Quantum Non-Locality and Wave Function Collapse

By George Shiber Tuesday, November 8, 2016Friday, October 26, 2018 In Mathematical Physics, Mathematics, Philosophy, Philosophy Of Physics, Philosophy Of Science, Physics, Time Heisenberg Uncertainty Principle, Quantum Entanglement

The foundational crises caused by the wave function collapse need no introduction. Here I will address the obvious need for a collapse theory that accounts simultaneously for quantum entanglement, time-symmetry and eliminability of any incorporation of a ‘decohering-observer’. The reason is clear, and not simply due to the universal wave functional collapse or the meta-entanglement Nakajima–Zwanzig-Wigner regress paradoxes. Here’s how to appreciate the acuteness of the problem. Look at the following collapse equation:

$\begin{array}{*{20}{l}}{d\left| {{\psi _t}} \right\rangle = \left[ { - \frac{i}{\hbar }} \right.Hdt + \sqrt \lambda \int {{d^3}} x\left( {N(\bar x) - {{\left\langle {N(\bar x)} \right\rangle }_t}} \right)}\\{ \cdot d{W_t}(\bar x) - \frac{\lambda }{2}\int {{d^3}} x\left. {{{\left( {N(\bar x) - {{\left\langle {N(\bar x)} \right\rangle }_t}} \right)}^2}dt} \right]\left| {{\psi _t}} \right\rangle }\end{array}$

ensuring the positivity of the reduced density matrix:

$\forall t\left\langle {\left. \psi \right|{{\hat \rho }_S}(t)\left| \psi \right.} \right\rangle \ge 0$

entails that the Lindblad quantum-jump equation:

$\begin{array}{l}{\rm{d}}\hat \rho _S^C = - i\left[ {{{\hat H}_S},\hat \rho _S^C} \right]{\rm{d}}t - \\\frac{1}{2}\sum\limits_\mu {{\kappa _\mu }} \left[ {{{\hat L}_\mu },\left[ {{{\hat L}_\mu },\hat \rho _S^C} \right]} \right]{\rm{d}}t + \\\sum\limits_\mu {\sqrt {{\kappa _\mu }} } W\left[ {{{\hat L}_\mu }} \right]\hat \rho _S^C{\rm{d}}{W_\mu }\end{array}$

is solvable iff ${L_k}$ commutes with the position-operator, with:

$\begin{array}{l}W\left[ {\hat L} \right]\hat \rho \equiv \hat L\hat \rho + \hat \rho {{\hat L}^\dagger } - \\\hat \rho {\rm{Tr}}\left\{ {\hat L\hat \rho + \hat \rho {{\hat L}^\dagger }} \right\}\end{array}$

and ${\rm{d}}{W_\mu }$ the Weiner-quantum-increments.

However:

$\tilde U: = - \frac{\lambda }{2}\int {{d^3}} x{\left( {N(\bar x) - {{\left\langle {N(\bar x)} \right\rangle }_t}} \right)^2}dt(...)$

commutes with the energy operator. By the Heisenberg energy-time uncertainty principle, the collapse equation cannot be integrated to get the collapse-localization double-integral in 4-D:

${\int {\int {\left| {\Psi _{{t_4}}^I\left( {x,X} \right) + \Psi _{{t_4}}^{II}\left( {x,X} \right)} \right|} } ^2}dxdX = 1$

Now applying the wave-particle duality to:

$\int { - \frac{\lambda }{2}} {d^3}xd{W_t}(\bar x) + \int {\frac{i}{\hbar }Hdt + \sqrt \lambda \int {{d^3}} xd{W_t}(\bar x)}$

we face the problem of decoherence-induced space-time localization, since the time-local non-Markovian master equation:

$\frac{{\rm{d}}}{{{\rm{d}}t}}{\hat \rho _S}(t) = \hat K(t){\hat \rho _S}(t)$

cannot be solvable consistent with the Lindblad quantum-jump equation also being solvable.

And here’s the paradox: the wave-particle duality is equivalent to the generalized Heisenberg uncertainty principle, and that would imply that the above collapse-localization double-integral in 4-D cannot have a solution and by entanglement entropy relation, that would contradict the Heisenberg time-energy uncertainty principle, thus the wave-particle duality ceases to make sense in the time-symmetry group representation of the corresponding Hilbert space of the quantum system.

Clearly the crux of any solution is deriving an ‘observer-independent’ collapse equation. Let me address some aspects of this here. I will simplify notation without any loss of generality.

Take an arbitrary system consisting of a quantum particle-‘object’ and the ‘active’-particles constitutive of the measuring instrument interacting with this particle-‘object’. And one-dimensional motion is only considered for ease of notation. Let $x$ refer to the coordinate of the particle-‘object’, and $X$ the generalized coordinates of the ‘active’ particles collection and $\iota$ the number of the generalized coordinate system and:

$dX = \prod\limits_{\iota = 1}^N {d{X^\iota }}$

the volume element of the configuration space of the ‘active’-particles.

