jaketodd said:
Doesn't it all come down to the math in the end? If the math says a particle takes infinitely many paths, then that's what is happening if it's correct regardless of the shrink wrap you put around it. However, a good question would be: "Is there a different way to mathematically reproduce the same results (and therefore the same accuracy) without the infinitely many paths being built into the math?" Or, am I wrong and that interpretation is not built into the math?
No - this is not true. Let me show how you to achieve exactly the same result using only one path, rather than an infinite number of them (this is an edited version of a post I made in a thread ages ago):
The main point about Feynman's theory is to calculate the propagator (essentially, the mathematical object that enables you to calculate the wave function at some space point and time in the future, given the wave function now).
Now if you subscribe to the view that electrons have trajectories (i.e. the de Broglie-Bohm view) and you use the obvious trajectory implied by the quantum formalism, then you can compute the propagator using only that single 'quantum' path rather than Feynman's infinite number of trajectories. Look at the interesting similarity between the following formulae for the propagators:
DE BROGLIE-BOHM
K^Q({\bf x}_1,t_1;{\bf x}_0,t_0) = \frac{1}{J(t)^ {\frac{1}{2}} } \exp\left[{\frac{i}{\hbar}}}\int_{t_0}^{t_1}L(t)\;dt\right]
FEYNMAN
K^F({\bf x}_1,t_1;{\bf x}_0,t_0) = N \sum_{all paths} \exp\left[{\frac{i}{\hbar}}\int_{t_0}^{t_1}L_{cl}(t)\;dt \right]
In the Feynman case the propagator linking two spacetime points is calculated by linearly superposing amplitudes e^{iS/\hbar} obtained by integrating the
classical Lagrangian L_{cl}(t)={\frac{1}{2}}mv^2-V associated with the infinite number of all possible paths connecting the points.
In the de Broglie-Bohm approach, you achieve the same effect by integrating the 'quantum' Lagrangian L(t)={\frac{1}{2}}mv^2-(V+Q) along precisely
one path (the one the electron actually follows). Here Q is the potential associated with the 'quantum force' (the particle is being pushed around or 'guided' by a physical field represented mathematically by the wave function).
So it's all a question of knowing the correct path/trajectory. Not a lot of people know this.. Note that this elevates the de Broglie-Bohm theory from being an 'interpretation' to a mathematical reformulation of quantum mechanics equivalent in status to Feynman's.
A couple of additional technical points:
(1) In the Feynman path integral method, computing the propagator by summing over all possible paths is only half of it. The Feynman propagator K^F is a many-to-many mapping i.e. all points are linked by all possible paths. So the full \Psi({\bf x}_1,t_1) is found from Huygen's principle by summing the contributions coming from all possible start points - you multiply the amplitude at {\bf x}_0,t_0 by the transition amplitude K^F for 'hopping' to {\bf x}_1,t_1. Then you sum (integrate) over all x_0:
\Psi({\bf x}_1,t_1)=\int K^F({\bf x}_1,t_1; {\bf x}_0,t_0) \Psi({\bf x}_0,t_0)\;d x_0
In the deBB approach you achieve the same end as in the path integral method - the computation of \Psi given the initial value - in a quite different and conceptually simpler manner with two spacetime points connected by at most a single path. The two steps in Feynman's approach (propagator then Huygens) are thus condensed into one (propagator). \Psi is generated from its initial form by a single-valued continuum of trajectories.
(2) You might think you need the wave function over all space to compute the propagator in the deBB case but this is not true. You just need the second derivative of the wave function - or more accurately the second derivative of its amplitude (for the quantum potential Q) and of its phase (for the \nabla\cdot{\bf v} in the Jacobian J) at the points along the track. In a practical numerical method, these can be calculated by sending a particle down the trajectory {\bf v} = \nabla S (i.e. following the streamlines of the quantum probability current) and then evaluating the required derivatives numerically using finite differencing or whatever. There is a whole community of physical chemists (believe it or not) who do precisely this to solve chemistry problems.So to conclude, Feynman's paths are mathematical tools for computing the evolution of \Psi, while (if it is the case that particles actually exist) one among the de Broglie-Bohm paths is the actual motion of the particle as deduced from the equations of QM, which exists in addition to the wave field \Psi. Keep finally in mind that path integrals are not exclusive to QM; one can write any linear field equation (e.g. Maxwell) in terms of path integrals.
OK - the last time I posted this observation it apparently caused one of the homework helpers to have a complete mental break down. Many apologies if this happens again!
