- #76

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But i will posed a new question : so why we are obliged to create a new mathematical measure for the path integrale (physically) please

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- #76

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But i will posed a new question : so why we are obliged to create a new mathematical measure for the path integrale (physically) please

- #77

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In other term, Why we don t use Lebesgue measure ?

What is the physical utility of feynmann measure?

What is the physical utility of feynmann measure?

- #78

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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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- #79

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http://www.vega.org.uk/video/series/5 or

http://www.vega.org.uk/video/subseries/8

...or else send them an email.

- #80

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It is interresant , thank you robphy.\

Do you think it is really an open problem>?

Do you think it is really an open problem>?

- #81

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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

- #82

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So no idea?

help me please

help me please

- #83

samalkhaiat

Science Advisor

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I am impressed, you created this thread 18 months ago!!:grumpy:Feynman said:So no idea?

help me please

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

- #84

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

- #85

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We start with an amplitude and identify it as a sum of paths weighted by a pure phase termFeynman said:

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

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

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

[tex]\sum_{\mbox{all paths}} \longrightarrow \int {\cal D}x[/tex]

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

[tex]\left.\frac{\delta(phase)}{\delta x}\right|_{x_{cl}}=0[/tex]

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

[tex]\left.\frac{\delta S}{\delta x}\right|_{x_{cl}}=0[/tex]

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

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

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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- #86

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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.Feynman said:What is the physical interpretation of Feynman path integral?

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.

- #87

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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?

- #88

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Feynman, I am jumping in here, but perhaps this will help.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

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

- ,
- "up",
- "horizontal" or
- "down".

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 "

1/3(up) + 1/3(horizontal) + 1/3(down) = 1,

but the

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

Ninki

- #89

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

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