Hence, the evolution-equation is given by the integral wave equation:

${\Psi _{{t_2}}}\left( {{x_2},{X_2}} \right) = \int {{K_{{t_2},{t_1}}}} \left( {{x_2}{X_2},{x_1},{X_2}} \right){\Psi _{{t_1}}}\left( {{x_1}{X_1}} \right)d{x_1}d{X_1}$

$\left\{ {\begin{array}{*{20}{c}}{{\Psi _{{t_2}}}\left( {{x_2},{X_2}} \right)}\\{{\Psi _{{t_1}}}\left( {{x_1},{X_1}} \right)}\end{array}} \right.$

are the wave functions of the particle at time ${t_2}$ and the initial time ${t_1}$ < ${t_2}$ , respectively. Now, let $\Gamma$ denote a virtual path in the configuration space of the system. So, the kernel of the integral evolution operator is given as:

${K_{{t_2},{t_1}}}\left( {{x_2},{x_1}} \right) = \int {\exp \frac{i}{\hbar }} {S_{1,2}}\left[ \Gamma \right]\left[ {d\Gamma } \right]$

Combining the last two expressions generates a transformation of the wave function, and it is clear that the Schrödinger equation cannot capture this possibility given wave-functional invariance for infinitesimal time-intervals.

To understand the cause of such jumps, take the action functional:

$\int {\exp \frac{i}{\hbar }} {S_{1,2}}\left[ \Gamma \right]\left[ {d\Gamma } \right] = {K_{{t_2},{t_1}}}\left( {{x_2},{x_1}} \right)$

and let ${U_A}(X)$ refer to the interaction potential energy of the active ‘particles’ with the particle-‘object’. So the action functional is:

${S_{1,2}}\left[ \Gamma \right] = \int\limits_{{t_1}}^{{t_2}} {\left( {{T^\Gamma }\left( {\dot X,\dot x} \right) - {U_q}\left( {x,X} \right)h\left( {t - {t_3}} \right) - U\left( {x,t} \right)} \right)} dt$

where ${T^\Gamma }\left( {\dot X,\dot x} \right)$ is the kinetic energy of all the particles of the system and $h\left( {t - {t_3}} \right)$ the Heaviside function that reflects the initiation of a macroscopic registering process in the measuring instrument at time ${t_3}$ .

That is what is meant by the potential energy for the paths of this new set to have a macroscopic value. Assume now that the registering process takes place in a small local spatial domain as this occurs when the particle coordinate is measured. Let ${\Omega ^I}$ be the region of the configuration space corresponding to this domain and ${\Omega ^{II}}$ the rest of the configuration space. The initiation of the registering process in this domain is expressed mathematically by the Lindblad quantum-jump equation corresponding ‘active’ particles in:

Now, consider the system just after this jump. Then from the above, we can derive:

$\begin{array}{l}K_{{t_4},{t_3}}^I\left( {{x_4},{X_4},{x_3},{X_3}} \right) = \\\exp \left( { - \frac{i}{\hbar }{U_a}\varepsilon } \right)\left( {\int_{{\Omega ^I}} {\exp \frac{i}{\hbar }{s_{3,4}}\left[ \gamma \right]\left[ {d\gamma } \right]} } \right)\delta \left( {{X_4} - {X_3}} \right)\end{array}$

for ${\Omega ^I}$ , and:

$\begin{array}{l}K_{{t_4},{t_3}}^{II}\left( {{x_4},{X_4},{X_4},{x_3},{X_3}} \right) = \\\int_{{\Omega ^{II}}} {\exp \frac{i}{\hbar }} {s_{3,4}}\left[ \gamma \right]\left( {\int \begin{array}{l}\exp \frac{i}{\hbar }{S_{3,4}}\\\left[ \Gamma \right]\left[ {d\Gamma } \right]\end{array} } \right)\left[ {d\Gamma } \right]\end{array}$

for ${\Omega ^{II}}$ . So, the quantum path integral can be reduced to the ‘real’ form of the path integral having the Wiener measure via a representation of the time variable in the complex form $t = \tau \exp \left( { - i\varphi } \right)$ and we take the transition amplitude for:

$\left\{ {\begin{array}{*{20}{c}}{\varphi = - \frac{\pi }{2}}\\{t = - i\tau }\end{array}} \right.$

Thus, we have time-reversal symmetry: however, this fails for the Lindblad quantum-jump due to its essential irreversibility.

