Assume the potential in question is(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

V = \left\{

\begin{matrix}

\infty, \qquad x<0 \\

-V_0, \qquad 0\leq x \leq a \\

0, \qquad x>a

\end{matrix}

\right.

[/tex]

where [tex]V_0[/tex] is positive.

if we need to find the bound state, we consider the energy is less than the potential. But the potential withn [0, a] is negative, is that mean the energy will be more negative (i.e. [tex]|E| > V_0[/tex]) ?

Generally, the Schrodinger equation will be written of the following form

[tex]

\frac{d^2\psi}{dx^2} + k^2\psi = 0

[/tex]

where

[tex]k = \sqrt{\frac{2m}{\hbar^2}(E-V)}[/tex]. The general solution is of the form

[tex]\psi = A\exp(ikx) + B\exp(-ikx)[/tex]

For x<0 or x>a region, the wavefunction must decay because the potential is larger than E, [tex]k=\sqrt{\frac{2m}{\hbar^2}(E-V)} = i \sqrt{\frac{2m}{\hbar^2}(V-E)} = i\kappa[/tex], the solutions in those region become

[tex]\psi = A\exp(\kappa x) + B\exp(-\kappa x)[/tex]

But within [0, a], if we want to find bound state, energy must be negative and [tex]|E|>V_0[/tex], so

[tex]k = \sqrt{\frac{2m}{\hbar^2}(-|E|-(-V_0))} = \sqrt{\frac{2m}{\hbar^2}(V_0-|E|)} [/tex]

but this also lead to k be imaginary number, that is, the solution within [0, a] is decaying again??? But I think the solution in that region should be oscillating. Where am I get wrong?

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# Bound state

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