When a particle in one dimension have discrete spectrum?

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SUMMARY

A system in one dimension with the potential V(x) = b|x| exhibits a discrete energy spectrum. The energy eigenvalues must satisfy the integral condition ∫x1x2 dx √{2m[E - λ|x|]} = (n + 1/2) πħ, which arises from the boundary conditions of the wave function in quantum mechanics. When the minimum potential exceeds the particle's energy, the solutions become complex, leading to a continuous spectrum. Conversely, when the energy surpasses the minimum potential, the solutions yield discrete values, indicating a bound state.

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What are the conditions for which it can be concluded that a system has discrete energy levels?
For example a system in one dimension with the potential
[itex]V(x)=b|x|[/itex]
has only a discrete spectrum. How I can prove it?
My book says moreover that the energy eigenvalues have to satisfy the condition
[itex]\lmoustache_{x_1}^{x_2} dx \sqrt{2m[E- \lambda |x|]} = (n+1/2) \pi \hbar[/itex]
why?

thanks for help.
 
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In general, in cases where the minimum potential is bigger than the energy of the particle, you find that the general solution will be complex, and the energy takes a continuous form. on the other hand when the energy is bigger than the minimum potential, the solutions take discrete values as we have a bound particle.
 

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