Driven simple harmonic oscillator

We want to solve the time-dependent Schrödinger equation for the full Hamiltonian \(\hat{H}=\hat{H}_0+\hat{V}(t)\) where \[\hat{H}_0=\hbar\omega\left(\hat{a}^\dagger\hat{a}+\frac{1}{2}\right) \quad \text{and}\quad\hat{V}(t)=f(t)\hat{a}^\dagger+f^*(t)\hat{a}\] Since \(\hat{a}+\hat{a}^\dagger\propto\hat{x}\), \(\hat{V}(t)\) acts like a driving force moving \(x\) as a function of time. Recall that \([\hat{a},\hat{a}^\dagger]=1\). Then, define \[\hat{A}=\hat{a}+\frac{f(t)}{\hbar\omega}\] One can easily verify that \[[\hat{A},\hat{A}^\dagger]=[\hat{a},\hat{a}^\dagger]=1\] Then, \[\hbar\omega\left(\hat{A}^\dagger\hat{A}+\frac{1}{2}\right)=\hbar\omega\left(\hat{a}^\dagger\hat{a}+\hat{a}^\dagger\frac{f(t)}{\hbar\omega}+\hat{a}\frac{f^*(t)}{\hbar\omega}+\frac{|f(t)|^2}{(\hbar\omega)^2}+\frac{1}{2}\right)=\hat{H}(t)+\frac{|f(t)|^2}{\hbar\omega}\] So, \[\hat{H}(t)=\hbar\omega\left(\hat{A}^\dagger\hat{A}+\frac{1}{2}\right)-\frac{|f(t)|^2}{\hbar\omega}\] We define the following: \[\hat{A}\left|0\right\rangle=0, \quad \left|n\right\rangle=\frac{(\hat{A}^\dagger)^n}{\sqrt{n!}}\left|0\right\rangle\] Then, \[\hat{H}\left|n\right\rangle=E_n(t)\left|n\right\rangle=\left[\hbar\omega\left(n+\frac{1}{2}\right)-\frac{|f(t)|^2}{\hbar\omega}\right]\left|n\right\rangle\] Expand \(\left|\psi(t)\right\rangle=\sum_nc_n(t)\left|n\right\rangle\). \[i\hbar \ \partial_t\left|\psi(t)\right\rangle=\sum_ni\hbar\left[\partial_tc_n(t)\left|n\right\rangle+c_n(t)\partial_t\left|n\right\rangle\right]=\hat{H}\left|\psi(t)\right\rangle=\sum_nc_n(t)E_n(t)\left|n\right\rangle.\] Multiply by \(\left\langle m\right|\) to get \[i\hbar \ \partial_tc_m(t)+i\hbar\sum_nc_n(t)\left\langle m\middle|\partial_t\middle|n\right\rangle=E_m(t)c_m(t)\] So we need to calculate \[\begin{aligned} \partial_t\left|n\right\rangle=\partial_t\frac{(\hat{A}^\dagger)^n}{\sqrt{n!}}\left|0\right\rangle=& \ \partial_t\left(\frac{\left (\hat{a}^\dagger+\frac{f^*(t)}{\hbar\omega}\right)^n}{\sqrt{n!}}\right)\left|0\right\rangle \\ =& \ n\frac{\left(\hat{a}^\dagger+\frac{f^*(t)}{\hbar\omega}\right)^{n-1}}{\sqrt{n!}}\frac{df^*(t)}{dt}\frac{1}{\hbar\omega}\left|0\right\rangle+\frac{(\hat{A}^\dagger)^n}{\sqrt{n!}}\partial_t\left|0\right\rangle \\ =& \ \frac{\sqrt{n}}{\hbar\omega}\frac{df^*(t)}{dt}\left|n-1\right\rangle+\frac{(\hat{A}^\dagger)^n}{\sqrt{n!}}\partial_t\left|0\right\rangle \end{aligned}\] But \(\hat{A}\left|0\right\rangle=0\) implies that \(\partial_t\left(\hat{a}+\frac{f(t)}{\hbar\omega}\right)\left|0\right\rangle=0\), which further implies that \[\frac{df}{dt}\frac{1}{\hbar\omega}\left|0\right\rangle+\left(\hat{a}+\frac{f(t)}{\hbar\omega}\right)\partial_t\left|0\right\rangle=0.