## Objective: connection between Euler and Schrödinger equations

Leonhard Euler

1707 - 1783

Erwin Schrödinger

1887 - 1961

## Why?

• Known since birth of QM, but often forgotten in modern textbooks

• Links to many active fields of mathematical research:
- Optimal transport
- Information theory
- PDE theory
- Stochastic differential equations
- etc.

• Illustrates "power" of geometry + physics

# The Players

## Player one: Euler fluid equations

\displaystyle\frac{\partial \mathbf v}{\partial t} + \mathbf v\cdot\nabla \mathbf v = -\nabla V-
$\displaystyle\frac{\partial \mathbf v}{\partial t} + \mathbf v\cdot\nabla \mathbf v = -\nabla V-$

incompressible

\displaystyle \frac{\partial \rho}{\partial t} + \mathrm{div}(\rho \mathbf v) = 0
$\displaystyle \frac{\partial \rho}{\partial t} + \mathrm{div}(\rho \mathbf v) = 0$
P(\rho) = e'(\rho)\rho^2
$P(\rho) = e'(\rho)\rho^2$
\mathbf x
$\mathbf x$
\mathbf v(\mathbf x)
$\mathbf v(\mathbf x)$
\displaystyle \frac{1}{\rho}\nabla P(\rho)
$\displaystyle \frac{1}{\rho}\nabla P(\rho)$
\displaystyle \frac{1}{\rho}\nabla p
$\displaystyle \frac{1}{\rho}\nabla p$
\displaystyle \mathrm{div}(\mathbf v) = 0
$\displaystyle \mathrm{div}(\mathbf v) = 0$

Pressure function:

\Omega
$\Omega$
\displaystyle E(\mathbf v,\rho) = \frac{1}{2}\int_\Omega |\mathbf v|^2 \rho \, d\mathbf x + \int_\Omega e(\rho)\rho\,d \mathbf x
$\displaystyle E(\mathbf v,\rho) = \frac{1}{2}\int_\Omega |\mathbf v|^2 \rho \, d\mathbf x + \int_\Omega e(\rho)\rho\,d \mathbf x$

Total energy:

w(\rho) = e'(\rho)\rho+e(\rho)
$w(\rho) = e'(\rho)\rho+e(\rho)$

Thermodynamic work:

\displaystyle \nabla w(\rho)
$\displaystyle \nabla w(\rho)$

Potential function:

V = V(\mathbf x)
$V = V(\mathbf x)$

## Player two: Schrödinger equation

\displaystyle\mathrm{i}\hbar\frac{\partial \psi}{\partial t} =\left[ \frac{-\hbar^2 }{2m}\Delta + V\right] \psi = 0
$\displaystyle\mathrm{i}\hbar\frac{\partial \psi}{\partial t} =\left[ \frac{-\hbar^2 }{2m}\Delta + V\right] \psi = 0$

Wave function:

\psi: \Omega \to \mathbb C
$\psi: \Omega \to \mathbb C$

Potential function:

V: \Omega \to \mathbb R
$V: \Omega \to \mathbb R$

Conservation laws

Hamiltonian operator:

\displaystyle E(\psi) = \langle \psi,\hat H \psi \rangle_{L^2} = \int_\Omega (\frac{\hbar^2 \lvert\nabla\psi \rvert^2}{2m} + V\lvert\psi\rvert^2)d\mathbf x
$\displaystyle E(\psi) = \langle \psi,\hat H \psi \rangle_{L^2} = \int_\Omega (\frac{\hbar^2 \lvert\nabla\psi \rvert^2}{2m} + V\lvert\psi\rvert^2)d\mathbf x$
\displaystyle \hat H = -\frac{\hbar^2}{2m}\Delta + V
$\displaystyle \hat H = -\frac{\hbar^2}{2m}\Delta + V$

Total energy:

Total probability:

\displaystyle 1=\langle \psi, \psi \rangle_{L^2} = \int_\Omega \lvert \psi\rvert^2 d\mathbf x
$\displaystyle 1=\langle \psi, \psi \rangle_{L^2} = \int_\Omega \lvert \psi\rvert^2 d\mathbf x$

## Connecting the dots

Euler equations

Schrödinger equation

Hamilton's equations on probabilities

???

