MUST the variation of the action be zero? [Archive]

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friend

Nov4-09, 06:41 PM

Feynman's path integral is:

\[
\int {Dx\,e^{{\textstyle{i \over \hbar }}\int {L(x,\dot x,t)dt} } }
\]

where the Action is:

\[
\int {L(x,\dot x,t)dt}
\]

and the Lagrangian is: \[
{L(x,\dot x,t)}
\]

Now we are told that as we functionally integrate the path integral in the region of far flung paths, contributions of the integral cancel out with other parts of other far flung paths, leaving only the classical path that contributes most to the final result. This classical part is also obtained by setting the functional derivative of the Action to zero and getting the Euler-Lagrange equations of motion.

My question is if the functional derivative of the Action integral does not have a zero for any path, no classical path, then is it still possible to evaluate the path integral at all? Or will everything cancel out? Thanks.

Atakor

Nov6-09, 02:36 PM

if you don't have a classical path..then it means you're still in the quantum domain. hence apply QM or QFT tools..

friend

Nov6-09, 03:44 PM

My question is if the functional derivative of the Action integral does not have a zero for any path, no classical path, then is it still possible to evaluate the path integral at all? Or will everything cancel out? Thanks.

The path integral is equated to <x1|x2> which is elsewhere also equated to the dirac delta function D(x1-x2). So the path integral itself equates to a Dirac delta function:

\[
< x|x_0 > \, = \,\,\delta (x - x_0 )\, = \int {{\rm{Dx}}\,{\rm{e}}^{\frac{{\rm{i}}}{\hbar }\int {{\rm{L(x,\dot x,t)dt}}} } }
\]

Now it occurs to me that the variation of the action in the exponential might be required to be zero if the functional derivative of the whole path integral were required to be zero. Taking the functional derivative of the whole path integral may result in a factor that is the functional derivative of the action integral in the exponential. So if the functional derivative of the whole thing were required to be zero, then this would require the action integral to have a variation of zero. So the question for me is what is the functional derivative of a Dirac delta function. Or what is,

\[
\frac{\delta }{{\delta (x)}}\,\delta (x - x_0 )
\]

My first instinct would be that it is 1. But when this is put in QM terms:

\[
\frac{\delta }{{\delta (x)}}\, < x|x_0 >
\]

Then perhaps it is zero if the states are eigenstates with discrete spectrum, for then there is no small variation of quantum states. Anyone have any idea what the meaning of this variation of an inner product would mean, physically or mathematically? Does this sound like a reasonable physical requirement? Thanks.

friend

Nov9-09, 07:24 PM

The path integral is equated to <x1|x2> which is elsewhere also equated to the dirac delta function D(x1-x2). So the path integral itself equates to a Dirac delta function:

\[
< x|x_0 > \, = \,\,\delta (x - x_0 )\, = \int {{\rm{Dx}}\,{\rm{e}}^{\frac{{\rm{i}}}{\hbar }\int {{\rm{L(x,\dot x,t)dt}}} } }
\]

Now it occurs to me that the variation of the action in the exponential might be required to be zero if the functional derivative of the whole path integral were required to be zero. Taking the functional derivative of the whole path integral may result in a factor that is the functional derivative of the action integral in the exponential. So if the functional derivative of the whole thing were required to be zero, then this would require the action integral to have a variation of zero.

OK, how does this sound? If we integrate the path integral one more time, it should equal 1 since it's a Dirac delta function. Then the functional derivative of that should be zero, since the functional derivative of a constant is zero. This much is necessarily True. But since the path integral is already an infinite number of integrations, I'm not sure that integrating it one more time would really affect its formulation. And then taking the functional derivative of that would probably create a factor consisting of the variation of the action. This would be required to be zero since nothing else could be set to zero.

The path integral written out more fully is:

\[
\mathop {\lim }\limits_{n \to \infty } \,\,\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } {...\int_{ - \infty }^{ + \infty } {\prod\nolimits_{j = 1}^n {dx_j \,\left( {{\textstyle{m \over {2\pi i\hbar \varepsilon }}}} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}} \,\exp \left( {{\textstyle{i \over \hbar }}\,S[x\left( t \right)]} \right)} } } }
\]

So if we integrate one more time the product would go from \[
\prod\nolimits_{j = 1}^n {}
\]
to \[
\prod\nolimits_{j = 1}^{n + 1} {}
\]

But what real difference would this make since they both go to infinity anyway?

