- #1
E92M3
- 68
- 0
I can solve the infinite will with length a. But what happens when the coordination is shifted, namely:
[tex]V(x)=0,\frac{ -a}{4}<x<\frac{3a}{4}[/tex]
I use the usual solution:
[tex]\psi=Asin(kx)+Bcos(kx)[/tex]
Now when I apply the first boundary condition:
[tex]\psi(\frac{ -a}{4})=\psi(\frac{3a}{4})=0[/tex]
I can't get rid of one of the terms (namely the B=0 like when the well was defined from 0 to a).
I know that A will be the same as if the well was defined from 0 to a and I can get k by applying the other boundary condition, but I can't do anything if I can't get rid of the cos term.
[tex]V(x)=0,\frac{ -a}{4}<x<\frac{3a}{4}[/tex]
I use the usual solution:
[tex]\psi=Asin(kx)+Bcos(kx)[/tex]
Now when I apply the first boundary condition:
[tex]\psi(\frac{ -a}{4})=\psi(\frac{3a}{4})=0[/tex]
I can't get rid of one of the terms (namely the B=0 like when the well was defined from 0 to a).
I know that A will be the same as if the well was defined from 0 to a and I can get k by applying the other boundary condition, but I can't do anything if I can't get rid of the cos term.