General Question About Path Integral Formulation of QM

[charlemagne94]

Hello all,

I will be learning about the path integral formulation, among other topics, in an advanced QM class during this upcoming semester, so I read ahead a little.

I understand that, essentially, the propagator between two points in spacetime is the normalized sum of exp(i*2pi*S/h) over all spacetime paths connecting the points, where S is the action of the path.

I became curious if there is a way of assigning a probability amplitude to paths in spacetime; in effect, finding an expression that says which paths are more likely than others, similar to the way the Schrödinger equation says which positions of a particle are more likely. My textbook seemed silent on this point, and I think it is a theoretically interesting question.

Thanks!

 

[lugita15]

Quantum mechanics tells you about the probabilities that you will get certain results if you measure a certain observable quantity, or observable for short. Like it will tell you the probability of measuring an electron to be in a spin-up state or a spin-down state when you measure the spin observable. Or it will tell you the probability of a particle to be over here or over there if you measure the position observable. But how can you ever measure the path a particle travels? You can, of course, find out where it is at any given time if you measure the position observable, but presumably you can't find out where it is when you DON'T measure the position observable. So quantum mechanics is silent on what path the particle takes when its position is not being measured, or whether it takes any path at all. When you're not measuring it, the only thing you know about the particle is its quantum state, from which you can find out the probabilities I mentioned before.

One thing you may be interested in is the Quantum Zeno Effect. You see, before you measure the position of particle, it has a certain quantum state, which continuously evolves in accordance with the Schrödinger equation. But once you measure the position, the quantum state becomes what is known as a "position eigenstate", basically a quantum state with definite position. This is what is known as wavefunction collapse. Then after that, the quantum state starts evolving from that. If you again make a position measurement a short time later, you'll probably get a position value that's close to the one you got before, because the state hasn't had time to evolve much. So if you measure the position once a second, it won't change much. If you measure it twice a second it will change even less, etc. So in the limit of measuring it infinitely often, i.e. continually measuring its position, the quantum state will never be able to change and thus the particle will never be able to move! It's like the expression "a watched pot never boils". Here a watched particle never moves.

 

[Chopin]

Since S is real, the exponential e^{i2\pi S/\hbar} is always a simple phase--that is, all paths have the same magnitude of probability. So no path has a greater probability than any other.

However, the phases of different paths will all interfere with each other, and the net result is that most of them will cancel each other out. The only paths which don't get canceled out by something else are those that occur where the action is stationary with respect to path variations. The path at that point is equal to the classical path, which explains why it is the one that actually seems to be taken.

 

[atyy]

Not directly, but there are correspondences between quantum mechanics and statistical mechanics.

http://www.arthurjaffe.com/Assets/pdf/CQFT.pdf (section 4 Quantum Theory as Statistical Physics)

http://arxiv.org/abs/cond-mat/0502068
"Given a generic quantum Hamiltonian construed as an abstract operator defined on some Hilbert space, we prove that there exists a continuous manifold of bases in which the representation of the quantum Hamiltonian is SMF decomposable, i.e., there is a (continuous) manifold of distinct stochastic classical systems related to the same quantum problem."