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Let us suppose we have a particle with energy $E$ and $E<U$ and the potential defined as

$U(x)=0$ for $x<0$ (I)

$U(x)=U_0$ for $x>L$ (III) and

$U(x)=U$ for $0<x<L$ (II)

I find the wavefunction for

region (II)

$$\psi(x)=Ae^{\beta x}+Be^{-\beta x}$$ for $\beta=\frac {\sqrt{2m(U-E)}}{\hbar}$

Is this true ? Because in the site of the hyperphysics it says it should be,

$$\psi(x)=Be^{-\beta x}$$ for $\beta=\frac {\sqrt{2m(U-E)}}{\hbar}$

I am not sure how we can derive this mathematically ? Why the

$Ae^{\beta x}$ term vanishes ?

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  • $\begingroup$ In each region you have different coefficients $A$ and $B$. In addition, the wave function should not grow to infinity; thus some growing exponents must have zero-coefficients. Finally, the wave function and its first derivatives at any point must be continuous. These conditions at $x=0$ and $x=L$ give you a unique wave function (up to a common coefficient). $\endgroup$ – Vladimir Kalitvianski Dec 8 '18 at 13:38
  • $\begingroup$ I can share the solutions for the other regions ? I dont need to find the coefficients but just write the general form of the equation. So you are saying its due to the conditions ? $\endgroup$ – Reign Dec 8 '18 at 13:51
  • $\begingroup$ In each region you have particular values of $A$, $B$, and $\beta$: $A_{I,II,III}$, $B_{I,II,III}$, and $\beta_{I,II,III}$ due to different values of $U(x)$. $\endgroup$ – Vladimir Kalitvianski Dec 8 '18 at 14:48
  • $\begingroup$ Yes I know that. $\endgroup$ – Reign Dec 8 '18 at 14:55
  • $\begingroup$ If $E$ is smaller than any particular value of $U$, then the difference $U-E$ is positive under the square root, so the exponential must decay at $x>0$ rather than grow. Thus $A=0$ to get rid of the growing exponential in the solution. $\endgroup$ – Vladimir Kalitvianski Dec 8 '18 at 18:09

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