Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New features!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

123 submissions , 104 unreviewed
3,547 questions , 1,198 unanswered
4,552 answers , 19,366 comments
1,470 users with positive rep
411 active unimported users
More ...

Time ordering and time derivative in path integral formalism and operator formalism

+ 8 like - 0 dislike
341 views

In operator formalism, for example a 2-point time-ordered Green's function is defined as

$\langle\mathcal{T}\phi(x_1)\phi(x_2)\rangle_{op}=\theta(x_1-x_2)\phi(x_1)\phi(x_2)+\theta(x_2-x_1)\phi(x_2)\phi(x_1),$

where the subscript "op" refers to operator formalism. Now if one is to take a time derivative of it, the result will be $\frac{\partial}{\partial x_1^0}\langle\mathcal{T}\phi(x_1)\phi(x_2)\rangle_{op}=\langle\mathcal{T}{\frac{\partial \phi(x_1)}{\partial x_1^0}}\phi(x_2)\rangle_{op}+\delta(x_1^0-x_2^0)[\phi(x_1),\phi(x_2)]$, the delta function comes from differentiating the theta functions. This means time derivative does not commute with time ordering.

If we consider path integral formalism, the time-ordered Green's function is defined as

$\langle\mathcal{T}\phi(x_1)\phi(x_2)\rangle_{pi}=\int\mathcal{D}\phi\phi(x_1)\phi(x_2)e^{iS(\phi)}$.

Of course $\langle\mathcal{T}\phi(x_1)\phi(x_2)\rangle_{op}=\langle\mathcal{T}\phi(x_1)\phi(x_2)\rangle_{pi},$ as is proved in any QFT textbook. However in path integral case time derivative commutes with time ordering, because we don't have anything like a theta function thus $\frac{\partial}{\partial x_1^0}\int\mathcal{D}\phi\phi(x_1)\phi(x_2)e^{iS(\phi)}=\int\mathcal{D}\phi\frac{\partial}{\partial x_1^0}\phi(x_1)\phi(x_2)e^{iS(\phi)}$. I did a bit googling and found out that for the path integral case the time-ordered product is called "$\mathcal{T^*}$ product" and operator case just "$\mathcal{T}$ product".

I am not that interested in what is causing the difference(still explanations on this are welcomed), because I can already vaguely see it's due to some sort of ambiguity in defining the product of fields at equal time. The question that interests me is, which is the right one to use when calculating Feynman diagrams?

I did find a case where both give the same result, i.e. scalar QED(c.f. Itzykson & Zuber, section 6-1-4), but is it always the case? If these two formulations are not effectively equivalent, then it seems every time we write down something like $\langle\partial_0\phi\cdots\rangle$, we have to specify whether it's in the sense of the path integral definition or operator definition.

EDIT: As much as I enjoy user1504's answer, after thinking and reading a bit more I don't think analytic continuation is all the mystery. In Peskin&Schroeder chap 9.6 they manage to use path integral to get a result equivalent to operator approach, without any reference to analytic continuation. It goes like this : Consider a T-product for free KG field $\langle T\{\phi(x)\phi(x_1)\}\rangle=\int\mathcal{D}\phi\phi(x)\phi(x_1)e^{iS(\phi)}$. Apply Dyson-Schwinger equation, we get $\int\mathcal{D}\phi(\partial^2+m^2)\phi(x)\phi(x_1)e^{iS}=-i\delta^4(x-x_1)$, then they just assume the $\partial^2$ commute with path integration(which is already weird according to our discussion) and they conclude

$(\partial^2+m^2)\int\mathcal{D}\phi\phi(x)\phi(x_1)e^{iS}=(\partial^2+m^2)\langle T\{\phi(x)\phi(x_1)\}\rangle=-i\delta^4(x-x_1)$.

This is just the right result given by operator approach, in which $\delta(x^0-x_1^0)$ comes from $\theta$ function. Given my limited knowledge on the issue, this consistency looks almost a miracle to me. What is so wicked behind these maths?

