ballzac
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Just a quick question. Solving the Schrodinger equation (time-independent) for an infinite potential barrier, I end up with two wavefunctions.
In region I,
V(x)=0
\Rightarrow\psi(x)=Acos(\frac{\sqrt{2mE}}{\hbar}x)+Bsin(\frac{\sqrt{2mE}}{\hbar}x)
In region II,
V(x)=\infty
\Rightarrow\psi(x)=Ce^\infty+De^{-\infty}=\infty+0
Clearly in the infinite potential region, the wavefunction must equal zero, so it is clear that the term with constant factor C must vanish. I interpret this as meaning that the D term is the wave that penetrates the barrier (in this case it does not because e^{-\infty}=0), and the C term is a wave coming from the right, that does not exist and therefore C=0. If I am right in assuming this, then how can one prove mathematically, that C=0, without resorting to non-mathematical common sense? If I'm wrong, then what is the explanation?
Thanks in advance, and sorry if there are any errors in the above...made a lot of typos the first time around :$
In region I,
V(x)=0
\Rightarrow\psi(x)=Acos(\frac{\sqrt{2mE}}{\hbar}x)+Bsin(\frac{\sqrt{2mE}}{\hbar}x)
In region II,
V(x)=\infty
\Rightarrow\psi(x)=Ce^\infty+De^{-\infty}=\infty+0
Clearly in the infinite potential region, the wavefunction must equal zero, so it is clear that the term with constant factor C must vanish. I interpret this as meaning that the D term is the wave that penetrates the barrier (in this case it does not because e^{-\infty}=0), and the C term is a wave coming from the right, that does not exist and therefore C=0. If I am right in assuming this, then how can one prove mathematically, that C=0, without resorting to non-mathematical common sense? If I'm wrong, then what is the explanation?
Thanks in advance, and sorry if there are any errors in the above...made a lot of typos the first time around :$