physical interpretation of Feynman path integral

[Feynman]

Hi
What is the physical interpretation of Feynman path integral?
Thanks :smile:

[marcus]

Hi
What is the physical interpretation of Feynman path integral?
Thanks :smile:

here is one view: it does not have a physical interpretation
because it is a way of calculating a sometimes very close approximation

but nature does not know or care how we calculate
she does something, we dont know exactly how
and we have this way of calculating that gets the right answer

it's like virtual particles---maybe they dont exist in nature but they are good to calculate with

it's like feynmann diagrams: maybe nature doesnt know about them---she doesnt need to because she has her own way of doing things without perturbation series----maybe nothing in nature corresponds to feynmann diagrams---they are useful, tho, for organizing the calculation of a perturbation expansion.

that is just one viewpoint :smile:
hopefully someone else will provide a contrasting one

BTW Feynmann, if you havent already I invite you to have a look in this Quantum Physics forum
at the electroweak conversation between turin and zephram.
if zephram answers on this thread it will probably be interesting

[Feynman]

But I know that the path integral have a physical interpretation but i don't know

[meldi]

hello marcus!
I'm interested by the subject but I didn't find the post of turin and zephram. Can you give me a link please?

[JohnDubYa]

Read Feynman and Hibbs' book "Path Integrals." The introduction should satisfy your curiosity.

[Feynman]

So i don't have the Feynman Hibbs book
So can u tell me about the physical interpretation of the path integrals :@

[Olias]

It relational to the many 'sum-over-history' interpretation also by Feynman.
For instance:The fundamental question in the path integral (PI) formulation of quantum mechanics is:If the particle is at a position q at time t = 0, what is the probability amplitude that it will be at some other position q0 at a later time t = T?

This I took from this good paper:http://arxiv.org/PS_cache/quant-ph/pdf/0004/0004090.pdf

If you read through it, you can form your own viewpoint, interpretation is open to a vast number of probibilities, every path has many (integral-able)(connecting) histories!

PS the original paper by Feynman was rejected!

[somy]

Dear Olias;
I see the link. can you tell me how did dirac observed the action plays a central role???
(also the difference between lagrangian and hamiltonian interpretation)
Tanks in advanced.

[robphy]

You might want to start off reading some of these introductory articles by EF Taylor
http://www.eftaylor.com/leastaction.html
and watching some of Feynman's lectures in Auckland (which became "QED")
http://www.vega.org.uk/series/lectures/feynman/

[Feynman]

But I think taht the Feynman path integral is a link between the quantum physics and the quatum field theory, but how????

[marcus]

hello marcus!
I'm interested by the subject but I didn't find the post of turin and zephram. Can you give me a link please?

hi meldi, sorry I didnt respond earlier, I will get a link to that
electroweak thread in the Quantum Physics forum

here's the part of my post that I think you are responding to:

...BTW Feynmann, if you havent already I invite you to have a look in this Quantum Physics forum
at the electroweak conversation between turin and zephram.
if zephram answers on this thread it will probably be interesting...

the thread begins here:
http://www.physicsforums.com/showthread.php?t=31857

first Duct Taper says something
then Zephram replies
then Turin says "I would like to hear a little more..."

it is really turins perceptive questioning that elicits the good exposition
that thread is a model of what I wish we had more of

[dlgoff]

...that thread is a model of what I wish we had more of

Yes indeed. I followed that thread and was fascinated by zephram. Thanks zephram.

Regards

[JohnDubYa]

RE: "So i don't have the Feynman Hibbs book "

Then buy it, or check it out from the library.

[meldi]

thank you very much Marcus!

[selfAdjoint]

RE: "So i don't have the Feynman Hibbs book "

Then buy it, or check it out from the library.

I bought Feynmann Hibbs about ten years ago. I can't really say it helped me understand field theory. The book that really made a breakthrough for me (YMMD) was Hatfield, Quantum Field Theory of Point Particles and Strings. This book is pretty useless as a general reference but it is very intuitive, somewhat in the style of Zefram_C's explanations, but with all the math, too.

[Haelfix]

Some of Dyson's original papers on the subject are also informative, and quite readable. It explains the link between canonical quantization and the path integral. So if you understand the former things are well off.

