- #1

Brewer

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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.