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Given two bosonic operators $A$, $B$ (in the Heisenberg picture) in a QFT, the time-ordered product of $A$ and $B$ is defined as $$ T\{A(t_1)B(t_2)\}=\theta(t_1-t_2)A(t_1)B(t_2)+\theta(t_2-t_1)B(t_2)A(t_1),\tag{I} $$ where $\theta(t)$ is the Heaviside step function.

However, assuming that our theory has time translational symmetry, and assuming we have $$ A(t)=e^{aH\cdot (t-t_0)}A(t_0)e^{-aH\cdot(t-t_0)}\tag{II}\\ (a=i~\text{for the Lorentzian signature, $a=1$ for the Euclidean signature}), $$ one might write down the following expression by substituting (II) into the l.h.s. of (I) $$ \begin{align} T\{A(t_1)B(t_2)\}=&T\{e^{aH(t_0)\cdot(t_1-t_0)}A(t_0)e^{-aH(t_0)\cdot(t_1-t_0)}B(t_2)\}\nonumber\\ =&\theta(t_0-t_2)e^{aH(t_0)\cdot(t_1-t_0)}A(t_0)e^{-aH(t_0)\cdot(t_1-t_0)}B(t_2)\\ +&\theta(t_2-t_0)B(t_2)e^{aH(t_0)\cdot(t_1-t_0)}A(t_0)e^{-aH(t_0)\cdot(t_1-t_0)}\tag{III}. \end{align} $$ Here, in the first line, I select a special time slice $t=t_0$ to compute the Hamiltonian, i.e., $$ H(t_0)=\int d^{d-1}\vec x \,\mathcal{H}(t_0,\vec x) $$ where $\mathcal{H}$ is the Hamiltonian density in the Heisenberg picture. It seems that (III) is incorrect, because we can use (II) again to rewrite (III) as $$ T\{A(t_1)B(t_2)\}=\theta(t_0-t_2)A(t_1)B(t_2)+\theta(t_2-t_0)B(t_2)A(t_1),\tag{IV} $$ which is obviously ridiculous. Where did I go wrong? What stops me from deriving eq. (III)

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TL;DR: It seems that to consistently apply a picture changing/symmetry operator [such as the time evolution operator $\hat{U}$], one should apply it to all operators in an expression democratically, e.g. in OP's case: Use eq. (II) on both operators $\hat{A}_H(t_1)$ and $\hat{B}_H(t_2)$.

  1. Recall the relationship $$\hat{A}_H(t)~=~\hat{U}(t)^{-1} \hat{A}_S(t)\hat{U}(t)\tag{A}$$ between the Schrödinger & Heisenberg pictures, where the time evolution operator is $$ \hat{U}(t)~:=~\left\{\begin{array}{rcl} T\exp\left[-\frac{i}{\hbar}\int_0^t\! dt^{\prime}~\hat{H}_S(t^{\prime})\right] &\text{for}& t ~\geq~0, \cr AT\exp\left[-\frac{i}{\hbar}\int_0^t\! dt^{\prime}~\hat{H}_S(t^{\prime})\right] &\text{for}& t ~\leq~0, \end{array}\right. \tag{B} $$ cf. e.g. this Phys.SE post.

  2. In eq. (B) we have generalized to Hamiltonians $\hat{H}_S(t)$ with explicit time dependence to make clear that the time evolution operator $\hat{U}(t)$ is more generally a Wilson-line-like operator that depends on a 1D curve worth of different times.

  3. Therefore, OP's eq. (II) is more generally$^1$ $$ \hat{A}_H(t)~=~\hat{U}(t)^{-1}\hat{U}(t_0)\hat{A}_H(t_0)\hat{U}(t_0)^{-1}\hat{U}(t) \tag{II}$$

  4. If we substitute the expression (II) for the $\hat{A}_H$ operator into OP's eq. (I), the argument $\hat{A}_H(t_1)\hat{B}_H(t_2)$ of the time-ordering operator $T$ does no longer depend on just 2 times $t_1$ & $t_2$, but rather infinitely many times, and thus it becomes more involved/less straightforward to apply the time-ordering operator $T$ in the way OP intends to do.

$^1$ Interestingly, the geometric interpretation of eq. (II) is not just a Wilson loop $t\to t_0 \to t$, but more involved.

Eq. (II) should be contrasted with the interaction picture where the analog $$ \hat{A}_I(t)~=~\hat{U}_I(t,t_0)\hat{A}_I(t_0)\hat{U}_I(t_0,t) \tag{C}$$ does have a geometric interpretation of a Wilson loop $t\to t_0 \to t$.

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  • $\begingroup$ Hi, @Qmechanic, thank you for your continuous attention to this problem. Your latest update looks correct and inspires me a lot. However, after accepting your logic, I have one more quick question: how do I return to the Schrodinger picture if the picture-changing operators I apply to every operator (in the Heisenberg picture) are the same? Anyway, I will accept this answer. $\endgroup$ Commented yesterday
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The l.h.s of expression $(III)$ you are time ordering with respect to times $t_1$ and $t_2$ whereas in the r.h.s you have changed this to $t_0$ and $t_2$ making them not equivalent. $t_0$ is an initial time you define your operators at, not a parameter.

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  • $\begingroup$ Hi, I'm confused about the point that when I write down $e^{a H(t_0)\cdot t} A(t_0)e^{-aH(t_0)\cdot t}$ in the first line of eq. (III), should I regard it as a composite operator defined at the time $t_0$, or an operator defined at the time $t_0+t$ $\endgroup$ Commented 2 days ago
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    $\begingroup$ The operator is defined at $t$ not $t_0$. The operator $A(t_0)$ which is evolved via the time evolution operator (to give $A(t)$) is just an initial operator, in the same way that a time dependent state in the Schrodinger picture is known at some initial time $t_0$, then given at $t$ by applying the time evolution operator. $\endgroup$ Commented 2 days ago
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The thime-ordering operator is applied to functions of time, not to the operators themselves. Your input are the two functions $t\mapsto A(t)$ and $t\mapsto B(t)$, and the output is a function of two arguments \begin{equation} (t_1,t_2)\mapsto \theta(t_1-t_2)A(t_1)B(t_2)+ \theta(t_2-t_1)A(t_2)B(t_1). \end{equation} In physical tradition one usually does not distinguish between the function $t\mapsto A(t)$ and its particular value $A(t)$ for some fixed value of $t$, which may lead to such confusions at the beginning.

Things get only worse in Quantum field theory, where, for example, by expression like \begin{equation} T\{\phi(x)^2\phi(y)\partial_{\mu}\phi(y)\} \end{equation} one actually means \begin{equation} \lim_{\substack{x_1\to x,\ x_2\to x,\\ y_1\to y,\ y_2 \to y}}\partial_{y_1^\mu}T\{\phi(x_1)\phi(x_2)\phi(y_1)\phi(y_2)\}, \end{equation} where, in particular, the derivative acts on the Heaviside functions, and it is a separate story how the limit should be rightly defined. The moral of this example is that in reality you should always think of the time-ordered product as an abbreivation of something more complicated, not just as a function of the arguments. With time you will get some intuition how to deal with them.

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  • $\begingroup$ Thank you, Budjum. I like your answer, and I think the spirit of my answer regarding this S.E. question(physics.stackexchange.com/questions/864388/…) somehow follows what you said, "Things get only worse in Quantum field theory...one actually means..." Do you know of any Refs that explicitly discuss this point of view? Thanks again : ) $\endgroup$ Commented 8 hours ago

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