- #1
T-7
- 64
- 0
Hi,
This is just a quick question -- I'm puzzled by the way this answer sheet represents the potential function.
The question asks us to determine the energy eigenvalues of the bound states of a well where the potential drops abruptly from zero to a depth Vo at x=0, and then increases linearly with position x until at x=a the potential is again zero.
They write:
[tex]V(x) = \frac{V_{0}x}{a}[/tex]
[tex]E = V(b) = \frac{V_{0}b}{a}[/tex]
where b is some point between x=0 and x=a.
But surely the correct representation of the potential function is
[tex]V(x) = V_{0}\left(\frac{x}{a}-1\right)[/tex]
so that V(0) = -Vo, and V(a) = 0. But, using my potential function, I end up with a somewhat different expression for the energy eigenvalues, in the end, than they do. Why do they do it that way? And what's wrong with my pot. function??
Cheers!
This is just a quick question -- I'm puzzled by the way this answer sheet represents the potential function.
The question asks us to determine the energy eigenvalues of the bound states of a well where the potential drops abruptly from zero to a depth Vo at x=0, and then increases linearly with position x until at x=a the potential is again zero.
They write:
[tex]V(x) = \frac{V_{0}x}{a}[/tex]
[tex]E = V(b) = \frac{V_{0}b}{a}[/tex]
where b is some point between x=0 and x=a.
But surely the correct representation of the potential function is
[tex]V(x) = V_{0}\left(\frac{x}{a}-1\right)[/tex]
so that V(0) = -Vo, and V(a) = 0. But, using my potential function, I end up with a somewhat different expression for the energy eigenvalues, in the end, than they do. Why do they do it that way? And what's wrong with my pot. function??
Cheers!