\] Multiply by \(\left\langle m\right|\) to get \[\frac{df}{dt}\frac{1}{\hbar\omega}\delta_{m0}+\left\langle m\middle|\hat{A}\partial_t\middle|0\right\rangle=0\] \[\implies\sqrt{m+1}\left\langle m+1\middle|\partial_t\middle|0\right\rangle=-\frac{df}{dt}\frac{1}{\hbar\omega}\delta_{m0}\] so, \[\left\langle m\middle|\partial_t\middle|0\right\rangle=-\frac{df}{dt}\frac{1}{\hbar\omega}\delta_{m1}\] \[\implies\partial_t\left|n\right\rangle=\frac{\sqrt{n}}{\hbar\omega}\frac{df^*}{dt}\left|n-1\right\rangle-\frac{\sqrt{n+1}}{\hbar\omega}\frac{df}{dt}\left|n+1\right\rangle.\] Substituting back into the Schrödinger equation, \[i\hbar \ \partial_tc_m(t)+i\hbar\frac{\sqrt{m+1}}{\hbar\omega}\frac{df^*}{dt}c_{m+1}(t)-\frac{i\hbar\sqrt{m}}{\hbar\omega}\frac{df}{dt}c_{m-1}(t)=E_m(t)c_m(t)\] So we have, \[\boxed{\frac{dc_m(t)}{dt}=-\frac{i}{\hbar}E_m(t)c_m(t)-\frac{\sqrt{m+1}}{\hbar\omega}\frac{df^*}{dt}c_{m+1}(t)+\frac{\sqrt{m}}{\hbar\omega}\frac{df}{dt}c_{m-1}(t)}\] This equation is a complicated coupled linear differential equation with no obvious solution when \(f\ne 0\). So, this attempt at a solution does not work.

Interaction representation

Let’s examine instead with the interaction representation picture. \[\left|\psi_S(t)\right\rangle=e^{-\frac{i}{\hbar}\hat{H}_0t}T\exp\left(-\frac{i}{\hbar}\int_0^tdt'\hat{V}_I(t')\right)\left|\psi_S(0)\right\rangle\] where \[\hat{V}_I(t)=e^{\frac{i}{\hbar}\hat{H}_0t}\hat{V}(t)e^{-\frac{i}{\hbar}\hat{H}_0t}.\] Note that \(V_I\) is different from \(V\), Not recognizing this is a common mistake and will lead to errors. In our case, we have \[\hat{V}_I(t)=e^{\frac{i}{\hbar}\hbar\omega\hat{a}^\dagger\hat{a}t}[f\hat{a}^\dagger+f^*\hat{a}]e^{-i\omega\hat{a}^\dagger\hat{a}t}.\] But \(e^{i\lambda\hat{a}^\dagger\hat{a}}\hat{a}^\dagger e^{-i\lambda\hat{a}^\dagger\hat{a}}\equiv g(\lambda)\) so \[ie^{i\lambda\hat{a}^\dagger\hat{a}}[\hat{a}^\dagger\hat{a},\hat{a}^\dagger]e^{-i\lambda\hat{a}^\dagger\hat{a}}=\frac{dg(\lambda)}{d\lambda}\] \[ig(\lambda)=\frac{dg(\lambda)}{d\lambda}\implies g(\lambda)=g(0)e^{i\lambda}.\] So, \[\boxed{\hat{V}_I(t)=f(t)e^{i\omega t}\hat{a}^\dagger+f^*(t)e^{-i\omega t}\hat{a}}.\] It could also be evaluated in our standard way to evaluate Hadamards as before, yielding the same result. It is useful to note that \[\begin{aligned} [\hat{V}_I(t),\hat{V}_I(t')]=& \ f(t)f^*(t')e^{i\omega t}e^{-i\omega t'}[\hat{a}^\dagger,\hat{a}]+f^*(t)f(t')e^{-i\omega t}e^{i\omega t'}[\hat{a},\hat{a}^\dagger] \\ =& \ -2i \ \text{Im}\left (f(t)f^*(t')e^{i\omega(t-t')}\right), \end{aligned}\] which is just a number. So, \([\hat{V}_I(t),\hat{V}_I(t')]\) commutes with all operators.
