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How do we define time-ordering operations in QFT?

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)

Answer*

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](https://en.wikipedia.org/wiki/Heisenberg_picture) 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](https://physics.stackexchange.com/q/682445/2451) 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](https://en.wikipedia.org/wiki/Wilson_loop)-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 eq. (II) 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](https://en.wikipedia.org/wiki/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 $t\to t_0 \to t$.

Edit Summary*

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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. $\endgroup$

Hezaraki
– Hezaraki

2025-12-06 09:46:12 +00:00
Commented 16 hours ago

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