I still love A Zee's book, Quantum field theory in a Nutshell. Its a very quick read, is wonderfully intuitive and quite deep. You pretty much have the path integral thrown at you right from the getgo, and its only a few chapters before its almost fully justified.

[Feynman]

So gentelmen,
How we can interprate for exemple the excat solution of Shrodinger equation for harmonic oscilator by Feynmanpath integral?
which is very complex

[Feynman]

OHHHHH
If i can't buy the book, what i do?

[Feynman]

what is the probabilisitic point view about path integral?

[Feynman]

Please what is the probabilisitic point view about path integral?

[Feynman]

What is the relation between Manifolds and the path integral?
Thanks

[Feynman]

What is the differential geometry role on path integral?

[jtolliver]

Please what is the probabilisitic point view about path integral?
The probability of travelling from one point to the other is the sum over all paths between them of the probability of travelling along that path; The path integral is basically just that sum over paths.

[Feynman]

So can we consider that the path integral is an integral over manifolds?

[Feynman]

So can we consider that the path integral is an integral over manifolds?

[Feynman]

So can we consider that the path integral is an integral over manifolds?

[Feynman]

?????????????????????????????????????????????????? ???????????????????????????????

[Mike2]

So can we consider that the path integral is an integral over manifolds?
Paths are manifold.

[Feynman]

Who and why Paths are manifold Mike?
I'm talking about path integrals

[selfAdjoint]

Mike means that a path (a one-dimensional continuum) is a manifold. A one-dimensional manifold. But I am sure that wasn't what you meant. But I can't figure out what you did mean.

[Mike2]

Mike means that a path (a one-dimensional continuum) is a manifold. A one-dimensional manifold. But I am sure that wasn't what you meant. But I can't figure out what you did mean.
I've taken to re-reading about path integrals to see if I can get a better intuition of what's going on. Hatfield's presentation seems to be the best I've seen so far - not so many things pulled out of the hat.

In (my) words, it is the integral over all possible paths of the exponential of "i" time the action. The "e" to the "i" of something is always less than or equal to one. This is the weight given to each path in the sum? So I wonder, what values can the Action take? Can it be anything from zero to infinity? Can it be negative? Thanks.

[selfAdjoint]

Mike, e to the power i times something is always a complex number of modulus (i.e. magnitude) 1. THe "something", a function of time, shows the angle the number makes with the real axis. As Feynmann says in QED, think of a "little arrow" of length 1 that can point anywhere in a 360o circle, like a dial or a compass needle. As the particle travels, the vector representing the exponential rotates around at a steady rate (because the exponent of e was a linear function of time). So at every point on the path it is pointing somewhere or other, and all the paths are like that they all have little arrow pointing at some angle in the plane. It's better to think of each path having a dial somewhere outside of the spacetime picture where the pointer can go around. This is all really metaphor for complex numbers.

So then you add up (integrate) all the complex values along each path and then integrate the sums across all the paths, and what happens is that all the different pointing arrows wash each other out and only the classical path comes out of the integration.

[Mike2]

Mike, e to the power i times something is always a complex number of modulus (i.e. magnitude) 1.
So each "path" is weighted by the same magnitude but different phase?

So then you add up (integrate) all the complex values along each path and then integrate the sums across all the paths, and what happens is that all the different pointing arrows wash each other out and only the classical path comes out of the integration.
I'm sure there is a little more to then that. The path integral does not result in a classical path; for then there would be no need for path integral in the first place. I think it means that the classical path simply contributes most to the path integral than for the far fetched paths, right?

[selfAdjoint]

First question, yes if it's just the e^{i\psi t}, but you can have multipliers of the whole exponential that change the modulus. The key is Euler's relation: me^{i\theta} = mcos\theta + imsin\theta.

Second question, no, there is AFAIK no reason prior to Feynmann's method that quantum amplitudes should give the classical (stationary action) path.

[Feynman]

So gentelman ,
we are taking about the Path wich mean (maybe) manifolds, and not complex .
My question is why the path can be consider that is a manifold and can be consider the path integral is an integral over a manifold?
Thanks

[selfAdjoint]

So gentelman ,
we are taking about the Path wich mean (maybe) manifolds, and not complex .
My question is why the path can be consider that is a manifold and can be consider the path integral is an integral over a manifold?
Thanks

I can't make head or tail of this. Clarify please?

