Given potential $V(x) = Asec(x)$ for $x > 0$. I want to calculate the ground-state energy $E_0$ via the Schrödinger equation.

I'm completely stuck on this one. I've set up the time-independent Schrödinger equation, but it can't be solved without using special functions. I don't see how I can calculate the energy without solving the schrodinger equation. Any hints?

  • $\begingroup$ Have you found the general shape of the curve? Have you found the minimum? Have you deduced the shape of the curve in the neighborhood of the minimum? How do you proceed? $\endgroup$ Commented Mar 21, 2012 at 3:04
  • 2
    $\begingroup$ This is exactly the point. As the particle is large, it is localized. The localization will be around the minimum of the potential. Could you write up a solution and post it, answering your own question? $\endgroup$
    – yohBS
    Commented Mar 21, 2012 at 7:27
  • 3
    $\begingroup$ It is really extremely confusing if you change the question completely so that previous comments suddenly don't make any sense at all. $\endgroup$
    – Lagerbaer
    Commented Mar 21, 2012 at 17:59

1 Answer 1


The energy spectrum of the problem(v3) with potential

$$\Phi(x)~=~\frac{A}{\cos x},\qquad\qquad x\in\mathbb{R}_{+}, $$

is unbounded from below, i.e., there is no ground state.

This can e.g. be seen using semiclassical methods a la this answer. Semiclassically, the reason is:

  1. because the potential $\Phi$ has infinitely many periods, and
  2. because the classically accessible length within one period is non-zero for any (potential) energy-level $V$.

The total accessible length $\ell(V)$ is therefore infinite for any (potential) energy-level $V$, no matter how negative $V$ is. In other words, the accessible region of phase space is always bigger than Planck constant $h$, and we can hence fit a semiclassical state, for any energy-level $E$, no matter how negative $E$ is.


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