Classical to Quantum in Hamiltonian Formalism

\begin{align*} H(q,p,t)\psi = i\hbar\frac{\partial}{\partial t}\psi \end{align*}

So, \(\lambda = \frac{h}{p}\to 0\).

So, \(\psi(q,t) \propto \exp\left(\frac{i}{\hbar}S(q,t)\right)\). Then,

\begin{align*} H(q,p,t)\psi &= \lim_{\hbar\to0}i\hbar\frac{\partial}{\partial t}\exp\left(\frac{i}{\hbar}S(q,t)\right) \\ &= i\hbar\frac{i}{\hbar}\psi\frac{\partial S}{\partial t} \\ &= \psi\left(-\frac{\partial S}{\partial t}\right) \\ \left[H(q,p,t) + \frac{\partial S}{\partial t}\right] &= 0 \\ H(q,p,t) + \frac{\partial S}{\partial t} &= 0. \end{align*}

For a seperable, periodic, system, we get quantized action,

\begin{align*} J_\sigma = \oint p_\sigma dq_\sigma = n_\sigma h \end{align*}

Consider a particle in a central force potential.

\(P_\theta = \ell_z = \mathbb{C}\). Then,

\begin{align*} J &= \oint P_\theta d\theta \\ &= 2\pi\ell_z = nh \\ \Rightarrow \ell_z &= \frac{nh}{2\pi} = n\hbar. \end{align*}

For a Hamiltonian not dependent on time,

\begin{align*} E &= H(J_1,\cdots,J_n) \\ &= H(n_1h,\cdots,n_nh). \end{align*}

Suppose \(n_\sigma\to n_\sigma+1\). Then,

\begin{align*} E' &= H(J_1,\cdots,J_\sigma',\cdots,J_n) \\ &= H(n_1h,\cdots,(n_\sigma+1)h,\cdots,n_nh) \\ \Delta E &= H(n_1h,\cdots,(n_\sigma+1)h,\cdots,n_nh) - H(n_1h,\cdots,n_nh) \\ &= (1h)\frac{\partial H}{\partial J_sigma} \\ &= h\nu_\sigma. \end{align*}

\begin{align*} [A,B] &= \sum \frac{\partial A}{\partial q}\frac{\partial B}{\partial p} - \frac{\partial A}{\partial p}\frac{\partial B}{\partial q} \\ \frac{dF}{dt} &= [F,H] + \frac{\partial F}{\partial t}. \end{align*}

For Quantum Mechanics, everything becomes an operator,

\begin{align*} [\hat{A},\hat{B}] &= \sum \frac{\partial A}{\partial q}\frac{\partial B}{\partial p} - \frac{\partial A}{\partial p}\frac{\partial B}{\partial q} \\ \frac{d\hat{F}}{dt} &= \frac{1}{i\hbar}[\hat{F},\hat{H}] + \frac{\partial \hat{F}}{\partial t} \\ &= \frac{i}{\hbar}[\hat{H},\hat{F}] + \frac{\partial \hat{F}}{\partial t}. \end{align*}