# Physical interpretation of Feynman path integral

#### Feynman

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

#### Feynman

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

#### Mike2

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

#### Feynman

Who and why Paths are manifold Mike?

Staff Emeritus
Gold Member
Dearly Missed
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.

Last edited:

Staff Emeritus
Gold Member
Dearly Missed
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?

Last edited:

Staff Emeritus
Gold Member
Dearly Missed
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

Staff Emeritus
Gold Member
Dearly Missed
Feynman said:
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

#### Mike2

Feynman said:
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

Homework Helper
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...

#### Feynman

Feynman path integral

Staff Emeritus
Gold Member
Dearly Missed
All path integrals are Feynmann path integrals. He invented 'em.

so?

so?

#### dextercioby

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

Staff Emeritus
Gold Member
Dearly Missed
Feynman said:
so?

*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

Feynman said:
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?

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving