We illustrate the transition from Liouville theorem in classical mechanics to Ehrenfest theorem in quantum mechanics, and then demonstrate a heuristic argument for Schrödinger equation.
We start from the fact that the evolution of a classical system in the phase space is incompressible, as thus:
Liouville:
\(0\)\(\;=\;\)\(\dfrac{\mathrm{d} \rho}{\mathrm{d} t}\)
\(\;=\;\)\(\dfrac{\partial \rho}{\partial t}\)\(\,+\,\)\( \displaystyle\sum\limits^{N}_{n \;=\; 1}\)\(\left(\vphantom{\dfrac{\partial \rho}{\partial q \vphantom{ q}_{n}} \dfrac{\mathrm{d} q \vphantom{ q}_{n}}{\mathrm{d} t} \,+\, \dfrac{\partial \rho}{\partial p \vphantom{ p}_{n}} \dfrac{\mathrm{d} p \vphantom{ p}_{n}}{\mathrm{d} t}}\right.\)\(\dfrac{\partial \rho}{\partial q \vphantom{ q}_{n}}\)\(\dfrac{\mathrm{d} q \vphantom{ q}_{n}}{\mathrm{d} t}\)\(\,+\,\)\(\dfrac{\partial \rho}{\partial p \vphantom{ p}_{n}}\)\(\dfrac{\mathrm{d} p \vphantom{ p}_{n}}{\mathrm{d} t}\)\(\left.\vphantom{\dfrac{\partial \rho}{\partial q \vphantom{ q}_{n}} \dfrac{\mathrm{d} q \vphantom{ q}_{n}}{\mathrm{d} t} \,+\, \dfrac{\partial \rho}{\partial p \vphantom{ p}_{n}} \dfrac{\mathrm{d} p \vphantom{ p}_{n}}{\mathrm{d} t}}\right)\)
By Hamiltonian equations, we have Liouville theorem:
\(\;=\;\)\(\dfrac{\partial \rho}{\partial t}\)\(\,+\,\)\( \displaystyle\sum\limits^{N}_{n \;=\; 1}\)\(\left(\vphantom{\dfrac{\partial \rho}{\partial q \vphantom{ q}_{n}} \dfrac{\partial H}{\partial p \vphantom{ p}_{n}} \,-\, \dfrac{\partial \rho}{\partial p \vphantom{ p}_{n}} \dfrac{\partial H}{\partial q \vphantom{ q}_{n}}}\right.\)\(\dfrac{\partial \rho}{\partial q \vphantom{ q}_{n}}\)\(\dfrac{\partial H}{\partial p \vphantom{ p}_{n}}\)\(\,-\,\)\(\dfrac{\partial \rho}{\partial p \vphantom{ p}_{n}}\)\(\dfrac{\partial H}{\partial q \vphantom{ q}_{n}}\)\(\left.\vphantom{\dfrac{\partial \rho}{\partial q \vphantom{ q}_{n}} \dfrac{\partial H}{\partial p \vphantom{ p}_{n}} \,-\, \dfrac{\partial \rho}{\partial p \vphantom{ p}_{n}} \dfrac{\partial H}{\partial q \vphantom{ q}_{n}}}\right)\)\(;\,\)
At this point, we have to imagine what this relation looks like in the matrix mechanics. Let \(\mathbf{R}\) be the density matrix, and \(\mathbf{H}\) the Hamiltonian matrix, and \(\mathbf{Q}\) and \(\mathbf{P}\) the position and momentum matrices. We have two things to scale here, \(\eth\)\(\rho\) and \(\eth\)\(H\)\(\,/\,\)\(\eth\)\(q \vphantom{ q}_{n}\)\(\eth\)\(p \vphantom{ p}_{n}\)\(\eth\)\(t\). We may scale both of their maximal values to \(1\), so \(\rho\) and \(H\) are interpreted as unitary evolutions; and we take that \(\eth\)\(q \vphantom{ q}_{n}\)\(\eth\)\(p \vphantom{ p}_{n}\)\(\;=\;\)\(\mathring{\imath}\)\(\hslash\) , the smallest unit of angular momentum, where the phase \(\mathring{\imath}\) is just a gauge.