Response to @drake:If $a$ is a positive infinitesimal, then $\int \dot A(t) B(t) \,e^{iS}\equiv\int D\phi\, {A(t+a)-A(t)\over a}B(t)\,e^{iS}=\frac{1}{a}\langle T\{A(t+a)B(t)\}\rangle-\frac{1}{a}\langle A(t)B(t)\rangle$, notice the second term has an ordering ambiguity from path integral(say $A=\dot{\phi},B=\phi$), and we can make it in any order we want by choosing an appropriate time discretization, c.f. Ron Maimon's post cited by drake. Keeping this in mind we proceed:

$\frac{1}{a}\langle T\{A(t+a)B(t)\}\rangle-\frac{1}{a}\langle A(t)B(t)\rangle\\=\frac{1}{a}\theta(a)\langle A(t+a)B(t)\rangle+\frac{1}{a}\theta(-a)\langle B(t)A(t+a)\rangle-\frac{1}{a}\langle A(t)B(t)\rangle\\=\frac{1}{a}\theta(a)\langle A(t+a)B(t)\rangle+\frac{1}{a}[1-\theta(a)]\langle B(t)A(t+a)\rangle-\frac{1}{a}\langle A(t)B(t)\rangle\\=\frac{\theta(a)}{a}\langle [A(t+a),B(t)]\rangle+\frac{1}{a}[\langle B(t)A(t+a)\rangle-\langle A(t)B(t)\rangle]$

Now taking advantage of ordering ambiguity of the last term to make it $\langle B(t)A(t)\rangle$(this amounts to defining A using backward discretization, say $A=\dot{\phi}(t)=\frac{\phi(t+\epsilon^-)-\phi(t)}{\epsilon^-}$), then the finally:

$\frac{\theta(a)}{a}\langle [A(t+a),B(t)]\rangle+\frac{1}{a}\langle B(t)[A(t+a)-A(t)\rangle]\to \frac{1}{2a}\langle [A(t),B(t)]\rangle+\langle B(t)\dot{A}(t)\rangle$.(Here again a very dubious step, to get $\frac{1}{2a}$ we need to assume $\theta(a\to 0^+)=\theta(0)=\frac{1}{2}$, but this is really not true because $\theta$ is discontinuous)

However on the other hand, since $a$ was defined to be a postive infinitesimal, at the very beginning we could've written $\frac{1}{a}\langle T\{A(t+a)B(t)\}\rangle-\frac{1}{a}\langle A(t)B(t)\rangle=\frac{1}{a}\langle A(t+a)B(t)\rangle-\frac{1}{a}\langle A(t)B(t)\rangle$, then all the above derivation doesn't work. I'm sure there are more paradoxes if we keep doing these manipulations.

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user Jia Yiyang
asked Jul 23, 2013 in Theoretical Physics by Jia Yiyang (2,410 points) [ no revision ]
Most voted comments show all comments
I think that the problem is that you cannot represent a commutator of 2 operators at equal time, by a path integral representation (it would be zero), but this does not mean that the derivative and the integral commute in the path integral formalism. These are equivalent formalism : $\int d\phi ~\phi(x)~\phi(y) ~e^{i\int_0^T dt L(\phi,t)} =\langle 0|T[\hat \phi(x),\hat \phi(y)]e^{-i \hat HT}|0\rangle$

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user Trimok
@Trimok: Maybe my example is not that good because $[\phi(x_1),\phi(x_2)]$ is indeed 0 at equal time. It's better to consider $[\phi(x_1),\dot{\phi}(x_2)]$, in this case path integral can also give non-zero result by this argument: en.wikipedia.org/wiki/…

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user Jia Yiyang
This may seem odd, but I believe this is actually the same ambiguity as in physics.stackexchange.com/q/69828. Its not very transparent, but see Sec II of prd.aps.org/pdf/PRD/v3/i10/p2486_1 for an example where the derivative should not commute with path integral. In terms of practical operations, the best advice seems to be "work the Hamiltonian formulation". There are no time derivatives in this case and then you can integrate out the momenta at the end.