So, on the one hand, for ${\Omega ^I}$ , we have:

$\begin{array}{l}K_{{\tau _4},{\tau _3}}^I\left( {{x_4},{X_4},{x_3},{X_3}} \right) = \\\exp \frac{1}{\hbar }{U_a}\varepsilon \left( {\int_{{\Omega ^I}} {{e^{\frac{1}{\hbar }{s_{3,4}}\left[ \gamma \right]\left[ {d\gamma } \right]}}} } \right)\delta \left( {{X_4} - {X_3}} \right)\end{array}$

while for ${\Omega ^{II}}$ , we have:

$\begin{array}{l}K_{{t_4},{t_3}}^{II}\left( {{x_4},{X_4},{x_3},{X_3}} \right) = \\\int_{{\Omega ^{II}}} {{e^{\left( { - \frac{1}{\hbar }{s_{3,4}}\left[ \gamma \right]} \right)}}} \left( {\int {{e^{\left( { - \frac{1}{\hbar }{S_{3,4}}\left[ \Gamma \right]} \right)}}} \left[ {d\Gamma } \right]} \right)\left[ {d\Gamma } \right]\end{array}$

Hence, the normalization of the wave function expressing the quantum particle integrity must be conserved in the localization process.

Therefore after the collapse we get

$\int {\int {\left| {\Psi _{{t_4}}^I} \right.} } \left( {x,X} \right) + \Psi _{{t_4}}^{II}{\left. {\left( {x,X} \right)} \right|^2}dxdX = 1$

in imaginary time:

$\int {\int {\left| {\Psi _{{\tau _4}}^I} \right.} } \left( {x,X} \right) + \Psi _{{\tau _4}}^{II}{\left. {\left( {x,X} \right)} \right|^2}d{x_4}d{X_4} = 1$

with:

$\begin{array}{l}\Psi _{{\tau _4}}^I\left( {{x_4},{X_4}} \right) = \\\int {\int {K_{{\tau _4},{\tau _3}}^I\left( {{x_4},{X_4},{x_3},{X_3}} \right)} } {\Psi _{\tau 3}}\left( {{x_3},{X_3}} \right)d{x_3}d{X_3}\end{array}$

and:

$\begin{array}{l}\Psi _{{\tau _4}}^{II}\left( {{x_4},{X_4}} \right) = \\\int {\int {K_{{\tau _4},{\tau _3}}^{II}\left( {{x_4},{X_4},{x_3},{X_3}} \right)} } {\Psi _{\tau 3}}\left( {{x_3},{X_3}} \right)d{x_3}d{X_3}\end{array}$

Since $K_{{\tau _4},{\tau _3}}^I$ strictly exceeds the amplitude $K_{{\tau _4},{\tau _3}}^{II}$ due to the macro-classical order of the magnitude of the exponent in this term, we can derive:

${\int {\int {\left| {\Psi _{{\tau _4}}^I\left( {{x_4},{X_4}} \right)} \right|} } ^2}d{x_4}d{X_4} \approx 1$

The corresponding volume of space occupied by the ‘active’ particles taking part in the registering process is infinitesimal and thus the interaction radius of the quantum particle with the ‘active’ particle is infinitesimal as well!

Let ${Y^I}$ be the space coordinate of this volume. And here is the key:

It then follows that all paths $\gamma$ at time ${t_4}$ pass through this point, and at time ${\tau _4}$ we have a localized state:

${\Psi _{{\tau _4}}}\left( {{x_4},{X_4}} \right) = \delta \left( {{x_4} - {Y^I}\left( {{\tau _4}} \right)} \right)\Phi \left( {{X_4}} \right)$

and the wave function of the active particles after the collapse has the form:

$\begin{array}{l}{\Phi _{{\tau _4}}}\left( {{X_4}} \right) = \delta \left( {X_4^k - X_I^k\left( {{\tau _4}} \right)} \right)\\{\Phi _{{\tau _4}}}\left( {X_4^k,...,X_4^{k - 1},X_4^{k + 1},...,X_4^N} \right)\end{array}$

and for ‘real’ time we have:

$\begin{array}{c}{\Psi _{{t_4}}}\left( {{x_4}.{X_4}} \right) = \delta \left( {{x_4} - {Y^I}({t_4})} \right)\\{\Phi _{{t_4}}}\left( {{X_4}} \right)\end{array}$

Thus, after the collapse, the entangled wave function of the total system transforms into the product of the wave function of the measuring apparatus and the wave function particle-‘object’ with the delta function form, thus maintaining time-symmetry. At time ${t_4}$ , the interaction of the particle-‘object’ and the measuring apparatus instantaneously vanishes everywhere with the exception of the domain ${\Omega ^I}$ , and the particle-‘object’ becomes a macro-classical projection interpretable as a single point. Moreover, there can be no violation of relativistic requirements since collapse phenomena are not spatio-temporal dynamical processes governed by local derivatives.

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