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Consider a quantum mechanical system with hamiltonian $\hat{H}$, and eigenstates $\psi_n$ with energies $E_n$. I want to `force' this system into a specific eigenstate $\psi_0$ that has energy $E_0$. My idea is to add a periodic perturbation $\epsilon \hat{V}(t)$ that is periodic in time with period $\frac{\hbar}{E_0}$; here $\epsilon$ is a small real number. The new system has Hamiltonian $$\hat{H}+\epsilon\hat{V}(t).$$ My intuition says that, regardless of the initial condition at $t=0$, the periodic perturbation would cause the whole system to be periodic with period $\frac{\hbar}{E_0}$ in the limit $t\rightarrow\infty$; and in the unperturbed system, the phase of eigenstate $\psi_0$ also oscillates with this period. So when $\epsilon$ is very small, the system would eventually move into a state close to $\psi_0$. However, I have not managed to prove this. Is this true in general?

Own attempts

Here are my attempts using variation of constants and Floquet's theorem. Firstly, write $\psi(t)=\sum_n c_n(t)\psi_n$ with all $c_n(t)\in\mathbb C$. Now we get from the Schrödinger equation: $$\frac{\partial c_n(t)}{\partial t} = \frac{1}{i\hbar}\left(E_nc_n(t)+\sum_m \langle \psi_n\mid\epsilon\hat{V}(t)\mid\psi_m\rangle c_m(t)\right).$$ This is a linear system of differential equations with time-periodic coefficients. So we can apply Floquet's theorem. This gives us \begin{equation} \begin{pmatrix} c_1(t)\\ c_2(t)\\ \vdots\\ c_N(t) \end{pmatrix} = P(t)e^{tB}\mathbf{v} \end{equation} where $P(t)$ is a time-periodic $N\times N$-matrix with period $\frac{\hbar}{E_0}$, $B$ is a constant $N\times N$-matrix, and $\mathbf{v}$ is a $N\times 1$-vector. We know that $|c_n(t)|^2\leq1$ for all $n,t$, so it follows that all eigenvalues of $B$ have real part at most $0$.

At this point I get confused. How can I prove that when $t\rightarrow\infty$, the system will converge to a state that is close to $\psi_0$? Or is that not true in general? With `close' I mean that when $\epsilon\downarrow0$, the limiting state will converge to $\psi_0$.

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  • $\begingroup$ I don't quite understand. What is the initial condition? Any vector or $\psi_0$ itself? $\endgroup$ Commented Oct 26 at 14:16
  • $\begingroup$ My intuition says that this will work for any initial condition, as long as the matrix for the dynamical system contains nonzero off-diagonal elements. I have now edited this into the question. $\endgroup$
    – Riemann
    Commented Oct 26 at 14:17
  • $\begingroup$ This is similar to the concept of forced vibration in classical mechanics: regardless of the initial condition, the system will eventually have the same periodicity as the driving force $\endgroup$
    – Riemann
    Commented Oct 26 at 14:18

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It will not in general converge to $\psi_0$ if the perturbation is periodic. Rather the system will oscillate between your initial state and other states in your Hilbert space. In general there is no guarantee that you can reach your target state $\psi_0$ with 100% probability at any time: this would depend on the details of your perturbation.

In the simplest model your $\hat V(t)$ would couple only your initial state with your target state $\psi_0$, and no other state. The problem is then equivalent to a 2-state Rabi model, where the populations oscillate between the two states.

There might be some specialized forms of $\hat V(t)$ that will get you to $\psi_0$ with 100% probability at some specific times, but in general the best you can do is to choose parameters in your $\hat V(t)$ to maximize the probability of getting $\psi_0$ at some specific times, but no guarantee that the population of all other states will be $0$.

As a side note, your approach to solution does not involve any perturbative expansion, but rather an exact solution to the time-dependent Schrödinger equation. A perturbative expansion might give you insights on how to choose parameters in your $\hat V(t)$ but any hope of getting a specific state at some specific time would require — as you set it up — an exact solution.

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  • $\begingroup$ Thanks for the answer! I've started reading about the Rabi model at Wikipedia, but there is something I don't understand. At some point they introduce a perturbation $H^1(t)=H^1e^{-i\omega t}$. But isn't this hamiltonian non-hermitian? en.wikipedia.org/wiki/… $\endgroup$
    – Riemann
    Commented Oct 26 at 15:40
  • $\begingroup$ Also, I found out the problem with my intuition: when the Hamiltonian is periodic with period $T$, that does not mean that the wavefunction is periodic with period $T$, or any period at all. This is true, for example, in case of the 2-state Rabi model. Equation 69 in the following pdf illustrates this phenomenon. bohr.physics.berkeley.edu/classes/221/notes/spinmagf.pdf $\endgroup$
    – Riemann
    Commented Oct 26 at 18:40
  • $\begingroup$ I asked about the wikipedia issue here: physics.stackexchange.com/questions/832265/… $\endgroup$
    – Riemann
    Commented Oct 26 at 20:24

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