This way, \(\left(\vphantom{\partial \rho \,/\, \partial q \vphantom{ q}_{n}}\right.\)\(\partial\)\(\rho\)\(\,/\,\)\(\partial\)\(q \vphantom{ q}_{n}\)\(\left.\vphantom{\partial \rho \,/\, \partial q \vphantom{ q}_{n}}\right)\)\(\eth\)\(q \vphantom{ q}_{n}\) can be considered as the component \(n\) of the density matrix under the \(q \vphantom{ q}_{n}\) basis, and \(\left(\vphantom{\partial H \,/\, \partial p \vphantom{ p}_{n}}\right.\)\(\partial\)\(H\)\(\,/\,\)\(\partial\)\(p \vphantom{ p}_{n}\)\(\left.\vphantom{\partial H \,/\, \partial p \vphantom{ p}_{n}}\right)\)\(\eth\)\(p \vphantom{ p}_{n}\), the component \(n\) of the Hamiltonian matrix under the \(p \vphantom{ p}_{n}\) basis. Unfortunately, we no longer have \(q \vphantom{ q}_{n}\)\(,\,\)\(p \vphantom{ p}_{n}\) as a compatible set of coordinates; rather, they are off by a change of basis. However, we can imagine \(\mathbf{R}\) to be written as a matrix from the position to the momentum. Hence from Liouville, according to the heuristic above, we have the relation:
\(0\)\(\;=\;\)\(\dfrac{\partial}{\partial t}\)\(\mathbf{R}\)\(\,+\,\)\(\dfrac{1}{\mathring{\imath} \hslash}\)\(\left(\vphantom{\mathbf{R} \mathbf{H} \,-\, \mathbf{H} \mathbf{R}}\right.\)\(\mathbf{R}\)\(\mathbf{H}\)\(\,-\,\)\(\mathbf{H}\)\(\mathbf{R}\)\(\left.\vphantom{\mathbf{R} \mathbf{H} \,-\, \mathbf{H} \mathbf{R}}\right)\)\(;\,\)
Switching to wave mechanics in Schrödinger picture, we let \(\rho\) be the density operator, and \(H\) the Hamiltonian.
\(0\)\(\;=\;\)\(\dfrac{\partial}{\partial t}\)\(\rho\)\(\,+\,\)\(\dfrac{1}{\mathring{\imath} \hslash}\)\(\left(\vphantom{\rho H \,-\, H \rho}\right.\)\(\rho\)\(H\)\(\,-\,\)\(H\)\(\rho\)\(\left.\vphantom{\rho H \,-\, H \rho}\right)\)\(;\,\)
Multiplying wave function \(\psi\) on both sides, and recalling \(\rho\)\(\psi\)\(\;=\;\)\(\psi\), we get:
\(0\)\(\;=\;\)\(\dfrac{\partial}{\partial t}\)\(\psi\)\(\,-\,\)\(\rho\)\(\dfrac{\partial}{\partial t}\)\(\psi\)\(\,+\,\)\(\dfrac{1}{\mathring{\imath} \hslash}\)\(\left(\vphantom{\rho H \psi \,-\, H \psi}\right.\)\(\rho\)\(H\)\(\psi\)\(\,-\,\)\(H\)\(\psi\)\(\left.\vphantom{\rho H \psi \,-\, H \psi}\right)\)\(;\,\)
Schrödinger:
\(\mathring{\imath}\)\(\hslash\)\(\dfrac{\partial}{\partial t}\)\(\psi\)\(\;=\;\)\(H\)\(\psi\)\(;\,\)
And we have recovered the Schrödinger equation.
❧ August 25, 2021