Solving Schrödinger's Equation for Cylindrical Boundaries

[fluidistic]

Homework Statement

I must get the first eigenvalues of the time independent Schrödinger's equation for a particle of mass m inside a cylinder of height h and radius a where $h \sim a$.
The boundary conditions are that psi is worth 0 everywhere on the surface of the cylinder.

Homework Equations

$-\frac{\hbar ^2}{2m} \triangle \psi =E \psi$.
Laplacian in cylindrical coordinates.

The Attempt at a Solution

I've used separation of variables on the PDE, seeking for the solutions of the form $\psi (\rho, \theta , z)=R(\rho) \Theta (\theta ) Z(z)$.
I reached that $\frac{Z''}{Z}=\text{constant}=-\lambda ^2$. Assuming that the Z function is periodic and worth 0 at the top and bottom of the cylinder, I reached that $Z(z)=B \sin \left ( \frac{n\pi n}{h} \right )$ where n=1,2, 3, etc.
Then I reached that $\frac{\Theta''}{\Theta} = -m^2$ where m=0, 1, 2, etc (because it must be periodic with period 2 pi). So that $\Theta (\theta )=C \cos (m \theta ) +D \sin (m \theta)$.
Then the last ODE remaining to solve is $\rho ^2 R''+\rho R'+R \{ \rho ^2 \left [ \frac{2mE}{\hbar ^2} - \left ( \frac{n\pi}{h} \right )^2 \right ] -m^2 \}=0$. This is where I'm stuck.
It's very similar to a Bessel equation and Cauchy-Euler equation but I don't think it is either. So I don't really know how to tackle that ODE. Any idea? Wolfram alpha does not seem to solve it either: http://www.wolframalpha.com/input/?i=x^2y%27%27%2Bxy%27%2By%28x^2*k-n^2%29%3D0.

 

[TSny]

Look's like Bessel's equation. See http://www.efunda.com/math/bessel/bessel.cfm

Of course, you'll need to rescale $\rho$ to simplify the expression inside your { }.

 

[fluidistic]

Look's like Bessel's equation. See http://www.efunda.com/math/bessel/bessel.cfm

Hmm ok.

Of course, you'll need to rescale $\rho$ to simplify the expression inside your { }.

Hmm what do you mean exactly? I have an equation of the form $\rho ^2 R''+\rho R' +R(\rho ^2 p^2 -m^2)$ where p is a constant for a given n.
Rescaling rho means to get $p^2=1$?

 

[TSny]

Yes. Define a new independent variable ($x$, say) in terms of $\rho$ such that you get the standard form of Bessel's equation.

 

[fluidistic]

Yes. Define a new independent variable ($x$, say) in terms of $\rho$ such that you get the standard form of Bessel's equation.

I try $x=\rho p$ so $\rho =x/p$ but then the ODE changes to $\frac{x^2R''}{p^2}+\frac{xR'}{p}+R(x^2-m^2)=0$. I could multiply by $p^2$ but I would not get the standard form of the Bessel equation. I don't see how I could rescale the factor in front of rho ^2 without rescaling the coefficients in front of R'' and R.

 

[TSny]

You need to take care of the rescaling in the derivatives, too. For example, $dR/d\rho = \left(dR/dx\right)\left(dx/d\rho\right)$

 

[fluidistic]

You need to take care of the rescaling in the derivatives, too. For example, $dR/d\rho = \left(dR/dx\right)\left(dx/d\rho\right)$

Oh right, I totally missed this!
So indeed now I recognize a Bessel equation!
Therefore I get that the solutions of the form $\psi =R ( \rho ) \Theta (\theta ) Z(z)=B_n \sin \left ( \frac{n\pi z}{h} \right ) [C_m \cos (m\theta ) + D_m \sin (m \theta )]J_m \left ( \rho \sqrt {\frac{2mE}{\hbar ^2} - \frac{n^2 \pi ^2}{h^2}} \right )$.
So the solution that satisfies the boundary condition is a linear combination of those.
I'm not 100% sure about what they mean by "eigenvalues". Eigenfrequencies? Lowest energies possible? (They only want the first 3 eigenvalues).
I'm pretty sure this will concern the cases (1) n=1 and m=0 and m=1. (2) n=2, m=0. But I'm not sure what they are asking me.

 

[TSny]

I think they want the three lowest energies. They are called eigenvalues because they are eigenvalues of the time independent Schrödinger equation $H|\psi> = E|\psi>$

 

[haruspex]

You don't seem to have used the boundary condition at ρ=a.

 

[fluidistic]

I think they want the three lowest energies. They are called eigenvalues because they are eigenvalues of the time independent Schrödinger equation $H|\psi> = E|\psi>$

Ok thanks!
Hmm I don't know how to get that information.
I have a feeling I should add a subscript "n" under "E" in $\psi =R ( \rho ) \Theta (\theta ) Z(z)=B_n \sin \left ( \frac{n\pi z}{h} \right ) [C_m \cos (m\theta ) + D_m \sin (m \theta )]J_m \left ( \rho \sqrt {\frac{2mE}{\hbar ^2} - \frac{n^2 \pi ^2}{h^2}} \right )$ and then isolate $E_n$ but not sure to what I should equate the equation.

 

[TSny]

Follow haruspex's lead.

 

[fluidistic]

Oh right guys sorry. And thanks for helping. I did not see haruspex's post.
So if $x_p$ is the p'th zero of the Bessel function then $E_n= \left ( \frac{\hbar ^2}{2m} \right ) \left [ \left ( \frac{x_p}{a} \right ) ^2 +\left ( \frac{n^2 \pi ^2}{h^2} \right ) \right ]$. I guess I'll have to check if I can replace "p" by "n". It's not obvious to me at a first glance.

 

[fluidistic]

Ok I've thought a bit on this. The first 3 lowest energy values are when $x_p=x_0$. So $E_1= \frac{\hbar ^2}{2m} \left [ \left ( \frac{x_0}{a} \right ) ^2 + \frac{\pi ^2}{h^2} \right ]$, $E_2= \frac{\hbar ^2}{2m} \left [ \left ( \frac{x_0}{a} \right ) ^2 + \frac{4\pi ^2}{h^2} \right ]$ and $E_3= \frac{\hbar ^2}{2m} \left [ \left ( \frac{x_0}{a} \right ) ^2 + \frac{9\pi ^2}{h^2} \right ]$.
I'm not very confident because I don't know if $\frac{\hbar ^2 }{2m} \left [ \left ( \frac{x_1}{a} \right ) ^2 + \frac{\pi ^2}{h^2} \right ] <\frac{\hbar ^2}{2m} \left [ \left ( \frac{x_0}{a} \right ) ^2 + \frac{9\pi ^2}{h^2} \right ]$ for example.

 

[TSny]

You'll need to consult a Table of Roots

The problem states that the height of the cylinder is approx. equal to the radius: $h\approx a$, which should help figure out the lowest three energies.

 

[fluidistic]

You'll need to consult a Table of Roots

The problem states that the height of the cylinder is approx. equal to the radius: $h\approx a$, which should help figure out the lowest three energies.

Great and thank you once more.
I get from lower to upper: $E_{1,0}$, $E_{1,1}$ and $E_{2,0}$ where the subscript are $E_{n,p}$.

 

[TSny]

I think that might be correct.

 

[fluidistic]

I think that might be correct.

Thanks for all.