Physical interpretation of Feynman path integral

Ok very good Hans de Vries , i undrstood you.
But i will posed a new question : so why we are obliged to create a new mathematical measure for the path integrale (physically) please

In other term, Why we don t use Lebesgue measure ?
What is the physical utility of feynmann measure?

robphy, the lectures at http://www.vega.org.uk/series/lectures/feynman/ [Broken] had some problem opening...! it was saying 'Failed to open stream. No Such File or Directory"

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It is interresant , thank you robphy.\
Do you think it is really an open problem>?

So, My question has no exact solution .
I replay my question : First if you should speak mathematiquelly Why we are obliged to create a new measure (wich it Feynman measure), Secondelly What this measure and this Feynman integrals represent physically (Energy , momentum ,.........?????)?
thanks

So no idea?

samalkhaiat
Feynman said:
So no idea?
I am impressed, you created this thread 18 months ago!!:grumpy:
You could have read three or four books on Path Intrgral in less than a year! You could have become an expert on the subject in 18 months!!!:surprised

sam

samalkhaiat It is an open problem!!!!!!!!!

Feynman said:
So, My question has no exact solution .
I replay my question : First if you should speak mathematiquelly Why we are obliged to create a new measure (wich it Feynman measure), Secondelly What this measure and this Feynman integrals represent physically (Energy , momentum ,.........?????)?
thanks
We start with an amplitude and identify it as a sum of paths weighted by a pure phase term

$$\langle x_1|e^{-iHT/\hbar}|x_2\rangle =\sum_{\mbox{all paths}}e^{i\cdot phase}$$

Since there are an infinity of paths between the end points we can convert the summation into a functional integral

$$\sum_{\mbox{all paths}} \longrightarrow \int {\cal D}x$$

To choose the phase term in the exponential we can identify the classical path $$x_{cl}$$ by the method of stationary phase

$$\left.\frac{\delta(phase)}{\delta x}\right|_{x_{cl}}=0$$

But the classical path is one that satisfies the principle of least action

$$\left.\frac{\delta S}{\delta x}\right|_{x_{cl}}=0$$

Where S is the action. Thus it seems sensible to use the action (over h-bar) as the phase term, for which one can gain some confidence in by using this approach in the double slit experiment. Thus our path integral becomes

$$\langle x_1|e^{-iHT/\hbar}|x_2\rangle =\int {\cal D}xe^{i\int d^4x{\cal L}/\hbar}$$

That the equality holds involves more justification that I'm not going to give.

The functional integral formalism can be made use of in calculating correlation functions and the Feynman rules for field theories, where we replace the measure Dx by D(field) and the integration now runs over the field configurations.

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Feynman said:
What is the physical interpretation of Feynman path integral?
In the beginning of the book "QFT in a Nutshell", by A. Zee, there is a description of the physical interpretation. I will give a poor outline of the discussion, but it is no substitute for reading it in Zee's words.

In the double slit experiment, the intensity of the light at a given point on the target screen is a sum of two products. The first product is the intensity of the light that went through one of the slits times the probability that the photon went through that slit. The second product is a similar one using the other slit.

If there were three slits, the the intensity at a given point would be a sum of three products and if you increase the number of slits, you increase the number of products to be summed.

If there is a second screen with slits in it placed between the first screen and the target, then you can calculate the intensity of the light reaching each slit in the second screen and use that to calculate the intensity of the light reaching the target. You can add more screens, and the calculations are of the same type.

Add more and more screens, each one with more and more holes. Add so many holes to each screen, that the screens aren't really there at all. That is, the intensity of light at each point on the target screen is the path integral from the source to that point.

jimmysnyder : thanks for your information,
But your are talking without the relativistic effect to the light on the paths.
So if we take the relativistic effect on the light, how can the geometric form of the paths become?

Feynman said:
So Hans de Vrie, i understood from you that this is a topology argument,
Ok i agrre this idea, but the problem is still here : What is the physical and not the mathematical signification of path integral
Feynman, I am jumping in here, but perhaps this will help.
When an electron is directed at the reverse side of your CRT it is intuitive that the electron takes but a single path, which is a straight line, to the spot targeted by the electronics manipulating the electromagetic elements that aim the electron.
Now simply draw other non-straight lines above and below the single straight line. These squiggly lines are those that the electron may take in arriving at the targeted spot on the screen of the CRT. However, these other lines all conveniently, "cancel" and voila, the electron goes where it was intended by those engineers that designed the CRT circuitry.
As far as I can tell Feynman, the one that is presently deceased, never intended that the electron in the CRT case actually take those squiggly paths, he had something else in mind.
Consider a stern-gerlach transition experiment (R. Feynman, "Lectures on Physics" Vol. III Chapter V.). A spin-1 particle can take one of three possible paths through the inhomogeneous magnetic field of the segment
1. ,
2. "up",
3. "horizontal" or
4. "down".
Now the particle, the atom, can only take one of the three possible trajectories, or paths. However, the magnetic field induces elements of the spin state that are not locally expressed, to expand out, and actually take each of the other two trajectories that the particle does not take (but would have taken if polarized as such). This is not as confusing as it may seem.
Consider the spin-1 particle as being able to generate each of the thee possible states very rapidly. When the particle, the atom, reaches the magnetic field the spin state that is currently generated becomes the polarized, or the observed state. The unexpressed spin states remain nonlocal (meaning that '0' does not mean 'off', like a light switch turns the room light off. 0 means nonlocal, or unobserved, and one cannot assign without more, any physical significance, meaning any observable significance, to the nonlocal state). While the spin generator is generating the spin states, there are always two states that are nonlocal and that simply wait their turn until they are generated. Being generated here means that the nonlocal state is made obervable.
Now, when one sums over the possible paths that the "spin states" may take, and if we assign a probability that the particle will be in one of these states, the probability function will always sum as,
1/3(up) + 1/3(horizontal) + 1/3(down) = 1,
but the particle state, the observed state will be in only one of these states at any time while in thje magnetic field. Before polarization the spin state is a rapidly changing function. The spin state time line history looks like the following (assume 100 = up, 010 = horizontal and 001 = down):
100 010 001 100 010 001 100 010 001 100 010 001 ...
and so on, where the '1' means the current observed state, the '0' the nonlocal (or nonobserved) state. After polarization the partcle spin state time line history looks like the following (assume the 100 state is the polarized state):
100 100 100 100 100 100 100 100 ....
and so on. Summing over the paths, the 1's and 0's, is merely summing the probabilities that the particle must be in one of the allowed number of possible states, which the sum will always equal one.
Ninki

So did the relativistic effect change the form and the physical sens of feynman path integral?