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Given a certain potential $ a^{x} $ with positive non-zero 'a' are there a discrete spectrum of energy state for the Schrodinger equation

$$ \frac{- \hbar ^{2}}{2m} \frac{d^{2}}{dx^{2}}f(x)+a^{x}f(x)=E_{n}f(x)$$

Is there an example of this potential in physics?

EDIT: what would happen if we put instead $ a^{|x|} $ so the potential is EVEN and tends to infinity as $ |x| \to \infty $

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after I read your question I immediately started working on $e^{|x|}$. Looks like we had the same idea :) You might want to try using the WKB approximation, that's what I am trying to do. – kηives Nov 10 '12 at 22:03
related: exponential potential $\exp(|x|)$. – Emilio Pisanty Aug 24 '13 at 2:50
up vote 6 down vote accepted

I) OP's potential

$$V(x)~=~a^x~=~e^{bx}, \qquad b~:=~\ln a ~\in~ \mathbb{R}, $$

is the so-called Liouville potential.

There are no (discrete) bound states. In scattering theory, an incoming wave at $x=-\infty$ gets reflected by the so-called "Liouville wall", and returns to $x=-\infty$.

This potential is used in e.g. Liouville theory, which is important in dilaton gravity theories and string theory.

II) On the other hand, even potentials $V(x)=V(-x)$ of e.g. the form




have discrete spectra.

III) Finally, let us mention that double Liouville potentials

$$V(x)~=~A_1e^{b_i x} + A_2e^{b_2 x} $$

(and multiple Liouville potentials) have also been studied in the literature. See also Toda field theory.

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oh it is clear however if we input that the wave function is $ y(0)=0=y(\infty) $ or modify the potential to ve $ V(x)=a^{|x|} $ what would happen then ?? – Jose Javier Garcia Nov 10 '12 at 21:56
I updated the answer. – Qmechanic Nov 10 '12 at 22:38

As $x \rightarrow -\infty$ the potential $V(x) = a^x$ will go to 0 so if you start with a particle as a wavepacket anywhere on that potential, it will eventually end up travelling to $x \rightarrow -\infty$. Even if the packet started traveling in the positive $x$ direction, it will bounce off the potential and go to $x \rightarrow -\infty$. So the only eigenvalues the potential will have would be the eigenvalues of a free particle in the region of $x \rightarrow -\infty$. Therefore there are no bound states and it will have a continuum of energy levels just like a free particle has.

I know of no potential like this in physics.

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