Propagators for Bounded Time-Dependent Hamiltonians

A time-dependent Hamiltonian calls for a two-time propagator rather than a one-parameter group. When the Hamiltonians are bounded and continuous in operator norm, this propagator can be constructed directly, with explicit convergence and approximation estimates. The physical treatment of drives, pulses, and energy balance belongs to Time-Dependent Hamiltonians. Unbounded-domain existence theorems are treated in Kato Theory.

Helpful background. Bounded Operators supplies norm convergence and operator-valued integration. Existence and Uniqueness of Time Evolution develops the autonomous variation-of-constants argument.

A bounded-Hamiltonian propagator theorem

Let $I=[a,b]$ be compact and let $H:I\to\mathcal B(\mathcal H)$ be continuous in operator norm, with $H(t)=H(t)^*$ at every time. Then there is a unique family of bounded operators $U(t,s)$ satisfying

$$i\hbar\partial_t U(t,s)=H(t)U(t,s),\qquad U(s,s)=I.$$

The derivative exists in operator norm. The family is unitary, jointly norm-continuous, and obeys

$$U(t,r)U(r,s)=U(t,s),\qquad U(t,s)^*=U(s,t).$$

It also has the initial-time derivative

$$i\hbar\partial_sU(t,s)=-U(t,s)H(s).$$

These claims hold for every ordering of the times in $I$. Generally $U(t,s)$ depends on both times, not just their difference.

The assumptions are deliberately strong enough for an elementary proof. Infinite-dimensional Hilbert space is allowed; finite-dimensional matrices are a special case. No uniform bound on the dimension is needed, only boundedness and norm continuity of the operator family on the chosen interval.

Convergent Volterra iteration

For $t\geq s$, start with $V_0(t,s)=I$ and recursively define

$$V_{n+1}(t,s) =-\frac{i}{\hbar}\int_s^t H(r)V_n(r,s)\,dr.$$

Put $M(t,s)=\hbar^{-1}\int_s^t\|H(r)\|\,dr$. Induction gives

$$\|V_n(t,s)\|\leq\frac{M(t,s)^n}{n!}.$$

For the induction step, integrate $\|H(r)\|M(r,s)^n/(\hbar n!)$ and use the derivative of $M(r,s)^{n+1}$. The series $U=\sum_{n\geq0}V_n$ therefore converges uniformly on the time triangle. Substitution under the integral gives the Volterra equation, and norm continuity of $H$ permits its differentiation.

If two solutions have the same initial value, their difference satisfies the homogeneous Volterra equation. Iterating its norm bound $n$ times produces a factorial denominator; taking $n\to\infty$ makes the difference zero. This proves uniqueness without first assuming unitarity.

The iterates are the ordered terms of the Dyson Expansion. Here the estimate upgrades that series to an existence proof. Its truncation after order $N$ has the sufficient bound

$$\left\|U-\sum_{n=0}^N V_n\right\| \leq e^M\frac{M^{N+1}}{(N+1)!}.$$

This bound controls a finite-interval operator error; it does not justify applying the same series to unbounded operators whose products may lack a common domain.

Unitarity requires invertibility as well as norm preservation

Differentiating in norm gives

$$\partial_t(U(t,s)^*U(t,s))=0.$$

Thus $U(t,s)^*U(t,s)=I$, so the constructed operator is an isometry. In infinite dimensions an isometry need not be onto. To finish the argument, solve the equation backward as well. Uniqueness gives the composition law and both identities

$$U(s,t)U(t,s)=I=U(t,s)U(s,t).$$

Hence the isometry is unitary. Differentiating the composition identity with respect to the intermediate initial time yields the displayed equation for $\partial_sU$; the Hamiltonian appears on the right and has the opposite Schrödinger-equation sign.

Frozen Hamiltonians give a controlled approximation

Assume in addition that

$$\|H(t)-H(s)\|\leq L|t-s|.$$

During a step of length $h>0$, freeze the Hamiltonian at the left endpoint and write $W(r)=e^{-irH(t)/\hbar}$. Differentiating $U(t+h,t+r)W(r)$ and integrating gives

$$U(t+h,t)-W(h) =-\frac{i}{\hbar}\int_0^h U(t+h,t+r) \times[H(t+r)-H(t)]W(r)\,dr.$$

Both surrounding factors are unitary. Therefore

$$\|U(t+h,t)-e^{-ihH(t)/\hbar}\| \leq\frac{Lh^2}{2\hbar}.$$

Telescoping products over $N$ equal steps, with $T=Nh$, gives a total error at most $LTh/(2\hbar)$. Each approximate step is exactly unitary, but the overall error is only first order in the step size. Norm preservation alone cannot certify accuracy.

For a bounded spin Hamiltonian $H(t)=(\hbar/2)[\omega_z\sigma_z+\Omega(t)\sigma_x]$ with $|\dot\Omega|\leq K$, one may take $L=\hbar K/2$. The bound becomes $KTh/4$. Rapid modulation demands smaller steps even when every matrix exponential preserves the state norm perfectly.

Exercises

1. Suppose all $H(t)$ commute pairwise. Explain why the solution is the ordinary exponential of the integral.

Solution

Set $K(t)=\int_s^t H(r)\,dr$. Pairwise commutation gives $[K(t),H(t)]=0$. Differentiating the norm-convergent exponential series yields $\partial_t e^{-iK(t)/\hbar}=-(i/\hbar)H(t)e^{-iK(t)/\hbar}$. The initial value is $I$, so uniqueness identifies this expression with $U(t,s)$.

2. Why does the bound $\|U(t,s)\|\leq e^{M(t,s)}$ obtained during iteration not indicate physical amplification?

Solution

The triangle inequality discards cancellations between ordered terms and does not use self-adjointness. It proves convergence for more general bounded generators. Self-adjointness and the inverse evolution subsequently imply the sharper equality $\|U(t,s)\|=1$ on a nonzero Hilbert space.

References

• A. Pazy, Semigroups of Linear Operators and Applications to Partial Differential Equations, Applied Mathematical Sciences 44, Springer, 1983, Chapter 5, publisher edition — time-dependent evolution equations.
• J. Schmid and M. Griesemer, “Kato’s theorem on the integration of non-autonomous linear evolution equations,” Mathematical Physics, Analysis and Geometry 17, 265–271, 2014, doi:10.1007/s11040-014-9154-5, author text — the additional analysis needed for common unbounded domains.