How to Derive Energy Expressions for Bound States in a 1D Potential Well?

[Brewer]

Question states:

Derive an expression for the energy of the bound states of a particle in the one-dimensional well defined by:

V(x) = \infinity x<=0 (Region I)
V(x) = 0 0<x<L (Region II)
V(x) = V_0 x>=L (Region III)

And its not been too bad. Up until a point.

I know for I \psi(x)=0.

In the region II \psi(x) = Acos(kx) + Bsin(kx)

In the region III \psi(x) = Ce^{-\alpha x}

where k = \sqrt{\frac{2mE}{\hbar^2}}
and \alpha = \sqrt{\frac{2m(V_0 - E)}{\hbar^2}}

I also know that at the boundary of II and III (x=L) that \psi_{II}(x) = \psi_{III}(x)
Solving for the even parity solutions I get tan(kL) = \frac{\alpha}{k} and for the odd parity solutions I get cot(kL) = -\frac{\alpha}{k}.

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 tan(kl) = - cot(kl), which on first glimpse seems ok, but after a bit of working I manage to get this to cancel to tan(kl) = i, 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.

 

[nrqed]

Question states:

Derive an expression for the energy of the bound states of a particle in the one-dimensional well defined by:

V(x) = \infinity x<=0 (Region I)
V(x) = 0 0<x<L (Region II)
V(x) = V_0 x>=L (Region III)

And its not been too bad. Up until a point.

I know for I \psi(x)=0.

In the region II \psi(x) = Acos(kx) + Bsin(kx)

In the region III \psi(x) = Ce^{-\alpha x}

where k = \sqrt{\frac{2mE}{\hbar^2}}
and \alpha = \sqrt{\frac{2m(V_0 - E)}{\hbar^2}}

I also know that at the boundary of II and III (x=L) that \psi_{II}(x) = \psi_{III}(x)
Solving for the even parity solutions I get tan(kL) = \frac{\alpha}{k} and for the odd parity solutions I get cot(kL) = -\frac{\alpha}{k}.

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 tan(kl) = - cot(kl), which on first glimpse seems ok,

no! You can not set them equal! Those are conditions for the energies of the even and odd solutions..Those are *different* solutions so those two equations should not be mixed! They correspond to different solutions of the Schrodinger equation.

What you must do is to solve separately

tan(kL) = \frac{\alpha}{k} and
cot(kL) = -\frac{\alpha}{k}.

Each is a transcendental solution. You must solve each graphically.

Hope this helps

Patrick

 

[Brewer]

Is there no way to solve either algebraicly?

 

[nrqed]

Is there no way to solve either algebraicly?

Unfortunately, no.

Patrick

 

[Brewer]

But this is the way to go about finding an expression for the bound states of the particle, correct?

It just seems odd to me that you have to do it this way, because while it is now a homework question, it is just a direct lift from an exam paper, where I wouldn't have time (or the capabilities) to draw these graphs.