# How can energy be negative in a finite square well?

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?

• 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. – aquirdturtle Jan 27 '16 at 7:23
• 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. – 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.