# Time-dependent propagators

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 4.3.

The fourth postulate of non-relativistic quantum mechanics concerns how states evolve with time. The postulate simply states that in non-relativistic quantum mechanics, a state satisfies the Schrödinger equation:

$\displaystyle i\hbar\frac{\partial}{\partial t}\left|\psi\right\rangle =H\left|\psi\right\rangle \ \ \ \ \ (1)$

where ${H}$ is the Hamiltonian, which is obtained from the classical Hamiltonian by means of the other postulates of quantum mechanics, namely that we replace all references to the position ${x}$ by the quantum position operator ${X}$ with matrix elements (in the ${x}$ basis) of

$\displaystyle \left\langle x^{\prime}\left|X\right|x\right\rangle =\delta\left(x-x^{\prime}\right) \ \ \ \ \ (2)$

and all references to classical momentum ${p}$ by the momentum operator ${P}$ with matrix elements

$\displaystyle \left\langle x^{\prime}\left|P\right|x\right\rangle =-i\hbar\delta^{\prime}\left(x-x^{\prime}\right) \ \ \ \ \ (3)$

In our earlier examination of the Schrödinger equation, we assumed that the Hamiltonian is independent of time, which allowed us to obtain an explicit expression for the propagator

$\displaystyle U\left(t\right)=e^{-iHt/\hbar} \ \ \ \ \ (4)$

The propagator is applied to the initial state ${\left|\psi\left(0\right)\right\rangle }$ to obtain the state at any future time ${t}$:

$\displaystyle \left|\psi\left(t\right)\right\rangle =U\left(t\right)\left|\psi\left(0\right)\right\rangle \ \ \ \ \ (5)$

What happens if ${H=H\left(t\right)}$, that is, there is an explicit time dependence in the Hamiltonian? The approach taken by Shankar is a bit hand-wavy, but goes as follows. We divide the time interval ${\left[0,t\right]}$ into ${N}$ small increments ${\Delta=t/N}$. To first order in ${\Delta}$, we can integrate 1 by taking the first order term in a Taylor expansion:

 $\displaystyle \left|\psi\left(\Delta\right)\right\rangle$ $\displaystyle =$ $\displaystyle \left|\psi\left(0\right)\right\rangle +\Delta\left.\frac{d}{dt}\left|\psi\left(t\right)\right\rangle \right|_{t=0}+\mathcal{O}\left(\Delta^{2}\right)\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|\psi\left(0\right)\right\rangle +-\frac{i\Delta}{\hbar}H\left(0\right)\left|\psi\left(0\right)\right\rangle +\mathcal{O}\left(\Delta^{2}\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(1-\frac{i\Delta}{\hbar}H\left(0\right)\right)\left|\psi\left(0\right)\right\rangle +\mathcal{O}\left(\Delta^{2}\right) \ \ \ \ \ (8)$

So far, we’ve been fairly precise, but now the hand-waving starts. We note that the term multiplying ${\left|\psi\left(0\right)\right\rangle }$ consists of the first two terms in the expansion of ${e^{-i\Delta H\left(0\right)/\hbar}}$, so we state that to evolve from ${t=0}$ to ${t=\Delta}$, we multiply the initial state ${\left|\psi\left(0\right)\right\rangle }$ by ${e^{-i\Delta H\left(0\right)/\hbar}}$. That is, we propose that

$\displaystyle \left|\psi\left(\Delta\right)\right\rangle =e^{-i\Delta H\left(0\right)/\hbar}\left|\psi\left(0\right)\right\rangle \ \ \ \ \ (9)$

[The reason this is hand-waving is that there are many functions whose first order Taylor expansion matches ${\left(1-\frac{i\Delta}{\hbar}H\left(0\right)\right)}$, so it seems arbitrary to choose the exponential. I imagine the motivation is that in the time-independent case, the result reduces to 4.]

In any case, if we accept this, then we can iterate the process to evolve to later times. To get to ${t=2\Delta}$, we have

 $\displaystyle \left|\psi\left(2\Delta\right)\right\rangle$ $\displaystyle =$ $\displaystyle e^{-i\Delta H\left(\Delta\right)/\hbar}\left|\psi\left(\Delta\right)\right\rangle \ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{-i\Delta H\left(\Delta\right)/\hbar}e^{-i\Delta H\left(0\right)/\hbar}\left|\psi\left(0\right)\right\rangle \ \ \ \ \ (11)$

The snag here is that we can’t, in general, combine the two exponentials into a single exponential by adding the exponents. This is because ${H\left(\Delta\right)}$ and ${H\left(0\right)}$ will not, in general, commute, as the Baker-Campbell-Hausdorff formula tells us. For example, the time dependence of ${H\left(t\right)}$ might be such that at ${t=0}$, ${H\left(0\right)}$ is a function of the position operator ${X}$ only, while at ${t=\Delta}$, ${H\left(\Delta\right)}$ becomes a function of the momentum operator ${P}$ only. Since ${X}$ and ${P}$ don’t commute, ${\left[H\left(0\right),H\left(\Delta\right)\right]\ne0}$, so ${e^{-i\Delta H\left(\Delta\right)/\hbar}e^{-i\Delta H\left(0\right)/\hbar}\ne e^{-i\Delta\left[H\left(0\right)+H\left(\Delta\right)\right]/\hbar}}$.

