Recall the time-dependent Schrödinger equation: \[i\hbar \ \partial_t\left|\psi(t)\right\rangle=\hat{H}(t)\left|\psi(t)\right\rangle\] Question: Suppose we are in the state \(\left|\psi(t_0)\right\rangle\) at time \(t_0\). What state are we in at some later time \(t\)? Note that this question is complicated by the fact that the Hamiltonian now changes with respect to time.
As a start, we suppose \(\hat{H}\) is independent of time. Introduce energy eigenstates \(\{\left|n\right\rangle\}\) such that \(\hat{H}\left|n\right\rangle=E_n\left|n\right\rangle\) and \(\left\langle m\middle|n\right\rangle=\delta_{mn}\). Then, expand \[\left|\psi(t_0)\right\rangle=\sum_nc_n(t_0)\left|n\right\rangle\] Inserting into the Schrödinger equation, \[\sum_{n}i\hbar \ \partial_tc_n(t)\left|n\right\rangle=\sum_nE_nc_n(t)\left|n\right\rangle\] Multiply by \(\left\langle m\right|\), \[i\hbar \ \partial_tc_m(t)=E_mc_m(t)\] which is solved by \[c_m(t)=c_m(t_0)e^{-iE_m(t-t_0)/\hbar}\] Hence, \[\boxed{\left|\psi(t)\right\rangle=\sum_n c_n(t_0)e^{-iE_n(t-t_0)/\hbar}\left|n\right\rangle}\] An alternate way to do this is as follows. \[\left|\psi(t)\right\rangle=e^{-i\hat{H}(t-t_0)/\hbar}\left|\psi(t_0)\right\rangle\] Since, \[e^{-i\hat{H}(t-t_0)/\hbar}\left|n\right\rangle=e^{-iE_n(t-t_0)/\hbar}\left|n\right\rangle\] because \(|n\rangle\) is an eigenstate of \(\hat{H}\), we then find that \[\left|\psi(t)\right\rangle=\sum_n c_n(t_0)e^{-iE_n(t-t_0)/\hbar}\left|n\right\rangle\] as before.
Now assume that \(\hat{H}(t)\) is
time-dependent \((\partial_t\hat{H}\ne
0)\).
Method A: Introduce any complete
orthonormal basis \(\{\left|n\right\rangle\}\). \[\left|\psi(t)\right\rangle=\sum_nc_n(t)\left|n\right\rangle\]
Then, \[i\hbar \
\partial_t\left|\psi(t)\right\rangle=\sum_ni\hbar \
\partial_tc_n(t)\left|n\right\rangle=\hat{H}(t)\left|\psi(t)\right\rangle=\sum_nc_n(t)\hat{H}(t)\left|n\right\rangle\]
Multiplying by \(\left\langle
m\right|\), \[i\hbar \
\partial_tc_m(t)=\sum_{n}c_n(t)\left\langle
m\middle|\hat{H}(t)\middle|n\right\rangle\] so \[i\hbar \
\partial_tc_m(t)=\sum_nH_{mn}(t)c_n(t)\] This is a matrix
differential equation that is hard to solve except for small-finite
sized problems. This is because there is nothing we know about the
matrix \(H_{mn}(t)\) and it can be a
complicated object to work with, especially if it is infinite
dimensional.
Method B: Introduce instantaneous eigenstates.