In any event, I have seen where the functional derivative commutes with integration. So the functional derivative of the path integral will commute with all the integrals (all infinity of them) and pass inside to the exponential. So functional differentiation commutes with functional integration, right?

Then I assume:

\[
\delta \left( {\exp \left( {\frac{{\rm{i}}}{\hbar }{\rm{S[x}}\left( {\rm{t}} \right){\rm{]}}} \right)} \right){\rm{ = }}\exp \left( {\frac{{\rm{i}}}{\hbar }{\rm{S}}\left[ {{\rm{x}}\left( {\rm{t}} \right)} \right]} \right){\rm{\cdot}}\left( {\frac{{\rm{i}}}{\hbar }\delta {\rm{S}}\left[ {{\rm{x}}\left( {\rm{t}} \right)} \right]} \right)
\]

And also that the only way this could be identically zero is if the variation of the action is zero. This might be where the principle of least action comes from.

Some of this functional calculus doesn't seem to be very well developed in the literature I've seen. I'd appreciate those with more experience could check the accuracy of these assumptions. Thank you.

friend

Nov13-09, 08:14 PM

OK, how does this sound? If we integrate the path integral one more time, it should equal 1 since it's a Dirac delta function. Then the functional derivative of that should be zero, since the functional derivative of a constant is zero. This much is necessarily True.
Yes, this much is true, but is it actually useful? Does it apply to any functional whatsoever, or just those that equal the Dirac delta function?

I do find it interesting that here we have the possibility of discovering the mathematical origin of the least action principle.

It's curious to note that the path integral can be developed from principle alone from the identiy operator:

\[
1 = \int_{ - \infty }^{ + \infty } {\left| {x_1 } \right\rangle \left\langle {x_1 } \right|dx_1 }
\]

And when this is applied an infinite number of times we get:

\[
1 = \int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } {...\int_{ - \infty }^{ + \infty } {\left| {x_1 } \right\rangle \left\langle {{x_1 }}
\mathrel{\left | {\vphantom {{x_1 } {x_2 }}}
\right. \kern-\nulldelimiterspace}
{{x_2 }} \right\rangle \left\langle {{x_2 }}
\mathrel{\left | {\vphantom {{x_2 } {x_3 }}}
\right. \kern-\nulldelimiterspace}
{{x_3 }} \right\rangle \left\langle {x_3 } \right|...\left| {x_n } \right\rangle \left\langle {x_n } \right|dx_1 dx_2 dx_3 ...dx_n } } } }
\]

where we can recognize \[
\left\langle {{x_i }}
\mathrel{\left | {\vphantom {{x_i } {x_j }}}
\right. \kern-\nulldelimiterspace}
{{x_j }} \right\rangle
\]
as the inner product between orthonormal vectors. And when this identity is applied to a general inner product we get:

\[
\left\langle {{x_F }}
\mathrel{\left | {\vphantom {{x_F } {x_0 }}}
\right. \kern-\nulldelimiterspace}
{{x_0 }} \right\rangle = \int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } {\int_{ - \infty }^{ + \infty } {...\int_{ - \infty }^{ + \infty } {\left\langle {{x_F }}
\mathrel{\left | {\vphantom {{x_F } {x_1 }}}
\right. \kern-\nulldelimiterspace}
{{x_1 }} \right\rangle \left\langle {{x_1 }}
\mathrel{\left | {\vphantom {{x_1 } {x_2 }}}
\right. \kern-\nulldelimiterspace}
{{x_2 }} \right\rangle \left\langle {{x_2 }}
\mathrel{\left | {\vphantom {{x_2 } {x_3 }}}
\right. \kern-\nulldelimiterspace}
{{x_3 }} \right\rangle \left\langle {x_3 } \right|...\left| {x_n } \right\rangle \left\langle {{x_n }}
\mathrel{\left | {\vphantom {{x_n } {x_0 }}}
\right. \kern-\nulldelimiterspace}
{{x_0 }} \right\rangle dx_1 dx_2 dx_3 ...dx_n } } } }
\]

If the vectors have finite components, then the \[
\left\langle {{x_i }}
\mathrel{\left | {\vphantom {{x_i } {x_j }}}
\right. \kern-\nulldelimiterspace}
{{x_j }} \right\rangle
\]
is the Kronecker delta, \[
\delta _j^i
\]
. But if the vectors have an infinte number of components, then the inner product is the Dirac delta function \[
\delta \left( {x_i - x_j } \right)
\]
.