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user BebopButUnsteady
@JiaYiyang : When you derive a time-ordered product, you have always the last term which looks like $ \langle0|\delta(x_0-y_0)[A(x), B(y)]e^{iS}|0\rangle$. And this term, no matter $A$ and $B$ are, cannot be represented by a path integral. But this term is not null, even if it cannot be represented by a path integral. So derivative and integral (in the path integral formalism) do not commute, even if the (not null) difference, cannot be represented by a path integral.

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user Trimok
@Trimok: so what's your comment on peskin$schroeder's approach as I described in the EDIT?

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user Jia Yiyang
Most recent comments show all comments
@drake: Just in case of any miscommunication, forward means $\frac{1}{a}(\phi(t+a)-\phi(t))$, backward means $\frac{1}{a}(\phi(t)-\phi(t-a))$, centered means $\frac{1}{2a}(\phi(t+a)-\phi(t-a))$

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user Jia Yiyang
Typo in my comment "This unnaturalness is just the same as in my derivation I defined $\dot{A}$˙ using backward discretization while $A$ is defined using forward discretization" , it should be $A$ is backward discretized and $\dot{A}$ is forward discretized.

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user Jia Yiyang

2 Answers

+ 7 like - 0 dislike

EDIT: I'm leaving this up as background reading to @drake's answer. (The point of the following is that the path integral does indeed give the correct time ordering, so it is producing the correct $\theta$-function weighted, time-ordered sums, which must be accounted for when differentiating its output.)

The two formalisms are equivalent; if they don't give the same result, something is wrong in the calculation. To see this you have to understand a subtlety which is not usually well-explained in textbooks, namely that the path integral is not defined merely by taking the limit of a bunch of integrals of the form $\int_{\mbox{lattice fields}} e^{iS(\phi)} d\phi$.

The problem is that these finite-dimensional integrals are not absolutely convergent, because $|e^{iS(\phi)}| = 1$. To define even the lattice path integral in Minkowski signature, you have to specify some additional information, to say exactly what is meant by the integral.

In QFT, the additional information you want is that the path integral should be calculating the kernel of the time evolution operator $e^{iH\delta t}$, which is an analytic function of $\delta t$. This fact is usually expressed by saying that the Minkowski signature path integral is the analytic continuation of a Euclidean signature path integral: The Euclidean $n$-point functions $E(y_1,...,y_n)$ defined by

$E(y_1,...,y_n) = \int \phi(y_1)...\phi(y_n) e^{-S_E(\phi)} d\phi$

are analytic functions of the Euclidean points $y_i \in \mathbb{R}^d$. This function $E$ can be continued to a function $A(z_1,...,z_n)$ of $n$ complex variables $z_i \in \mathbb{C}^d$. This analytic function $A$ does not extend to the entire plane; it has singularities, and several different branches. Each branch corresponds to a different choice of time-ordering. One branch is the correct choice, another choice is the 'wrong sign' time-ordering. Other choices have wrong signs on only some subsets of the points. If you restrict $A$ to the set $B$ of boundary points of the correct branch, you'll get the Minkowski-signature $n$-point functions $A|_B = M$, where $M(x_1,...,x_n) = \langle \hat{\phi}(x_1)...\hat{\phi}(x_n)\rangle_{op}$ and the $x_i$ are points in Minkowski space.

In perturbation theory, most of this detail is hidden, and the only thing you need to remember is that the $+i\epsilon$ prescription selects out the correct time-ordering.

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user user1504
answered Jul 23, 2013 by user1504 (1,100 points) [ no revision ]
+1. I didn't realize this subtlety. So I guess in path integral the issue is $\partial_0$ does not commute with analytic continuation? And it implies when differentiating w.r.t time it's safer to use operator definition of T-ordering, while in path integral one has to take the trouble first differentiating in Euclidean signature then analytically continue it? It would be nicer if you can talk more about the time derivative issue.

