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.

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