Then if we use the gaussian from of the delta function, \[
\delta \left( {x - x_0 } \right) = \mathop {\lim }\limits_{\Delta \to 0} {\textstyle{1 \over {\left( {\pi \Delta ^2 } \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}} }}}e^{ - \left( {x - x_0 } \right)^2 /\Delta ^2 }
\]
, then set \[
\Delta = i^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}} \Delta t
\], the exponential becomes \[
i\left( {\left( {x - x_0 } \right)/\Delta t} \right)^2
\], which can be recognized as the square of the velocity when t is considered the time variable. With this substitution for each of the \[
\left\langle {{x_i }}
\mathrel{\left | {\vphantom {{x_i } {x_j }}}
\right. \kern-\nulldelimiterspace}
{{x_j }} \right\rangle
\]
, the above integral becomes:

\[
\int_{ - \infty }^{ + \infty } {...\int_{ - \infty }^{ + \infty } {\prod\nolimits_{1 = 1}^n {dx_i } } } \,e^{ - i\int_0^t {\dot x^2 \,\,dt} }
\]

where each of the inner products contributes infinitesimally to the integral in the exponent. And apart from the constant m/2 in the exponent, this is the lagrangian for the kinetic energy of a free particle. And this is very similar to the way I've seen the path integral derived is some text books.

If we think more generally, we can substitute x or even x' in the exponent with some other function or field. The requirement would be that these functions be square-integrable so that the the exponent exists. This is all part of the definition of a Hilbert space of functions that seem to be the only functions applicable. And then we are talking about quantum field theory when functions replace x, etc. (not necessarily physical fields). Wouldn't it be interesting if ALL of physics could be derived from nothing more than the Identity? Perhaps there are some interesting philosophical implications that can be gained from that.

So it seems the path integral can be derived apart from physical concerns, and it may be that physics is just a part of these results.

So one question in this context is if the least action principle can be required on principle alone as well. Or is this purely a requirement of physics?

In any event, I have seen where the functional derivative commutes with integration. So the functional derivative of the path integral will commute with all the integrals (all infinity of them) and pass inside to the exponential. So functional differentiation commutes with functional integration, right?

Then I assume:

\[
\delta \left( {\exp \left( {\frac{{\rm{i}}}{\hbar }{\rm{S[x}}\left( {\rm{t}} \right){\rm{]}}} \right)} \right){\rm{ = }}\exp \left( {\frac{{\rm{i}}}{\hbar }{\rm{S}}\left[ {{\rm{x}}\left( {\rm{t}} \right)} \right]} \right){\rm{\cdot}}\left( {\frac{{\rm{i}}}{\hbar }\delta {\rm{S}}\left[ {{\rm{x}}\left( {\rm{t}} \right)} \right]} \right)
\]

And also that the only way this could be identically zero is if the variation of the action is zero. This might be where the principle of least action comes from.

I've seen in the book, Quantum Field Theory, by Lowell S. Brown, page 13 that:

\[
{\textstyle{1 \over i}}{\textstyle{\delta \over {\delta f\left( t \right)}}}\left\langle {q',t_2 |q'',t_1 } \right\rangle ^f = \int {[dq]\,e^{i\int_{t1}^{t2} {dt'L_0 } } \,{\textstyle{1 \over i}}{\textstyle{\delta \over {\delta f\left( t \right)}}}\,e^{i\int_{t1}^{t2} {dt'\,f\left( {t'} \right)q\left( {t'} \right)} } \, = \,} \int {[dq]\,q\left( t \right)} \,e^{i\int_{t1}^{t2} {dt'\,\left[ {L_0 + f\left( {t'} \right)q\left( {t'} \right)} \right]} }
\]

