- #1
ballzac
- 104
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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,
[tex]V(x)=0[/tex]
[tex]\Rightarrow\psi(x)=Acos(\frac{\sqrt{2mE}}{\hbar}x)+Bsin(\frac{\sqrt{2mE}}{\hbar}x)[/tex]
In region II,
[tex]V(x)=\infty[/tex]
[tex]\Rightarrow\psi(x)=Ce^\infty+De^{-\infty}=\infty+0[/tex]
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 [tex]e^{-\infty}=0[/tex]), and the C term is a wave coming from the right, that does not exist and therefore [tex]C=0[/tex]. If I am right in assuming this, then how can one prove mathematically, that [tex]C=0[/tex], 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,
[tex]V(x)=0[/tex]
[tex]\Rightarrow\psi(x)=Acos(\frac{\sqrt{2mE}}{\hbar}x)+Bsin(\frac{\sqrt{2mE}}{\hbar}x)[/tex]
In region II,
[tex]V(x)=\infty[/tex]
[tex]\Rightarrow\psi(x)=Ce^\infty+De^{-\infty}=\infty+0[/tex]
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 [tex]e^{-\infty}=0[/tex]), and the C term is a wave coming from the right, that does not exist and therefore [tex]C=0[/tex]. If I am right in assuming this, then how can one prove mathematically, that [tex]C=0[/tex], 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 :$