- #1

JohanL

- 158

- 0

A particle of mass m is contained in a right circular cylinder(pillbox) of radius R and height H. The particle is described by a wave function satisfying the Schrödinger wave equation

[tex]

-\frac{\hbar^2}{2m}\nabla^2\Psi(r,\phi,z) = E\Psi(r,\phi,z)

[/tex]

and the condition that the wave function go to zero over the surface of the pillbox. Find the lowest(zero point) permitted energy.

Solution:

First i express the laplace operator in cylindrical coordinates and use separation of variables. This gives three differential equations.

[tex]

\frac {d^2Z(z)} {dz^2} = \lambda^2z

[/tex]

[tex]

\frac {d^2\Phi(\phi)} {d\phi^2} = -m^2\Phi

[/tex]

[tex]

r\frac {d} {dr}(r\frac {dR(r)} {dr}) + (\mu^2r^2-m^2)R(r) = 0

[/tex]

with

[tex]

k^2 + \lambda^2 = \mu^2

[/tex]

and [tex]\lambda, \mu, m [/tex] are the separation constants

The periodicity and the boundary conditions Z(0)=0 and Z(H)=0 gives the solutions of the first two D.E are

[tex]

Z(z) = C_1sinh(\pi lz/H)

[/tex]

[tex]

\Phi(\phi) = C_2exp(im\phi)

[/tex]

where l is an integer and m can take on the positive and negative integers and zero.

The third differential equation is bessels equation and because R(0) should be finite the solutions are bessels functions of the first kind.

The wave function is then...i think.

[tex]

\Psi(r,\phi,z) = const.*J_m(\sqrt{\frac{2mE}{\hbar^2}}r)exp(im\phi)sinh(\pi lz/H)

[/tex]

Usually you then use the other boundary condition R(r=R)=0 to get the energies. If the wave function above didnt include the sinh-term (the z dependence) i know that the energies are given by the condition:

[tex]

J_m(\sqrt{\frac{2mE}{\hbar^2}}r) = 0

[/tex]

which gives

[tex]

E(m,n) = \frac{\hbar^2}{2mR^2}r_m_n

[/tex]

where r_mn is the n:th root of the m:th Bessel function.

But with this problem's wave function i don't know how to calculate the energies.

Any ideas?

The answer is:

[tex]

E(m,n) = \frac{\hbar^2}{2m}[(\frac{r_m_n}{R})^2+(\frac{l\pi}{H})^2]

[/tex]