Link between hydrodynamics and quantum mechanics

Klas Modin

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-
vt+vv=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
ρt+div(ρv)=0\displaystyle \frac{\partial \rho}{\partial t} + \mathrm{div}(\rho \mathbf v) = 0
P(\rho) = e'(\rho)\rho^2
P(ρ)=e(ρ)ρ2P(\rho) = e'(\rho)\rho^2
\mathbf x
x\mathbf x
\mathbf v(\mathbf x)
v(x)\mathbf v(\mathbf x)
\displaystyle \frac{1}{\rho}\nabla P(\rho)
1ρP(ρ)\displaystyle \frac{1}{\rho}\nabla P(\rho)
\displaystyle \frac{1}{\rho}\nabla p
1ρp\displaystyle \frac{1}{\rho}\nabla p
\displaystyle \mathrm{div}(\mathbf v) = 0
div(v)=0\displaystyle \mathrm{div}(\mathbf v) = 0

Pressure function:

\Omega
Ω\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
E(v,ρ)=12Ωv2ρ dx+Ωe(ρ)ρ dx\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(ρ)=e(ρ)ρ+e(ρ)w(\rho) = e'(\rho)\rho+e(\rho)

Thermodynamic work:

\displaystyle \nabla w(\rho)
w(ρ)\displaystyle \nabla w(\rho)

Potential function:

V = V(\mathbf x)
V=V(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
iψt=[22mΔ+V]ψ=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
ψ:ΩC\psi: \Omega \to \mathbb C

Potential function:

V: \Omega \to \mathbb R
V:ΩRV: \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
E(ψ)=ψ,H^ψL2=Ω(2ψ22m+Vψ2)dx\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
H^=22mΔ+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
1=ψ,ψL2=Ωψ2dx\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

Madelung transform

???

Special solution

Review of analytical mechanics

(Classical mechanics in the framework of differential geometry)

\displaystyle\dot{\mathbf{q}} =\frac{\partial H}{\partial \textbf p}
q˙=Hp\displaystyle\dot{\mathbf{q}} =\frac{\partial H}{\partial \textbf p}
\displaystyle\dot{\mathbf{p}} =-\frac{\partial H}{\partial \mathbf q}
p˙=Hq\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 \}
P(Ω)={ρC(Ω)Ωρ dx=1,  ρ>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)
M=P(Ω)    TP(Ω)=P(Ω)×C0(Ω)\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
H(ρ,S)=Ω(S22m+c4log(ρ)2+V)ρ dx\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}
ρ˙=δHδS\displaystyle\dot\rho = \frac{\delta H}{\delta S}
\displaystyle\dot S = -\frac{\delta H}{\delta \rho}
S˙=δHδρ\displaystyle\dot S = -\frac{\delta H}{\delta \rho}

\(\Rightarrow\)

\displaystyle\dot\rho + \mathrm{div}(\rho\nabla S/m ) = 0
ρ˙+div(ρS/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
S˙+S22mcΔρρ+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
S˙+1mS2ScΔρρ+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
v=1mS\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 -
v˙+vv=V\displaystyle\dot\mathbf v + \mathbf v\cdot\nabla \mathbf v = -\nabla V -
\displaystyle\dot\rho + \mathrm{div}(\rho\mathbf v ) = 0
ρ˙+div(ρv)=0\displaystyle\dot\rho + \mathrm{div}(\rho\mathbf v ) = 0
\displaystyle\mathbf v = \frac{1}{m}\nabla S
v=1mS\displaystyle\mathbf v = \frac{1}{m}\nabla S

Thermodynamic work:

\displaystyle w(\rho) = -\frac{c\Delta\sqrt{\rho}}{m\sqrt{\rho}}
w(ρ)=cΔρmρ\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
e(ρ)=c4mlogρ2\displaystyle e(\rho) = \frac{c}{4m}\lvert \nabla \log \rho \rvert^2
\surd
\surd
\nabla w(\rho)
w(ρ)\nabla w(\rho)

Connecting the dots

Euler equations

Schrödinger equation

Hamilton's equations on probabilities

Madelung transform

Special solution

Madelung transform

Erwin Madelung

1881 - 1972

T^*P(\Omega) \to L^2(\Omega,\mathbb C)
TP(Ω)L2(Ω,C)T^*P(\Omega) \to L^2(\Omega,\mathbb C)
(\rho,S) \mapsto \psi = \sqrt{\rho}\mathrm e^{\mathrm i S/\hbar}
(ρ,S)ψ=ρeiS/(\rho,S) \mapsto \psi = \sqrt{\rho}\mathrm e^{\mathrm i S/\hbar}
\lvert\psi\rvert^2 = \rho
ψ2=ρ\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
ψ˙=(12ρ˙ρ+iS˙)ψ\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
ψ=(ρρ+iS)ψ\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
Δψ=(ρρ+iS)2ψ+(ρρ+iΔS)ψ\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
=(Δρρ12S2+idiv(ρS)ρ)ψ\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
iψ˙=(S˙+i2ρ˙ρ)ψ\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
(22mΔρρ+S22m+Vi2mdiv(ρS)ρ)ψ\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
iψ˙=[22mΔ+V]ψ=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 =
22mΔψ+Vψ=\displaystyle \frac{-\hbar^2 }{2m}\Delta\psi + V\psi =

Schrödinger equation

\displaystyle\dot\rho + \mathrm{div}(\rho\nabla S/m ) = 0
ρ˙+div(ρS/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
S˙+S22m22mΔρρ+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}
c=22m\displaystyle c= \frac{\hbar^2}{2m}
\surd
\surd

Connecting the dots

Euler equations

Schrödinger equation

Hamilton's equations on probabilities

Madelung transform

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

Further reading about the geometry of the Madelung transform:

Geometric Hydrodynamics via Madelung Transform, PNAS 2018

The fluid-quantum connection

By Klas Modin

The fluid-quantum connection

A popular 20 min seminar about the connection between hydrodynamics and quantum mechanics.

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