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-
∂t∂v+v⋅∇v=−∇V−
incompressible
\displaystyle \frac{\partial \rho}{\partial t} + \mathrm{div}(\rho \mathbf v) = 0
∂t∂ρ+div(ρv)=0
P(\rho) = e'(\rho)\rho^2
P(ρ)=e′(ρ)ρ2
\mathbf x
x
\mathbf v(\mathbf x)
v(x)
\displaystyle \frac{1}{\rho}\nabla P(\rho)
ρ1∇P(ρ)
\displaystyle \frac{1}{\rho}\nabla p
ρ1∇p
\displaystyle \mathrm{div}(\mathbf v) = 0
div(v)=0
Pressure function:
\Omega
Ω
\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)=21∫Ω∣v∣2ρdx+∫Ωe(ρ)ρdx
Total energy:
w(\rho) = e'(\rho)\rho+e(\rho)
w(ρ)=e′(ρ)ρ+e(ρ)
Thermodynamic work:
\displaystyle \nabla w(\rho)
∇w(ρ)
Potential function:
V = V(\mathbf x)
V=V(x)
Player two: Schrödinger equation
\displaystyle\mathrm{i}\hbar\frac{\partial \psi}{\partial t} =\left[ \frac{-\hbar^2 }{2m}\Delta + V\right] \psi = 0
iℏ∂t∂ψ=[2m−ℏ2Δ+V]ψ=0
Wave function:
\psi: \Omega \to \mathbb C
ψ:Ω→C
Potential function:
V: \Omega \to \mathbb R
V:Ω→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^ψ⟩L2=∫Ω(2mℏ2∣∇ψ∣2+V∣ψ∣2)dx
\displaystyle \hat H = -\frac{\hbar^2}{2m}\Delta + V
H^=−2mℏ2Δ+V
Total energy:
Total probability:
\displaystyle 1=\langle \psi, \psi \rangle_{L^2} = \int_\Omega \lvert \psi\rvert^2 d\mathbf x
1=⟨ψ,ψ⟩L2=∫Ω∣ψ∣2dx
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˙=∂p∂H
\displaystyle\dot{\mathbf{p}} =-\frac{\partial H}{\partial \mathbf q}
p˙=−∂q∂H
Properties
- Conservation of total energy H
- Conservation of phase space volume (symplecticity)
Evolves on phase space T∗M≃R2n
Examples:
Celestial mechanics
M=R3N
Rigid body
M=SO(3)
Hamiltonian dynamics on probability densities
Space of probability densities on Ω :
\displaystyle P(\Omega) = \{ \rho \in C^\infty(\Omega)\mid \int_\Omega \rho\, d\mathbf x = 1, \; \rho > 0 \}
P(Ω)={ρ∈C∞(Ω)∣∫Ωρdx=1,ρ>0}
Now take as configuration manifold
\displaystyle M = P(\Omega) \;\Rightarrow\; T^* P(\Omega) = P(\Omega) \times C^\infty_0(\Omega)
M=P(Ω)⇒T∗P(Ω)=P(Ω)×C0∞(Ω)
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(2m∣∇S∣2+4c∣∇log(ρ)∣2+V)ρdx
Hamilton's equations on T∗P(Ω)
\displaystyle\dot\rho = \frac{\delta H}{\delta S}
ρ˙=δSδH
\displaystyle\dot S = -\frac{\delta H}{\delta \rho}
S˙=−δρδH
⇒
\displaystyle\dot\rho + \mathrm{div}(\rho\nabla S/m ) = 0
ρ˙+div(ρ∇S/m)=0
\displaystyle\dot S + \frac{\lvert\nabla S\rvert^2}{2m} - c \frac{\Delta\sqrt{\rho}}{\sqrt{\rho}} + V = 0
S˙+2m∣∇S∣2−cρΔρ+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˙+m1∇S⋅∇2S−c∇ρΔρ+∇V=0
Now set:
\displaystyle\mathbf v = \frac{1}{m}\nabla S
v=m1∇S
Which equation does v fulfill?
We recover the Euler equations!
\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\mathbf v = \frac{1}{m}\nabla S
v=m1∇S
Thermodynamic work:
\displaystyle w(\rho) = -\frac{c\Delta\sqrt{\rho}}{m\sqrt{\rho}}
w(ρ)=−mρcΔρ
Internal energy:
\displaystyle e(\rho) = \frac{c}{4m}\lvert \nabla \log \rho \rvert^2
e(ρ)=4mc∣∇logρ∣2
\surd
√
\nabla w(\rho)
∇w(ρ)
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)
T∗P(Ω)→L2(Ω,C)
(\rho,S) \mapsto \psi = \sqrt{\rho}\mathrm e^{\mathrm i S/\hbar}
(ρ,S)↦ψ=ρeiS/ℏ
\lvert\psi\rvert^2 = \rho
∣ψ∣2=ρ
Notice that:
Key calculations for ψ= ρei S/ℏ
\displaystyle \dot\psi = \left(\frac{1}{2}\frac{\dot\rho}{\rho} + \frac{\mathrm i}{\hbar} \dot S\right)\psi
ψ˙=(21ρρ˙+ℏiS˙)ψ
\displaystyle \nabla\psi = \left(\frac{\nabla\sqrt\rho}{\sqrt\rho} + \frac{\mathrm i}{\hbar} \nabla S\right)\psi
∇ψ=(ρ∇ρ+ℏi∇S)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
Δψ=(ρ∇ρ+ℏi∇S)2ψ+(∇⋅ρ∇ρ+ℏiΔS)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
=(ρΔρ−ℏ21∣∇S∣2+ℏiρdiv(ρ∇S))ψ
\displaystyle \mathrm{i}\hbar\dot\psi = \left(- \dot S + \frac{\mathrm{i\hbar}}{2}\frac{\dot\rho}{\rho} \right)\psi
iℏψ˙=(−S˙+2iℏρρ˙)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
(−2mℏ2ρΔρ+2m∣∇S∣2+V−2miℏρdiv(ρ∇S))ψ
\displaystyle\mathrm{i}\hbar\dot\psi =\left[ \frac{-\hbar^2 }{2m}\Delta + V\right] \psi = 0
iℏψ˙=[2m−ℏ2Δ+V]ψ=0
Left-hand side
Right-hand side
\displaystyle \frac{-\hbar^2 }{2m}\Delta\psi + V\psi =
2m−ℏ2Δψ+Vψ=
Schrödinger equation
\displaystyle\dot\rho + \mathrm{div}(\rho\nabla S/m ) = 0
ρ˙+div(ρ∇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˙+2m∣∇S∣2−2mℏ2ρΔρ+V=0
\displaystyle c= \frac{\hbar^2}{2m}
c=2mℏ2
\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:
Link between hydrodynamics and quantum mechanics Klas Modin