How to prove that there is a bound state in the potential $U(x) = -A e^{-a |x|}$, where for all $a \in \mathbb{R}$ and $A>0$. I heard that we can say something to the minimum of this form $ \left( \psi \right| H \left| \psi \right)$ for some vector of hilbert space, but that it will give us?

So i want to know, why if there is $\psi$ such that $\left( \psi \right| H \left| \psi \right) < 0$ then there is bound state?

Thank you!

  • $\begingroup$ Your a>0 if you are to have a sensible potential bounded below. So you accept the RR principle, no? $\endgroup$ – Cosmas Zachos Feb 13 '18 at 2:24
  • $\begingroup$ @Cosmas Zachos, I understand that any self-adjoint operator has an expansion of unity, that is, a complete set of functions, but we do not know if there are bound states. Therefore, I can not understand why it is not proved the existence of a minimum of the functional in RR principle. $\endgroup$ – Ann Feb 13 '18 at 6:41
  • $\begingroup$ $a$ can be negative?? Then the potential (and spectrum) is unbounded from below, i.e. there is no stable ground state. $\endgroup$ – Qmechanic Feb 13 '18 at 7:04
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    $\begingroup$ Anyway, if we overlook this pathology, it is essentially a duplicate of physics.stackexchange.com/q/143630/2451. I gave a general 1D proof in my answer. $\endgroup$ – Qmechanic Feb 13 '18 at 7:11

The potential energy can have any reference energy without change in the outcome. Thus a negative energy is relative to the assumed reference energy. If you take a large enough negative reference, you will only have positive bound energy states.

  • $\begingroup$ Hi, i found Rayleigh–Ritz variational theorem: If $U(x) < 0$ for any $x \in \mathbb{R}$ and there is $\psi \in H$ such that $ \left(\psi, \psi \right) = 1$ and $ \left( \psi \right| H \left| \psi \right) < 0$. Then there is at least 1 solution of schroedinger equation. But I could not find proof of this. $\endgroup$ – Ann Feb 13 '18 at 1:31
  • $\begingroup$ It is important for me to understand why it exists. physics.stackexchange.com/q/143630 $\endgroup$ – Ann Feb 13 '18 at 1:42

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