10.4: Kinetic equation (in classical mechanics and quantum mechanics)

In this section, we consider the simplest kinetic equation in classical system and quantum system.

Consider the state space $\Omega$ such that $\Omega ={\mathbb R}^2$, that is,

\begin{align} {\mathbb R}^2 = {\mathbb R}_q\times{\mathbb R}_p {{=}} \{ (q,p)=(\mbox{position , momentum }) \; | \; q,p\in{\mathbb R}\} \tag{10.14} \end{align}

Hamiltonian ${\mathcal H}(q,p)$ is defined by the total energy, for example, as the typical case ($m$: particle mass), we consider that

\begin{align} & [{\mbox{Hamiltonian }}(={\mathcal H}(q,p))] \nonumber \\ = & [\mbox{kinetic energy}(=\frac{p^2}{2m})] + [\mbox{potential energy}(=V(q))] \tag{10.15} \end{align}
Newtonian equation(=Hamilton's canonical equation)

Concerning Hamiltonian ${\cal H}(q,p)$, Hamilton's canonical equation is defined by

\begin{align} \mbox{Hamilton's canonical equation}= \left\{\begin{array}{ll} \frac{dp}{dt}=-\frac{{\cal H}(q,p)}{\partial q} \\ \\ \frac{dq}{dt}=\frac{{\cal H}(q,p)}{\partial p} \end{array}\right. \tag{10.16} \end{align}

And thus, in the case of (10.15}), we get

\begin{align} \mbox{Hamilton's canonical equation} = \left\{\begin{array}{ll} \frac{dp}{dt}=-\frac{{\cal H}(q,p)}{\partial q}=-\frac{\partial V(q,p)}{\partial q} \\ \\ \frac{dq}{dt}=\frac{{\partial \mathcal H}(q,p)}{\partial p}=\frac{p}{m} \end{array}\right. \tag{10.17} \end{align}

which is the same as Newtonian equation. That is,

\begin{align} m \frac{d^2 q}{dt^2}=[\mbox{Mass}] \times [\mbox{Acceleration}]= -\frac{\partial V(q,p)}{\partial q}(=\mbox{Force}) \end{align}

Now, let us describe the above (10.17) in terms of quantum language. For each $t \in T={\mathbb R}$, define the state space $\Omega_t$ by

\begin{align} \Omega_t= \Omega = {\mathbb R}^2 = {\mathbb R}_q\times{\mathbb R}_p {{=}} \{ (q,p)=(\mbox{position , momentum }) \; | \; q,p\in{\mathbb R}\} \tag{10.18} \end{align}

and assume Lebesgue measure $\nu$.

Then, we have the classical basic structure:

\begin{align} [ C_0(\Omega_t ) \subseteq L^\infty(\Omega_t ) \subseteq B(L^2(\Omega_t ) ) ] \qquad (\forall t \in T ={\mathbb R}) \end{align}

The solution of the canonical equation (10.17) is defined by

\begin{align} \Omega_{t_1} \ni \omega_{t_1} \mapsto \phi_{t_1,t_2}(\omega_{t_1}) = \omega_{t_2} \in \Omega_{t_2} \tag{10.19} \end{align}

Since (10.19) determines the deterministic causal map, we have the deterministic sequential causal operator $\{ \Phi_{t_1, t_2} : L^\infty (\Omega_{t_2} ) \to L^\infty (\Omega_{t_1} )$ $\}_{(t_1, t_2 )\in T^2_{\le}}$ such that

\begin{align} [\Phi_{t_1, t_2} ( f_{t_2})](\omega_{t_1})= f_{t_2} (\phi_{t_1,t_2}(\omega_{t_1}) ) \qquad (\forall f_{t_2} \in L^\infty (\Omega_2), \forall \omega_{t_1} \in \Omega_{t_1}, t_1 \le t_2 ) \tag{10.20} \end{align}

10.4.3: Schrödinger equation (quantizing Hamiltonian)

The quantization is the following procedure:

\begin{align} \color{blue}{\mbox{quantization}} \left\{\begin{array}{ll} \mbox{total energy} E & \xrightarrow[\mbox{ quantumization}]{} \frac{\hbar \sqrt{-1} \partial}{\partial t} \\ \\ \mbox{momentum }p & \xrightarrow[\mbox{ quantumization}]{} \frac{\hbar { \partial}}{ \sqrt{-1} \partial q} \\ \\ \mbox{position }q & \xrightarrow[\mbox{ quantumization}]{} q \end{array}\right. \tag{10.21} \end{align} Substituting the quantumization (10.21) to the classical Hamiltonian: \begin{align} E={\mathcal H}(q,p)=\frac{p^2}{2m} + V(q) \end{align} we get \begin{align} { \hbar \sqrt{-1}} \frac{\partial }{\partial t } = {\cal H}(q, \frac{\hbar}{\sqrt{-1}} \frac{ \partial}{ {} \partial q }) = - \frac{\hbar^2}{2 m} \frac{ \partial^2 }{ \partial q^2 } + V(q) \tag{10.22} \end{align}