Special solution

## Review of analytical mechanics

(Classical mechanics in the framework of differential geometry)

\displaystyle\dot{\mathbf{q}} =\frac{\partial H}{\partial \textbf p}
$\displaystyle\dot{\mathbf{q}} =\frac{\partial H}{\partial \textbf p}$
\displaystyle\dot{\mathbf{p}} =-\frac{\partial H}{\partial \mathbf q}
$\displaystyle\dot{\mathbf{p}} =-\frac{\partial H}{\partial \mathbf q}$

Properties

• Conservation of total energy $$H$$
• Conservation of phase space volume (symplecticity)

Evolves on phase space $$T^*M \simeq \mathbb{R}^{2n}$$

Examples:

Celestial mechanics

$$M=\mathbb{R}^{3N}$$

Rigid body

$$M = SO(3)$$

## Hamiltonian dynamics on probability densities

Space of probability densities on $$\Omega$$ :

\displaystyle P(\Omega) = \{ \rho \in C^\infty(\Omega)\mid \int_\Omega \rho\, d\mathbf x = 1, \; \rho > 0 \}
$\displaystyle P(\Omega) = \{ \rho \in C^\infty(\Omega)\mid \int_\Omega \rho\, d\mathbf x = 1, \; \rho > 0 \}$

Now take as configuration manifold

\displaystyle M = P(\Omega) \;\Rightarrow\; T^* P(\Omega) = P(\Omega) \times C^\infty_0(\Omega)
$\displaystyle M = P(\Omega) \;\Rightarrow\; T^* P(\Omega) = P(\Omega) \times C^\infty_0(\Omega)$

Take as Hamiltonian

\displaystyle H(\rho,S) = \int_\Omega \left(\frac{\lvert\nabla S\rvert^2}{2m} + \frac{c}{4}\lvert\nabla\log(\rho)\rvert^2 + V\right)\rho\, d\mathbf x
$\displaystyle H(\rho,S) = \int_\Omega \left(\frac{\lvert\nabla S\rvert^2}{2m} + \frac{c}{4}\lvert\nabla\log(\rho)\rvert^2 + V\right)\rho\, d\mathbf x$

## Hamilton's equations on $$T^*P(\Omega)$$

\displaystyle\dot\rho = \frac{\delta H}{\delta S}
$\displaystyle\dot\rho = \frac{\delta H}{\delta S}$
\displaystyle\dot S = -\frac{\delta H}{\delta \rho}
$\displaystyle\dot S = -\frac{\delta H}{\delta \rho}$

$$\Rightarrow$$

\displaystyle\dot\rho + \mathrm{div}(\rho\nabla S/m ) = 0
$\displaystyle\dot\rho + \mathrm{div}(\rho\nabla S/m ) = 0$
\displaystyle\dot S + \frac{\lvert\nabla S\rvert^2}{2m} - c \frac{\Delta\sqrt{\rho}}{\sqrt{\rho}} + V = 0
$\displaystyle\dot S + \frac{\lvert\nabla S\rvert^2}{2m} - c \frac{\Delta\sqrt{\rho}}{\sqrt{\rho}} + V = 0$

Apply gradient to second equation, yields:

\displaystyle\nabla\dot S + \frac{1}{m}\nabla S\cdot\nabla^2 S - c \nabla\frac{\Delta\sqrt{\rho}}{\sqrt{\rho}} + \nabla V = 0
$\displaystyle\nabla\dot S + \frac{1}{m}\nabla S\cdot\nabla^2 S - c \nabla\frac{\Delta\sqrt{\rho}}{\sqrt{\rho}} + \nabla V = 0$

Now set:

\displaystyle\mathbf v = \frac{1}{m}\nabla S
$\displaystyle\mathbf v = \frac{1}{m}\nabla S$

Which equation does $$\mathbf v$$ fulfill?

## We recover the Euler equations!

\displaystyle\dot\mathbf v + \mathbf v\cdot\nabla \mathbf v = -\nabla V -
$\displaystyle\dot\mathbf v + \mathbf v\cdot\nabla \mathbf v = -\nabla V -$
\displaystyle\dot\rho + \mathrm{div}(\rho\mathbf v ) = 0
$\displaystyle\dot\rho + \mathrm{div}(\rho\mathbf v ) = 0$
\displaystyle\mathbf v = \frac{1}{m}\nabla S
$\displaystyle\mathbf v = \frac{1}{m}\nabla S$

Thermodynamic work:

\displaystyle w(\rho) = -\frac{c\Delta\sqrt{\rho}}{m\sqrt{\rho}}
$\displaystyle w(\rho) = -\frac{c\Delta\sqrt{\rho}}{m\sqrt{\rho}}$

Internal energy:

\displaystyle e(\rho) = \frac{c}{4m}\lvert \nabla \log \rho \rvert^2
$\displaystyle e(\rho) = \frac{c}{4m}\lvert \nabla \log \rho \rvert^2$
\surd
$\surd$
\nabla w(\rho)
$\nabla w(\rho)$

## Connecting the dots

Euler equations

Schrödinger equation

Hamilton's equations on probabilities

Special solution

1881 - 1972

T^*P(\Omega) \to L^2(\Omega,\mathbb C)
$T^*P(\Omega) \to L^2(\Omega,\mathbb C)$
(\rho,S) \mapsto \psi = \sqrt{\rho}\mathrm e^{\mathrm i S/\hbar}
$(\rho,S) \mapsto \psi = \sqrt{\rho}\mathrm e^{\mathrm i S/\hbar}$
\lvert\psi\rvert^2 = \rho
$\lvert\psi\rvert^2 = \rho$