 
To be concrete, choose \(f(t)=Ce^{i\Omega t}\) with \(C\in\mathbb{R}\). Then, \[\hat{V}_I(t)=C(e^{i(\omega+\Omega)t}\hat{a}^\dagger+e^{-i(\omega+\Omega)t}\hat{a})\] and \[[\hat{V}_I(t),\hat{V}_I(t')]=-2iC^2 \ \text{Im}(e^{i\Omega}(t-t')e^{i\omega(t-t')})=-2iC^2\sin((\omega+\Omega)(t-t')).\] Gottfried says to consider \[\hat{W}(t)=\int_0^t\hat{V}_I(t') \ dt'=\frac{C}{i(\omega+\Omega)}(e^{i(\Omega+\omega)t}-1)\hat{a}^\dagger-\frac{C}{i(\omega+\Omega)}(e^{-i(\Omega+\omega)t}-1)\hat{a}\] Then, \[\begin{aligned} [\hat{W}(t),\hat{V}_I(t)]=& \ \frac{C^2}{i(\omega+\Omega)}(1-e^{-i(\omega+\Omega)t})(-1)-\frac{C^2}{i(\omega+\Omega)}(1-e^{i(\omega+\Omega)t}) \\ =& \ -\frac{2C^2}{i(\omega+\Omega)}(1-\cos(\omega+\Omega)t), \end{aligned}\] which is just a number again. Now make the unitary transformation \(\left|\psi_I(t)\right\rangle=e^{-\frac{i}{\hbar}\hat{W}(t)}\left|\psi_{II}(t)\right\rangle\). But \[\left[i\hbar \ \partial_t-\hat{V}_I(t)\right]\left|\psi_I(t)\right\rangle=0\] So we have \[e^{\frac{i}{\hbar}\hat{W}(t)}\left[i\hbar \ \partial_t-\hat{V}_I(t)\right]e^{-\frac{i}{\hbar}\hat{W}(t)}e^{\frac{i}{\hbar}\hat{W}(t)}\left|\psi_I(t)\right\rangle=0\] Hence, \[e^{\frac{i}{\hbar}\hat{W}(t)}\left[i\hbar \ \partial_t-\hat{V}_I(t)\right]e^{-\frac{i}{\hbar}\hat{W}(t)}\left|\psi_{II}(t)\right\rangle=0\] Since \(\hat{W}(t)\) does not commute with \(\hat{V}_I(t)\), we cannot easily evaluate the derivative term. So let us expand the exponentials in a power series and then differentiate.