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user Jia Yiyang
I don't have much to add on the time derivatives. Take them on the operator side, and be aware that naively passing them through the path integral is hazardous.

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user user1504
I updated my question, since it's too long for a comment.

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user Jia Yiyang
Hi. I upvoted your question because I think everything you say is correct, however I think it doesn't answer the question. Do you have any reason why passing time derivatives through the path integral is so dangerous? Can you have a look at my answer?

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user drake
+ 6 like - 0 dislike

Good question. It has made me think.

  • Strictly speaking, it is not possible to compute $\theta(t-t')\langle \dot\phi(t)\phi(t')\rangle+\theta (t'-t)\langle \phi (t')\dot\phi (t)\rangle$ ( shorthand notation: $\langle\,\equiv \langle 0 |\,$. Also note that I'm omitting the spatial arguments of the fields ) using the Lagrangian version of the path integral, because to derive this, we are to assume that the insertions (factors multiplying $e^{iS}$ in the integrand of the path integral) are functionals of the fields at a given time (i.e., they are independent of momenta $\pi$). Thus,

$$\partial_t \,\left[\theta(t-t')\langle \phi(t)\phi(t')\rangle+\theta (t'-t)\langle \phi (t')\phi (t)\rangle\right]=\lim_{a\to 0}{1\over a}\left[\theta(t+a-t')\langle \phi(t+a)\phi(t')\rangle+\theta (t'-t-a)\langle \phi (t')\phi (t+a)\rangle - \theta(t-t')\langle \phi(t)\phi(t')\rangle+\theta (t'-t)\langle \phi (t')\phi (t)\rangle\right]=\lim_{a\to 0}{1\over a}\int D\phi\, (\phi(t+a)\phi(t')-\phi(t)\phi(t'))\,e^{iS}=\int D\phi\, \dot\phi(t)\phi(t')\,e^{iS}$$

which agrees with $\theta(t-t')\langle \dot\phi(t)\phi(t')\rangle+\theta (t'-t)\langle \phi (t')\dot\phi (t)\rangle$ only if $t\neq t'$.$^1$

  • Example. Let $A(t)$ and $B(t)$ two functional at a given time, then you can check that

$$\int \dot A(t) B(t) \,e^{iS}=\langle\dot A(t)B(t)\rangle+\lim_{a\to 0}{1\over 2a}\langle [A(t),B(t)]\rangle$$

The last term represents the Dirac delta you found after deriving the step function.

  • While your question is very interesting, I think that your example is unfortunate since $\delta (t-t')[\phi(x),\phi(x')]=\delta (t-t')[\phi(t,\vec x),\phi(t,\vec x')]=0$. Let me choose an example in which this last commutator is different from zero to show how the time derivative of a correlation function splits in the two terms you mention: $\partial_t \langle \, T\, \phi(t)\,A(t')\,\rangle$, where $A$ is a functional of fields and momenta at a given time. Let's make a general variation (which doesn't modified the path integral measure) of the Hamiltonian or phase-space path integral ($S=\int dt\, \dot \phi\,\pi-H \, $ ). Since the momentum is an integration variable, the integral may not change:

$$0={\delta\over \delta \pi (t)}\int D\phi D\pi \, A(t')\,e^{iS}\,\delta \pi=\\ \int D\phi D\pi\, \left( {\delta A(t')\over \delta \pi (t)}+A(t')i\dot\phi (t)-A(t')(-i){\delta H (t)\over \delta \pi (t)}\right)\,e^{iS}\,\delta \pi$$

As $\delta \pi$ is a general variation and:

$${\delta A(t')\over \delta \pi (t)}=\delta(t-t')(-i)\,[\phi(t),A(t')]$$ $${\delta H (t)\over \delta \pi (t)}=-i\,[\phi(t), H]$$

we obtain:

$$\partial_t \langle \, T\, \phi(t)\,A(t')\,\rangle=i\langle \, T\, [\phi(t),H]\,A(t')\,\rangle+\delta (t-t')\,\langle \, [\phi(t),A(t')]\,\rangle \\ =\theta(t-t')\langle \, \dot\phi(t)\,A(t')\,\rangle+\theta(t'-t)\langle \, A(t')\,\dot\phi(t)\,\rangle+\delta (t-t')\,\langle \, [\phi(t),A(t')]\,\rangle$$

If, for example, $A(t')=\pi (t',\vec x')$, the last term gives $i\,\delta ^4 (x-x')$.

