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We are considering the instantaneous eigenstates of an analytically time-dependent hamiltonian and I would like to know how legitimate it is to extend them to the complex plane.

Specifically, our hamiltonian describes an atom or a molecule in an external laser pulse. Thus it includes kinetic energy terms, all Coulomb potentials, and a laser interaction, and it can be written $$H(t)=H_0+\sum_{i=1}^N\mathbf{F}(t)\cdot\mathbf{r}_i.$$ Here $H_0$ can be assumed to be as well-behaved as necessary as long as one preserves the basic structure, but in general it's outside our capacity for exact solutions. $\mathbf{F}(t)$ is a vector-valued analytic (and if necessary entire) function of the time $t$ which is real for real times. We define the instantaneous eigenstates to obey the relation $$H(t)|n(t)\rangle=E_n(t)|n(t)\rangle$$ in Dirac notation.

My question: Since $\mathbf{F}(t)$ can be extended to complex times, can the eigenstates $|n(t)\rangle$ be extended as well?

My specific worry is the following. One can always write, in this setting, $$H(t)=T+V_1(t)+iV_2(t),$$ where $T$ is the kinetic energy and all three of $T$, $V_1$ and $V_2$ are hermitian and well-behaved. However, for nonzero $V_2$ it appears that $H(t)$ is no longer normal: $$[H,H^\dagger]=[T+V_1+iV_2,T+V_1-iV_2]=2i[V_2,T],$$ which is in general nonzero since $T$ contains derivatives. It would appear then that formally none of the $|n(t)\rangle$ exist outside $t\in\mathbb{R}$. However, I would find such a strict cutoff somewhat strange - i.e., what if $V_2$ is nonzero but can somehow be neglected w.r.t. the other terms? surely there is some kind of a gray area around the real axis of some sort. Further, the existence and analyticity of the eigenstates for complex times, and in particular matrix elements involving them, are crucial to our results and physical intuition seems to indicate they should exist.

For completeness, we are considering in particular matrix elements of the form $$\langle m(t_1)|U(t_1,t_2)A(t_2)U(t_2,t_3)|n(t_3\rangle,$$ where $A(t)$ is analytic in $t$ and hermitian for real times, $U(t,t')$ is the propagator and obeys $$i\frac{\partial}{\partial t}U(t,t')=H(t)U(t,t'),\quad U(t,t)=\textrm{id},$$ and all three of the times involved, $t_1, t_2$ and $t_3$, may be complex.

Any insights will be deeply appreciated.

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    $\begingroup$ In the following simple case the answer is yes : $H(t) = -\frac{d^{2}}{dx^{2}} + (x-F(t))^{2}$ $\endgroup$
    – jjcale
    Commented Aug 17, 2012 at 18:43

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An operator can have eigenvectors without being normal. Matters are complicated somewhat by having unbounded operators, but under some reasonable conditions you do have analyticity of eigenvectors for isolated eigenvalues. See Kato, "Perturbation Theory for Linear Operators", http://books.google.ca/books?id=8ji2kN_D3BwC

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