Koopman–von Neumann classical mechanics

In classical statistical mechanics, the probability density (with respect to Liouville measure) obeys the Liouville equation^[12][13] $${\displaystyle i{\frac {\partial }{\partial t}}\rho (x,p,t)={\hat {L}}\rho (x,p,t)}$$ with the self-adjoint Liouvillian $${\displaystyle {\hat {L}}=-i{\frac {\partial H(x,p)}{\partial p}}{\frac {\partial }{\partial x}}+i{\frac {\partial H(x,p)}{\partial x}}{\frac {\partial }{\partial p}},}$$ where ${\displaystyle H(x,p)}$ denotes the classical Hamiltonian (i.e. the Liouvillian is ${\displaystyle i}$ times the Hamiltonian vector field considered as a first order differential operator). The same dynamical equation is postulated for the KvN wavefunction $${\displaystyle i{\frac {\partial }{\partial t}}\psi (x,p,t)={\hat {L}}\psi (x,p,t),}$$ thus $${\displaystyle {\frac {\partial }{\partial t}}\psi (x,p,t)=\left[-{\frac {\partial H(x,p)}{\partial p}}{\frac {\partial }{\partial x}}+{\frac {\partial H(x,p)}{\partial x}}{\frac {\partial }{\partial p}}\right]\psi (x,p,t),}$$ and for its complex conjugate $${\displaystyle {\frac {\partial }{\partial t}}\psi ^{*}(x,p,t)=\left[-{\frac {\partial H(x,p)}{\partial p}}{\frac {\partial }{\partial x}}+{\frac {\partial H(x,p)}{\partial x}}{\frac {\partial }{\partial p}}\right]\psi ^{*}(x,p,t).}$$ From $${\displaystyle \rho (x,p,t)=\psi ^{*}(x,p,t)\psi (x,p,t)}$$ follows using the product rule that $${\displaystyle {\frac {\partial }{\partial t}}\rho (x,p,t)=\left[-{\frac {\partial H(x,p)}{\partial p}}{\frac {\partial }{\partial x}}+{\frac {\partial H(x,p)}{\partial x}}{\frac {\partial }{\partial p}}\right]\rho (x,p,t)}$$ which proves that probability density dynamics can be recovered from the KvN wavefunction.

Remark

The last step of this derivation relies on the classical Liouville operator containing only first-order derivatives in the coordinate and momentum; this is not the case in quantum mechanics where the Schrödinger equation contains second-order derivatives.

^{[12] [13]}

The above axioms (i) to (iv), with the inner product written in the bra–ket notation, are

1. ${\displaystyle \langle \psi (t)|\psi (t)\rangle =1}$,
2. The expectation value of an observable ${\displaystyle {\hat {A}}}$ at time ${\displaystyle t}$ is ${\displaystyle \langle A(t)\rangle =\langle \Psi (t)|{\hat {A}}|\Psi (t)\rangle .}$
3. The probability that a measurement of an observable ${\displaystyle {\hat {A}}}$ at time ${\displaystyle t}$ yields ${\displaystyle A}$ is ${\displaystyle \left|\langle A|\Psi (t)\rangle \right|^{2}}$, where ${\displaystyle {\hat {A}}|A\rangle =A|A\rangle }$. (This axiom is an analogue of the Born rule in quantum mechanics.^[15])
4. (see Tensor product of Hilbert spaces).

