13 3D Problems Separable in Cartesian Coordinates
We will now look at the case ofpotentials that separate in Cartesian coordinates. These will
be of the form.
V(~r) =V 1 (x) +V 2 (y) +V 3 (z)
In this case, we can solve the problem byseparation of variables.
H=Hx+Hy+Hz
(Hx+Hy+Hz)u(x)v(y)w(z) =Eu(x)v(y)w(z)
[Hxu(x)]v(y)w(z) +u(x) (Hy+Hz)v(y)w(z) = Eu(x)v(y)w(z)
Hu(x)
u(x)=E−
(Hy+Hz)v(y)w(z)
v(y)w(z)
= ǫxThe left hand side of this equation depends only onx, while the right side depends onyandz. In
order for the two sides to be equal everywhere, theymust both be equal to a constantwhich
we callǫx.
Thexpart of the solution satisfies the equation
Hxu(x) =ǫxu(x).Treating the other components similarly we get
Hyv(y) =ǫyv(y)
Hzw(z) =ǫzw(z)and the total energy is
E=ǫx+ǫy+ǫz.
There are only a few problems which can be worked this way but they are important.
13.1 Particle in a 3D Box
An example of a problem which has a Hamiltonian of the separable form istheparticle in a 3D
box. The potential is zero inside thecube of sideLand infinite outside. It can be written as a
sum of terms.
H=Hx+Hy+Hz
The energies are
Enx,ny,nz=π^2 ̄h^2
2 mL^2
(n^2 x+n^2 y+n^2 z).They depend onthree quantum numbers, (since there are 3 degrees of freedom).
unx,ny,nz(~r) = sin(n
xπx
L)
sin(n
yπy
L)
sin(n
zπz
L)
For a cubic box like this one, there will often be degenerate states.