# Continuity condition box 2D

1. Jun 25, 2014

### PeteSampras

Hello, i had studied the problem in 1D, but i thinking the problem in 2d, an i have the following question:

in a potential -V between (-a,a) an 0 otherwise.

One dimensional case:

One of the boundary condition are :

$\phi_I \in (-a,a)$, and $\phi_{II} \in (a,\infty)$

$\phi_I(a)=\phi_{II}(a)$

and continuity condition

$\phi_I'(a)=\phi_{II}'(a)$

in two dimensional case , for example with separation variables:

$X_I(x)Y_I(y) \in (x,y) \in (-a,a)$, and $X_{II}(x) Y_{II}(y) \in (x,y) \in (a,\infty)$

how are the boundary and continuity condition?

I think that

$X_I(x=a)Y_I(y=a)=X_{II}(x=a)Y_{II}(x=a)$

but, ¿how i write the continuity condition?,

2. Jun 25, 2014

### ChrisVer

You mean $x \in (-a,a)$ and $y \in (-a,a)$
You don't have to do much, just match the corresponding solutions. Y,X are independent of x,y respectively, so you can deal with the x,y axis separately as having the 1D case (only twice cause now you have 2 -x,y- 1D )
So
$X_I (a)= X_{II} (a)$

$Y_I (a)= Y_{II} (a)$

The same argument:
The only other way to have the continuity of $\Psi$ at those points would be to say:
$X_{I}(a)= Y_{II}(a)$
$Y_{I}(a)= X_{II}(a)$
But this wouldn't make any sense, since that's the reason of applying the separation of solutions- to treat each variable independently of the other.

Last edited: Jun 25, 2014
3. Jun 26, 2014

### Jilang

ϕI(a,a)=ϕII(a,a) ?