- #1
Brewer
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Question states:
Derive an expression for the energy of the bound states of a particle in the one-dimensional well defined by:
[tex]V(x) = \infinity x<=0[/tex] (Region I)
[tex]V(x) = 0 0<x<L[/tex] (Region II)
[tex]V(x) = V_0 x>=L[/tex] (Region III)
And its not been too bad. Up until a point.
I know for I [tex]\psi(x)=0[/tex].
In the region II [tex]\psi(x) = Acos(kx) + Bsin(kx)[/tex]
In the region III [tex]\psi(x) = Ce^{-\alpha x}[/tex]
where [tex]k = \sqrt{\frac{2mE}{\hbar^2}}[/tex]
and [tex]\alpha = \sqrt{\frac{2m(V_0 - E)}{\hbar^2}}[/tex]
I also know that at the boundary of II and III (x=L) that [tex]\psi_{II}(x) = \psi_{III}(x)[/tex]
Solving for the even parity solutions I get [tex]tan(kL) = \frac{\alpha}{k}[/tex] and for the odd parity solutions I get [tex]cot(kL) = -\frac{\alpha}{k}[/tex].
Now in order to find an expression for E I want to solve these equations for k (as k is a function of E).
Equating the 2 equations I end up with [tex]tan(kl) = - cot(kl)[/tex], which on first glimpse seems ok, but after a bit of working I manage to get this to cancel to [tex]tan(kl) = i[/tex], and correct me if I'm wrong, but I'm sure that I can't do the inverse tan of i can I? Does this look like the right way to go about attempting this question? Have I just messed up the maths somewhere? Is there anything you suggest that might point me in the right direction?
Thanks guys.
Derive an expression for the energy of the bound states of a particle in the one-dimensional well defined by:
[tex]V(x) = \infinity x<=0[/tex] (Region I)
[tex]V(x) = 0 0<x<L[/tex] (Region II)
[tex]V(x) = V_0 x>=L[/tex] (Region III)
And its not been too bad. Up until a point.
I know for I [tex]\psi(x)=0[/tex].
In the region II [tex]\psi(x) = Acos(kx) + Bsin(kx)[/tex]
In the region III [tex]\psi(x) = Ce^{-\alpha x}[/tex]
where [tex]k = \sqrt{\frac{2mE}{\hbar^2}}[/tex]
and [tex]\alpha = \sqrt{\frac{2m(V_0 - E)}{\hbar^2}}[/tex]
I also know that at the boundary of II and III (x=L) that [tex]\psi_{II}(x) = \psi_{III}(x)[/tex]
Solving for the even parity solutions I get [tex]tan(kL) = \frac{\alpha}{k}[/tex] and for the odd parity solutions I get [tex]cot(kL) = -\frac{\alpha}{k}[/tex].
Now in order to find an expression for E I want to solve these equations for k (as k is a function of E).
Equating the 2 equations I end up with [tex]tan(kl) = - cot(kl)[/tex], which on first glimpse seems ok, but after a bit of working I manage to get this to cancel to [tex]tan(kl) = i[/tex], and correct me if I'm wrong, but I'm sure that I can't do the inverse tan of i can I? Does this look like the right way to go about attempting this question? Have I just messed up the maths somewhere? Is there anything you suggest that might point me in the right direction?
Thanks guys.