Time Ordering Operator for Fermions & Bosons

[kanato]

As I understand it, the time ordering operator works as follows (for t<0):
T c^\dagger(t) c(0) = -c(0) c^\dagger(t) for fermions and
T c^\dagger(t) c(0) = c(0) c^\dagger(t) for bosons.

Now suppose instead of these creation/annihilation operators, I had a more general commutation relation, ie [d,d^\dagger] = S, how does the time ordering operator behave?

Edit: After rereading that, I should be more specific. I'm trying to formulate DMFT equations in a non-orthogonal basis. So the creation/annihilation operators anti-commute to give [d_a,d_b^\dagger]_+ = S_{a,b}. The Green's function is usually defined as G(\tau) = \langle T c(\tau) c^\dagger(0) \rangle and I need to understand exactly what the time ordering operator does, but I'm not even totally sure how its really defined, because as I understand it above, it works differently for fermions and bosons.

 

[vanhees71]

The time-ordering operator just orders time-dependent operators in a product according to their time arguments (from right to left in ascending order). They don't care about any commutators.

Your definition of the time-ordered Green's function is correct with this definition. The time-ordering comes into the game when solving the time-evolution equation for an explicitly time-dependent Hamiltonian. This happens, e.g., in the usual interaction picture, where the operators time-evolove according to the free Hamiltonian $\hat{H}_0$, and the states with the interaction Hamiltonian $\hat{H}_I$, and $\hat{H}_I=\hat{H}_I(t)$ is usually time-dependent. The time-evolution operator $\hat{C}$ for the states obeys the equation of motion
$$\mathrm{i} \hbar \dot{\hat{C}}(t)=\hat{H}_I(t) \hat{C}(t),$$
which you cannot so easily integrate as it might look, because the Hamiltonian $\hat{H}_I(t)$ may not be commuting at different times.

The key to a (formal) solution is to rewrite the equation of motion (eom) in terms of an integral equation, working in the initial condition $\hat{C}(t_0)=\hat{1}$. Integrating the eom then leads to
$$\hat{C}(t)=\hat{1} -\mathrm{i}/\hbar \int_{t_0}^t \mathrm{d} t' \hat{H}_I(t') \hat{C}(t').$$
Now you can solve this equation iteratively, i.e., you start with the (very crude) approximation $\hat{C}_0(t)=\hat{1}$ and plug this approximation into the right-hand side of the integral eom, giving you
$$\hat{C}_1(t)=\hat{1} - \mathrm{i}/\hbar \int_{t_0}^t \mathrm{d} t_1 \hat{H}_I(t_1).$$
This solution you plug again into the right-hand side of the integral eom giving
$$\hat{C}_2(t)=\hat{1} - \mathrm{i}/\hbar \int_{t_0}^t \mathrm{d} t_1 \hat{H}_I(t_1) + (-\mathrm{i}/\hbar)^2 \int_{t_0}^t \mathrm{d} t_2 \int_{t_0}^{t_2} \mathrm{d} t_1 \hat{H}_I(t_2) \hat{H}_I(t_1).$$
Now you can rewrite the final integral by reading it as a surface integal in the $(t_1,t_2)$ plane. Instead of integrating over $t_2$ first you can as well integrate over $t_1$ first (just draw the triangular integration region!)
$$\int_{t_0}^t \mathrm{d} t_2 \int_{t_0}^{t_2} \mathrm{d} t_1 \hat{H}_I(t_2) \hat{H}_I(t_1) = \int_{t_0}^t \mathrm{d} t_1 \int_{t_1}^{t} \mathrm{d} t_2 \hat{H}_I(t_2) \hat{H}_I(t_1).$$
We always have to keep to Hamiltonian with the smaller time argument to the right. Now we rename the integration variables on the right-hand side of the equation and in another step use the time-ordering operator:
$$\int_{t_0}^t \mathrm{d} t_2 \int_{t_0}^{t_2} \mathrm{d} t_1 \hat{H}_I(t_2) \hat{H}_I(t_1) = \int_{t_0}^t \mathrm{d} t_2 \int_{t_2}^{t} \mathrm{d} t_1 \hat{H}_I(t_1) \hat{H}_I(t_2)= \int_{t_0}^t \mathrm{d} t_2 \int_{t_2}^{t} \mathrm{d} t_1 \mathcal{T} \hat{H}_I(t_2) \hat{H}_I(t_1).$$
Since we can also write a time ordering operator in front of the operator product on the left-hand side, we can just add the two equal integrals and divide by $2$, leading to
$$\int_{t_0}^t \mathrm{d} t_2 \int_{t_0}^{t_2} \mathrm{d} t_1 \hat{H}_I(t_2) \hat{H}_I(t_1) = \frac{1}{2} \int_{t_0}^t \mathrm{d} t_2 \int_{t_0}^t \mathrm{d} t_1 \mathcal{T} \hat{H}_I(t_2) \hat{H}_I(t_1).$$
This argument you can now iterate further, and finally you get as a formal solution of the eom
$$\hat{C}(t)=\mathcal{T} \exp \left [-\frac{\mathrm{i}}{\hbar} \int_{t_0}^t \mathrm{d} t' \hat{H}_I(t') \right].$$
Here you have to expand the exponential in its power series, giving each integral in the power another name of the time-integration variable, then the time-ordering symbol makes sense. The general correction of $N$th order thus reads
$$\frac{1}{N!} \left (-\frac{\mathrm{i}}{\hbar} \right)^N \int_{t_0}^t \mathrm{d} t_N \int_{t_0}^t \mathrm{d} t_{N-1} \cdots \int_{t_0}^t \mathrm{d} t_1 \mathcal{T} \hat{H}_I (t_N) \hat{H}_I(t_{N-1}) \cdot \hat{H}(t_1).$$
The Green's function becomes important in evaluating these integrals because of Wick's theorem for vacuum expectation values (see any QFT textbook).

 

[Rahzan]

In the penultimate expression by Vanhees71 the integral depends only on one time variable, so T becomes superfluous; simply integrate then exponentiate. Why keep T then?

 

[otennert]

Because this is an abstract notation only and not a simple integral. It is the result of a limiting procedure as vanhees71 has explained where actually an infinite number of in-between integrations have a fixed lower limit, but an upper limit which in turn is an integration variable in the next integration step, just as exemplified in the expression before.

 

[yucheng]

Now we rename the integration variables on the right-hand side of the equation and in another step use the time-ordering operator:
$$\int_{t_0}^t \mathrm{d} t_2 \int_{t_0}^{t_2} \mathrm{d} t_1 \hat{H}_I(t_2) \hat{H}_I(t_1) = \int_{t_0}^t \mathrm{d} t_2 \int_{t_2}^{t} \mathrm{d} t_1 \hat{H}_I(t_1) \hat{H}_I(t_2)= \int_{t_0}^t \mathrm{d} t_2 \int_{t_2}^{t} \mathrm{d} t_1 \mathcal{T} \hat{H}_I(t_2) \hat{H}_I(t_1).$$

Hi Vanhees. Is this a typo in your book?， https://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf

 

[vanhees71]

Thanks a lot of pointing this out. In (1.24) the time arguments have to be exchanged. It's always time-ordered, i.e., the operators are time-ordered from right to left. In (1.24) $\tau_2>\tau_1$, i.e.,
$$A^{(2)}=\int_{t_0}^t \mathrm{d} \tau_1 \int_{\tau_1}^t \mathrm{d} \tau_2 \mathbf{X}(\tau_2) \mathbf{X}(\tau_1).$$
I also corrected the typo in the corresponding figure ;-).