A Do Time-ordering and Time Integrals commute? Peskin(4.22)(4.31)(4.44)

George Wu
Messages
6
Reaction score
3
TL;DR Summary
In Peskin's QFT textbook, according to the explanation below(4.22), the Time-ordered exponential is just a notation, however in the derivation of (4.31),the Time-ordered exponential seems more than just a notation.
In Peskin P85:
1683566820582.png

It says the Time-ordered exponential is just a notation,in my understanding, it means
$$\begin{aligned}
&T\left\{ \exp \left[ -i\int_{t_0}^t{d}t^{\prime}H_I\left( t^{\prime} \right) \right] \right\}\\
&\ne T\left\{ 1+(-i)\int_{t_0}^t{d}t_1H_I\left( t_1 \right) +\frac{(-i)^2}{2!}\int_{t_0}^t{d}t_1dt_2H_I\left( t_1 \right) H_I\left( t_2 \right) +\cdots \right\}\\
\end{aligned}$$
However in the derivation of (4.31):
1683567804588.png

A way to derivate this is: Jens Wagemaker (https://physics.stackexchange.com/users/103623/jens-wagemaker), Why can the time-ordered exponentials be brought to the right?, URL (version: 2023-05-06): https://physics.stackexchange.com/q/762829
Here is his way of derivation:
Note that the time ordering operator is like a sorting algorithm, hence it doesn't matter if we permute something before applying the time ordering operator. In particular we can perform some additional time-ordering by inserting an additional time-ordering operator:
$$\begin{aligned}
A&=T\left\{ \phi _I(x)\phi _I(y)\exp \left[ -i\int_{-T}^T{d}tH_I(t) \right] \right\}\\
&=T\left\{ \phi _I(x)\phi _I(y)T\left\{ \exp \left[ -i\int_{-T}^T{d}tH_I(t) \right] \right\} \right\}\\
&=T\left\{ \phi _I(x)\phi _I(y)U(T,-T) \right\}\\
\end{aligned}$$
We show the case ##x_0>y_0##. By 4.26 we get
$$U(T,-T)=U\left(T, x_0\right) U\left(x_0, y_0\right) U\left(y_0,-T\right)$$
which we substitute.
$$A=T\left\{ \phi _I(x)\phi _I(y)U\left( T,x_0 \right) U\left( x_0,y_0 \right) U\left( y_0,-T \right) \right\} $$
Now we want to apply the time ordering. For this we note that the ##U\left(T, x_0\right)## contains only operators with the with in the interval ##\left[T, x_0\right]##, and similar for the terms ##U\left(x_0, y_0\right)## and##U\left(y_0,-T\right)##. Hence, if we apply the time ordering we get.
##A=U\left(T, x_0\right) \phi_I(x) U\left(x_0, y_0\right) \phi_I(y) U\left(y_0,-T\right)##, which occurs in the numerator of your second expression.
This step seems straightforward:
$$\begin{aligned}
A&=T\left\{ \phi _I(x)\phi _I(y)\exp \left[ -i\int_{-T}^T{d}tH_I(t) \right] \right\}\\
&=T\left\{ \phi _I(x)\phi _I(y)T\left\{ \exp \left[ -i\int_{-T}^T{d}tH_I(t) \right] \right\} \right\}\\
\end{aligned}$$
However this means:
$$\begin{aligned}
A&=T\left\{ \phi _I(x)\phi _I(y)\exp \left[ -i\int_{-T}^T{d}tH_I(t) \right] \right\}\\
&=T\left\{ \phi _I(x)\phi _I(y)\left[ 1+(-i)\int_{t_0}^t{d}t_1H_I\left( t_1 \right) +\frac{(-i)^2}{2!}\int_{t_0}^t{d}t_1dt_2H_I\left( t_1 \right) H_I\left( t_2 \right) +\cdots \right] \right\}\\
&=T\left\{ \phi _I(x)\phi _I(y)T\left[ 1+(-i)\int_{t_0}^t{d}t_1H_I\left( t_1 \right) +\frac{(-i)^2}{2!}\int_{t_0}^t{d}t_1dt_2H_I\left( t_1 \right) H_I\left( t_2 \right) +\cdots \right] \right\}\\
&=T\left\{ \phi _I(x)\phi _I(y)\left[ 1+(-i)\int_{t_0}^t{d}t_1H_I\left( t_1 \right) +\frac{(-i)^2}{2!}\int_{t_0}^t{d}t_1dt_2T\left\{ H_I\left( t_1 \right) H_I\left( t_2 \right) \right\} +\cdots \right] \right\}\\
&=T\left\{ \phi _I(x)\phi _I(y)T\left\{ \exp \left[ -i\int_{-T}^T{d}tH_I(t) \right] \right\} \right\}\\
&=T\left\{ \phi _I(x)\phi _I(y)U(T,-T) \right\}\\
\end{aligned}$$
In this process we use:
$$\begin{aligned}
&T\left\{ 1+(-i)\int_{t_0}^t{d}t_1H_I\left( t_1 \right) +\frac{(-i)^2}{2!}\int_{t_0}^t{d}t_1dt_2H_I\left( t_1 \right) H_I\left( t_2 \right) +\cdots \right\}\\
&=1+(-i)\int_{t_0}^t{d}t_1H_I\left( t_1 \right) +\frac{(-i)^2}{2!}\int_{t_0}^t{d}t_1dt_2T\left\{ H_I\left( t_1 \right) H_I\left( t_2 \right) \right\} +\cdots\\
\end{aligned}$$
which means :
$$
T\int_{t_0}^t{d}t_1\cdots dt_nH_I\left( t_1 \right) \cdots H_I\left( t_n \right) =\int_{t_0}^t{d}t_1\cdots dt_nT\left\{ H_I\left( t_1 \right) \cdots H_I\left( t_n \right) \right\} $$
What's more, in the derivation of (4.44):
1683568909241.png

