Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The Path Integral

  1. Dec 21, 2014 #1
    What's the main logic of path integral formulation ? (Feyman path integral formulation)
    I mean what's the reason to think this way ?
    Thanks
     
  2. jcsd
  3. Dec 21, 2014 #2

    Nugatory

    User Avatar

    Staff: Mentor

    I've asked Quarlep to narrow his question down a bit.
     
  4. Dec 21, 2014 #3
    Example special relativity based on two postulates or ideas so Path Integration based on what
     
  5. Dec 21, 2014 #4

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    The main goal of path integrals is to compute transition probabilities. If a particle is known at time [itex]t_i[/itex] to be at a location [itex]x_i[/itex], then you want to know what the probability is that the particle will be found at postion [itex]x_f[/itex] at a later time [itex]t_f[/itex]. Quantum mechanically, this is computed in terms of probability amplitudes, and you square the amplitude to get the probability.

    The two facts about this amplitude, [itex]\Psi(x_f, t_f, x_i, t_i)[/itex] that are important for computing path integrals (I'm only doing single-particle nonrelativistic quantum mechanics here, but it should give you the general idea) are:
    1. [itex]\Psi(x_f, t_f, x_i, t_i) = \int d x_1 \Psi(x_f, t_f, x_1, t_1) \Psi(x_1, t_1, x_i, t_i)[/itex], where [itex]t_1[/itex] is any time between [itex]t_i[/itex] and [itex]t_f[/itex]. The idea behind this is that in order to go from [itex]x_i[/itex] at time [itex]t_i[/itex] to location [itex]x_f[/itex] at time [itex]t_f[/itex], the particle must be somewhere, call it [itex]x_1[/itex], at time [itex]t_1[/itex]. The amplitude for the entire path is the amplitude to get from [itex]x_i[/itex] to [itex]x_1[/itex] times the amplitude to get from [itex]x_1[/itex] to [itex]x_f[/itex]. To include all possibilities for the intermediate location, [itex]x_1[/itex] we integrate over all possibilities.
    2. In the limit as [itex]\delta t \rightarrow 0[/itex], the amplitude [itex]\Psi(x_f, t_i + \delta t, x_i, t_i)[/itex] approaches the value [itex]K(\delta t) e^{\frac{i}{\hbar}} L(x, v, t) \delta t[/itex] where [itex]x = \frac{x_i + x_f}{2}[/itex] (the "average" position), and [itex]t = t_i + \frac{\delta t}{2}[/itex] (the "average" time), and [itex]v = \frac{x_f - x_i}{\delta t}[/itex] (the "average" velocity), and where [itex]L(x,v,t)[/itex] is the classical Lagrangian, which for nonrelativistic single-particle motion is given by: [itex]L = \frac{m v^2}{2} - V(x)[/itex], where [itex]m[/itex] is the mass and [itex]V[/itex] is the potential energy.
     
  6. Dec 21, 2014 #5

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    Given those facts, we can write the amplitude [itex]\Psi(x_f, t_f, x_i, t_i)[/itex] in the approximate form: (Picking [itex]N[/itex] to be a large number, and letting [itex]\delta t = \frac{t_f - t_i}{N+1}[/itex])

    [itex]\Psi(x_f, t_f, x_i, t_i) = (K(\delta t))^N \int dx_1 dx_2 ... dx_N e^{\frac{i A}{\hbar}}[/itex]

    where [itex]A = \sum_j L(x_j, v_j, t_j) \delta t[/itex], with [itex]v_j = \frac{x_{j+1}-x_j}{\delta t}[/itex]

    When [itex]N \rightarrow \infty[/itex], [itex]A \rightarrow \int_{t_i}^{t_f} L(x, v, t) dt[/itex], and we get (in some sense) the path integral for the amplitude [itex]\Psi(x_f, t_f, x_i, t_i)[/itex]. I say "in some sense" because it's difficult to give a rigorous meaning to the concept of an integral over an infinite number of variables.
     
  7. Dec 21, 2014 #6

    bhobba

    User Avatar
    Science Advisor
    Gold Member

    Just to expand a bit on the above here is something I wrote on it a while back.

    You start out with <x'|x> then you insert a ton of ∫|xi><xi|dxi = 1 in the middle to get ∫...∫<x'|x1><x1|......|xn><xn|x> dx1.....dxn. Now <xi|xi+1> = ci e^iSi so rearranging you get
    ∫.....∫c1....cn e^ i∑Si.

    Focus in on ∑Si. Define Li = Si/Δti, Δti is the time between the xi along the jagged path they trace out. ∑ Si = ∑Li Δti. As Δti goes to zero the reasonable physical assumption is made that Li is well behaved and goes over to a continuum so you get ∫L dt.

    Now Si depends on xi and Δxi. But for a path Δxi depends on the velocity vi = Δxi/Δti so its very reasonable to assume when it goes to the continuum L is a function of x and the velocity v.

    Its a bit of fun working through the math with Taylor approximations seeing its quite a reasonable process.

    In this way you see the origin of the Lagrangian. And by considering close paths we see most cancel and you are only left with the paths of stationary action.

    Go and get copy of Landau - Mechanics where all of Classical mechanics is derived from this alone - including the existence of mass and that its positive. Strange but true. Actually some other assumptions are also made, but its an interesting exercise first seeing what they are, and secondly their physical significance. Then from that going through Chapter 3 of Ballentine: QM - A Modern Development.

