B.4 The Evolution Operator $\hat{S}$

To solve the equations of motion in the interaction picture (B.7), a unitary operator $\hat{S}(t,t_0)$ that determines the state vector at time $t$ in terms of the state vector at time $t_0$ is introduced

$$\begin{array}{l} \displaystyle \vert\Psi_\mathrm{I}(t)\rangle... ...\hat{S}(t,t_0) \vert\Psi_\mathrm{I}(t_0)\rangle \ , \end{array}$$ (B.15)

$\hat{S}$ satisfies the initial condition $\hat{S}(t_0,t_0)=1$. For finite times $\hat{S}(t,t_0)$ can be constructed explicitly by employing the SCHRÖDINGER picture

$$\begin{array}{ll} \displaystyle \vert\Psi_\mathrm{I}(t)\rangl... ..._{0}t_0/\hbar} \vert\Psi_\mathrm{I}(t_0)\rangle \ , \end{array}$$ (B.16)

which therefore identifies

$$\begin{array}{l} \hat{S}(t,t_0) \ = \ e^{i\hat{H}_{0}t/\hbar}... ...hat{H}(t-t_0)/\hbar} e^{-i\hat{H}_{0}t_0/\hbar} \ . \end{array}$$ (B.17)

Since $\hat{H}$ and $\hat{H}_0$ do not commute with each other, the order of the operators must be carefully maintained. Equation (B.17) immediately yields several general properties of $\hat{S}$ [189]

• $\hat{S}^\dagger(t,t_{0})\hat{S}(t,t_{0}) = \hat{S}(t,t_{0})\hat{S}^\dagger(t,t_{0}) = 1$, implying that $\hat{S}$ is unitary $\hat{S}^\dagger(t,t_{0})=\hat{S}^{-1}(t,t_{0})$,
• $\hat{S}(t_{1},t_{2})\hat{S}(t_{2},t_{3}) = \hat{S}(t_{1},t_{3})$, which shows that $\hat{S}$ has the group property, and
• $\hat{S}(t,t_{0})\hat{S}(t_{0},t) = 1$, implying that $\hat{S}(t_{0},t)=\hat{S}^\dagger(t,t_{0})$ .

Although (B.17) is the formal solution to the problem posed by (B.15), it is not very useful for computational purposes. Instead one can construct an integral equation for $\hat{S}$, which can then be solved by iteration. It follows from (B.7) and (B.15) that $\hat{S}$ satisfies the differential equation

$$\begin{array}{l} \displaystyle i\hbar \partial_t\hat{S}(t,t_0... ...at{H}^\mathrm{int}_\mathrm{I}(t) \hat{S}(t,t_0) \ . \end{array}$$ (B.18)

Integrating both sides of the (B.18) with respect to time with the initial condition $\hat{S}(t_0,t_0)=1$ yields

$$\begin{array}{ll} \displaystyle \hat{S}(t,t_0) & \displaystyl... ...athrm{int}_\mathrm{I}(t_{1}) \hat{S}(t_{1},t_0) \ . \end{array}$$ (B.19)

By iterating this equation repeatedly one gets

$$\begin{array}{lll} \displaystyle \hat{S}(t,t_0) &\ = \ & \dis... ...{2})\ldots \hat{H}^\mathrm{int}_\mathrm{I}(t_{n})\ .\end{array}$$ (B.20)

Equation (B.20) has the characteristic feature that the operator containing the latest time stands farthest to the left. At this point it is convenient to introduce the time-ordering operator denoted by the symbol $T_\mathrm{t}$

$$\displaystyle T_\mathrm{t}\{\hat{A}(t_{1})\hat{B}(t_{2})\}\ = \ \... ...}(t_{1})\hat{B}(t_{2})\ + \ \theta(t_{2}-t_{1})\hat{B}(t_{2})\hat{A}(t_{1}) \ .$$ (B.21)

where $\theta(t)$ is the step function^B.1. Each time two FERMIons are interchanged, the resulting expression changes its sign. By rearranging the integral using $T_\mathrm{t}$

$$\begin{array}{l}\displaystyle \frac{1}{2!}\int_{t_0}^{t} dt_{... ...}(t_{2}) \hat{H}^\mathrm{int}_\mathrm{I}(t_{1}) \ . \end{array}$$ (B.22)

The second term on the right hand-side is equal to the first, which is easy to see by just redefining the integration variables $t_{1}\rightarrow t_{2}$, $t_{2}\rightarrow t_{1}$. Thus one gets

$$\frac{1}{2!}\int_{t_0}^{t} dt_{1} \int_{t_0}^{t} dt_{2}\ T_\mathr... ...at{H}^\mathrm{int}_\mathrm{I}(t_{1}) \hat{H}^\mathrm{int}_\mathrm{I}(t_{2}) \ .$$ (B.23)

Thus for the expansion of the $\hat{S}(t,t_0)$ one obtains

$$\begin{array}{ll} \displaystyle \hat{S}(t,t_0) & \displaystyl... ...t'\hat{H}^\mathrm{int}_\mathrm{I}(t') \right) \} \ .\end{array}$$ (B.24)

M. Pourfath: Numerical Study of Quantum Transport in Carbon Nanotube-Based Transistors