# Amplitude and phase of a Feynman path

1. Feb 7, 2004

### Mike2

Can the amplitude and phase associated with each of the paths in the Feynman path integral be connected to geometric attributes of that path? For example, is the amplitude and phase connected to how long the path is or how much it curves or how much it deviates from the geodesic?

2. Feb 7, 2004

### lethe

these are paths through the space of all field configurations, which is not a Riemannian manifold, so the words "length" and "curvature" and "geodesic" are out of place here.

3. Feb 7, 2004

Staff Emeritus
The phase increases at a steady rate along each path. That is the only criterion. Then when the paths are summed up in the path integral, the phases of paths far from the stationary one cancel.

4. Feb 7, 2004

### Mike2

Re: Re: Amplitude and phase of a Feynman path

So it would seem that phase is related to the length of the path. What about amplitude?

5. Feb 8, 2004

### MathNerd

1) Each separate path has an equal amplitude

2) The phase for each path is the action along the path in units of Plank’s constant

6. Feb 8, 2004

### lethe

Re: Re: Re: Amplitude and phase of a Feynman path

you're not listening. you are integrating over a space that does not have a metric, so you cannot ask the length of the path, nor the curvature.

7. Feb 8, 2004

Staff Emeritus
Re: Re: Re: Re: Amplitude and phase of a Feynman path

Excuse me Lethe, but isn't there a nuance here? Each of the paths in the space-of-paths is a PATH, in euclidean space, and in that space it does have a length. And indeed it is in that space that the physics happens that is attributed to the path when you integrate over the space-of-paths. N'est-ce pas?

8. Feb 8, 2004

### lethe

Re: Re: Re: Re: Re: Amplitude and phase of a Feynman path

the space of all field configurations is a Euclidean space? why do you say that? for starters, it is certainly infinite dimensional, and Euclidean space is finite dimensional.

tell me what expression you want to represent the length of a path through the space of field configurations. the action?

9. Feb 8, 2004

### Mike2

Re: Re: Re: Re: Re: Re: Amplitude and phase of a Feynman path

Isn't the phase determined from the Action Integral, and isn't the action integral proportional to the length?

10. Feb 8, 2004

### lethe

Re: Re: Re: Re: Re: Re: Re: Amplitude and phase of a Feynman path

the action for a relativistic point particle travelling on a Lorentzian manifold is proportional to the length of its worldline.

but the action for, say, a scalar field? as far as i can tell, this not the length.

11. Feb 8, 2004

### Loren Booda

Is the amplitude of related Feynman paths normalized over actions which generally contrast between their domain magnitudes ("lengths"), but retain a constant phase difference?

12. Feb 8, 2004

### MathNerd

Re: Re: Re: Re: Re: Re: Re: Amplitude and phase of a Feynman path

Action

$$S = \int_{t_0}^{t_1} L \ dt$$

where S = the classical action
L = the lagrangian

Action of a scalar field

$$S = \int_{t_0}^{t_1} \int_{V} L \ d^3V \ dt$$

but in this case L = the lagrangian density.

The probability amplitude for a given path is $$A e^{i S}$$, where A is some constant that is chosen so that when you sum over all paths the total probability adds up to one and S is the action along that particular path. Remember that A is the same for ALL paths.

Last edited by a moderator: Feb 8, 2004
13. Feb 9, 2004

### Haelfix

Ahh you forget the all important summation sign, in front of that expression, I insist it makes no sense physically to talk about one path, even in principle! Unless you are in a very restricted world =)

Its the coupling of all possible paths, in exactly that way, that is the fundamental quantum *thing*!
.

Last edited: Feb 9, 2004
14. Feb 16, 2004

### Mike2

What is the "coupling" between all possible paths?

15. Feb 16, 2004

Staff Emeritus
Integration over. The paths then occupy the role that points do in elementary integration.

16. Feb 16, 2004

### Mike2

Re: Re: Re: Re: Re: Re: Re: Re: Amplitude and phase of a Feynman path

So it would seem that it is the Lagrangian that chooses the classical path out of all the possible paths by determining how the phases will add to produce the classical result. And the Lagrangian must comply with the Euler-Lagrange equation which some interpret as a vector always normal to the path. What does this all prove?