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I'm struggling with the following problem:

Consider the semi-infinite potential well with equation: $$ V(x) = \begin{cases} +\infty & x<0 \\ -V_0 & 0<x<a \\ 0 & x>a \end{cases}$$ Show, by solving TISE, that for $E<-V_0 \implies \psi(x)=0$ everywere. $-V_0 < 0$ and $a>0$.

Note: I know that $E$ cannot be less than the minimum value of the potential but here I'm asked to explicitly make calculations and show what happens.

I know that, in theory, the problem that I should encounter is a normalization problem that will force me to set all normalization constant to $0$.

Here's what I did to solve the problem. First I divided the problem in $3$ regions (I for $x<0$, II for $0<x<a$, III for $x>a$).

For region I the wave function will be $0$ $$\psi_I(x)=0$$ For region II I have: $$-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} - V_0 \psi = E\psi \implies \psi_{II}(x)=Ae^{kx}+Be^{-kx} \ \ \ \ \ \ k=\frac{\sqrt{-2m(E+V_0)}}{\hbar}$$ For region III I have: $$-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} = E\psi \implies \psi_{III}(x)=Ce^{lx}+De^{-lx} \ \ \ \ \ \ l=\frac{\sqrt{-2mE}}{\hbar}$$ Now $\psi(0)=0 \implies A=-B$ and $C=0$ since the function has to go to zero at infinity. I'm left with the following wave function: $$ \psi(x) = \begin{cases} 0 & x<0 \\ A(e^{kx}-e^{-kx}) & 0<x<a \\ De^{-lx} & x>a \end{cases}$$

Here I have the problem: this function appears to be normalizable since I can impose continuity at $a$ to get: $$A(e^{ka}-e^{-ka}) = De^{-la} \implies D=A(e^{ka}-e^{-ka})e^{la}$$ therefore I can solve the usual integral to get the value of A. This integral in my calculations does not diverge at all.

Finally, I tried to use additional condition imposing continuity of the first derivative at $a$ but still could not extract any useful realtion from this condition.

Sorry for the length of this post but I really tried to solve this and I wanted to show everything I did.

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  • $\begingroup$ For "usual integral" i mean $\int_0^{\infty} |\psi(x)|^2 dx = 1$. Both integral parts (between 0 and a and between a and infinity) yelds a finite value. $\endgroup$ Jan 15 at 19:24
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    $\begingroup$ Enforcing the continuity of the derivative at $a$ should work. You will find a relation between $k$ and $l$ that is incompatible with other assumptions that you have made. $\endgroup$
    – Rishi
    Jan 16 at 0:28
  • $\begingroup$ Thanks, I tried again today and I found the solution using your suggestion, I will post the complete follow-up as an answer to this question. $\endgroup$ Jan 16 at 17:27

1 Answer 1

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As has been noticed in your post, the solution in region II is given by: $$\psi_{II}(x)=Ae^{\alpha x}+Be^{-\alpha x}.$$ Where $\alpha$ is a pure imaginary number given by: $$\bigg(-{2m(V_0+E)\over{\hbar}^2}\bigg)^{1/2}.$$ For region III we have that: $$\psi_{III}(x)=Ce^{\beta x}+De^{\beta x},$$ with $\beta$, a real number given by: $$\bigg(-{2mE\over\hbar}\bigg)^{1/2}=\bigg({2m|E|\over\hbar}\bigg)^{1/2}.$$ Since $\psi(x)$ must go to zero as $x$ goes to infinity, we have that $C=0$, thus: $$\psi_{III}(x)=De^{-\beta x}.$$ Applying the boundary condition at $x=0$ yields $A=-B$ so that $$\psi_{II}(x)=A(e^{\alpha x}-e^{-\alpha x}).$$ Applying the boundary condition at $x=a$ yields: $$A(e^{\alpha a}-e^{-\alpha a})=De^{\beta a}.$$ Note that the boundary condition then implies that $D/A$ is a pure imaginary number since: $$e^{\alpha a}-e^{-\alpha a}=2i\;Im(e^{\alpha a}).$$ Now, making use of the fact that $\psi(x)$ must have a continuous first derivative at $x=a$, we compute: $${d\psi_{II}\over dx}(x)\bigg|_{x=a}=A\alpha(e^{\alpha a}+e^{-\alpha a})$$ and also that: $${d\psi_{III}\over dx}(x)\bigg|_{x=a}=-D\beta e^{-\beta a}.$$ By continuity, the above two relations are equal so that: $$A\alpha(e^{\alpha a}+e^{-\alpha a})=-D\beta e^{\beta a}.$$ From this relation we note that $D/A$ is a real number since: $$e^{\alpha a}+e^{-\alpha a}=2\;Re(e^{\alpha a}).$$ However, this is a contradiction since the we earlier determined that $D/A$ was a pure imaginary number. The only choice that satisfies all the boundary conditions is therefore $A=D=0$. Hence, $\psi(x)=0$ for all $x$ in the three regions of interest.

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