But it's just given as is, and they don't say how they derive this, or how the functional derivate commutes with integration and passes onto the exponential and then to the action. But it is as if to say,

\[
{\textstyle{1 \over i}}{\textstyle{\delta \over {\delta f\left( t \right)}}}\int {[dq]} \,e^{i\int_{t1}^{t2} {dt'\,\left[ {L_0 + f\left( {t'} \right)q\left( {t'} \right)} \right]} } \, = \,\int {[dq]} \,{\textstyle{1 \over i}}{\textstyle{\delta \over {\delta f\left( t \right)}}}\left( {e^{i\int_{t1}^{t2} {dt'\,\left[ {L_0 + f\left( {t'} \right)q\left( {t'} \right)} \right]} } } \right)
\]

\[
= \int {[dq]} \,e^{i\int_{t1}^{t2} {dt'\,\left[ {L_0 + f\left( {t'} \right)q\left( {t'} \right)} \right]} } {\textstyle{1 \over i}}{\textstyle{\delta \over {\delta f\left( t \right)}}}\left( {i\int_{t1}^{t2} {dt'\,\left[ {L_0 + f\left( {t'} \right)q\left( {t'} \right)} \right]} } \right)
\]

\[
= \int {[dq]} \,e^{i\int_{t1}^{t2} {dt'\,\left[ {L_0 + f\left( {t'} \right)q\left( {t'} \right)} \right]} } {\textstyle{\partial \over {\partial f\left( t \right)}}}\,\left[ {L_0 + f\left( {t'} \right)q\left( {t'} \right)} \right]\, = \,\int {[dq]} \,e^{i\int_{t1}^{t2} {dt'\,\left[ {L_0 + f\left( {t'} \right)q\left( {t'} \right)} \right]} } q\left( {t'} \right)
\]

as I suspect is the case. So perhaps my suspicions are correct. Does anyone want to comment?

It might be that physicists do not concern themselves too much with the details of functional calculus. They seem to only use it in passing to get the least action principle and move on from there. I feel that the question I raise here should be of interest to physicists. But perhaps this is more of a technical math question and should be moved to the appropriate forum. I'm not sure which forum would be most interested in functional derivatives and integrals.

Fra

Nov14-09, 05:45 AM

Here are some random ramblings.

So one question in this context is if the least action principle can be required on principle alone as well. Or is this purely a requirement of physics?

It seems you partly search for a deeper understanding of things, but you seek to seek a deduction of some of the laws of physics from deductive logic, is that right?

I can related to your thinking, I seek something similar but in a different way where I replace the logical system with a more general one. I seek an inductive type of inference of the laws of physics from the a more general (not generally deductive) logic.

As I see it, it's a sort of parallell to your vision, but we use different logic. You seem to seek a closed fixed logical system, mine is an open evolving one.

In my think, the "least action principle" is treated on the same basis as "the second law". So Max ent rules and least action rules are two special cases of a more general idea. Conceptually I think this belongs in evolving inductive logic, rather than deductive.

The connection is that the action take by a system, relates to it's expectations of the future. And the least action then suggest that each possible future has a "weight" according to it's plausability. The relation between least action and entropy is that least action is loosely a form of "state relative entropy". This unifies the two principles.

The weighted possibilities, is quite analogous to a risk analysis, where the action takes the form of minimum speculation.

A hint of this is described here
http://en.wikipedia.org/wiki/Kullback%E2%80%93Leibler_divergence

Although that is still a simplification, the full evolving idea is not incorporated there.

So if we only can described the expected encoding information microstructure of a system, I think, it's expected actions should follow from such a intrinsic principle of minimum information divergence - close to - from "principle alone".

minimum information divergence is in effect a principle of minimum speculation or least action.

/Fredrik

friend

Nov14-09, 09:29 AM

It seems you partly search for a deeper understanding of things, but you seek to seek a deduction of some of the laws of physics from deductive logic, is that right?
This seems like the reasonable thing to do.

I can related to your thinking, I seek something similar but in a different way where I replace the logical system with a more general one. I seek an inductive type of inference of the laws of physics from the a more general (not generally deductive) logic.

As I see it, it's a sort of parallell to your vision, but we use different logic. You seem to seek a closed fixed logical system, mine is an open evolving one.

Inductive logic comes from deductive logic. Deductive logic is concerned with whether a proposition is true or false. Inductive logic counts the relative number of samples in various sets. But you have to be able to acknowledge whether it is true or false that a particular sample is included in your set before you can count how many there are.