[Mike2]

So gentelman ,
we are taking about the Path wich mean (maybe) manifolds, and not complex .
My question is why the path can be consider that is a manifold and can be consider the path integral is an integral over a manifold?
Thanks
String theory uses a type of path integral, only it sums up 2D "paths", on 1D paths. There they do say that the Feynman path integral is "summed over manifolds". I suppose the same thing can be said of the 1D case.

[Feynman]

So do you have some idea about contruction of this path integral?
thanks

[dextercioby]

Which path integral ?For each physical system u have a path integral that gives u the amplitude of probability of transition from one quantum state to another...
:confused:

[Feynman]

Feynman path integral

[selfAdjoint]

All path integrals are Feynmann path integrals. He invented 'em.

[Feynman]

so?
what do you mean selfAdjoint

[Feynman]

so?
what do you mean selfAdjoint

[dextercioby]

He means "equal to his Adjoint". :tongue2: :rofl:

[selfAdjoint]

so?
what do you mean selfAdjoint


*You were asked, "Which path integral?" meaning for which observable.

*You replied Feynmann path integral.

*I pointed out that all the path integrals under discussion are Feynmann path integrals, as a gentle hint to look at what the question meant. Sigh...

[Mike2]

If we are really going to discuss path integrals, let me write some down so we can discuss what the variables are:
(These equations were taken out of Hatfield's book)


\psi (x,t)\,\, = \,\, G(x,t;xo,to)\,\, = \,\,\int D \hat x(\hat t) \,\, {\rm exp(}i{\rm }\int_{t_o }^t {L[\dot {\hat x},\hat x,\hat t]d\hat t} ) \,\, = \,\,\,\int D \hat x(\hat t) \,\, {\rm exp(}i{\rm }\, S[\hat x(\hat t)])


My question is what is \psi (x,t)? Isn't this just the normal wave function solved for with the regular Schrodinger equation? Where does h or h-bar go in these equations?


\Psi [\phi (\vec x,t)]\,\, = \,\, G[\phi ,t;\phi o,to]\,\, = \,\,\int D \hat \phi \,\,{\rm exp(}i{\rm }\int_{t_o }^t {d\hat t\int {d^3 x} \,\,L[\dot {\hat \phi} ,\hat \phi ,\hat t]} )\,\, = \,\,\,\int D \hat \phi \,\, {\rm exp(}i{\rm }\,\, S[\hat \phi (\hat t)])


What is \Psi [\phi (\vec x,t)] called?
Is it possible to do a 3rd quantization? What would that be? Would that be the field of all possible fields? Would this be the field from which any kind of particle field would emerge?



\psi (x,t)\,\, = \,\,\,\int D \hat x(\hat t) \,\,{\rm exp(}i\,\int\limits_{to}^t {d\hat t\,S_o [\hat x,\hat t]} \, - \,\lambda V(\hat x))\,\, = \,\,\sum\limits_{n = 0}^\infty {{{( - i\lambda )^n } \over {n!}}\,\,\int D \hat x(\hat t) \,\,{\,\,\,(\int\limits_{to}^t {dt'\,\,{\rm V}[x(t)])^n }\, \,\,\,{\rm exp(}i\,\,\int\limits_{to}^t {d\hat t\,S_o [\hat x,\hat t]} } )\,\,}


Is this correct, or should it be the time integral over the lagrangian minus the potential times lambda? Is it true that lambda is the charge giving rise to the potential V? Or is it the charge subject to the potential V?

Are there symmetries involved with So that make it easy to solve for? Does the addition of lambda time V always a form of symmetry breaking process? Does this mean that any time there is a symmetry breaking process that there will be a lambda that is a quantized value? Does this mean any time there is a quantized value there is a process of symmetry breaking responsible for it? Is quantization equal to symmetry breaking?

Thanks.

PS: It took me an hour and a half to construct the equations and write this post.

[Mike2]

I took this from Hatfield's book, Quantum Field Theory Of Point Particles and Strings, page 307, eq 13.12.


