Review of the time-ordered product

Last time we developed an expansion for \(\hat{U}(t, t_{0})\) in powers of \(\hat{H}(t)\): \[\begin{aligned} |\psi(t)\rangle&=\hat{U}\left(t, t_{0})\left|\psi\left(t_{0}\right)\right\rangle\right. \\ \hat{U}\left(t, t_{0}\right)&=T \exp \left[-\frac{i}{\hbar} \int_{t_{0}}^{t} d t^{\prime} \ \hat{H}\left(t^{\prime}\right)\right] \\ & =\sum_{n=0}^{\infty}\left(-\frac{i}{\hbar}\right)^{n} \frac{1}{n!} \int_{t_{0}}^{t} d t_{1} \cdots \int_{t_{0}}^{t} d t_{n} \ T\left(\hat{H}\left(t_{1}\right) \cdots H\left(t_{n}\right)\right) \\ & =\sum_{n=0}^{\infty }\left(-\frac{i}{\hbar}\right)^{n} \int_{t_{0}}^{t} d t_{1} \int_{t_{0}}^{t_{1}} d t_{2} \cdots \int_{t_{0}}^{t_{n-1}} d t_{n} \ \hat{H}\left(t_{1}\right) \hat{H}\left(t_{2}\right) \cdots H\left(t_{n}\right) \end{aligned}\] Note also that the equation of motion for \(\hat{U}\) is \[i \hbar \frac{\partial}{\partial t} \hat{U}\left(t, t_{0}\right)=\hat{H}(t) \hat{U}(t, t_{0})\] which follows from the last form for \(\hat{U}\). Now, note that \(t\) only appears in the upper limit of the last integral, which allows us to take the derivative with respect to \(t\) as follows: \[\begin{aligned} i \hbar \frac{\partial}{\partial t} \hat{U}\left(t, t_{0}\right) & =\sum_{n=0}^{\infty}-\left(\frac{i}{\hbar}\right)^{n} i \hbar \frac{\partial}{\partial t} \int_{t_{0}}^{t} d t_{1} \int_{t_{0}}^{t_{1}} d t_{2} \cdots \int_{t_{0}}^{t_{n-1}} d t_{n} \hat{H}\left(t_{1}\right) \hat{H}\left(t_{2}\right) \cdots \hat{H}\left(t_{n}\right) \\ & =\frac{-i}{\hbar} \left (i \hbar \hat{H}(t) \right )\sum_{n=0}^{\infty}\left(-\frac{i}{\hbar}\right)^{n} \int_{t_{0}}^{t} d t_{1} \int_{t_{0}}^{t_{2}} d t_{2} \cdots \int_{t_{0}}^{t_{n-1}} d t_{n} \hat{H}\left(t_{1}\right) \cdots H\left(t_{n}\right) \\ & =\hat{H}(t) \hat{U}\left(t_{1}\left(t_{0}\right)\right.. \end{aligned}\]

But in many cases, the Hamiltonian separates into a time independent piece and a small time-dependent piece as \(\hat{H}=\hat{H}_{0}+\hat{V}(t)\). In those cases, we want an expansion in \(\hat{V}\) not \(\hat{H}\).

