Quantum hamiltonian with an expoenntial potetial.

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given the Schroedinger equation with an exponential potential

-D^{2}y(x)+ae^{bx}y(x)-E_{n}y(x)= 0

with the boudnary conditons y(0)=0=y(\infty)

is this solvable ?? what would be the energies and eigenfunction ? thanks.
 
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##V(x)=ae^{bx}## represents an infinite barrier when approached from the left ... if b>0.
Where did you get those boundary conditions from?

##V(x=0)=ae^{0}=a## if E>a, then y(0) need not be zero.
You do need y(x) to be continuous where it crosses the barrier.
 
um i forgot .. y(0)=0 assume there is an infinite potential barrier so the wave function must be 0 at the origin.
 
Oh well, that one will have solutions that look a lot like the 1/x potential - at least, for the lower energies.

For simplicity, measure energy from the bottom of the well so ##V(0<x)=a(e^{bx}-1)## ... positive values of E will include bound states - so the answer to your question is: yes - solutions exist, and the SE for this potential should be solvable.

If you want a rigorous proof of solvability you'll have to ask a mathematician ;)
Getting the analytical solution will probably be a bit of a pain... but it usually is.
How did this come up?
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!

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