This means that the best we can usually do is to write

 $\displaystyle \left|\psi\left(t\right)\right\rangle$ $\displaystyle =$ $\displaystyle \left|\psi\left(N\Delta\right)\right\rangle \ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \prod_{n=0}^{N-1}e^{-i\Delta H\left(n\Delta\right)/\hbar}\left|\psi\left(0\right)\right\rangle \ \ \ \ \ (13)$

The propagator then becomes, in the limit

$\displaystyle U\left(t\right)=\lim_{N\rightarrow\infty}\prod_{n=0}^{N-1}e^{-i\Delta H\left(n\Delta\right)/\hbar} \ \ \ \ \ (14)$

This limit is known as a time-ordered integral and is written as

$\displaystyle T\left\{ \exp\left[-\frac{i}{\hbar}\int_{0}^{t}H\left(t^{\prime}\right)dt^{\prime}\right]\right\} \equiv\lim_{N\rightarrow\infty}\prod_{n=0}^{N-1}e^{-i\Delta H\left(n\Delta\right)/\hbar} \ \ \ \ \ (15)$

One final note about the propagators. Since each term in the product is the exponential of ${i}$ times a Hermitian operator, each term is a unitary operator. Further, since the product of two unitary operators is still unitary, the propagator in the time-dependent case is a unitary operator.

We’ve defined a propagator as a unitary operator that carries a state from ${t=0}$ to some later time ${t}$, but we can generalize the notation so that ${U\left(t_{2},t_{1}\right)}$ is a propagator that carries a state from ${t=t_{1}}$ to ${t=t_{2}}$, that is

$\displaystyle \left|\psi\left(t_{2}\right)\right\rangle =U\left(t_{2},t_{1}\right)\left|\psi\left(t_{1}\right)\right\rangle \ \ \ \ \ (16)$

We can chain propagators together to get

 $\displaystyle \left|\psi\left(t_{3}\right)\right\rangle$ $\displaystyle =$ $\displaystyle U\left(t_{3},t_{2}\right)\left|\psi\left(t_{2}\right)\right\rangle \ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U\left(t_{3},t_{2}\right)U\left(t_{2},t_{1}\right)\left|\psi\left(t_{1}\right)\right\rangle \ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U\left(t_{3},t_{1}\right)\left|\psi\left(t_{1}\right)\right\rangle \ \ \ \ \ (19)$

Therefore

$\displaystyle U\left(t_{3},t_{1}\right)=U\left(t_{3},t_{2}\right)U\left(t_{2},t_{1}\right) \ \ \ \ \ (20)$

Since the Hermitian conjugate of a unitary operator is its inverse, we have

$\displaystyle U^{\dagger}\left(t_{2},t_{1}\right)=U^{-1}\left(t_{2},t_{1}\right) \ \ \ \ \ (21)$

We can combine this with 20 to get

 $\displaystyle \left|\psi\left(t_{1}\right)\right\rangle$ $\displaystyle =$ $\displaystyle I\left|\psi\left(t_{1}\right)\right\rangle \ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U^{-1}\left(t_{2},t_{1}\right)U\left(t_{2},t_{1}\right)\left|\psi\left(t_{1}\right)\right\rangle \ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U^{\dagger}\left(t_{2},t_{1}\right)U\left(t_{2},t_{1}\right)\left|\psi\left(t_{1}\right)\right\rangle \ \ \ \ \ (24)$

Therefore

 $\displaystyle U^{\dagger}\left(t_{2},t_{1}\right)U\left(t_{2},t_{1}\right)$ $\displaystyle =$ $\displaystyle U\left(t_{1},t_{1}\right)=I\ \ \ \ \ (25)$ $\displaystyle U^{\dagger}\left(t_{2},t_{1}\right)$ $\displaystyle =$ $\displaystyle U\left(t_{1},t_{2}\right) \ \ \ \ \ (26)$

That is, the Hermitian conjugate (or inverse) of a propagator carries a state ‘backwards in time’ to its starting point.