Computing the time-ordered product directly

First, operate the derivative on the RHS. \[\left(1+\frac{i}{\hbar}\hat{W}+\frac{1}{2!}\left(\frac{i}{\hbar}\right)^2\hat{W}^2+\frac{1}{3!}\left(\frac{i}{\hbar}\right)^3\hat{W}^3+\cdots\right)\left[i\hbar \ \partial_t-\hat{V}_I\right]\left(1-\frac{i}{\hbar}\hat{W}+\frac{1}{2!}\left(\frac{-i}{\hbar}\right)^2\hat{W}^2+\frac{1}{3!}\left(\frac{-i}{\hbar}\right)^3\hat{W}^3+\cdots\right)\] \[\begin{gathered} =\left(1+\frac{i}{\hbar}\hat{W}+\frac{1}{2!}\left(\frac{i}{\hbar}\right)^2\hat{W}^2+\frac{1}{3!}\left(\frac{i}{\hbar}\right)^3\hat{W}^3+\cdots\right)\left\{i\hbar \ \partial_t-\hat{V}_I-i\hbar\left(i\hbar\dot{\hat{W}}+\hat{W}i\hbar \ \partial_t-\hat{V}_I\hat{W}\right) \right. \\ +\frac{1}{2!}\left(\frac{-i}{\hbar}\right)^2\left[i\hbar\dot{\hat{W}}\hat{W}+i\hbar\hat{W}\dot{\hat{W}}+\hat{W}^2i\hbar \ \partial_t-\hat{V}_I\hat{W}^2\right] \\ \left.+\frac{1}{3!}\left(\frac{-i}{\hbar}\right)^3\left[i\hbar\dot{\hat{W}}\hat{W}^2+i\hbar\hat{W}\dot{\hat{W}}\hat{W}+i\hbar\hat{W}^2\dot{\hat{W}}+\hat{W}^3i\hbar \ \partial_t-\hat{V}_I\hat{W}^3 \right]+\cdots\right\} \end{gathered}\] \[\begin{gathered} =i\hbar \ \partial_t-\hat{V}_I+\dot{\hat{W}}+\hat{W}\partial_t+\frac{i}{\hbar}\hat{V}_I\hat{W}-\hat{W}\partial_t-\frac{i}{\hbar}\hat{W}\hat{V}_I-\frac{1}{2}\frac{i}{\hbar}\left(\dot{\hat{W}}\hat{W}+\hat{W}\dot{\hat{W}}+\hat{W}^2\partial_t-\frac{i}{\hbar}\hat{V}_I\hat{W}^2\right) \\ +\frac{i}{\hbar}\left(\hat{W}\dot{\hat{W}}+\hat{W}^2\partial_t+\frac{i}{\hbar}\hat{W}\hat{V}_I\hat{W}\right)-\frac{1}{2}\frac{i}{\hbar}\hat{W}^2\partial_t+\frac{1}{2}\frac{1}{\hbar^2}\hat{W}^2\hat{V}_I+\frac{1}{6}\left(\frac{i}{\hbar}\right)^2\left(\dot{\hat{W}}\hat{W}^2+\hat{W}\dot{\hat{W}}\hat{W}+\hat{W}^2\dot{\hat{W}}\right) \\ -\frac{1}{2}\left(\frac{i}{\hbar}\right)^2\left(\hat{W}\dot{\hat{W}}\hat{W}+\hat{W}^2\dot{\hat{W}}\right)+\frac{1}{2}\left(\frac{i}{\hbar}\right)^2(\hat{W}^2\dot{\hat{W}})+\cdots \end{gathered}\] But \(\dot{\hat{W}}=\hat{V}_I(t)\) so we get \[\begin{gathered} =i\hbar \ \partial_t-\hat{V}_I+\hat{V}_I-\frac{i}{\hbar}[\hat{W},\hat{V}_I]-\frac{i}{2\hbar}\left(\hat{V}_I\hat{W}+\hat{W}\hat{V}_I-2\hat{W}\hat{V}_I\right)+\frac{1}{\hbar^2}\hat{V}_I\hat{W}^2-\frac{1}{\hbar^2}\hat{W}\hat{V}_I\hat{W}+\frac{1}{2\hbar^2}\hat{W}^2\hat{V}_I \\ -\frac{1}{6\hbar^2}\left(\hat{V}_I\hat{W}^2+\hat{W}\hat{V}_I\hat{W}+\hat{W}^2\hat{V}_I\right)+\frac{1}{2\hbar^2}\left(\hat{W}\hat{V}_I\hat{W}+\hat{W}^2\hat{V}_I\right)-\frac{1}{2\hbar^2}\hat{W}^2\hat{V}_I+\cdots \end{gathered}\] \[=i\hbar \ \partial_t-\frac{i}{2\hbar}[\hat{W},\hat{V}_I]+\frac{1}{\hbar^2}\left(\frac{1}{3}\hat{V}_I\hat{W}^2-\frac{2}{3}\hat{W}\hat{V}_I\hat{W}+\frac{1}{3}\hat{W}^2\hat{V}_I\right)+\cdots\] \[=i\hbar \ \partial_t-\frac{i}{2\hbar}[\hat{W},\hat{V}_I]+\frac{1}{3\hbar^2}\left(-[\hat{W},\hat{V}_I]\hat{W}+\hat{W}[\hat{W},\hat{V}_I]\right)+\cdots\] \[=i\hbar \ \partial_t-\frac{i}{2\hbar}[\hat{W},\hat{V}_I]+\frac{1}{3\hbar^2}[\hat{W},[\hat{W},\hat{V}_I]]+\cdots\] All higher-order terms are multiple commutators, but \([\hat{W},\hat{V}_I]\) commutes with everything so \[=i\hbar \ \partial_t-\frac{i}{2\hbar}[\hat{W},\hat{V}_I]=i\hbar \ \partial_t+\frac{C^2}{\hbar(\omega+\Omega)}(1-\cos{(\Omega+\omega)t}).\] So \[\left[i\hbar \ \partial_t+\frac{C^2}{\hbar(\omega+\Omega)}(1-\cos{(\Omega+\omega)t}\right]\left|\psi_{II}(t)\right\rangle=0\] \[\implies\left|\psi_{II}(t)\right\rangle=e^{\frac{i}{\hbar}\int_0^t\frac{C^2}{\hbar(\omega+\Omega)}(1-\cos{(\Omega+\omega)t} \ dt'}\left|\psi_{II}(0)\right\rangle\] \[=e^{\frac{iC^2}{\hbar^2(\omega+\Omega)}\left(t-\frac{\sin{(\omega+\Omega)t}}{\omega+\Omega}\right)}\left|\psi_{II}(0)\right\rangle.\] Hence, \[\begin{aligned} \left|\psi_I(t)\right\rangle=& \ e^{-\frac{i}{\hbar}\hat{W}(t)}\left|\psi_{II}(t)\right\rangle \\ =& \ \exp\left(-\frac{C}{\hbar(\omega+\Omega)}(e^{i(\omega+\Omega)t}-1)\hat{a}^\dagger+\frac{C}{\hbar(\omega+\Omega)}(e^{-i(\omega+\Omega)t}-1)\hat{a}\right)\\ \times& \ \exp\left(\frac{iC^2}{\hbar^2(\omega+\Omega)}(t-\frac{\sin{(\omega+\Omega)t}}{\omega+\Omega})\right)\left|\psi_{II}(0)\right\rangle \end{aligned}\] and \[\left|\psi_S(t)\right\rangle=e^{-i\omega t(\hat{a}^\dagger\hat{a}+\frac{1}{2})}\left|\psi_{I}(t)\right\rangle\] To check, we could compute overlaps with states \(\left\langle m\right|\) and verify the differential equation for \(c_m(t)\) holds, but I won’t do that.
 
In summary, when \([\hat{V}_I(t),\hat{V}_I(t')]\) is a number, the time-ordered product simplifies. In particular, we have \[\fbox{ $\begin{array}{c} T\exp(-\frac{i}{\hbar}\int_0^t\hat{V}_I(t') \ dt')=T\exp\left(-\frac{i}{\hbar}C\int_0^t\left[e^{i(\omega+\Omega)t'}\hat{a}^\dagger+e^{-i(\omega+\Omega)t'}\hat{a}\right] \ dt'\right) \\ =\exp\left(-\frac{C}{\hbar(\omega+\Omega)}(e^{i(\omega+\Omega)t}-1)\hat{a}^\dagger+\frac{C}{\hbar(\omega+\Omega)}(e^{-i(\omega+\Omega)t}-1)\hat{a}\right)\ \exp\left(\frac{iC^2}{\hbar^2(\omega+\Omega)}(t-\frac{\sin{(\omega+\Omega)t}}{\omega+\Omega})\right) \end{array}$ }\] for this case.