\Psi [\phi (\vec x,t)]\,\, = \,\, G[\phi ,t;\phi_o,t_o]\,\, = \,\,\int D \hat \phi \,\,{\rm exp(}i{\rm }\int_{t_o }^t {d\hat t\int {d^3 x} \,\,L[\dot {\hat \phi} ,\hat \phi ,\hat t]} )\,\, = \,\,\,\int D \hat \phi \,\, {\rm exp(}i{\rm }\,\, S[\hat \phi (\hat t)])


My question is what is \Psi (x,t) called? Are there alternative names for this? How is it described? How does it differ from \phi (x,t)? Where does h or h-bar go in these equations? Are there "limits" to the integration over D \hat \phi ? Or would this be considered some type of indefinate integral requiring some sort of initial conditions to determine a constant of integration?

Thanks.

[Feynman]

Thank you Mike2, \Psi (x,t) is the solution of SCHROD equation's,
But the problem is HOW WE CA DEFINE D \hat \phi ? mathematically and physically
thx

[Feynman]

???????????????????

[Mike2]

Thank you Mike2, \Psi (x,t) is the solution of SCHROD equation's,
But the problem is HOW WE CAN DEFINE D \hat \phi ? mathematically and physically
thx
As I recall, \phi (x,t) is the wave function of 1st quantization. And I believe \Psi (x,t) is called the "field" of second quantization. Are there any other names for \Psi (x,t), for example, amplitude of something, field of something? Thanks.

The integration over D \hat \phi is over all of "functional space", over all possible changes in the function \phi (x,t) that gets you from the starting \phi_i (x,t) to the ending \phi_f (x,t). It would appear that there is no geometry involved with this space, right? I mean, there would have to be a metric associated with this functional space in order to have geometry, right?

[Feynman]

But integration over a sheme or path (i think) is a deduction from homotopy theory and algebic topology?

[selfAdjoint]

But integration over a sheme or path (i think) is a deduction from homotopy theory and algebic topology?

??? Integration derives from measure theory. There is a link to homotopy, but I don't think homotopy is prior.

[Mike2]

As I recall, \phi (x,t) is the wave function of 1st quantization. And I believe \Psi (x,t) is called the "field" of second quantization. Are there any other names for \Psi (x,t), for example, amplitude of something, field of something? Thanks.

The integration over D \hat \phi is over all of "functional space", over all possible changes in the function \phi (x,t) that gets you from the starting \phi_i (x,t) to the ending \phi_f (x,t). It would appear that there is no geometry involved with this space, right? I mean, there would have to be a metric associated with this functional space in order to have geometry, right?
Just a moment... There is an "inner product" defined on these spaces, aren't there? This inner product determines amplitudes, right? So would this inner product define a metric for this space? What is this space called, Hilbert space, Fock space, ...? Thanks

[Feynman]

But Mr selfAdjoint I have an article wich talk about the importance of the homotopy to construct the Feynman path integral

[Feynman]

But Mr Mike2 How we can define this measure?
And i think that the differential and algebric geometry has an important role on construction of these measure

[selfAdjoint]

But Mr selfAdjoint I have an article wich talk about the importance of the homotopy to construct the Feynman path integral

Citation please? Link if possible?

[Feynman]

ok Mr selfAdjoint But this article is on french!

[dextercioby]

That's irrelevant.Please make a link to the article...

[Feynman]

ok dextercioby , but i don't know how i send this article.

[selfAdjoint]

ok dextercioby , but i don't know how i send this article.

If the article is online, go to it, copy the URL (in the address box at the top of your screen), then come here and paste the url into the reply window. PF doesn't even need all the link apparatus, it automatically puts url tags before and after every url it recognizes.

[Feynman]

This is the article

[Feynman]

so?please help me

[Feynman]

Who can i browse this article ?

[selfAdjoint]

Who can i browse this article ?

What article? Did you have a link?

[Feynman]

Finnally this is the adrres of the article:
lpt1.u-strasbg.fr/kenoufi/MEMOIRES/magistere.pdf

[Feynman]

so no answer?

[Hans de Vries]

Finnally this is the adrres of the article:
http://lpt1.u-strasbg.fr/kenoufi/MEMOIRES/magistere.pdf

The paper discusses the path-integral in so-called "multiple connected"
spaces, as opposed to "single connected" spaces. The homotopy aspect
here is not important to understand the principle of the path integral.



