Say if the potential $V(x) < 0$ in the well but the sides or the scattered states its zero potential, anyways

  • How is that the energy in the well is less than zero?

  • Is it because the potential is less than zero?

  • $\begingroup$ I'm not sure what "but the sides or the scattered states its zero potential" means. Energy being zero is just a convention, you could raise the entire potential by some constant (making E positive) and none of the physics would change. I'm not sure where exactly the confusion is, but if the potential is negative and a particle has less kinetic energy than potential energy, this immediately results in a negative energy. $\endgroup$ – aquirdturtle Jan 27 '16 at 7:23
  • $\begingroup$ Before trying to understand negative energy in quantum mechanics, note that this also happens in classical mechanics. E.g. it is well-known that the mechanical energy $E$ of a satellite is negative when its orbit is bounded to the planet. $\endgroup$ – Qmechanic Jan 27 '16 at 9:29

Energy or the value of $V(x)$ negative means it is a bound system. Think of it in this way, if a particle is free and has no kinetic energy and potential energy then it's total energy is zero. If this particle is not free or otherwise is bounded by a negative potential well then it's potential energy is $-V$. You have to give the same amount of energy, in this case $+V$ to make it free. Then it's total energy will be $-V+V=0$ and it will go free. To summarize a particle in a potential well $-V(x)$ means it is bounded by the well and has $V$ amount of energy less than of what it would need to become free.

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    $\begingroup$ You should make it clear that there is a convention here, that you are choosing the zero of potential energy. There exist positive energy bound states if you shift the potential energy up enough. In fact, there are potentials for which there are an infinite number of positive-energy bound states, no matter how you shift the energy: e.g. the harmonic oscillator. Negative energy on its own means nothing until you know what the exact structure of the potential is. $\endgroup$ – march Jan 27 '16 at 7:56
  • $\begingroup$ Yes @march, I should have made it clear. Thanks for the comment, I appreciate it. $\endgroup$ – Taskin Jan 27 '16 at 8:48

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