Pictures for time evolution

We start by looking at different pictures for quantum mechanizes.
We are all familiar with the Schrodinger representation. \[\hat{H}|\psi ( t)\rangle=i \hbar \frac{\partial}{\partial t}|\psi(t)\rangle \quad \text { assume } \hat{H} \text { independent of time here. }\] Now consider the expectation value of an operator \(\hat{A}\) with no time dependence, where the expectation value can have time dependence from the dime dependence of the states: \[A(t)=\langle\psi(t)|\hat{A}|\psi(t)\rangle.\] Then, taking the derivative gives us \[\begin{aligned} \frac{d}{d t} A(t) & =\left(\frac{\partial}{\partial t}\langle\psi(t)|\right) \hat{A}|\psi(t)\rangle+\langle\psi(t)| \hat{A} \frac{\partial}{\partial t}|\psi(t)\rangle \\ & =\frac{i}{\hbar}\langle\psi(t)| \hat{H} \hat{A}|\psi(t)\rangle-\frac{i}{\hbar}\langle\psi(t)| \hat{A} \hat{H}|\psi(t)\rangle \\ & =-\frac{i}{\hbar}\langle\psi(t)|[\hat{A}, \hat{H}] | \psi(t)\rangle \\ i\hbar \frac{d}{d t} A(t) & =\langle\psi(t)|[\hat{A}, \hat{H}]|\psi(t)\rangle. \end{aligned}\] All time dependence comes from the wave functions in the Schrodinger representation when \(\frac{\partial H}{\partial t}=0\).
Heisenberg representation
Write \(\left|\psi_{s}(t)\right\rangle=\hat{U}(t)\left|\psi_{H}\right\rangle \quad \text{with}\quad\hat{U}(t)=e^{-\frac{i}{\hbar} \hat{H} t}\) Then define \[\hat{A}_{H}(t)=\hat{U}^{\dagger}(t) \hat{A} \hat{U}(t).\] Then \[\begin{aligned} A(t)&=\left\langle\psi_{s}(t)\right| \hat{A}\left|\psi_{s}(t)\right\rangle\\ & =\left\langle\psi_{H}\right| \hat{U}^{\dagger}(t) \hat{A} \hat{U}(t)\left|\psi_{H}\right\rangle \\ & =\left\langle\psi_{H}\right| \hat{A}_{H}(t)\left|\psi_{H}\right\rangle, \end{aligned}\] and all time dependence is in the operators now. We find that \[\begin{aligned} & i \hbar \frac{d}{\partial t} \hat{A}_{H}(t)=i \hbar \frac{\partial}{\partial t} \hat{U}^{\dagger}\hat{A} \hat{U}+\hat{U}^{\dagger} \hat{A} i \hbar \frac{\partial}{\partial t} \hat{U} \\ &=-\hat{U}^{\dagger} \hat{H} \hat{A} \hat{U}+\hat{U}^{\dagger} \hat{A} H\hat{U} \\ &=\hat{U}^{\dagger}[\hat{A}, \hat{H}] U \\ &=\left[\hat{A}_{H}(t), \hat{H}\right] \quad \text {since }\left[\hat{U}, H\right]=0 \text { for time-independent } \hat{H}. \\ \end{aligned}\] Summarizing, we have that \[\boxed{ i \hbar \frac{d}{d t} \hat{A}_{H}(t)=\left[\hat{A}_{H}(t), \hat{H}\right]}\] which is called the equation of motion. All time dependence is now in the operators, which evolve like Poisson brackets in classical mechanics.

Pictures for time-dependent Hamiltonians

If \(\hat{H}\) depends on time, \(\hat{H}_{s}(t)\), we proceed similarly \[\begin{aligned} & \left|\psi_{s}(t)\right\rangle=\hat{U}\left(t, t_{0}\right)\left|\psi_{H}\left(t_{0}\right)\right\rangle \\ & \hat{A}_{H}(t)=\hat{U}^{\dagger}\left(t, t_{0}\right)\ \hat{A}_{s}(t)\ \hat{U}\left(t, t_{0}\right) \\ & \hat{H}_{H}(t)=\hat{U}^{\dagger}\left(t, t_{0}\right) \ \hat{H}_{s}(t) \ \hat{U}\left(t,t_{0}\right) \end{aligned}\] \[i \hbar \frac{\partial}{\partial t} \hat{U}\left(t, t_{0}\right)=\hat{H}_{s}(t) \hat{U}\left(t, t_{0}\right),\] so \[\begin{aligned} \frac{\partial}{\partial t} \hat{A}_{H}(t)=&i \hbar \frac{\partial}{\partial t} \hat{U}^{\dagger}\left(t,t_0\right) \hat{A}_{s}(t) U(t, t_0)\\ & +i \hbar \hat{U}^{\dagger}(t, t_0) \frac{\partial}{\partial t} \hat{A}_{s}(t) \hat{U}(t, t_0)+i \hbar \hat{U}^{\dagger}(t, t_0) \hat{A}_{s}(t) \frac{\partial}{\partial t} \hat{U}(t, t) \\ =&-\hat{U}^{\dagger}\left(t, t_{0}\right) \hat{H}_{s}(t) \hat{A}_{s}(t) \hat{U}\left(t, t_{0}\right) \\ & +i \hbar \hat{U}^{\dagger}(t, t_0) \frac{\partial}{\partial t} \hat{A}_{s}(t) \hat{U}(t, t_0)+\hat{U}^{\dagger}\left(t, t_{0}\right) \hat{A}_{s}(t) \hat{H}_{s}(t) \hat{U}{(t,t_0)}. \end{aligned}\] but \[\hat{U}^{\dagger} \hat{H}_{s} \hat{A}_{s} \hat{U}=\hat{U}^{\dagger} \hat{H}_{s} \hat{U}^{\dagger} \hat{U}^{\dagger} \hat{A}_{s} \hat{U}=\hat{H}_{H}(t) \hat{A}_{H}(t),\] so \[\boxed{i \hbar \frac{\partial}{\partial t} \hat{A}_{H}(t)=\left[\hat{A}_{H}(t), \hat{H}_{H}(t)\right]+i \hbar \frac{\partial \hat{A}_{H}}{\partial t}(t)}.\]