\[\hat{H}(t)\left|n(t)\right\rangle=E_n(t)\left|n(t)\right\rangle\]
So now, everything depends on time. But, we still have orthonormality
and completeness: \[\left\langle
m(t)\middle|n(t)\right\rangle=\delta_{mn}\quad\text{and}\quad\sum_{m}\left|m(t)\right\rangle\left\langle
m(t)\right|=\mathbb{I}.\] Then, let \[\left|\psi(t)\right\rangle=\sum_nc_n(t)\left|n(t)\right\rangle.\]
Inserting into the Schrödinger equation, \[\sum_ni\hbar \
\partial_tc_n(t)\left|n(t)\right\rangle+\sum_ni\hbar
c_n(t)\partial_t\left|n(t)\right\rangle=\sum_nc_n(t)E_n(t)\left|n(t)\right\rangle\]
Multiply by \(\left\langle m\right|\),
\[i\hbar \
\partial_tc_m(t)+i\hbar\sum_nc_n(t)\left\langle
m(t)\middle|\partial_t\middle|n(t)\right\rangle=c_m(t)E_m(t).\]
In general, it is hard to calculate the middle term. But note, \[\left\langle
m(t)\middle|n(t)\right\rangle=\delta_{mn}\implies \partial_t\left\langle
m(t)\middle|n(t)\right\rangle+\left\langle
m(t)\middle|\partial_t\middle|n(t)\right\rangle=0\] and \[\hat{H}(t)\left|n(t)\right\rangle=E_n(t)\left|n(t)\right\rangle\implies\partial_t\hat{H}(t)\left|n(t)\right\rangle+\hat{H}(t)\partial_t\left|n(t)\right\rangle=\partial_tE_n(t)\left|n(t)\right\rangle+E_n(t)\partial_t\left|n(t)\right\rangle\]
Multiply by \(\left\langle m\right|\),
\[\left\langle
m(t)\middle|\partial_t\hat{H}(t)\middle|n(t)\right\rangle+E_m(t)\left\langle
m(t)\middle|\partial_t\middle|n(t)\right\rangle=\partial_tE_n(t)\delta_{mn}+E_n(t)\left\langle
m(t)\middle|\partial_t\middle|n(t)\right\rangle\] Then, break out
into cases:
If \(m=n\), then we find \(\left\langle n(t)\middle|\partial_t\hat{H}(t)\middle|n(t)\right\rangle=\partial_tE_n(t)\).
If \(m\ne n\), then we find \(\frac{\left\langle m(t)\middle|\partial_t\hat{H}(t)\middle|n(t)\right\rangle}{E_n(t)-E_m(t)}=\left\langle m(t)\middle|\partial_t\middle|n(t)\right\rangle\)
So, \[i\hbar \ \partial_tc_m(t)+i\hbar c_m(t)\left\langle m(t)\middle|\partial_t\middle|m(t)\right\rangle+i\hbar\sum_{m\ne n}c_n(t)\frac{\left\langle m(t)\middle|\partial_t\hat{H}(t)\middle|n(t)\right\rangle}{E_n(t)-E_m(t)}=c_m(t)E_m(t)\] The derivative matrix element is still hard to deal with, but at least now there is only one such term.
This approach can be used to develop what is called adiabatic perturbation theory, where we assume the changes of the Hamiltonian in time are slow. Griffiths textbook has an excellent treatment of this problem.
Define \(\hat{U}(t,t_0)\) by \[\left|\psi(t)\right\rangle=\hat{U}(t,t_0)\left|\psi(t_0)\right\rangle.\]
\(U\) takes us from time \(t_0\) to \(t\). If \(\hat{H}(t)\) is time-independent then \[\hat{U}(t,t_0)=\exp\left
(-\frac{i}{\hbar}\hat{H}(t-t_0)\right)\] as we saw earlier. Here
are some general properties of \(\hat{U}\).
1.) \(\hat{U}(t_0,t_0)=\mathbb{I}\)
2.) \(\hat{U}(t,t')\hat{U}(t',t_0)=\hat{U}(t,t_0)\)
3.) Due to the Hermiticity of \(\hat{H}\), \(\hat{H}^\dagger=\hat{H}\). \[\begin{aligned}
&i\hbar \
\partial_t\left|\psi(t)\right\rangle=\hat{H}(t)\left|\psi(t)\right\rangle
\\
&\left\langle\psi(t)\right|(-i\hbar\partial_t)=\left\langle\psi(t)\right|\hat{H}(t),
\end{aligned}\] where the operator acts to the left. Therefore,
\[\begin{aligned}
\partial_t\left\langle\psi(t)\middle|\psi(t)\right\rangle=& \
\left\langle\psi(t)\middle|\partial_t\middle|\psi(t)\right\rangle+\left\langle\psi(t)\middle|\partial_t\middle|\psi(t)\right\rangle
\\
=& \
\frac{i}{\hbar}\left\langle\psi(t)\middle|\hat{H}(t)\middle|\psi(t)\right\rangle-\frac{i}{\hbar}\left\langle\psi(t)\middle|\hat{H}(t)\middle|\psi(t)\right\rangle
\\
=& \ 0.