In my think, the "least action principle" is treated on the same basis as "the second law". So Max ent rules and least action rules are two special cases of a more general idea. Conceptually I think this belongs in evolving inductive logic, rather than deductive.
If you don't know how your system of reasoning is evolving, then you can't say you have a valid system that will give you reliable answers. You might be using the "old" system which no longer applies. But if you do know how your system of reasoning is evolving, then this is because you know the rules by which it evolves. But then these rules by which it evolves are now more basic and fixed and serve as a rigid logic to be used.

The connection is that the action take by a system, relates to it's expectations of the future. And the least action then suggest that each possible future has a "weight" according to it's plausability. The relation between least action and entropy is that least action is loosely a form of "state relative entropy". This unifies the two principles.

The weighted possibilities, is quite analogous to a risk analysis, where the action takes the form of minimum speculation.

/Fredrik

It's interesting to note that the Dirac delta function is also considered a probability distributions. It's a gaussian distribution which represents completely random processes. In the development of the path integral (post 5) the inner product was a dirac delta function. The inner product also describes a transition from one state to the next. This is a kind of implication where the previous state necessarily results in the next state. This also may be a hint that deductive logic can be used to derive the path integral.

So perhaps this structure of physics ( the path integral) from completely random processes (the Dirac delta gaussian distribution) is the evolving law that you're looking for.

friend

Nov14-09, 07:08 PM

I've seen in the book, Quantum Field Theory, by Lowell S. Brown, page 13 that:

\[
{\textstyle{1 \over i}}{\textstyle{\delta \over {\delta f\left( t \right)}}}\left\langle {q',t_2 |q'',t_1 } \right\rangle ^f = \int {[dq]\,e^{i\int_{t1}^{t2} {dt'L_0 } } \,{\textstyle{1 \over i}}{\textstyle{\delta \over {\delta f\left( t \right)}}}\,e^{i\int_{t1}^{t2} {dt'\,f\left( {t'} \right)q\left( {t'} \right)} } \, = \,} \int {[dq]\,q\left( t \right)} \,e^{i\int_{t1}^{t2} {dt'\,\left[ {L_0 + f\left( {t'} \right)q\left( {t'} \right)} \right]} }
\]

But it's just given as is, and they don't say how they derive this, or how the functional derivate commutes with integration and passes onto the exponential and then to the action. But it is as if to say,

\[
{\textstyle{1 \over i}}{\textstyle{\delta \over {\delta f\left( t \right)}}}\int {[dq]} \,e^{i\int_{t1}^{t2} {dt'\,\left[ {L_0 + f\left( {t'} \right)q\left( {t'} \right)} \right]} } \, = \,\int {[dq]} \,{\textstyle{1 \over i}}{\textstyle{\delta \over {\delta f\left( t \right)}}}\left( {e^{i\int_{t1}^{t2} {dt'\,\left[ {L_0 + f\left( {t'} \right)q\left( {t'} \right)} \right]} } } \right)
\]

\[
= \int {[dq]} \,e^{i\int_{t1}^{t2} {dt'\,\left[ {L_0 + f\left( {t'} \right)q\left( {t'} \right)} \right]} } {\textstyle{1 \over i}}{\textstyle{\delta \over {\delta f\left( t \right)}}}\left( {i\int_{t1}^{t2} {dt'\,\left[ {L_0 + f\left( {t'} \right)q\left( {t'} \right)} \right]} } \right)
\]

\[
= \int {[dq]} \,e^{i\int_{t1}^{t2} {dt'\,\left[ {L_0 + f\left( {t'} \right)q\left( {t'} \right)} \right]} } {\textstyle{\partial \over {\partial f\left( t \right)}}}\,\left[ {L_0 + f\left( {t'} \right)q\left( {t'} \right)} \right]\, = \,\int {[dq]} \,e^{i\int_{t1}^{t2} {dt'\,\left[ {L_0 + f\left( {t'} \right)q\left( {t'} \right)} \right]} } q\left( {t'} \right)
\]

This somewhat confuses me because I thought that any variation of the action integral was considered to be identically zero. And this seems to be what he's doing. He then goes on to interate this process in order to form the definition of time ordered products. How can he do this?

Is the general variation, \[
\delta S
\]
, somehow different than a specific variation with respect to a function, \[
\frac{\delta }{{\delta f\left( t \right)}}S
\]
such that the first is zero when the second is not zero?

friend

Nov16-09, 01:30 PM

This somewhat confuses me because I thought that any variation of the action integral was considered to be identically zero. And this seems to be what he's doing. He then goes on to interate this process in order to form the definition of time ordered products. How can he do this?

