Derivation of Dyson Series

  • #1

Main Question or Discussion Point

Hi,

I'm looking at this wikipedia entry ( http://en.wikipedia.org/wiki/Dyson_series ) for the derivation of the Dyson series and I'm having a great deal of difficulty with:

[itex]S_n=\int_{t_0}^t{dt_1\int_{t_0}^{t_1}{dt_2\cdots\int_{t_0}^{t_{n-1}}{dt_nK(t_1, t_2,\dots,t_n)}}}.[/itex]

If K is symmetric in its arguments, we can define (look at integration limits):

[itex]K_n=\int_{t_0}^t{dt_1\int_{t_0}^{t}{dt_2\cdots\int_{t_0}^t{dt_nK(t_1, t_2,\dots,t_n)}}}.[/itex]

And so it is true that:

[itex]S_n=\frac{1}{n!}K_n.[/itex]

I simply can't get my head around this. How can the S integral lead to the K integral (up to an n factorial)? I don't understand how this integral can be performed. I mean S is a series of coupled integrals that must be performed in succession, K is a product of independent integrals...


Thanks in Advance
 
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  • #2
A. Neumaier
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Hi,

I'm looking at this wikipedia entry ( http://en.wikipedia.org/wiki/Dyson_series ) for the derivation of the Dyson series and I'm having a great deal of difficulty with:

[itex]S_n=\int_{t_0}^t{dt_1\int_{t_0}^{t_1}{dt_2\cdots\int_{t_0}^{t_{n-1}}{dt_nK(t_1, t_2,\dots,t_n)}}}.[/itex]

If K is symmetric in its arguments, we can define (look at integration limits):

[itex]K_n=\int_{t_0}^t{dt_1\int_{t_0}^{t}{dt_2\cdots\int_{t_0}^t{dt_nK(t_1, t_2,\dots,t_n)}}}.[/itex]

And so it is true that:

[itex]S_n=\frac{1}{n!}K_n.[/itex]

I simply can't get my head around this. How can the S integral lead to the K integral (up to an n factorial)? I don't understand how this integral can be performed. I mean S is a series of coupled integrals that must be performed in succession, K is a product of independent integrals...
Well, you assumed that K is symmetric in the t_i. In the Dyson series, it isn't. Thats why you need the time-ordering to make it look symmetric....
 
  • #3
Well, you assumed that K is symmetric in the t_i. In the Dyson series, it isn't. Thats why you need the time-ordering to make it look symmetric....
Those equations aren't me they're from the wikipedia article. However, that doesn't answer how those two can be equal (up to a permutation constant).
 
  • #4
A. Neumaier
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Those equations aren't me they're from the wikipedia article. However, that doesn't answer how those two can be equal (up to a permutation constant).
OK, I misunderstood what you were asking for. In context it becomes clear that the formulation in wikipedia is a bit misleading. The ''If'' should be ''Since'', since K is already defined as the time-ordered version.

In the first expression that you quoted, the integration region is over the simplex t_1>...>t_n. It is a geometric fact that the n-dimensional box with sides [t_0,t_n] can be partitioned into the n! simplices obtained by reordering the t_i in all possible ways. Because of symmetry, each of these integrals has the same value. Hence the second integral is n! times the first.
 
  • #5
I appreciate the response but I'm afraid that's way over my head. To me I see this S (for say the third term) as

[itex] \int^{t}_{t_0} \int^{t_1}_{t_0} \int^{t_2}_{t_0} T [ V(t_1)V(t_2)V(t_3) ] dt_1 dt_2 dt_3 [/itex]

and the first integral would be

[itex] \int^{t}_{t_0} \int^{t_1}_{t_0} T [ V(t_1)V(t_2) (V^{(1)}(t_2)-V^{(1)}(t_0) ) ] dt_1 dt_2 [/itex]

where [itex]V^{(1)}(t)[/itex] represents the first integral of V with respect to t. Which leads to a second integral

[itex] \int^{t}_{t_0} V(t_1) \int^{t_1}_{t_0} \left( V(t_2) (V^{(1)}(t_2)-V^{(1)}(t_0) ) ] dt_1 dt_2 [/itex]

and who knows what [itex] \int^{t_1}_{t_0} V(t_2) V^{(1)}(t_2) [/itex] is but you get the picture and this sequence of integrals cascades up until you finally get to the integral from t_0 to t. I don't see how this could possible be equal to

[itex] \frac{1}{n!} \left( \int^t_{t_0} V(t') dt' \right)^n [/itex]
 
  • #6
A. Neumaier
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I appreciate the response but I'm afraid that's way over my head.
Forget about the T, and work with K, where the essence of the argument is. To begin understanding what is going on, consider first the case n=2 and t_0=0, t=1. Take the difference K_2-S_2 and simplify, to get a double integral where the second integral goes from t_1 to 1. Now interchange the two integrals and figure out how the integration interval change, by looking at the possible pairs (t_1,t_2) in the plane. Then use the symmetry of K to see that the integral reduces again to K_2. Then do the same with n=3 and n=4, until you reach enlightenment.
 
  • #7
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Draw a square. Now draw a diagonal line connecting the lower left corner to the upper right corner, and shade in the lower triangle. The S integral is the lower triangle. The K integral is the entire square.
 
  • #8
tiny-tim
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hi maverick_starstrider! :smile:

let's take a specific set of values for t1 t2 and t3

consider t1 = 9, t2 = 5, t3 = 7

then ∫∫∫ T( V(t1),V(t2),V(t3) ) = ∫∫∫ ( V(5),V(7),V(9))

similarly, ∫∫∫ T( V(t1),V(t3),V(t2) ) = ∫∫∫ ( V(5),V(7),V(9))

∫∫∫ T( V(t2),V(t3),V(t1) ) = ∫∫∫ ( V(5),V(7),V(9))

∫∫∫ T( V(t2),V(t1),V(t3) ) = ∫∫∫ ( V(5),V(7),V(9))

∫∫∫ T( V(t2),V(t1),V(t2) ) = ∫∫∫ ( V(5),V(7),V(9))

∫∫∫ T( V(t3),V(t2),V(t1) ) = ∫∫∫ ( V(5),V(7),V(9))

(or is it ( V(9),V(7),V(5)) ? … i cant remember :redface:)

(i really like copy-and-pasting!! :biggrin:)

in other words: they're all the same!

now add them all up and divide by 3! … still the same! :wink:

ok? :smile:
 

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