By Programmer


2017-11-20 15:37:25 8 Comments

When solving the Schrödinger equation for finite potential well, the solution outside of the well is $$\psi _{1}=Fe^{{-\alpha x}}+Ge^{{\alpha x}}\,\!$$ and $$\psi _{3}=He^{{-\alpha x}}+Ie^{{\alpha x}}\,\!$$ However, when solving the Schrödinger equation for the quantum barrier, the solution of the regions are \begin{align} \psi _{L}(x)&=A_{r}e^{{ik_{0}x}}+A_{l}e^{{-ik_{0}x}}\quad x<0\\ \psi _{C}(x)&=B_{r}e^{{ik_{1}x}}+B_{l}e^{{-ik_{1}x}}\quad 0<x<a\\ \psi_{R}(x)&=C_{r}e^{{ik_{2}x}}+C_{l}e^{{-ik_{2}x}}\quad x>a \end{align}

The solutions for quantum barrier have the imaginary $i$ on the exponent of $e$ for example $A_{l}e^{{-ik_{0}x}}$. But the solution for the finite potential well does not have the imaginary $i$ on the exponent $e$ for example $Fe^{{-\alpha x}}$.

So why is there an imaginary $i$ for quantum barrier problem while there isn't an imaginary $i$ for potential well problem when both solution of the problem is derived by solving the Schrödinger equation in the form of $(\left[{\frac {d^{2}}{dx^{2}}}\psi (x)\right]=b\psi (x))$ ?

Thank you for the reply, and I understand the quantum barrier and potential well a bit more. However, I have a follow up question: Why is the region with potential $V=0$ in the finite potential well have a wavefunction of the form $$\psi _{2}=Asinkx+Bsinkx\,\!$$ But for the quantum barrier, the regions with potential $V=0$ have a wavefunction of $$\psi _{L}(x)=A_{r}e^{{ik_{0}x}}+A_{l}e^{{-ik_{0}x}}\quad x<0\\ $$ So why does the Schrodinger equation produce different result for the region of $V=0$?

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