We begin from the following equations for expectation values of the coordinate x and momentum p

${\displaystyle m{\frac {d}{dt}}\langle x\rangle =\langle p\rangle ,\qquad {\frac {d}{dt}}\langle p\rangle =\langle -U'(x)\rangle ,}$

aka, Newton's laws of motion averaged over ensemble. With the help of the operator axioms, they can be rewritten as

${\displaystyle {\begin{aligned}m{\frac {d}{dt}}\langle \Psi (t)|{\hat {x}}|\Psi (t)\rangle &=\langle \Psi (t)|{\hat {p}}|\Psi (t)\rangle ,\\{\frac {d}{dt}}\langle \Psi (t)|{\hat {p}}|\Psi (t)\rangle &=\langle \Psi (t)|-U'({\hat {x}})|\Psi (t)\rangle .\end{aligned}}}$

Notice a close resemblance with Ehrenfest theorems in quantum mechanics. Applications of the product rule leads to

${\displaystyle {\begin{aligned}\langle d\Psi /dt|{\hat {x}}|\Psi \rangle +\langle \Psi |{\hat {x}}|d\Psi /dt\rangle &=\langle \Psi |{\hat {p}}/m|\Psi \rangle ,\\\langle d\Psi /dt|{\hat {p}}|\Psi \rangle +\langle \Psi |{\hat {p}}|d\Psi /dt\rangle &=\langle \Psi |-U'({\hat {x}})|\Psi \rangle ,\end{aligned}}}$

into which we substitute a consequence of Stone's theorem ${\displaystyle i|d\Psi (t)/dt\rangle ={\hat {L}}|\Psi (t)\rangle }$ and obtain

${\displaystyle {\begin{aligned}im\langle \Psi (t)|[{\hat {L}},{\hat {x}}]|\Psi (t)\rangle &=\langle \Psi (t)|{\hat {p}}|\Psi (t)\rangle ,\\i\langle \Psi (t)|[{\hat {L}},{\hat {p}}]|\Psi (t)\rangle &=-\langle \Psi (t)|U'({\hat {x}})|\Psi (t)\rangle .\end{aligned}}}$

Since these identities must be valid for any initial state, the averaging can be dropped and the system of commutator equations for the unknown ${\displaystyle {\hat {L}}}$ is derived

${\displaystyle im[{\hat {L}},{\hat {x}}]={\hat {p}},\qquad i[{\hat {L}},{\hat {p}}]=-U'({\hat {x}}).}$

commutator eqs for L

Assume that the coordinate and momentum commute ${\displaystyle [{\hat {x}},{\hat {p}}]=0}$. This assumption physically means that the classical particle's coordinate and momentum can be measured simultaneously, implying absence of the uncertainty principle.

The solution ${\displaystyle {\hat {L}}}$ cannot be simply of the form ${\displaystyle {\hat {L}}=L({\hat {x}},{\hat {p}})}$ because it would imply the contractions ${\displaystyle im[L({\hat {x}},{\hat {p}}),{\hat {x}}]=0={\hat {p}}}$ and ${\displaystyle i[L({\hat {x}},{\hat {p}}),{\hat {p}}]=0=-U'({\hat {x}})}$. Therefore, we must utilize additional operators ${\displaystyle {\hat {\lambda }}_{x}}$ and ${\displaystyle {\hat {\lambda }}_{p}}$ obeying

${\displaystyle [{\hat {x}},{\hat {\lambda }}_{x}]=[{\hat {p}},{\hat {\lambda }}_{p}]=i,\quad [{\hat {x}},{\hat {p}}]=[{\hat {x}},{\hat {\lambda }}_{p}]=[{\hat {p}},{\hat {\lambda }}_{x}]=[{\hat {\lambda }}_{x},{\hat {\lambda }}_{p}]=0.}$

KvN algebra

The need to employ these auxiliary operators arises because all classical observables commute. Now we seek ${\displaystyle {\hat {L}}}$ in the form ${\displaystyle {\hat {L}}=L({\hat {x}},{\hat {\lambda }}_{x},{\hat {p}},{\hat {\lambda }}_{p})}$. Utilizing KvN algebra, the commutator eqs for L can be converted into the following differential equations:^[14][16]

${\displaystyle mL'_{\lambda _{x}}(x,\lambda _{x},p,\lambda _{p})=p,\qquad L'_{\lambda _{p}}(x,\lambda _{x},p,\lambda _{p})=-U'(x).}$