Is the general variation, \[
\delta S
\]
, somehow different than a specific variation with respect to a function, \[
\frac{\delta }{{\delta f\left( t \right)}}S
\]
such that the first is zero when the second is not zero?

Yes, this seems to be the case because what I found is that if

\[
S\left( {x,y\left( x \right),z\left( x \right)} \right) = \int_a^b {L\left( {x,y\left( x \right),z\left( x \right)} \right)} \,dx
\]

then

\[
\delta S = \frac{{\partial L}}{{\partial y}}\delta y + \frac{{\partial L}}{{\partial z}}\delta z
\]

but

\[
\frac{{\delta S}}{{\delta z}} = \frac{{\partial L}}{{\partial z}}
\]

So there are extra terms in \[
\delta S
\]
that are not in \[
\frac{{\delta S}}{{\delta z}}
\]
which can be used to make first zero when the second is not.

This makes me wonder if when we start adding fields in the Lagrangian to account for other kinds of particles and interactions, is there any care taken to make sure that the whole variation of the action remains zero? From what I've seen in my limited experience, they add terms to the lagrangian in a somewhat ad-hoc way to account for extra fields and self-interactions. I don't remember seeing any procedure where they take the time to check. Perhaps the coupling constants or the exponent of the fields are restricted in such a procedure. As just a wild quess, maybe this is where the cosmological constant problem comes in.

friend

Nov20-09, 03:43 PM

OK, how does this sound? If we integrate the path integral one more time, it should equal 1 since it's a Dirac delta function. Then the functional derivative of that should be zero, since the functional derivative of a constant is zero. This much is necessarily True... And then taking the functional derivative of that would probably create a factor consisting of the variation of the action. This would be required to be zero since nothing else could be set to zero.

As a brief introduction for purposes of context, functional derivatives are developed as follows:

Let \[
S\left[ {x,y\left( x \right),z\left( x \right)} \right]
\]
be a functional of the functions y(x) and z(x).

Then this functional, S, can be expanded in a taylor series

\[
S\left[ {x ,y_0(x) + h_y ,z_0(x) + h_z } \right] = \,S\left[ {x,y_0 ,z_0 } \right] + \frac{{\partial S}}{{\partial y}}|_{y_0 ,z_0 } h_y + \frac{{\partial S}}{{\partial z}}|_{y_0 ,z_0 } h_z + \frac{1}{{2!}}\left( {\frac{{\partial ^2 S}}{{\partial y^2 }}|_{y_0 ,z_0 } h_y ^2 + 2\frac{{\partial S}}{{\partial y}}\frac{{\partial S}}{{\partial z}}|_{y_0 ,z_0 } h_y h_z + \frac{{\partial ^2 S}}{{\partial z^2 }}|_{y_0 ,z_0 } h_z ^2 } \right) + ...
\]

Or the finite difference is

\[
\Delta S = S\left[ {x,y + h_y ,z + h_z } \right] - S\left[ {x,y,z} \right] = \frac{{\partial S}}{{\partial y}}h_y + \frac{{\partial S}}{{\partial z}}h_z + \frac{1}{{2!}}\left( {\frac{{\partial ^2 S}}{{\partial y^2 }}h_y ^2 + 2\frac{{\partial S}}{{\partial y}}\frac{{\partial S}}{{\partial z}}h_y h_z + \frac{{\partial ^2 S}}{{\partial z^2 }}h_z ^2 } \right) + ...
\]

where it is understood that we evaluate S using the functions \[
,y_0(x) ,z_0(x)
\]

We can change notation so that \[
h_y = \delta y
\]
and \[
h_z = \delta z
\]
and get

\[
\begin{array}{l}\\
\Delta S = S\left[ {x,y + \delta y,z + \delta z } \right] - S\left[ {x,y,z} \right] = \frac{{\partial S}}{{\partial y}}\delta y + \frac{{\partial S}}{{\partial z}}\delta z + \frac{1}{{2!}}\left( {\frac{{\partial ^2 S}}{{\partial y^2 }}\left( {\delta y} \right)^2 + 2\frac{{\partial S}}{{\partial y}}\frac{{\partial S}}{{\partial z}}\delta y\delta z + \frac{{\partial ^2 S}}{{\partial z^2 }}\left( {\delta z} \right)^2 } \right) + ... \\
\end{array}
\]

