Now we’ll try to solve the case of a time-dependent velocity field, , which will lead us to introduce the “time-ordering” operator .

Table of Contents

• A Time-Dependent Velocity Field
• The Time-Ordered Exponential for a Pullback
• The Time-Ordered Exponential for a Pushforward
  • The Interaction Picture
  • The Pushforward Continued
• In Summary

A Time-Dependent Velocity Field

Return again to our original ODE in one dimension, but now with a time-dependent velocity field:

If we discretize time into intervals of size , the solution should have the form of a sequence of time translation operators , where the translation operator itself varies with time:

The true trajectory should look like a limit of this expression, with each computed at time and applying the vector field at , the result of the path so far.

We could express the discretized solution on functions or densities instead:

But we cannot simply generalize the solutions

by replacing (note I’ve suppressed the dependence, but this is really ); this would only depend on the velocity at the final time and not any of the intermediates. We might try to write these with an integral over time in the exponent:

but this would only produce the correct sequence of infinitesimal time-evolution operators if the identity holds for these integrals. If we try to work out the pullback , say, with just with time-steps,

a term appears at second order

with a to the right of a . The correct expression would have all its s to the right of the s.

At this point we could guess at the right answer. It should be some kind of “product integral” like

appropriately ordered so that the largest times are on the right. Likewise for and . And we should be able to approximate the exponential by its first-order term since its argument is small:

which can be rewritten as a series in powers of :

It’s starting to look like an exponential series, but with an argument which varies by term.

From here it’s hard to see where to go. Let us try to approach the problem a different way.

The Time-Ordered Exponential for a Pullback

If we were to integrate directly we would get something like

and then we could imagine expanding the integrant

The time derivative would be , and repeating this procedure would produce something like a Taylor series for , involving increasingly-complicated time derivatives of . This might work, but it’s hard to see how it would be enlightening, and it will only really work for anyway without further machinery.

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We might then try to perform the same kind of “Taylor series” on itself, starting with the operator diff. eq.

we found in the previous post, which will still hold for . But we would, once again, run into the problem that an arbitrary manifold will not come with a notion of addition with which to build a series expansion.

Instead we should try to work with either the pullback or pushforward . The former obeys a simpler differential equation, and avoids all of the Jacobian details we worked out last time, but at composes in opposite order and will give a less-natural final result. I’ll take this tradeoff—we’ll start with the pullback, and then work out the pushforward afterwards.

We start with the diff eq.

which should now be written with ,

It will be useful to give the combined operator a name of its own:

Note that operators at different times do not commute:

With this, the operator diff. eq. is

The being on the right is important. The equivalent derivation for would wind up with an operator on the left, corresponding to the fact that pushforwards composes in the normal order.

We can now try to solve this operator diff. eq. to find an explicit form of . We recursively applying the fundamental theorem of calculus for the operator , using liberally and the fact that , the identity operator:

This is a fairly tidy infinite-sum-of-nested-integrals, at least. Note that the integration variables obey , and the total integration volume of term is therefore that of a -simplex of side length , .

Observe also that each product occurs in reverse time-order (), with the smallest time on the left. Also note that each participating only in a single of the integrals.

Then, if we simultaneously…

• expand integral to cover the full time interval , which will multiple the overall integration volume by , but will then include all possible products of s, most of which have their times out of order,
• replace the products-of-s with a “reverse time-ordered product” , such that the operator argument always appear in the order of original time, even when the time parameters range into the new region of integration,
• and divide by to undo the overcounting introduced by the first two steps…

… we should get an equivalent expression which even better resembles a true exponential series,:

Now each is independent of the others, so we can write them as a power of a single term, relying on the symbol to order them properly:

We see that the argument to is simply the exponential of the integral , suggesting a notation as a “(reverse) time-ordered exponential”:

This, finally, is a general solution to time evolution, albeit an unwieldy one.

The reverse time-ordering operator must be used because time evolution composes in reverse order for pullbacks. We will see that the forward will appear insted for both flows and pushforwards.

The Time-Ordered Exponential for a Pushforward

I chose to work out the time-evolution of a function because the equivalent derivation for is a little more complicated. However, it will be useful to write both.

We start with the diff. eq. for ,

The r.h.s. operator we name because it, technically, is the adjoint to in integrals like

At this point we can repeat the derivation from before, which would lead to the obvious time-ordered exponential:

Here we use a normal time-ordering because is acting on from the left.

But the operator

is, again, made complicated by the two non-commuting terms.

We can repeat some of the analysis from the previous post. The second term is exactly the negative of , and alone must implement the pullback part , but via a forward time-ordered exponential:

The divergence term against must be responsible for the Jacobian term in :

Therefore there must be some way to factor the time-ordered exponential of the first term of out of its exponential. Writing , with , we have

We should now rerun the derivation from the previous post where we considered . But it will be cleaner to work out the general case first.

The Interaction Picture

For an operator diff. eq.

with time-dependent coefficients , we know a derivation like one above would give a general solution as a time-ordered exponential:

At this point the move is to cleverly choose some with a simpler time-derivative than itself. For we chose, effectively, , but that won’t work here; we need something time-ordered at a minimum since its argument is a function of .

Specifically, we want the prefactor to be the equivalent operator which “propagates backwards in time with alone for a time ”. The operation which propagates forwards with would be

The inverse of this need to obey

We can guess its form. , as a time-ordered exponential, is like a series of factors

with the smallest time on the right. Its inverse need to hvae the same factors but in the opposite order, and each with a negative exponential, such that each factor pair cancels outwards from the middle. That is, it should be reverse-time-ordered:

Note that the time derivative of this expression has a on the right:

(Compare with the earlier pullback , which obeyed .)

Then is

and its time derivative is

We’ve defined

which “conjugates” by . This, it will turn out, is exactly the “interaction picture” of quantum mechanics, though we do not have the context to see it that way now.

The solution for , by analogy to our previous solutions, is

and the solution for will be found by

though there’s not much we can do with this in the general case.

The Pushforward Continued

Now plug in

with and . The analog of is

so becomes

as before.

The transformed density itself has solution

where has analog

It is not obvious to me looking at this, but this can be simplified greatly. We allow the pullbacks to act on a test function:

Therefore, as operators,

Putting it together, we get

and, going through the same steps from before to turn back into , we get

… which is exactly what we got before, except with a time-ordered exponential.

The pushforward operator itself can therefore be written as any of:

In Summary

The table again:

Was this worth it? The interesting thing, to me, is that the operator arises here, in the study of the most generic first-order ODE. I first encountered it in quantum mechanics and later in field theory, but it is in fact quite fundamental to dynamical systems.

The time-ordered exponentials like

are, as we saw, more like a “product integral”, but the time-ordering symbol allows us to express them as a normal integral.

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