And therefore, we get the Schrödinger equation :

\begin{align} { \hbar \sqrt{-1}} \frac{\partial u(t,q)}{\partial t } = {\cal H}(q, \frac{\hbar}{\sqrt{-1}} \frac{ \partial}{ {} \partial q })u(t,q) = - \frac{\hbar^2}{2 m} \frac{ \partial^2 }{ \partial q^2 }u(t,q) + V(q) u(t,q) \tag{10.23} \end{align}

Putting $u(t, \cdot )=u_t \in L^2({\mathbb R} )$ $(\forall t \in T={\mathcal R})$ we denote the Schrödinger equation (10.23) by

\begin{align} u_t = \frac{1}{\hbar \sqrt{-1}}{\mathcal H}u_t \end{align}

Solving this formally, we see

\begin{align} u_t = e^{\frac{\mathcal H}{\hbar \sqrt{-1}}t}u_0 \quad \mbox{ (Thus, the state representation is $|u_t \rangle \langle u_t| = |e^{\frac{\mathcal H}{\hbar \sqrt{-1}}t}u_0 \rangle \langle e^{\frac{\mathcal H}{\hbar \sqrt{-1}}t}u_0|$ } ) \tag{10.24} \end{align}

where,$u_0 \in L^2({\mathbb R} )$ is an initial condition.

Now,put Hilbert space $H_t=L^2({\mathbb R})$ $(\forall t \in T={\mathbb R})$, and consider the quantum basic structure:

\begin{align} [ {\mathcal C}(L^2({\mathbb R})) \subseteq B(L^2({\mathbb R})) \subseteq B(L^2({\mathbb R})) ] \end{align}

The dual sequential causal operator $\{ \Phi_{t_1, t_2}^* : {\mathcal Tr}(H_{t_1}) \to {\mathcal Tr}(H_{t_2}) \}_{(t_1, t_2 )\in T^2_{\le}}$ is defined by

\begin{align} \Phi_{t_1, t_2}^* (\rho)=e^{\frac{\mathcal H}{\hbar \sqrt{-1}}(t_2-t_1)} \rho e^{\frac{-{\mathcal H}}{\hbar \sqrt{-1}}(t_2-t_1)} \quad (\forall \rho \in {\mathcal Tr}(H_{t_1}) = (B(H_{t_1}))_* = {\mathcal C}(H_{t_1})^*) \tag{10.25} \end{align}

And therefore, { the sequential causal operator $\{ \Phi_{t_1, t_2} : B(H_{t_2}) \to B(H_{t_1}) \}_{(t_1, t_2 )\in T^2_{\le}}$ is defined by

\begin{align} \Phi_{t_1, t_2} (A)=e^{\frac{-{\mathcal H}}{\hbar \sqrt{-1}}(t_2-t_1)} A e^{\frac{{\mathcal H}}{\hbar \sqrt{-1}}(t_2-t_1)} \quad (\forall A \in B(H_{t_2})) \tag{10.26} \end{align}

} Also, since

\begin{align} \Phi_{t_1, t_2}^*( {\frak S}^p({\mathcal C}(H_{t_1})^* ) ) \subseteq {\frak S}^p({\mathcal C}(H_{t_2})^* ), \end{align}

the sequential causal operator $\{ \Phi_{t_1, t_2} : B(H_{t_2}) \to B(H_{t_1}) \}_{(t_1, t_2 )\in T^2_{\le}}$ is deterministic. Since we deal with the time-invariant system, putting $t=t_2-t_1$, we see that (10.26) is equal to

\begin{align} A_t=\Phi_t(A_0)= e^{\frac{-{\mathcal H}}{\hbar \sqrt{-1}}t} A_0 e^{\frac{{\mathcal H}}{\hbar \sqrt{-1}}t} \tag{10.27} \end{align}

And thus, we get the differential equation:

\begin{align} \frac{dA_t}{dt} &= \frac{-{\mathcal H}}{\hbar \sqrt{-1}} e^{\frac{-{\mathcal H}}{\hbar \sqrt{-1}}t} A_0 e^{\frac{{\mathcal H}}{\hbar \sqrt{-1}}t} + \frac{-{\mathcal H}}{\hbar \sqrt{-1}} e^{\frac{-{\mathcal H}}{\hbar \sqrt{-1}}t} A_0 e^{\frac{{\mathcal H}}{\hbar \sqrt{-1}}t}\frac{{\mathcal H}}{\hbar \sqrt{-1}} \nonumber \\ &= \frac{-{\mathcal H}}{\hbar \sqrt{-1}}A_t + A_t\frac{{\mathcal H}}{\hbar \sqrt{-1}} = \frac{1}{\hbar \sqrt{-1}} \Big( A_t {\mathcal H} - {\mathcal H}A_t \Big) \tag{10.28} \end{align}

which is just Heisenberg's kinetic equation. In quantum lanuage, we say that

 $(\sharp)$ Heisenberg's kinetic equation is formal, and Schrödinger equation is makeshift.
though the two are usually said to be equivalent.