The first variation of S, labeled \[
\delta S
\]
, which is called the first functional derivative of S, is the linear part of the Taylor expansion, or

\[
\delta S = \frac{{\partial S}}{{\partial y}}\delta y + \frac{{\partial S}}{{\partial z}}\delta z
\]

and the term with second power in \[
\delta y
\]
and \[
\delta z
\]
is called the second variation or second functional derivative, or

\[
\delta ^2 S = \frac{{\partial ^2 S}}{{\partial y^2 }}\left( {\delta y} \right)^2 + 2\frac{{\partial S}}{{\partial y}}\frac{{\partial S}}{{\partial z}}\delta y\delta z + \frac{{\partial ^2 S}}{{\partial z^2 }}\left( {\delta z} \right)^2
\]

and similarly for higher order variations, \[
\delta ^3 S,\,\delta ^4 S,\,...
\]

And with this notation,

\[
\Delta S = \delta S + \frac{1}{{2!}}\delta ^2 S + \frac{1}{{3!}}\delta ^3 S + ...
\]

What is \[
\delta \left( {\delta S} \right)
\]
?

\[
\delta \left( {\delta S} \right) = \delta \left( {\frac{{\partial S}}{{\partial y}}\delta y + \frac{{\partial S}}{{\partial z}}\delta z} \right) = \frac{\partial }{{\partial y}}\left( {\frac{{\partial S}}{{\partial y}}\delta y + \frac{{\partial S}}{{\partial z}}\delta z} \right)\delta y + \frac{\partial }{{\partial z}}\left( {\frac{{\partial S}}{{\partial y}}\delta y + \frac{{\partial S}}{{\partial z}}\delta z} \right)\delta z
\]

\[
= \left( {\frac{{\partial ^2 S}}{{\partial y^2 }}\delta y + \frac{{\partial S}}{{\partial y}}\frac{\partial }{{\partial y}}\left( {\delta y} \right) + \frac{{\partial ^2 S}}{{\partial y\partial z}}\delta z + \frac{{\partial S}}{{\partial z}}\frac{\partial }{{\partial y}}\left( {\delta z} \right)} \right)\delta y + \left( {\frac{{\partial ^2 S}}{{\partial z\partial y}}\delta y + \frac{{\partial S}}{{\partial y}}\frac{\partial }{{\partial z}}\left( {\delta y} \right) + \frac{{\partial ^2 S}}{{\partial z^2 }}\delta z + \frac{{\partial S}}{{\partial z}}\frac{\partial }{{\partial z}}\left( {\delta z} \right)} \right)\delta z
\]

But y(x) and z(x) do not depend on one another, so terms like \[
{\frac{\partial }{{\partial y}}\left( {\delta z} \right)}
\]
are obviously zero.

And with terms like \[
{\frac{\partial }{{\partial y}}\left( {\delta y} \right)}
\]
we have that \[
{\delta y}
\]
is a function of x and is independent of y. So they go to zero as well. What is left is

\[
\delta \left( {\delta S} \right) = \frac{{\partial ^2 S}}{{\partial y^2 }}\left( {\delta y} \right)^2 + \frac{{\partial ^2 S}}{{\partial y\partial z}}\delta y\delta z + \frac{{\partial ^2 S}}{{\partial z\partial y}}\delta y\delta z + \frac{{\partial ^2 S}}{{\partial z^2 }}\left( {\delta z} \right)^2 = \frac{{\partial ^2 S}}{{\partial y^2 }}\left( {\delta y} \right)^2 + 2\frac{{\partial ^2 S}}{{\partial y\partial z}}\delta y\delta z + \frac{{\partial ^2 S}}{{\partial z^2 }}\left( {\delta z} \right)^2
\]

But this last part was exactly how \[
\delta ^2 S
\]
was defined. So we have

\[
\delta \left( {\delta S} \right) = \delta ^2 S
\]

