is time-reparametrization invariant - changing $\tau$ to $\xi(\tau)$ transforms $$ n(q(\tau))\mapsto n(q(\xi)),\quad \dot{q}\mapsto \frac{\mathrm{d}\xi}{\mathrm{d}\tau}q', \quad \mathrm{d}\tau\mapsto \frac{\mathrm{d}\tau}{\mathrm{d}\xi}\mathrm{d}\xi$$ and the action is invariant under this ...


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The quick answer is: no. The Hamiltonian operator is a unitary operator that maps state vectors to other state vectors in a given Hilbert Space, regardless of time. Lubos's answer in this thread discusses this distinction very clearly: Why $\displaystyle i\hbar\frac{\partial}{\partial t}$ can not be considered as the Hamiltonian operator? Another point ...


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In general, $\frac{\partial L}{\partial \dot{q}}$ is the canonical (or generalized or conjugate*) momentum, and $m\dot x$, for $x$ the actual position, is kinetic momentum. Likewise, the cross product of the former with the generalized coordinate vector $q$ might be called "canonical angular momentum", and the cross product of the latter "kinetic angular ...


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This is more or less an exercise in chasing definitions. The adiabatic invariant $I$ is defined as $$ I\equiv \oint p \frac{\mathrm{d}q}{2\pi}\tag{49.7}$$ where the integral is taken over the path for given $E$ and $\lambda$. The external parameter $\lambda(t)$ is a slowly varying function of time $t$ in $\S49$, but is assumed to be a constant in $\S50$. ...



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