I'm having this problem for a while, but I'm still failing to realize what is bothering me in this, so here is my attempt at explaining my doubt:

I have the time-independent Schrodinger equation: $$-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}u(x)+V(x)u(x)=Eu(x).$$ If, for example, I want to find solutions for $E<0$, in an infinite well $V(x)=0$, I have: $$\frac{\partial^2}{\partial x^2}u(x)-k^2u(x)=0$$ and specify $k^2=\frac{2m}{\hbar^2}|E|$.

But now consider the potential barrier, with $E<V_0$, $$V(x)= \begin{cases} 0, \hspace{0.5cm} x<-a\\ V_0, \hspace{0.5cm} -a<x<a\\ 0, \hspace{0.5cm} x>a \end{cases}$$

And therefore I would know define $\alpha^2=\frac{2m|E-V_0|}{\hbar^2}$ and $$\frac{\partial^2}{\partial x^2}u(x)-\alpha^2u(x)=0.$$

Do these parameters $k,\alpha, etc...$, have any special conditions? Must they be always positive? For example, Gasiorowicz, 3rd edition, page.71, specifies, for the potential barrier, $\alpha^2=-\frac{2m(E-V_0)}{\hbar^2}$ and writes: $$\frac{\partial^2}{\partial x^2}u(x)-\alpha^2u(x)=0,$$ but now that would be $$\frac{\partial^2}{\partial x^2}u(x)-(-\frac{2m(E-V_0)}{\hbar^2})u(x)=0\rightarrow\frac{\partial^2}{\partial x^2}u(x)+\frac{2m(E-V_0)}{\hbar^2}u(x)=0.$$ Why not only $$\frac{\partial^2}{\partial x^2}u(x)-\frac{2m(E-V_0)}{\hbar^2}2u(x)=0,$$ instead?

I'm sorry if I was unclear!


1 Answer 1


There's no requirement that $\alpha^2>0$, although your differential equation will be either $u''+u=0$ or $u''-u=0$ depending on your choice of the sign of $\alpha^2$. The author chooses to define $\alpha^2$ in the way he does probably for one of two reasons:

  1. he is cleverly anticipating the final result, a real-valued exponential, inside the barrier; or
  2. he wants to treat $\alpha$ as an observable, hence it must be real-valued.

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