The last term is \(\hat{U}^{\dagger}\left(t, t_{0}\right) \frac{\partial A_{s}(t)}{\partial t} \hat{U}\left(t, t_{0}\right)=i \hbar \frac{\partial \hat{A}_{H}(t)}{\partial t}.\)

Interaction representation

The interaction representation is halfway between Schrodinger and Heisenberg. We have the same break-up into a time-independent and a small time-dependent piece: \[\hat{H}(t)= \hat{H}+\hat{V}(t).\] So we have \[\hat{H}_{0}|n\rangle=E_{n}^{0}|n\rangle.\] Define \[\begin{aligned} &\left|\psi_{I}(t)\right\rangle=e^{\frac{i}{\hbar} \hat{H_{0}}\left(t-t_{0}\right)}\left|\psi_{s}(t)\right\rangle\\ & \hat{A}_{I}(t)=e^{\frac{i}{\hbar} \hat{H}_{0}\left(t-t_{0}\right)} A_{s}(t) e^{-\frac{i}{\hbar} \hat{H}_{0}\left(t-t_{0}\right)} \\ & \hat{V}_{I}(t)=e^{\frac{i}{\hbar} \hat{H}_{0}\left(t-t_{0}\right)} \hat{V}_{s}(t) e^{-\frac{i}{\hbar} \hat{H}_{0}\left(t-t_{0}\right)}. \end{aligned}\] Note that \(H_{0I} (t) = H_{0s}(t)=H_0(t)\) since \[\left[\hat{H}_{0}, e^{\frac{i}{\hbar}\hat{H}_{0}\left(t-t_{0}\right)} \right]=0 .\] So, we find that \[\begin{aligned} i \hbar \frac{\partial}{\partial t}\left|\psi_{I}(t)\right\rangle&=i \hbar \frac{d}{d t}\left[e^{\frac{i}{\hbar} \hat{H}_{0}\left(t-t_{0}\right)}\left|\psi_{s}(t)\right\rangle\right] \\ & =-\hat{H}_{0} e^{\frac{i}{\hbar} \hat{H}_{0}\left(t-t_{0}\right)}\left|\psi_{s}(t)\right\rangle+e^{\frac{i}{\hbar} \hat{H}_{0}\left(t-t_{0}\right)} \hat{H}_{s}(t)\left|\psi_{s}(t)\right\rangle \\ & =e^{\frac{i}{\hbar} \hat{H}_{0}\left(t-t_{0}\right)}\left(\underbrace{-\hat{H}_{0}+\hat{H}}_0+\hat{V}_{s}(t)\right)\left|\psi_{s}(t)\right\rangle \\ & =e^{\frac{i}{\hbar} \hat{H}_{0}\left(t-t_{0}\right)} \hat{V}_{s}(t) e^{-\frac{i}{\hbar} \hat{H}_{0}\left(t-t_{0}\right)} e^{\frac{i}{\hbar} \hat{H}_{0}\left(t-t_{0}\right)}\left|\psi_{s}(t)\right\rangle \\ & =\hat{V}_{I}(t)\left|\psi_{I}(t)\right\rangle. \end{aligned}\] Hence, if \(\hat{U}_{I}\left(t_{1} t_{0}\right)\left|\psi_{I}\left(t_{0}\right)\right\rangle=\left|\psi_{I}(t)\right\rangle\), then we have

\[i\hbar\frac{\partial}{\partial t} \hat{U}_{I}\left(t, t_{0}\right)=\hat{V}_{I}(t) \hat{U}_{I}\left(t, t_{0}\right).\]