\end{aligned}\] Hence, normalized states remain normalized for
all \(t\). This tells us that \[\left\langle\psi(t)\middle|\psi(t)\right\rangle=\left\langle\psi(t_0)\middle|\hat{U}^\dagger(t,t_0)\hat{U}(t,t_0)\middle|\psi(t_0)\right\rangle=\left\langle\psi(t_0)\middle|\psi(t_0)\right\rangle.\]
Hence \(\hat{U}^\dagger(t,t_0)\hat{U}(t,t_0)=\mathbb{I}\)
meaning \(\hat{U}\) is unitary.
4.) Plugging \(\left|\psi(t)\right\rangle=\hat{U}(t,t_0)\left|\psi(t_0)\right\rangle\)
into the Schrödinger equation, \[i\hbar \
\partial_t\hat{U}(t,t_0)\left|\psi(t_0)\right\rangle=\hat{H}(t)\hat{U}(t,t_0)\left|\psi(t_0)\right\rangle\]
Since this is true for all \(t_0\),
\[\boxed{i\hbar \
\partial_t\hat{U}(t,t_0)=\hat{H}(t)\hat{U}(t,t_0)}\] which is
called the equation of motion.
Now, we wish to integrate the equation of motion to obtain an expression for \(\hat{U}(t,t_0)\). \[i\hbar(\hat{U}(t,t_0)-\hat{U}(t_0,t_0))=\int_{t_0}^t\hat{H}(t')U(t',t_0) \ dt'\] \[\implies \hat{U}(t,t_0)=\mathbb{I}-\frac{i}{\hbar}\int_{t_0}^t\hat{H}(t')\hat{U}(t',t_0) \ dt'.\] If we iterate, \[\begin{aligned} \hat{U}(t,t_0)=& \ \mathbb{I}-\frac{i}{\hbar}\int_{t_0}^t\hat{H}(t') \ dt'+\left(\frac{-i}{\hbar}\right)^2\int_{t_0}^tdt'\int_{t_0}^{t_1}dt_2\hat{H}(t_1)\hat{H}(t_2)+\cdots \\ =& \ \mathbb{I}-\frac{i}{\hbar}\int_{t_0}^tdt_1 T(\hat{H}(t_1))+\left(\frac{-i}{\hbar}\right)^2\frac{1}{2!}\int_{t_0}^tdt_1\int_{t_0}^tT(\hat{H}(t_1)\hat{H}(t_2))+\cdots \end{aligned}\] where the time-ordered product \(T\) is defined by \[\begin{aligned} &T(\hat{A}_1(t_1))=\hat{A}_1(t_1) \\ &T(\hat{A}_1(t_1)\hat{A}_2(t_2))=\Theta(t_1-t_2)\hat{A}_1(t_1)\hat{A}_2(t_2)+\Theta(t_2-t_1)\hat{A}_2(t_2)\hat{A}_1(t_1). \end{aligned}\] The rule is to “put Later times to the Left”. The proof of this formula is simple. Each time-ordered piece gives the same result as the original series. There are \(n!\) different orderings of \(n\) operators so this cancels out the \(\frac{1}{n!}\). We write \[\hat{U}(t,t_0)=T\exp(-\frac{i}{\hbar}\int_{t_0}^tdt'\hat{H}(t'))=\sum_{n=0}^\infty\left(\frac{-i}{\hbar}\right)^n\frac{1}{n!}\int_{t_0}^t\cdots\int_{t_0}^tdt_nT(\hat{H}(t_1)\hat{H}(t_2)\cdots\hat{H}(t_n))\] This is a power series in \(\hat{H}\) not in some time-dependent perturbation \(\hat{V}(t)\), hence it often is not useful for applications. Note that property 4 says \[i\hbar \ \partial_tT\exp(-\frac{i}{\hbar}\int_{t_0}^tdt'\hat{H}(t'))=\hat{H}(t)T\exp(-\frac{i}{\hbar}\int_{t_0}^tdt'\hat{H}(t'))\] So time-ordered products generalize to operators the fundamental property of an exponential function - namely that the derivative of an exponential is the derivative of the argument times the exponential. This is nontrivial because the operators may not commute at different times.