S.E in cylinder - Bessel's equation

[JohanL]

Im trying to solve the following problem:

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

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

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.

<br /> \frac {d^2Z(z)} {dz^2} = \lambda^2z<br />

<br /> \frac {d^2\Phi(\phi)} {d\phi^2} = -m^2\Phi<br />

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

with

<br /> k^2 + \lambda^2 = \mu^2<br />

and \lambda, \mu, m 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

<br /> Z(z) = C_1sinh(\pi lz/H)<br />
<br /> \Phi(\phi) = C_2exp(im\phi) <br />

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.

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

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:


<br /> J_m(\sqrt{\frac{2mE}{\hbar^2}}r) = 0<br />

which gives

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

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:

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

 

[OlderDan]

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

Perhaps \lambda needs to be imaginary to satisfy your boundary conditions??

 

[HallsofIvy]

OlderDan's point is that sinh(x) is a one to one function and is not 0 at two different points. If &lambda; is imaginary (equivalently: if &lamda;² is negative), you get sine and cosine solutions which are periodic.

 

[JohanL]

Thank you.
ok...lets see if i understand.

we have

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

My book says that if \lambda &gt; 0 the general solution is

C_1cosh (\lambda z) + C_2sinh (\lambda z)

But in the problem we have Z=0 at two different points

"OlderDan's point is that sinh(x) is a one to one function and is not 0 at two different points."

so \lambda must be imaginary (or am i something wrong when i eliminate C1cosh because of the boundary condition Z(0)=0. Either way the main problem remains.)

"If λ is imaginary you get sine and cosine solutions which are periodic."

and the general solution is

C_1cos (\lambda z) + C_2sin (\lambda z)



Then you get

Z(z) = C_1sin(\pi lz/H)

\Psi(r,\phi,z) = R(r) \Phi(\phi)Z(z)

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

sin instead of sinh in the wave function.
where

R(r) = C*J_m(\sqrt{\frac{2mE}{\hbar^2}}r)

One problem solved...i hope.
But i still don't know how to continue.

Now you should use the other boundary condition
R(r=R)=0 to get the energies. Is this correct:

const.*J_m(\sqrt{\frac{2mE}{\hbar^2}}R)exp(im\phi) sin(\pi lz/H) = 0

and then you solve for E and get the energies?

I don't really know how to do this tho.

 

[OlderDan]

Assuming you now have the correct eigenfunction, if you put it back into the original equation

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

and operate on the left hand side you should get a constant times the original function. The const. in your function divides out. You just need to associate the constant that comes from doing the operation on the left hand side with the E on the right hand side.

 

[JohanL]

I solved it. Thank you!