Boundary conditions in finite potential well

[Ritorufon]

Homework Statement

Hi guys, I'm having trouble understanding the finite potential well, in particular the boundary conditions

The well under scrutiny has potential

$$V(x)= 0$$ for $$|x|<a$$

and

$$V(x)=V_0$$ for $$>a$$

Homework Equations

$$\frac{d^2\psi}{dx^2}=-\sqrt{\frac{2mE}{\hbar^2}}\psi=-\alpha^2\psi$$ (1)

$$\frac{d^2\psi}{dx^2}=-\sqrt{\frac{2m(V_0-E)}{\hbar^2}}\psi=\beta^2\psi$$ (2)

with solutions to (1)

$$Asin\alpha x+Bcos\alpha x$$

and (2)

$$Ce^{-\beta x}+De^{+\beta x}$$

The Attempt at a Solution

Know that $$\psi(x)$$ is single valued, finite and continuous so at the boundary x = a

$$\psi_{inside} (a)=\psi_{outside} (a)$$

and

$$\frac{d^2 \psi_{inside} (a)}{dx^2}=\frac{d^2 \psi_{outside} (a)}{dx^2}$$

so for odd parity at the border x = a

$$Asin\alpha a = Ce^{-\beta a}$$ for $$\psi(a)$$ continuity

and

$$\alpha Acos\alpha a = -\beta Ce^{-\beta a}$$ for $$\frac{d^2 \psi (a)}{dx^2}$$ continuity

so if you divide the differentials by the functions you get

$$tan \alpha a = -\alpha / \beta$$

The same can be done with the even parity but if done the same as above you get

$$cot \alpha a= \beta/\alpha$$

but this incorrect the correct answer is

$$tan \alpha a=\beta/\alpha$$

but to get this answer you would have to divide the differental by the function and not the other way around like what was done for odd solutions.

My question is how do you know whether to divide the differental by the function or the function by the differential, also what physical meaning do these solutions have in relation to the well?

 

[Klockan3]

$$\frac{d^2 \psi_{inside} (a)}{dx^2}=\frac{d^2 \psi_{outside} (a)}{dx^2}$$

Just to mention, this one is really wrong. The fun thing is though that you calculate as if you did write the correct thing though, so it seems like you either slipped up everywhere in your post or you don't understand this part.

The same can be done with the even parity but if done the same as above you get

$$cot \alpha a= \beta/\alpha$$

but this incorrect the correct answer is

$$tan \alpha a=\beta/\alpha$$

but to get this answer you would have to divide the differental by the function and not the other way around like what was done for odd solutions.

Try to read what you wrote here again slowly, try to remember where you got everything from. No it doesn't matter which one you divide with which, I leave it to you to figure out why. Once you do it all properly you should understand why you got the wrong answer.

 

[Ritorufon]

oh yea that should be a first order differential, not a second order, okay ill take a look at them again thanks a lot! :)