    Thanks
    Bill
     
    Last edited: Dec 21, 2014
  8. Dec 22, 2014 #7
    I understand that every trajectory is a wavefunction.Some of them cancels or some of them makes interferance and it makes the real trajectory.(In double slit experiment every slit)
     
  9. Dec 22, 2014 #8

    atyy

    User Avatar
    Science Advisor

    No, every trajectory is not a wave function. In the full quantum formalism, the trajectory is not a fundamental quantity, only the wave function in an abstract space and the measurements performed on it. However the path integral formalism (for bosons) is a way to calculate using trajectories, and gives us an especially nice intuition for how the classical limit arises, since the classical trajectory minimzes the action and is the "saddle point" approximation of the path integral.
     
  10. Dec 22, 2014 #9
    If theres a good lecture video can you send me I didnt understand it
     
  11. Dec 22, 2014 #10
    Lets suppose I am in London I will go to Berlin but Before go there I must be somewhere between this two points.Lets call it Paris so The probability to go Berlin is London-Paris probability ( wave function squared) multiply by Paris Berlin probability. Is that true ?
     
  12. Dec 22, 2014 #11

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes. That's right. The assumption is that you make a random choice of what city to go to from London, picking Paris with probability [itex]A[/itex] and then you make an independent random choice of where to go from Paris, picking Berlin with probability [itex]B[/itex]. Then the probability that you will go from London to Berlin via Paris is just [itex]A \cdot B[/itex]

    Let's make it a tiny bit more complicated. Suppose that you're flying out of London, and the way that you fly is to go to the airport, and pick a random destination. When you get to that airport, you pick another random destination. Etc. Now, if that's the way you're choosing your trip, then you can ask the probability that you will get to Berlin on your second flight.

    So here's how you figure that out: You start in London. You pick your first destination at random. So you have a list of cities that are reachable from London on nonstop flight. To simplify, let's suppose that from London (L) you can get to Paris (P) or Amsterdam (A). From Paris, you can get to either London or Berlin (B). From Amsterdam, you can get to London or Berlin. Let [itex]Pr_1(X,Y)[/itex] be the probability that you will go from [itex]X[/itex] to [itex]Y[/itex] in one hop.

    Then there are two ways to get from London to Berlin in two hops:
    1. London to Paris to Berlin.
    2. London to Amsterdam to Berlin.
    The probability of going the first route is [itex]Pr_1(L, P) Pr_1(P, B)[/itex]. The probability of going by the second route is [itex]Pr_1(L, A) Pr_1(A, B)[/itex]. To compute the probability of getting to Berlin on the second hop, you just add those probabilities together:

    [itex]Pr_2(L, B) = Pr_1(L, P) Pr_1(P, B) + Pr_1(L, A) Pr_1(A, B)[/itex]
    If we generalize to starting at any city [itex]X[/itex] and getting to any city [itex]Y[/itex], we conclude similarly:

    [itex]Pr_2(X, Y) = \sum_Z Pr_1(X, Z) Pr_1(Z, Y)[/itex]
    Similarly, for [itex]N+1[/itex] hops, instead of just two:

    [itex]Pr_{N+1}(X, Y) = \sum_{Z_1} \sum_{Z_2} ... \sum_{Z_N} Pr_1(X, Z_1) Pr_1(Z_1, Z_2) ... Pr_1(Z_N, Y)[/itex]
    So a trip itenerary is just a list of cities in the order that they are visited: [itex]X, Z_1, Z_2, ..., Z_N, Y[/itex]. The probability of randomly choosing that itenerary is [itex]Pr_1(X, Z_1) Pr_1(Z_1, Z_2) ... Pr_1(Z_N, Y)[/itex]. The total probability of getting from [itex]X[/itex] to [itex]Y[/itex] in [itex]N+1[/itex] steps is [itex]Pr_{N+1}(X,Y)[/itex] given above.

    The idea behind a path integral is the same, except that instead of the fundamental quantity being probabilities, they are instead probability amplitudes. You have to square the total probability amplitude to get the probability.
     
    Last edited: Dec 22, 2014
  13. Dec 22, 2014 #12
    I am understand thank you very much.However I want to ask one little question whats the difference between probability and probability amplitude
     
  14. Dec 22, 2014 #13

    dx

    User Avatar
    Homework Helper
    Gold Member

    The laws for probabilities like

    P(E) = ∑ P(i)

    where the i are microstates that lead to the event E are true only on the assumption that if E happens, then the system must have had one of the microstates which lead to that event. The difference between amplitudes and probabilities are that the analogous rules for amplitudes do not presuppose this assumption. You can think of the amplitudes as an appropriate generalization of the concept of probability which allows a logical representation of the relationships without this assumption being made.
     
  15. Dec 22, 2014 #14
    Ψ(x1,t1) is (London) to Ψ(x2,t2) (Berlin) you said
    Ψ(x1,t1,x2,t2)=∑i Ψ(x1,t1,xi,ti)Ψ(xi,ti,x2,t2) i→∞ isn't it.This is probability amplitude and the square of this is probability.
     
  16. Dec 26, 2014 #15
    Why the square? Because it is 2 events with the same probability happen in a row? So you just multiply one event by itself which is basicly the same as multiplying the 2 events?

    Sorry if it is a stupid question.

    TIA
     
  17. Dec 26, 2014 #16

    Vanadium 50

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    It's not a stupid question, but it's a question covered about two years before what's being discussed here. The short answer is that the probability corresponds to the wavefunction's intensity, and this is a mechanism for calculating the amplitude. The intensity is the square of the amplitude.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: The Path Integral
  1. Path integral (Replies: 3)

  2. Path integrals (Replies: 5)

  3. The Path Integral (Replies: 0)

  4. Path integral (Replies: 2)

Loading...