# Confusion with time ordering

I am thinking about Proof of correlation function formula in quantum field theory and have realized there is a deeper confusion underpinning that. Consider:

$$T\{U_I(T, t_2)\Phi_I(x_1)\}$$

where $t_2>t_1$ where these are a field operator and the time evolution operator, respectively, both in the Interaction picture.$A=U_I(T, t_2)=T\big\{exp\big(-i\int_{t_2}^{T} dt'\, H_I(t')\big)\big\} = e^{iH_0 (T-t_0)}e^{iH (T-t_2)}e^{iH_0 (t_2-t_0)}$ See here for derivation.

Now, the first representation with an integral in the exponent, only has operators (in interaction picture) with time in the interval $(T, t_2) > t_1$, so we find:

\begin{align*} T\{U_I(T, t_2)\Phi_I(x_1)\} &=T\{T\big\{\exp\big(-i\int_{t_2}^{T} dt'\, H_I(t')\big)\big\}\Phi_I(x_1)\} \\ &=T\big\{\exp\big(-i\int_{t_2}^{T} dt' H_I(t')\big)\big\}\Phi_I(x_1)=U_I(T, t_2)\Phi_I(x_1) \end{align*}

However, the second representation has only operators in the Schrodinger picture, which are equal to those operators in the interaction picture, at the arbitrary time $t_0$, which may well be chosen to be less than $t_1$, and so we get:

\begin{align*} T\{U_I(T, t_2)\Phi_I(x_1)\} &= T\{e^{i(T-t_0)\cdot H_{I,0}(t_0)}e^{i(T-t_2)\cdot H_I(t_0)}e^{i(t_2-t_0)\cdot H_{I,0}(t_0)}\Phi_I(x_1)\} \\ &=\Phi_I(x_1)e^{i(T-t_0)\cdot H_{I,0}(t_0)}e^{i(T-t_2)\cdot H_I(t_0)}e^{i(t_2-t_0)\cdot H_{I,0}(t_0)} \\ &=\Phi_I(x_1)U_I(T, t_2) \neq U_I(T, t_2)\Phi_I(x_1) \end{align*}

• Operators like $D$ simply do not occur! It maps from one time-mark to two different, unrelated time-marks. The time arguments stem (usually) from the interaction-picture time evolved field operators, so you can at most define $D = AB$ in the Schrödinger picture, then $D(t) = A(t)B(t)$ (i.e. they receive the same time argument). – Sebastian Riese Dec 14 '15 at 17:45
• But in the example given, $A = U(T, t_2)$ has two time variables not one, and $t_1$ does not appear at all! How should I compare it with the original question? I believe your example is not so illustrative. I think you should revise the question before an answer is possible to provide. Do you want to prove $U_I(T, t_2)= T\big\{exp\big(-i\int_{t_2}^{T} dt' H_I(t')\big)\big\} = e^{iH_0 (T-t_0)}e^{iH (T-t_2)}e^{iH_0 (t_2-t_0)}$? – AlQuemist Dec 14 '15 at 18:51
• I still fail to see the problem. The correct way to apply time ordering is always in the Taylor expansion of the exponential, this way you can show, that the $T \exp {\int \cdots}$ expression fulfils the time-evolution equation for $U_I$ and then you can show that $U_I$ has the second form stated by evolving a state. – Sebastian Riese Dec 14 '15 at 18:53
• @guillefix As I've said: There is no time ordering defined for $U$ only for time-marks on field operators. You have to expand the $\exp$ into a power series and then the time ordering will place $\Phi$ where it belongs. The "outer" time ordering still acts on all individual field operators in the expansion of $U$. – Sebastian Riese Dec 14 '15 at 19:06
• @SebastianRiese: could you modify the question accordingly, and provide an answer? – AlQuemist Dec 14 '15 at 19:11

The first result is correct. Time ordering acts on interaction picture field operators. $H_I(t)$ is to be understood as a function of interaction picture field operators at time $t$.
The representation $$U(t, t') = e^{iH_0t}e^{-iH(t - t')}e^{-iH_0t}$$ does not represent $U$ in terms of interaction picture field operators and is therefore not permissible. Only because the operators are the same at that instant does not mean you can use them as interaction picture operators. The point here is, that the time-marks occur also explicitly, while the time dependence has to be implicit as the time dependence of the field operators for the technique to work.
To correctly evaluate expressions involving time ordering one has to expand the exponential and time order every term: $$T\left\{\exp \left(\int_{t_1}^{t_2}dt\,H(t) \right) \phi(t_3) \right\} = \sum_{n=0}^\infty \frac{(-i)^n}{n!} \int_{t_1}^{t_2} dt'_1 \cdots \int_{t_1}^{t_2} dt'_n T H(t'_1) \cdots H(t'_n)\phi(t_3).$$ If now $t_3 < t_1 < t_3$ the operator $\phi(t_3)$ will always be ordered to the right, and therefore can be factored out to the right and so the result is the time evolution operator followed by $\phi(t_3)$.