- #1
KFC
- 488
- 4
Assume the potential in question is
[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?
[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?