Elementary introduction to grasp the ideas:

"QED, the strange theory of light and matter from Feynman"
http://www.amazon.fr/exec/obidos/ASIN/0691024170/qid=1133806210/sr=1-2/ref=sr_1_8_2/403-4743641-8596453



Feynman deriving Schrödingers equation from the Path-integral:

"Space-time approach to non-relativistic Quantum Mechanics"
(Rev.Mod Phys. 20, 367-387 (1948)

Which you can find reprinted here in "Feynman's Thesis"
http://www.amazon.fr/exec/obidos/ASIN/9812563806/qid=1133806672/sr=1-1/ref=sr_1_0_1/403-4743641-8596453

This brand new book which was published just last month is edited by
Laurie M. Brown who wrote a breakthrough paper together with Feynman
in 1952 (!) "Radiative Corrections to Compton Scattering"



The important follow-up papers from Feynman are:

"The theory of positrons (1949)" , start of the relativistic path integral.

Followed by:

"Space-time approach to Quantum Electro Dynamics, (1949)"



All the above papers can be found in the collection:

"Selected Papers of Richard Feynman"
http://www.amazon.fr/exec/obidos/ASIN/9810241313/qid=1133807324/sr=1-2/ref=sr_1_0_2/403-4743641-8596453
(Again with Laurie M. Brown as editor)


Regards, Hans

[straycat]

Hi
What is the physical interpretation of Feynman path integral?
Thanks :smile:

Here's a little one-page summary of the FPI that I wrote after studying the Feynman and Hibbs text for a while. Go to the files section of my yahoo group and download the file: FPI.pdf.

http://groups.yahoo.com/group/QM_from_GR/files/

:biggrin:

David

[Feynman]

For Hans de Vries , so wath is the physical difference between the multiple connected and single connected and where the path integral entered in this case?:grumpy:

[Hans de Vries]

For Hans de Vries , so wath is the physical difference between the multiple connected and single connected and where the path integral entered in this case?:grumpy:


You can find the meaning of "simply connected" and "multiply connected"
in any textbook on complex analysis. Simply connected means that you
can deform each possible path between two points into any other.
Otherwise they are multiply connected

However, This is hardly relevant for studying the path-integral in a four
dimensional world. Maybe you should ask the author here:

http://dcwww.camp.dtu.dk/~kenoufi

Since the paper is his first years master's thesis.

(When he talks about multiply connected paths, he means paths in 3D space)

-

Regards, Hans

:grumpy:..... Grumpy?

[Feynman]

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]

And where is the signification between the topological argument and the physical interpretation?

[Feynman]

So no ideas please?

[Feynman]

mr Hans de Vries, i have looked the memory of Dr Knoufi on path integrals,
But in this memory , it is a mathematical study for path integrals,
I still search a physical interpretation

[Hans de Vries]

mr Hans de Vries, i have looked the memory of Dr Knoufi on path integrals,
But in this memory , it is a mathematical study for path integrals,
I still search a physical interpretation


It's a bit difficult to respond to the somewhat vague questions without
knowing your background.


The simplest idea behind the path-integral is to do the same in general
what Huygens principle did for the propagation of light. This states that
light goes every but most paths cancel each other out. Leaving the
path in the direction of the wave front. This is the idea which Feynman
brings forward in his public lectures on QED.

The most important physical idea behind the path-integral is the principle
of least action. The classic Lagrangian becomes the measure of the
phase changes over the trajectory of the particle in Quantum Mechanics.

A important offshoot comes from the way how the functional integrals
are solved in the path-integral formalism. They are not solved directly
but expanded into series of which the individual terms can be solved.

What Feynman did was to associate to each such term an interaction
with virtual particles. Each term is associated with a so-called Feynman
diagram which represents the interaction.


Regards, Hans

[Feynman]

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

[Feynman]

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

[starfield]

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

[robphy]

I don't know what is wrong... try entering by
http://www.vega.org.uk/video/series/5 or
http://www.vega.org.uk/video/subseries/8
...or else send them an email.

[Feynman]

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

[Feynman]

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

[Feynman]

:confused: So no idea?
help me please

[samalkhaiat]

:confused: So no idea?
help me please

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!:mad: You could have become an expert on the subject in 18 months!!!:surprised


sam

[Feynman]

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

[Perturbation]

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.

[Jimmy Snyder]

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.

[Feynman]

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?

[ninki]

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,
"up", "horizontal" or "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

[Feynman]

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