In order to apply wick's theorm, Time-ordering must go inside the integral.
So my question is:
Does Time-ordering and Time Integral commute?
Or in another word:
Can Time-ordering can go inside the integral?
 
Last edited:
Physics news on Phys.org
I have found a great answer:
Prahar (https://physics.stackexchange.com/users/8821/prahar), Time ordering of integral, URL (version: 2022-12-17): https://physics.stackexchange.com/q/741494
$$T\int_{t_0}^t{d}t_1\cdots dt_nH_I\left( t_1 \right) \cdots H_I\left( t_n \right)$$
is defined to equal to
$$\int_{t_0}^t{d}t_1\cdots dt_nT\left\{ H_I\left( t_1 \right) \cdots H_I\left( t_n \right) \right\}$$
Otherwise the left side does mean anything.
This definition can be extended to cases like##T\left\{ \phi \left( t\prime_1 \right) \cdots \phi \left( t\prime_m \right) \int_{t_0}^t{d}t_1\cdots dt_n\phi \left( t_1 \right) \cdots \phi \left( t_n \right) \right\} ##
$$\begin{aligned}
T\left\{ \phi \left( t\prime_1 \right) \cdots \phi \left( t\prime_m \right) \int_{t_0}^t{d}t_1\cdots dt_n\phi \left( t_1 \right) \cdots \phi \left( t_n \right) \right\} :&=T\left\{ \phi \left( t\prime_1 \right) \cdots \phi \left( t\prime_m \right) \int_{t_0}^t{d}t_1\cdots dt_nT\left\{ \phi \left( t_1 \right) \cdots \phi \left( t_n \right) \right\} \right\}\\
&=\int_{t_0}^t{d}t_1\cdots dt_nT\left\{ \phi \left( t\prime_1 \right) \cdots \phi \left( t\prime_m \right) \phi \left( t_1 \right) \cdots \phi \left( t_n \right) \right\}\\
\end{aligned}$$
 
Yes, they commute and it can go inside the integral. Your first inequality should be the equality.
 
I think it's a definition of the time-ordering symbol to apply it to the integrand, i.e., to commute it with the integral. It's defined to get a compact notation of the Dyson series of time-dependent perturbation theory in terms of a "time-ordered exponential". If you look at the derivation of this formula, that'll become clear. See, e.g., Sect. 1.9, where it's derived for potential scattering in non-relativistic QM:

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
According to recent podcast between Jacob Barandes and Sean Carroll, Barandes claims that putting a sensitive qubit near one of the slits of a double slit interference experiment is sufficient to break the interference pattern. Here are his words from the official transcript: Is that true? Caveats I see: The qubit is a quantum object, so if the particle was in a superposition of up and down, the qubit can be in a superposition too. Measuring the qubit in an orthogonal direction might...
Back
Top