This is called the Dyson expansion for the evolution operator.
Note that \[\begin{aligned} \left.|\psi_{I}(t)\right\rangle & =e^{\frac{i}{\hbar} \hat{H}_{0}\left(t-t_{0}\right)}\left|\psi_{s}(t)\right\rangle=e^{\left.\frac{i}{\hbar} \right. \hat{H}_{0}\left(t-t_{0}\right)} \hat{U}_{s}\left(t, t_{0}\right)\left|\psi_{s}\left(t_{0}\right)\right\rangle \\ & =\hat{U}_{I}\left(t,\left.t_{0}\right)\left|\psi_{I}\left(t_{0}\right)\right\rangle\right. \end{aligned}\] but \(\left|\psi_{I}\left(t_{0}\right)\right\rangle=\left|\psi_{S}\left(t_{0}\right)\right\rangle=\left|\psi_{H}\left(t_{0}\right)\right\rangle\) since \(U\left(t_{0}, t_{0}\right)=1\) in all pictures.
So \(\boxed{U_{s}\left(t, t_{0}\right)=e^{-\frac{i}{\hbar} \hat{H}_{0}(t-t)} T \exp \left[-\frac{i}{\hbar} \int_{t_{0}}^{t} d t^{\prime} \hat{V}_{ I}\left(t^{\prime}\right)\right]}\)
This is the desired expansion of the evolution operator in a power series in \(\hat{V}\) and it will be used for time dependent perturbation theory in a few lectures.
We can derive this result directly though. Recall \(i \hbar\frac{\partial}{\partial t} T e^{-\frac{i}{\hbar} \int_{t_{0}}^{t} d t^{\prime} \hat{A}\left(t^{\prime}\right)}=\hat{A}(t) T e^{-\frac{i}{\hbar} \int_{t_{0}}^{t} d t^{\prime} \hat{A}\left(t^{\prime}\right)}\) so \[\begin{aligned} & i \hbar \frac{d}{d t}\left[e^{-\frac{i}{\hbar} \hat{H}_{0}\left(t-t_{0}\right)} T \exp \left[-\frac{i}{\hbar} \int_{t_{0}}^{t} d t^{\prime} \ \hat{V}_{I}\left(t^{\prime}\right)\right]\right] \\ & =\hat{H}_{0} e^{-\frac{i}{\hbar} \hat{H}_{0}\left(t-t_{0}\right)} T \exp \left[-\frac{i}{\hbar} \int_{t_{0}}^{t} \hat{V}_{I}\left(t^{\prime}\right) d t^{\prime}\right] +e^{-\frac{i}{\hbar} \hat{H}_{0}(t-t_0)} \hat{V}_{I}(t) T \exp \left[-\frac{i}{\hbar} \int_{t_{0}}^{t} \hat{V}_{I}\left(t^{\prime}\right) d t^{\prime}\right]. \end{aligned}\] But \(e^{-\frac{i}{\hbar} \hat{H}_{0}(t-t_0)} \hat{V}_{I}(t)=\hat{V}_{s}(t) e^{-\frac{i}{\hbar} \hat{H}_{0}\left(t - t_{0}\right)}\) so \[\begin{gathered} =\left[\underbrace{\hat{H}_{0}+\hat{V}_{s}(t)}_{\hat{H}(t)}\right] e^{-\frac{i}{\hbar} \hat{H}_{0}(t-t)} T e^{-\frac{i}{\hbar} \int_{t_{0}}^{t} d t^{\prime} \hat{V}_{I}(t)} . \end{gathered}\] So \(i \hbar\frac{\partial}{\partial t} O_{p}(t, t_0)=\hat{H}(t) \ O_{p}(t, t_0)\) and \(O_{p}\left(t_{0}, t_{0}\right)=\mathbb{I}\), which implies that \(O_p(t, t_0)=\hat{U}_{s}\left(t, t_{0}\right),\) so we have proved the result directly.