Imagine that we have a particle in a cylinder of finite length and neglible radius. We can then assume that the system is axisymmetric and can be solved in one dimension.

Let us consider a time varying Hamiltonian $$ \hat H (x,t) = \frac{\hat p^2}{2m} + \begin{cases} V_0x &: t \in \left[0,\frac{T}{2}\right) \\ -V_0x &: t \in \left[\frac{T}{2}, T\right) \end{cases}. $$ where $x \in [0,L]$. We may extend it to all $t>0$ by identifying $T$ with $0$, so that we essentially have a discrete cosine wave. I quote the following from Sakurai (2nd Edition) Section 5.6 on page 345.

This section is devoted to time-dependent Hamiltonians, with some "obvious" approximations in the case of very fast or very slow time dependence. A careful look, though, points out some interesting phenomena, one of which did not come to light until very late in the twentieth century.

Sudden Approximation

If a Hamiltonian changes very quickly, then the system "doesn't have time" to adjust to the change. This leaves the system in the same state it was in before the change and is the essence of the so-called "sudden approximation." Rewrite the Schrodinger equation for the time evolution operator (2.1.25) as $$i\frac{\partial}{\partial s} \mathcal U (t, t_0) = \frac{H}{\hbar / T} \mathcal U (t, t_0 = \frac{H}{\hbar \Omega} \mathcal U (t, t_0)$$ where we have written time $t=sT$ in terms of a dimensionless parameter $s$ and a time scale $T$, and defined $\Omega = 1/T$. In the sudden approximation, the time scale $T\to 0$, which means that $\hbar \Omega$ will be much larger than the energy scale represented by $H$. Assuming we can redefine $H$ by adding or subtracting an arbitrary constant, introducing some overall phase factor in the state vectors, we see that $$\lim_{T\to0} \mathcal U (t, t_0) = 1$$

In the above, $\mathcal U$ is defined as a unitary time evolution operator such that $$|\alpha,t_0;t_0+dt\rangle = \mathcal U(t_0 + dt, t_0)|\alpha, t_0\rangle$$ and $$ \lim_{dt \to 0} \mathcal U (t_0 + dt, t_0) = 1. $$

Claim 1: Given that the potential changes sign every half of a period, I believe that the energy eigenstates of the system during $t\in[0,T/2)$ are the same as the energy eigenstates of the system during $t\in[T/2,T)$ up to a constant phase shift. (Thus, each eigenstate will be unitary throughout time.)

Claim 2: The potential in question may be thought of as gluing together Heaviside step function. From a loose distributional sense, the partial derivatives of the potential in time will be Dirac delta functions. Then, to ascribe meaning to the second derivative, I must integrate the system over a small epsilon neighborhood about $T/2$ and $T$.

Claim 3 (Difficult): If we were to make the magnitude of $V_0$ sufficiently large, could relativistic effects become non negligible without jeopardizing the validity of the Sudden Approximation.

Question: Are my three claims correct?

  • $\begingroup$ @Qmechanic♦, Thanks for the edits. My main confusion is about Claim 3. If we define the potential to be so large that the particle could reach a relativistic speed in a small time scale, would the Sudden Approximation still be valid? Or, would their be higher order nonlinearities that would ruin the model? Could I try solving the Dirac equation instead? Thanks again. $\endgroup$
    – Talmsmen
    Commented Dec 31, 2022 at 20:26
  • $\begingroup$ Is $V_0$ supposed to be a constant, i.e. independent of $x$ (and/or $p$)? If this is the case, the time dependent extra term (periodic or not) has no physically observable consequences. The time evolutions due to the Hamiltonians $H_0$ and $H_0 + f(t) \mathbf{1}$, where $f(t)$ is a time-dependent c-number function, differ only by a time-dependent phase factor $\exp(-i \int_0^t d\tau f(\tau) / \hbar)$, which cancels in all observable quantities (expectation values). On the other hand, if you intend to have an $x$-dependent potential $V_0(x)$, you should state this explicitly in your question. $\endgroup$
    – Hyperon
    Commented Jan 1, 2023 at 0:49
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    $\begingroup$ That linear (instantaneous) potential has no ground state, which will probably cause serious problems. In particular, the energy eigenstates are not normalizable (and badly so), which means you can't just take matrix elements like you normally would. If you want to get something sensible, you probably need a better behaved $V(x)$. $\endgroup$
    – Buzz
    Commented Jan 1, 2023 at 6:29
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    $\begingroup$ If it's a finite well, then there will certainly be normalizable instantaneous energy eigenstates. In fact, generally the only way to force $\psi=0$ at the endpoints is to make the potential go to infinity at those points, since otherwise the wave functions that do vanish there do not form a complete set. However, if you want to keep the instantaneous energy spectrum identical before and after the switches, you presumably want to have your well boundaries symmetrically placed around $x=0$, at $\pm \frac{L}{2}$. $\endgroup$
    – Buzz
    Commented Jan 1, 2023 at 6:47
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    $\begingroup$ Instead of having hard boundaries, you could instead start with a harmonic trap (zeros of Airy functions vs ladder operators). It has the advantage of making everything tractable analytically, while keeping the qualitative picture. Also, perhaps change your notation $V_0\to F$ since it has the dimension of a force. $\endgroup$
    – LPZ
    Commented Jan 2, 2023 at 16:07


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