Now if,

\[
S\left[ {x,y\left( x \right),z\left( x \right)} \right] = dF\left[ {x,y\left( x \right),z\left( x \right)} \right]
\]

then,

\[
\Delta S\left[ {x,y\left( x \right),z\left( x \right)} \right] = S\left[ {x,y + \delta y,z + \delta z} \right] - S\left[ {x,y,z} \right] = \Delta dF\left[ {x,y\left( x \right),z\left( x \right)} \right]
\]

\[
= dF\left[ {x,y + \delta y,z + \delta z} \right] - dF\left[ {x,y\left( x \right),z\left( x \right)} \right] = d\left( {F\left[ {x,y + \delta y,z + \delta z} \right] - F\left[ {x,y\left( x \right),z\left( x \right)} \right]} \right) = d\Delta F\left[ {x,y\left( x \right),z\left( x \right)} \right]
\]

So that,

\[
\Delta d = d\Delta
\]

\[
\Delta
\]
commutes with \[
d
\]

Then,

\[
\Delta dF = \delta \left( {dF} \right) + \frac{1}{{2!}}\delta ^2 \left( {dF} \right) + \frac{1}{{3!}}\delta ^3 \left( {dF} \right) + ... = d\left( {\delta F + \frac{1}{{2!}}\delta ^2 F + \frac{1}{{3!}}\delta ^3 F + ...} \right) = d\delta F + \frac{1}{{2!}}d\delta ^2 F + ...
\]

And keeping only the first approximation linear terms on both sides of the equation,

\[
\delta d = d\delta
\]

\[
\delta
\]
commutes with \[
d
\]

And if

\[
S\left[ {x,y\left( x \right),z\left( x \right)} \right] = \int_a^b {L\left[ {x,y\left( x \right),z\left( x \right)} \right]} \,dx
\]

Then

\[
\Delta S = S\left[ {x,y + \delta y,z + \delta z} \right] - S\left[ {x,y,z} \right] = \int_a^b {L\left[ {x,y + \delta y,z + \delta z} \right]} \,dx - \int_a^b {L\left[ {x,y,z} \right]} \,dx = \Delta \int_a^b {L\left[ {x,y,z} \right]} \,dx
\]

\[
= \int_a^b {L\left[ {x,y + \delta y,z + \delta z} \right] - L\left[ {x,y,z} \right]} \,dx = \int_a^b {\Delta L\left[ {x,y,z} \right]} \,dx
\]

So we see here that \[
\Delta
\]
commutes with \[
\int {}
\]

But also,

\[
\Delta S = \delta S + \frac{1}{{2!}}\delta ^2 S + ... = \delta \left( {\int_a^b L \,dx} \right) + \frac{1}{{2!}}\delta ^2 \left( {\int_a^b L \,dx} \right) + ...
\]

\[
= \int_a^b {\Delta L} \,dx = \int_a^b {\delta L + \frac{1}{{2!}}\delta ^2 L + ...} \,dx = \int_a^b {\delta L\,dx + \frac{1}{{2!}}\int_a^b {\delta ^2 L\,dx + ...} }
\]

Then keeping first approximation linear terms on each side of the equation we see

\[
\delta \int_a^b L \,dx = \int_a^b {\delta L\,dx}
\]

Or, \[
\delta \,\,\,{\rm{commutes}}\,\,{\rm{with}}\,\,\,\int {}
\]

Many text use the integral definition of a functional, \[
S = \int {L\,dx}
\]
in its development. I suppose they do this because it makes it easier to justify taking only the linear terms inside the integral since differentials approach zero more naturally in the process of integration.

It's easy to see that variation commutes with any number of integral signs since a difference outside the integral is translated to a difference inside the integral signs. Or,

\[
\Delta \int {\int {...\int {L\left[ {x,y,z} \right]} } } dx_1 dx_2 ...dx_n = \Delta \int {d[x]} L\left[ {x,y,z} \right]
\]

\[
= \int {d[x]} L\left[ {x,y + \delta y,z + \delta z} \right] - \int {d[x]} L\left[ {x,y,z} \right] = \int {d[x]\left( {L\left[ {x,y + \delta y,z + \delta z} \right] - L\left[ {x,y,z} \right]} \right)} = \int {d[x]\Delta L\left[ {x,y,z} \right]}
\]

So following a similar procedure of keeping only linear terms,

\[
\delta \int {d[x]\,L = \int {d[x]\delta L} }
\]

This means that the variation of the integration of the path integral gets passed inside all the infinite number of integrations to taking the variation of the exponent of the action.

friend

Nov22-09, 11:50 AM