$\\$

Just in case it is not clear enough, let me remark that derivatives do commute with the path integral measure. The key point is that

$$\partial_t \,\left[\theta(t-t')\langle \phi(t)\phi(t')\rangle+\theta (t'-t)\langle \phi (t')\phi (t)\rangle\right]=\int D\phi\, {\phi(t+\epsilon^+)-\phi (t)\over \epsilon ^+}\phi(t')\,e^{iS}\neq\theta(t-t')\langle \dot\phi(t)\phi(t')\rangle+\theta (t'-t)\langle \phi (t')\dot\phi (t)\rangle$$

In addition, I would like to emphasize that ${\delta H (t)\over \delta \pi (t)}$ is a functional at a given time, while $\dot\phi (t)$ in the integrand of a path integral must be interpreted as ${\phi(t+\epsilon^+)-\phi (t)\over \epsilon ^+}$, that is, as a difference between fields evaluated at different times.


$^1$ Note that the derivative can be defined in different ways giving rise to different operator orderings, see Maimon's excellent answer Path integral formulation of quantum mechanics, be careful with some typo in the expressions: where says $x(t)p(t)$ should say $p(t)x(t)$.


EDIT: To derive some of the results above, one needs to take $\theta (0)=1/2$. However, one can proceed in a slightly different manner to avoid such choice (whi, in my opinion, is totally right). For example,

$$\int \dot A(t) B(t) \,e^{iS}=\int {A(t+a)-A(t)\over a } B(t+a/2) \,e^{iS}=\langle\dot A(t)B(t)\rangle+\lim_{a\to 0}{1\over a}\langle [A(t),B(t)]\rangle$$

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user drake
answered Jul 27, 2013 by drake (880 points) [ no revision ]
Most voted comments show all comments
+1. (Also: stray time derivatives inside the expectation values of line 1 of your first equation.)

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user user1504
Two questions:1. Even if it's $\int D\phi\, {\dot{\phi}(t+\epsilon^+)-\dot{\phi (t)}\over \epsilon ^+}\phi(t)\,e^{iS}$, according to my understanding of Ron Maimon's thread, will be equal to $\langle\ddot{\phi}(t)\phi(t)\rangle$(as long as we also use forward discretization to define $\dot{\phi}(t)$. 2. Again, the only way I could get a $\delta(0)\langle[A(t),B(t)]\rangle$ is to first get back to the operator definition where we have step functions, then differentiate, but this would be a circular logic. I want a calculation without any reference to operator definition, and compare the result

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user Jia Yiyang
with the operator one, see if they are consistent.

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user Jia Yiyang
@JiaYiyang I don't know where you are stuck. $\epsilon^{+}$ or $a$ are positive infinitesimals. Apply the standard relation and use $\theta (0)=1/2$, $\theta (a)=1$, $\theta (-a)=0$. You can also get a similar result if you use $\dot f(t)={f(t+a/2)-f(t-a/2)\over a}$ with $a$ a positive infinitesimal.

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user drake
@drake: OK done, check it out.

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user Jia Yiyang
Most recent comments show all comments
I realize my question is poorly written because it actually contains three questions: 1)where does the ambiguity come from. 2)why is Dyson-Schwinger "safe" in the sense it reproduces the operator result. 3)if we don't carefully define conventions, the feynman rules are different, which one to use? I think your answer at least clarifies the first question, thanks!@drake

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user Jia Yiyang
Please continue this discussion by moving it to Physics Chat .Thank you.

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user Manishearth

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverf$\varnothing$ow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...