Notice that:

## Key calculations for $$\psi = \sqrt{\rho}\mathrm e^{\mathrm i S/\hbar}$$

\displaystyle \dot\psi = \left(\frac{1}{2}\frac{\dot\rho}{\rho} + \frac{\mathrm i}{\hbar} \dot S\right)\psi
$\displaystyle \dot\psi = \left(\frac{1}{2}\frac{\dot\rho}{\rho} + \frac{\mathrm i}{\hbar} \dot S\right)\psi$
\displaystyle \nabla\psi = \left(\frac{\nabla\sqrt\rho}{\sqrt\rho} + \frac{\mathrm i}{\hbar} \nabla S\right)\psi
$\displaystyle \nabla\psi = \left(\frac{\nabla\sqrt\rho}{\sqrt\rho} + \frac{\mathrm i}{\hbar} \nabla S\right)\psi$
\displaystyle \Delta\psi = \left(\frac{\nabla\sqrt\rho}{\sqrt\rho} + \frac{\mathrm i}{\hbar} \nabla S\right)^2\psi + \left(\nabla\cdot\frac{\nabla\sqrt\rho}{\sqrt\rho} + \frac{\mathrm i}{\hbar} \Delta S\right)\psi
$\displaystyle \Delta\psi = \left(\frac{\nabla\sqrt\rho}{\sqrt\rho} + \frac{\mathrm i}{\hbar} \nabla S\right)^2\psi + \left(\nabla\cdot\frac{\nabla\sqrt\rho}{\sqrt\rho} + \frac{\mathrm i}{\hbar} \Delta S\right)\psi$
\displaystyle = \left(\frac{\Delta\sqrt\rho}{\sqrt\rho} - \frac{1}{\hbar^2}\lvert\nabla S\rvert^2 + \frac{\mathrm i }{\hbar} \frac{\mathrm{div}(\rho\nabla S)}{\rho} \right)\psi
$\displaystyle = \left(\frac{\Delta\sqrt\rho}{\sqrt\rho} - \frac{1}{\hbar^2}\lvert\nabla S\rvert^2 + \frac{\mathrm i }{\hbar} \frac{\mathrm{div}(\rho\nabla S)}{\rho} \right)\psi$
\displaystyle \mathrm{i}\hbar\dot\psi = \left(- \dot S + \frac{\mathrm{i\hbar}}{2}\frac{\dot\rho}{\rho} \right)\psi
$\displaystyle \mathrm{i}\hbar\dot\psi = \left(- \dot S + \frac{\mathrm{i\hbar}}{2}\frac{\dot\rho}{\rho} \right)\psi$
\displaystyle \left( - \frac{\hbar^2}{2m}\frac{\Delta\sqrt\rho}{\sqrt\rho} + \frac{\lvert\nabla S\rvert^2}{2m} + V - \frac{\mathrm i \hbar}{2m} \frac{\mathrm{div}(\rho\nabla S)}{\rho} \right)\psi
$\displaystyle \left( - \frac{\hbar^2}{2m}\frac{\Delta\sqrt\rho}{\sqrt\rho} + \frac{\lvert\nabla S\rvert^2}{2m} + V - \frac{\mathrm i \hbar}{2m} \frac{\mathrm{div}(\rho\nabla S)}{\rho} \right)\psi$
\displaystyle\mathrm{i}\hbar\dot\psi =\left[ \frac{-\hbar^2 }{2m}\Delta + V\right] \psi = 0
$\displaystyle\mathrm{i}\hbar\dot\psi =\left[ \frac{-\hbar^2 }{2m}\Delta + V\right] \psi = 0$

Left-hand side

Right-hand side

\displaystyle \frac{-\hbar^2 }{2m}\Delta\psi + V\psi =
$\displaystyle \frac{-\hbar^2 }{2m}\Delta\psi + V\psi =$

Schrödinger equation

\displaystyle\dot\rho + \mathrm{div}(\rho\nabla S/m ) = 0
$\displaystyle\dot\rho + \mathrm{div}(\rho\nabla S/m ) = 0$
\displaystyle\dot S + \frac{\lvert\nabla S\rvert^2}{2m} - \frac{\hbar^2}{2m} \frac{\Delta\sqrt{\rho}}{\sqrt{\rho}} + V = 0
$\displaystyle\dot S + \frac{\lvert\nabla S\rvert^2}{2m} - \frac{\hbar^2}{2m} \frac{\Delta\sqrt{\rho}}{\sqrt{\rho}} + V = 0$
\displaystyle c= \frac{\hbar^2}{2m}
$\displaystyle c= \frac{\hbar^2}{2m}$
\surd
$\surd$

## Connecting the dots

Euler equations

Schrödinger equation

Hamilton's equations on probabilities

Special solution

## Summary

• Link between compressible Euler equations (with "strange" internal energy) and Schrödinger equation

• Schrödinger equation is a classical Hamiltonian system on the infinite-dimensional space of probability densities