Whence, we conclude that the classical KvN wave function ${\displaystyle |\Psi (t)\rangle }$ evolves according to the Schrödinger-like equation of motion

${\displaystyle i{\frac {d}{dt}}|\Psi (t)\rangle ={\hat {L}}|\Psi (t)\rangle ,\qquad {\hat {L}}={\frac {\hat {p}}{m}}{\hat {\lambda }}_{x}-U'({\hat {x}}){\hat {\lambda }}_{p}.}$

KvN dynamical eq

Let us explicitly show that KvN dynamical eq is equivalent to the classical Liouville mechanics.

Since ${\displaystyle {\hat {x}}}$ and ${\displaystyle {\hat {p}}}$ commute, they share the common eigenvectors

${\displaystyle {\hat {x}}|x,p\rangle =x|x,p\rangle ,\quad {\hat {p}}|x,p\rangle =p|x,p\rangle ,\quad A({\hat {x}},{\hat {p}})|x,p\rangle =A(x,p)|x,p\rangle ,}$

xp eigenvec

with the resolution of the identity ${\textstyle 1=\int dx\,dp\,|x,p\rangle \langle x,p|.}$ Then, one obtains from equation (KvN algebra) $${\displaystyle \langle x,p|{\hat {\lambda }}_{x}|\Psi \rangle =-i{\frac {\partial }{\partial x}}\langle x,p|\Psi \rangle ,\qquad \langle x,p|{\hat {\lambda }}_{p}|\Psi \rangle =-i{\frac {\partial }{\partial p}}\langle x,p|\Psi \rangle .}$$ Projecting equation (KvN dynamical eq) onto ${\displaystyle \langle x,p|}$, we get the equation of motion for the KvN wave function in the xp-representation

${\displaystyle \left[{\frac {\partial }{\partial t}}+{\frac {p}{m}}{\frac {\partial }{\partial x}}-U'(x){\frac {\partial }{\partial p}}\right]\langle x,p|\Psi (t)\rangle =0.}$

KvN dynamical eq in xp

The quantity ${\displaystyle \langle x,\,p|\Psi (t)\rangle }$ is the probability amplitude for a classical particle to be at point ${\displaystyle x}$ with momentum ${\displaystyle p}$ at time ${\displaystyle t}$. According to the axioms above, the probability density is given by ${\displaystyle \rho (x,p;t)=\left|\langle x,p|\Psi (t)\rangle \right|^{2}}$. Utilizing the identity $${\displaystyle {\frac {\partial }{\partial t}}\rho (x,p;t)=\langle \Psi (t)|x,p\rangle {\frac {\partial }{\partial t}}\langle x,p|\Psi (t)\rangle +\langle x,p|\Psi (t)\rangle \left({\frac {\partial }{\partial t}}\langle x,p|\Psi (t)\rangle \right)^{*}}$$ as well as (KvN dynamical eq in xp), we recover the classical Liouville equation

${\displaystyle \left[{\frac {\partial }{\partial t}}+{\frac {p}{m}}{\frac {\partial }{\partial x}}-U'(x){\frac {\partial }{\partial p}}\right]\rho (x,p;t)=0.}$

Liouville eq

Moreover, according to the operator axioms and (xp eigenvec), $${\displaystyle {\begin{aligned}\langle A\rangle &=\langle \Psi (t)|A({\hat {x}},{\hat {p}})|\Psi (t)\rangle =\int dx\,dp\,\langle \Psi (t)|x,p\rangle A(x,p)\langle x,p|\Psi (t)\rangle \\&=\int dx\,dp\,A(x,p)\langle \Psi (t)|x,p\rangle \langle x,p|\Psi (t)\rangle =\int dx\,dp\,A(x,p)\rho (x,p;t).\end{aligned}}}$$ Therefore, the rule for calculating averages of observable ${\displaystyle A(x,p)}$ in classical statistical mechanics has been recovered from the operator axioms with the additional assumption ${\displaystyle [{\hat {x}},{\hat {p}}]=0}$. As a result, the phase of a classical wave function does not contribute to observable averages. Contrary to quantum mechanics, the phase of a KvN wave function is physically irrelevant. Hence, nonexistence of the double-slit experiment^[13][17][18] as well as Aharonov–Bohm effect^[19] is established in the KvN mechanics.

