Time-evolution operator $U_B[q]\equiv U_B(t)$

This is given by the solution of the Schrödinger equation,

$$i\frac{\partial}{\partial t} U_B(t) = H_B(t)U_B(t),\quad U_B(0)=1,$$ (172)

the formal solution of which is

$$U_B(t) = {\rm T} e^{-i\int_0^tdt' H_B(t')}$$ (173)

with the time-ordering operator T. Now, $U_B(t)$ can't be directly calculated from Eq. (7.177) because the $H_B(t')$ do not commute with each other at different times . One solution is to calculate $U_B(t)$ by direct evaluation of the path integral which is tedious but can be done. Here, we show an alternative solution: introduce the interaction picture and write

$$H_B(t) = H_0 + g(t) x \equiv H_0 + V(t)$$

$$U_B(t) = e^{-iH_0t} \tilde{U}(t),\quad i{\partial_t} \tilde{U}(t) = \tilde{V}(t)\tilde{U}(t)$$

$$\quad \tilde{V}(t) = e^{iH_0t} V(t) e^{-iH_0t}= g(t)\left( \hat{x} \cos \Omega t + \frac{\hat{p}}{M\Omega} \sin \Omega t \right).$$ (174)

We solve for $\tilde{U}(t)$ by making the general ansatz

$$\tilde{U}(t) = e^{-iA(t)}e^{-iB(t)\hat{x}}e^{-iC(t)\hat{p}}$$ (175)

with functions $A(t)$ etc to be determined by taking the time-derivative of $\tilde{U}(t)$. This yields

$$i\frac{\partial}{\partial t} \tilde{U}(t) = \dot{A}(t) \tilde{U}(t) + \hat{x} \dot{B}(t) \tilde{U}(t) + \dot{C}(t) e^{-iA(t)} \underline{e^{-iB(t)\hat{x}} \hat{p}} e^{-iC(t)\hat{p}}$$

$$= \dot{A}(t) \tilde{U}(t) + \hat{x} \dot{B}(t) \tilde{U}(t) + \dot{... ...B(t)\hat{x}} +\left[e^{-iB(t)\hat{x}},\hat{p}\right]} \right) e^{-iC(t)\hat{p}}$$

$$= [e^{-i\alpha x},p] = i {\partial_x} e^{-i\alpha x}= \alpha e^{-i\alpha x}$$

$$= \left( \dot{A}(t) + \hat{x} \dot{B}(t) + \hat{p} \dot{C}(t) + {B}(t) \dot{C}(t)\right) \tilde{U}(t) \equiv \tilde{V}(t)\tilde{U}(t).$$ (176)

Therefore, comparing with the expression for $\tilde{V}(t)$ yields

$$B(t) = \int_{0}^{t}dt' g(t') \cos \Omega t',\quad C(t) = \frac{1}{M\Omega}\int_{0}^{t}dt' g(t') \sin \Omega t'$$

$$A(t) = -\frac{1}{M\Omega} \int_{0}^{t}dt'\int_{0}^{t'}ds g(t') g(s) \cos \Omega s \sin \Omega t'$$ (177)

and therefore,

$$\langle x \vert U_B(t) \vert x' \rangle = \langle x \vert e^{-iH_0 t} e^{-iA(t)}e^{-iB(t)\hat{x}}e^{-iC(t)\hat{p}}\vert x' \rangle$$

$$= e^{-iA(t)} %%\int dy \langle x\vert e^{-iH_0 t} \vert y\rangle \l... ...B(t)\hat{x}} \langle x\vert e^{-iH_0 t} e^{-iB(t)\hat{x}} \vert x'+C(t) \rangle$$

$$= e^{-iA(t)}\langle x\vert e^{-iH_0 t}\vert x'+C(t) \rangle e^{-iB(t)\left[ x'+C(t)\right]}$$ (178)

In order to get explicit results here, we now need the propagator matrix elements for the harmonic oscillator,

$$\fbox{$ \displaystyle \langle x\vert e^{-iH_0 t}\vert x'\rangle =... ...a t} \left [ \left( x^2+x'^2\right) \cos \Omega t - 2 x x' \right] \right\}. $}$$ (179)

These again can either be obtained by direct evaluation of the single path integral for the harmonic oscillator or (somewhat simpler) by using the stationary eigenstates. The matrix element for the driven harmonic oscillator, $\langle x \vert U_B(t) \vert x' \rangle$, can then after some transformations (straightforward algebra with trig functions) be written as

$$\langle x \vert U_B(t) \vert x' \rangle = \sqrt{\frac{M\Omega}{2\pi i \sin \Omega t}} \exp \left\{ i S(x,t;x') \right\}$$

$$S(x,t;x') \equiv \frac{iM\Omega}{2 \sin \Omega t} \Big [ \left( x^2+x'^2\right) \cos \Omega t - 2 x x'$$

$$- \frac{2x}{M\Omega}\int_{0}^{t}dt' g(t') \sin \Omega t' -\frac{2x'}{M\Omega}\int_{0}^{t}dt' g(t') \sin \Omega (t-t')$$

$$- \frac{2}{M^2\Omega^2}\int_{0}^{t}\int_{0}^{t'}dt'ds g(t') g(s)\sin \Omega (t-t')\sin \Omega s \Big].$$ (180)

This coincides with the result given in L. S. Schulman, Techniques and Applications of Path Integration, Wiley (1981).