Projecting KvN dynamical eq onto the common eigenvector of the operators ${\displaystyle {\hat {x}}}$ and ${\displaystyle {\hat {\lambda }}_{p}}$ (i.e., ${\displaystyle x\lambda _{p}}$-representation), one obtains classical mechanics in the doubled configuration space,^[20] whose generalization leads ^{[20] [21] [22] [23] [24]} to the phase space formulation of quantum mechanics.

As in the derivation of classical mechanics, we begin from the following equations for averages of coordinate x and momentum p

${\displaystyle m{\frac {d}{dt}}\langle x\rangle =\langle p\rangle ,\qquad {\frac {d}{dt}}\langle p\rangle =\langle -U'(x)\rangle .}$

With the help of the operator axioms, they can be rewritten as

${\displaystyle {\begin{aligned}m{\frac {d}{dt}}\langle \Psi (t)|{\hat {x}}|\Psi (t)\rangle &=\langle \Psi (t)|{\hat {p}}|\Psi (t)\rangle ,\\{\frac {d}{dt}}\langle \Psi (t)|{\hat {p}}|\Psi (t)\rangle &=\langle \Psi (t)|-U'({\hat {x}})|\Psi (t)\rangle .\end{aligned}}}$

These are the Ehrenfest theorems in quantum mechanics. Applications of the product rule leads to

${\displaystyle {\begin{aligned}\langle d\Psi /dt|{\hat {x}}|\Psi \rangle +\langle \Psi |{\hat {x}}|d\Psi /dt\rangle &=\langle \Psi |{\hat {p}}/m|\Psi \rangle ,\\\langle d\Psi /dt|{\hat {p}}|\Psi \rangle +\langle \Psi |{\hat {p}}|d\Psi /dt\rangle &=\langle \Psi |-U'({\hat {x}})|\Psi \rangle ,\end{aligned}}}$

into which we substitute a consequence of Stone's theorem

${\displaystyle i\hbar |d\Psi (t)/dt\rangle ={\hat {H}}|\Psi (t)\rangle ,}$

where ${\displaystyle \hbar }$ was introduced as a normalization constant to balance dimensionality. Since these identities must be valid for any initial state, the averaging can be dropped and the system of commutator equations for the unknown quantum generator of motion ${\displaystyle {\hat {H}}}$ are derived

${\displaystyle im[{\hat {H}},{\hat {x}}]=\hbar {\hat {p}},\qquad i[{\hat {H}},{\hat {p}}]=-\hbar U'({\hat {x}}).}$

Contrary to the case of classical mechanics, we assume that observables of the coordinate and momentum obey the canonical commutation relation ${\displaystyle [{\hat {x}},{\hat {p}}]=i\hbar }$. Setting ${\displaystyle {\hat {H}}=H({\hat {x}},{\hat {p}})}$, the commutator equations can be converted into the differential equations ^[14][16]

${\displaystyle mH'_{p}(x,p)=p,\qquad H'_{x}(x,p)=U'(x),}$

whose solution is the familiar quantum Hamiltonian

${\displaystyle {\hat {H}}={\frac {{\hat {p}}^{2}}{2m}}+U({\hat {x}}).}$

Whence, the Schrödinger equation was derived from the Ehrenfest theorems by assuming the canonical commutation relation between the coordinate and momentum. This derivation as well as the derivation of classical KvN mechanics shows that the difference between quantum and classical mechanics essentially boils down to the value of the commutator ${\displaystyle [{\hat {x}},{\hat {p}}]}$.